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Physics

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Introductory Physics I
Elementary Mechanics

by

Robert G. Brown
Duke University Physics Department
Durham, NC 27708-0305 rgb@phy.duke.edu Copyright Notice
Copyright Robert G. Brown 1993, 2007, 2013

Notice
This physics textbook is designed to support my personal teaching activities at Duke University, in particular teaching its Physics 141/142, 151/152, or 161/162 series (Introductory Physics for life science majors, engineers, or potential physics majors, respectively). It is freely available in its entirety in a downloadable PDF form or to be read online at: http://www.phy.duke.edu/∼rgb/Class/intro physics 1.php
It is also available in an inexpensive (really!) print version via Lulu press here: http://www.lulu.com/shop/product-21186588.html where readers/users can voluntarily help support or reward the author by purchasing either this paper copy or one of the even more inexpensive electronic copies.
By making the book available in these various media at a cost ranging from free to cheap, I enable the text can be used by students all over the world where each student can pay (or not) according to their means.
Nevertheless, I am hoping that students who truly find this work useful will purchase a copy through Lulu or a bookseller (when the latter option becomes available), if only to help subsidize me while I continue to write inexpensive textbooks in physics or other subjects.
This textbook is organized for ease of presentation and ease of learning. In particular, they are hierarchically organized in a way that directly supports efficient learning. They are also remarkably complete in their presentation and contain moderately detailed derivations of many of the important equations and relations from first principles while not skimping on simpler heuristic or conceptual explanations as well.
As a “live” document (one I actively use and frequently change, adding or deleting material or altering the presentation in some way), this textbook may have errors great and small, “stub” sections where I intend to add content at some later time but haven’t yet finished it, and they cover and omit topics according to my own view of what is or isn’t important to cover in a one-semester course. Expect them to change with little warning or announcement as I add content or correct errors. Purchasers of the paper version should be aware of its probable imperfection and be prepared to either live with it or mark up their copy with corrections or additions as need be. The latest (and hopefully most complete and correct) version is always available for free online anyway, and people who have paid for a paper copy are especially welcome to access and retrieve it.
I cherish good-hearted communication from students or other instructors pointing out errors or suggesting new content (and have in the past done my best to implement many such corrections or suggestions). Books by Robert G. Brown
Physics Textbooks
• Introductory Physics I and II
A lecture note style textbook series intended to support the teaching of introductory physics, with calculus, at a level suitable for Duke undergraduates.
• Classical Electrodynamics
A lecture note style textbook intended to support the second semester (primarily the dynamical portion, little statics covered) of a two semester course of graduate Classical Electrodynamics.

Computing Books
• How to Engineer a Beowulf Cluster
An online classic for years, this is the print version of the famous free online book on cluster engineering. It too is being actively rewritten and developed, no guarantees, but it is probably still useful in its current incarnation.

Fiction
• The Book of Lilith
ISBN: 978-1-4303-2245-0
Web: http://www.phy.duke.edu/∼rgb/Lilith/Lilith.php
Lilith is the first person to be given a soul by God, and is given the job of giving all the things in the world souls by loving them, beginning with Adam. Adam is given the job of making up rules and the definitions of sin so that humans may one day live in an ethical society.
Unfortunately Adam is weak, jealous, and greedy, and insists on being on top during sex to
“be closer to God”.
Lilith, however, refuses to be second to Adam or anyone else. The Book of Lilith is a funny, sad, satirical, uplifting tale of her spiritual journey through the ancient world soulgiving and judging to find at the end of that journey – herself.

Poetry
• Who Shall Sing, When Man is Gone
Original poetry, including the epic-length poem about an imagined end of the world brought about by a nuclear war that gives the collection its name. Includes many long and short works on love and life, pain and death.
Ocean roaring, whipped by storm in damned defiance, hating hell with every wave and every swell, every shark and every shell and shoreline.
• Hot Tea!
More original poetry with a distinctly Zen cast to it. Works range from funny and satirical to inspiring and uplifting, with a few erotic poems thrown in.

Chop water, carry wood. Ice all around, fire is dying. Winter Zen?

All of these books can be found on the online Lulu store here: http://stores.lulu.com/store.php?fAcctID=877977 The Book of Lilith is available on Amazon, Barnes and Noble and other online bookseller websites.

Contents
Preface

xi

Textbook Layout and Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xii

I: Getting Ready to Learn Physics

3

Preliminaries

3

See, Do, Teach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

Other Conditions for Learning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8

Your Brain and Learning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13

How to Do Your Homework Effectively . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19

The Method of Three Passes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

22

Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23

Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

26

Homework for Week 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

27

II: Elementary Mechanics

31

Week 1: Newton’s Laws

33

Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

1.1: Introduction: A Bit of History and Philosophy . . . . . . . . . . . . . . . . . . . . . .

38

1.2: Dynamics

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

39

1.3: Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41

1.4: Newton’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

46

1.5: Forces

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47

1.5.1: The Forces of Nature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47

1.5.2: Force Rules

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49

1.6: Force Balance – Static Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . .

51

Example 1.6.1: Spring and Mass in Static Force Equilibrium

. . . . . . . . . . . . .

51

1.7: Simple Motion in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

52

i

ii

CONTENTS
Example 1.7.1: A Mass Falling from Height H

. . . . . . . . . . . . . . . . . . . . .

53

Example 1.7.2: A Constant Force in One Dimension . . . . . . . . . . . . . . . . . .

58

1.7.1: Solving Problems with More Than One Object . . . . . . . . . . . . . . . . . .

61

Example 1.7.3: Atwood’s Machine . . . . . . . . . . . . . . . . . . . . . . . . . . . .

61

Example 1.7.4: Braking for Bikes, or Just Breaking Bikes? . . . . . . . . . . . . . . .

63

1.8: Motion in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

64

1.8.1: Free Flight Trajectories – Projectile Motion . . . . . . . . . . . . . . . . . . .

66

Example 1.8.1: Trajectory of a Cannonball . . . . . . . . . . . . . . . . . . . . . . .

66

1.8.2: The Inclined Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

69

Example 1.8.2: The Inclined Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . .

69

1.9: Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

71

1.9.1: Tangential Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

72

1.9.2: Centripetal Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

73

Example 1.9.1: Ball on a String . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

74

Example 1.9.2: Tether Ball/Conic Pendulum . . . . . . . . . . . . . . . . . . . . . .

75

1.9.3: Tangential Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

76

1.10: Conclusion: Rubric for Newton’s Second Law Problems . . . . . . . . . . . . . . . .

77

Homework for Week 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

78

Week 2: Newton’s Laws: Continued

95

Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

95

2.1: Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

97

Example 2.1.1: Inclined Plane of Length L with Friction . . . . . . . . . . . . . . . .

98

Example 2.1.2: Block Hanging off of a Table . . . . . . . . . . . . . . . . . . . . . . . 100
Example 2.1.3: Find The Minimum No-Skid Braking Distance for a Car . . . . . . . 102
Example 2.1.4: Car Rounding a Banked Curve with Friction . . . . . . . . . . . . . . 104
2.2: Drag Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
2.2.1: Stokes, or Laminar Drag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
2.2.2: Rayleigh, or Turbulent Drag . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
2.2.3: Terminal velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
Example 2.2.1: Falling From a Plane and Surviving . . . . . . . . . . . . . . . . . . . 112
Example 2.2.2: Solution to Equations of Motion for Stokes’ Drag . . . . . . . . . . . 113
2.2.4: Advanced: Solution to Equations of Motion for Turbulent Drag . . . . . . . . 114
Example 2.2.3: Dropping the Ram . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
2.3: Inertial Reference Frames – the Galilean Transformation . . . . . . . . . . . . . . . . 117
2.3.1: Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
2.3.2: Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

CONTENTS

iii

2.4: Non-Inertial Reference Frames – Pseudoforces . . . . . . . . . . . . . . . . . . . . . . 121
2.4.1: Identifying Inertial Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
Example 2.4.1: Weight in an Elevator . . . . . . . . . . . . . . . . . . . . . . . . . . 124
Example 2.4.2: Pendulum in a Boxcar . . . . . . . . . . . . . . . . . . . . . . . . . . 125
2.4.2: Advanced: General Relativity and Accelerating Frames . . . . . . . . . . . . . 127
2.5: Just For Fun: Hurricanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
Homework for Week 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
Week 3: Work and Energy

141

Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
3.1: Work and Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
3.1.1: Units of Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
3.1.2: Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
3.2: The Work-Kinetic Energy Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
3.2.1: Derivation I: Rectangle Approximation Summation . . . . . . . . . . . . . . . 146
3.2.2: Derivation II: Calculus-y (Chain Rule) Derivation . . . . . . . . . . . . . . . . 148
Example 3.2.1: Pulling a Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
Example 3.2.2: Range of a Spring Gun . . . . . . . . . . . . . . . . . . . . . . . . . . 150
3.3: Conservative Forces: Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 151
3.3.1: Force from Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
3.3.2: Potential Energy Function for Near-Earth Gravity . . . . . . . . . . . . . . . . 154
3.3.3: Springs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
3.4: Conservation of Mechanical Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
3.4.1: Force, Potential Energy, and Total Mechanical Energy . . . . . . . . . . . . . 157
Example 3.4.1: Falling Ball Reprise . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
Example 3.4.2: Block Sliding Down Frictionless Incline Reprise . . . . . . . . . . . . 158
Example 3.4.3: A Simple Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
Example 3.4.4: Looping the Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
3.5: Generalized Work-Mechanical Energy Theorem . . . . . . . . . . . . . . . . . . . . . 161
Example 3.5.1: Block Sliding Down a Rough Incline . . . . . . . . . . . . . . . . . . 161
Example 3.5.2: A Spring and Rough Incline . . . . . . . . . . . . . . . . . . . . . . . 162
3.5.1: Heat and Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . 162
3.6: Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
Example 3.6.1: Rocket Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
3.7: Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
3.7.1: Energy Diagrams: Turning Points and Forbidden Regions . . . . . . . . . . . . 168
Homework for Week 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

iv

CONTENTS

Week 4: Systems of Particles, Momentum and Collisions

181

Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
4.1: Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
4.1.1: Newton’s Laws for a System of Particles – Center of Mass . . . . . . . . . . . 186
Example 4.1.1: Center of Mass of a Few Discrete Particles . . . . . . . . . . . . . . . 188
4.1.2: Coarse Graining: Continuous Mass Distributions . . . . . . . . . . . . . . . . . 189
Example 4.1.2: Center of Mass of a Continuous Rod . . . . . . . . . . . . . . . . . . 191
Example 4.1.3: Center of mass of a circular wedge . . . . . . . . . . . . . . . . . . . 192
Example 4.1.4: Breakup of Projectile in Midflight . . . . . . . . . . . . . . . . . . . . 193
4.2: Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
4.2.1: The Law of Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . 194
4.3: Impulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
Example 4.3.1: Average Force Driving a Golf Ball . . . . . . . . . . . . . . . . . . . 198
Example 4.3.2: Force, Impulse and Momentum for Windshield and Bug . . . . . . . 198
4.3.1: The Impulse Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
4.3.2: Impulse, Fluids, and Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
4.4: Center of Mass Reference Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
4.5: Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
4.5.1: Momentum Conservation in the Impulse Approximation . . . . . . . . . . . . 204
4.5.2: Elastic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
4.5.3: Fully Inelastic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
4.5.4: Partially Inelastic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
4.5.5: Dimension of Scattering and Sufficient Information . . . . . . . . . . . . . . . 205
4.6: 1-D Elastic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
4.6.1: The Relative Velocity Approach . . . . . . . . . . . . . . . . . . . . . . . . . . 208
4.6.2: 1D Elastic Collision in the Center of Mass Frame . . . . . . . . . . . . . . . . 209
4.6.3: The “BB/bb” or “Pool Ball” Limits . . . . . . . . . . . . . . . . . . . . . . . . 211
4.7: Elastic Collisions in 2-3 Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
4.8: Inelastic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
Example 4.8.1: One-dimensional Fully Inelastic Collision (only) . . . . . . . . . . . . 215
Example 4.8.2: Ballistic Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
Example 4.8.3: Partially Inelastic Collision . . . . . . . . . . . . . . . . . . . . . . . 218
4.9: Kinetic Energy in the CM Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
Homework for Week 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
Week 5: Torque and Rotation in One Dimension

235

Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

CONTENTS

v

5.1: Rotational Coordinates in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . 236
5.2: Newton’s Second Law for 1D Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . 238
5.2.1: The r-dependence of Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
5.2.2: Summing the Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . 242
5.3: The Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243
Example 5.3.1: The Moment of Inertia of a Rod Pivoted at One End . . . . . . . . . 243
5.3.1: Moment of Inertia of a General Rigid Body . . . . . . . . . . . . . . . . . . . . 243
Example 5.3.2: Moment of Inertia of a Ring . . . . . . . . . . . . . . . . . . . . . . . 244
Example 5.3.3: Moment of Inertia of a Disk . . . . . . . . . . . . . . . . . . . . . . . 245
5.3.2: Table of Useful Moments of Inertia . . . . . . . . . . . . . . . . . . . . . . . . 246
5.4: Torque as a Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
Example 5.4.1: Rolling the Spool . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247
5.5: Torque and the Center of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248
Example 5.5.1: The Angular Acceleration of a Hanging Rod . . . . . . . . . . . . . . 249
5.6: Solving Newton’s Second Law Problems Involving Rolling . . . . . . . . . . . . . . . 249
Example 5.6.1: A Disk Rolling Down an Incline . . . . . . . . . . . . . . . . . . . . . 250
Example 5.6.2: Atwood’s Machine with a Massive Pulley . . . . . . . . . . . . . . . . 252
5.7: Rotational Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
5.7.1: Work Done on a Rigid Object . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
5.7.2: The Rolling Constraint and Work . . . . . . . . . . . . . . . . . . . . . . . . . 255
Example 5.7.1: Work and Energy in Atwood’s Machine . . . . . . . . . . . . . . . . 256
Example 5.7.2: Unrolling Spool . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257
Example 5.7.3: A Rolling Ball Loops-the-Loop . . . . . . . . . . . . . . . . . . . . . 258
5.8: The Parallel Axis Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259
Example 5.8.1: Moon Around Earth, Earth Around Sun . . . . . . . . . . . . . . . . 261
Example 5.8.2: Moment of Inertia of a Hoop Pivoted on One Side . . . . . . . . . . 261
5.9: Perpendicular Axis Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
Example 5.9.1: Moment of Inertia of Hoop for Planar Axis . . . . . . . . . . . . . . 264
Homework for Week 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265
Week 6: Vector Torque and Angular Momentum

277

Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277
6.1: Vector Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
6.2: Total Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279
6.2.1: The Law of Conservation of Angular Momentum . . . . . . . . . . . . . . . . . 280
6.3: The Angular Momentum of a Symmetric Rotating Rigid Object . . . . . . . . . . . . 281
Example 6.3.1: Angular Momentum of a Point Mass Moving in a Circle . . . . . . . 283

vi

CONTENTS
Example 6.3.2: Angular Momentum of a Rod Swinging in a Circle . . . . . . . . . . 283
Example 6.3.3: Angular Momentum of a Rotating Disk . . . . . . . . . . . . . . . . 284
Example 6.3.4: Angular Momentum of Rod Sweeping out Cone . . . . . . . . . . . . 285
6.4: Angular Momentum Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
Example 6.4.1: The Spinning Professor . . . . . . . . . . . . . . . . . . . . . . . . . . 285
6.4.1: Radial Forces and Angular Momentum Conservation . . . . . . . . . . . . . . 286
Example 6.4.2: Mass Orbits On a String . . . . . . . . . . . . . . . . . . . . . . . . . 287
6.5: Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289
Example 6.5.1: Fully Inelastic Collision of Ball of Putty with a Free Rod . . . . . . . 291
Example 6.5.2: Fully Inelastic Collision of Ball of Putty with Pivoted Rod . . . . . . 294
6.5.1: More General Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296
6.6: Angular Momentum of an Asymmetric Rotating Rigid Object . . . . . . . . . . . . . 296
Example 6.6.1: Rotating Your Tires . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
6.7: Precession of a Top . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300
Example 6.7.1: Finding ωp From ∆L/∆t (Average) . . . . . . . . . . . . . . . . . . . 302
Example 6.7.2: Finding ωp from ∆L and ∆t Separately . . . . . . . . . . . . . . . . 302
Example 6.7.3: Finding ωp from Calculus . . . . . . . . . . . . . . . . . . . . . . . . 303
Homework for Week 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305

Week 7: Statics

313

Statics Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313
7.1: Conditions for Static Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313
7.2: Static Equilibrium Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
Example 7.2.1: Balancing a See-Saw . . . . . . . . . . . . . . . . . . . . . . . . . . . 316
Example 7.2.2: Two Saw Horses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317
Example 7.2.3: Hanging a Tavern Sign . . . . . . . . . . . . . . . . . . . . . . . . . . 318
7.2.1: Equilibrium with a Vector Torque . . . . . . . . . . . . . . . . . . . . . . . . . 319
Example 7.2.4: Building a Deck . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320
7.3: Tipping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321
Example 7.3.1: Tipping Versus Slipping . . . . . . . . . . . . . . . . . . . . . . . . . 321
Example 7.3.2: Tipping While Pushing . . . . . . . . . . . . . . . . . . . . . . . . . . 323
7.4: Force Couples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
Example 7.4.1: Rolling the Cylinder Over a Step . . . . . . . . . . . . . . . . . . . . 325
Homework for Week 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327

III: Applications of Mechanics
Week 8: Fluids

339
339

CONTENTS

vii

Fluids Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339
8.1: General Fluid Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340
8.1.1: Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
8.1.2: Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
8.1.3: Compressibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344
8.1.4: Viscosity and fluid flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345
8.1.5: Properties Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345
Static Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346
8.1.6: Pressure and Confinement of Static Fluids . . . . . . . . . . . . . . . . . . . . 346
8.1.7: Pressure and Confinement of Static Fluids in Gravity . . . . . . . . . . . . . . 348
8.1.8: Variation of Pressure in Incompressible Fluids . . . . . . . . . . . . . . . . . . 350
Example 8.1.1: Barometers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350
Example 8.1.2: Variation of Oceanic Pressure with Depth . . . . . . . . . . . . . . . 353
8.1.9: Variation of Pressure in Compressible Fluids . . . . . . . . . . . . . . . . . . . 353
Example 8.1.3: Variation of Atmospheric Pressure with Height . . . . . . . . . . . . 354
8.2: Pascal’s Principle and Hydraulics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355
Example 8.2.1: A Hydraulic Lift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356
8.3: Fluid Displacement and Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357
8.3.1: Archimedes’ Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359
Example 8.3.1: Testing the Crown I . . . . . . . . . . . . . . . . . . . . . . . . . . . 360
Example 8.3.2: Testing the Crown II . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
8.4: Fluid Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
8.4.1: Conservation of Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364
8.4.2: Work-Mechanical Energy in Fluids: Bernoulli’s Equation . . . . . . . . . . . . 367
Example 8.4.1: Emptying the Iced Tea . . . . . . . . . . . . . . . . . . . . . . . . . . 368
8.4.3: Fluid Viscosity and Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . 368
8.4.4: A Brief Note on Turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371
8.5: The Human Circulatory System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372
Example 8.5.1: Atherosclerotic Plaque Partially Occludes a Blood Vessel . . . . . . . 376
Example 8.5.2: Aneurisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
Example 8.5.3: The Giraffe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378
Homework for Week 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
Week 9: Oscillations

389

Oscillation Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389
9.1: The Simple Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390
9.1.1: The Archetypical Simple Harmonic Oscillator: A Mass on a Spring . . . . . . 391

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CONTENTS
9.1.2: The Simple Harmonic Oscillator Solution . . . . . . . . . . . . . . . . . . . . . 396
9.1.3: Plotting the Solution: Relations Involving ω . . . . . . . . . . . . . . . . . . . 397
9.1.4: The Energy of a Mass on a Spring . . . . . . . . . . . . . . . . . . . . . . . . . 398
9.2: The Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398
9.2.1: The Physical Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400
9.3: Damped Oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402
9.3.1: Properties of the Damped Oscillator . . . . . . . . . . . . . . . . . . . . . . . . 404
Example 9.3.1: Car Shock Absorbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 406
9.4: Damped, Driven Oscillation: Resonance

. . . . . . . . . . . . . . . . . . . . . . . . . 407

9.4.1: Harmonic Driving Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409
9.4.2: Solution to Damped, Driven, Simple Harmonic Oscillator . . . . . . . . . . . . 411
9.5: Elastic Properties of Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414
9.5.1: Simple Models for Molecular Bonds . . . . . . . . . . . . . . . . . . . . . . . . 415
9.5.2: The Force Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417
9.5.3: A Microscopic Picture of a Solid . . . . . . . . . . . . . . . . . . . . . . . . . . 418
9.5.4: Shear Forces and the Shear Modulus . . . . . . . . . . . . . . . . . . . . . . . 420
9.5.5: Deformation and Fracture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421
9.6: Human Bone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423
Example 9.6.1: Scaling of Bones with Animal Size . . . . . . . . . . . . . . . . . . . 425
Homework for Week 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427
Week 10: The Wave Equation

435

Wave Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435
10.1: Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436
10.2: Waves on a String . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437
10.3: Solutions to the Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439
10.3.1: An Important Property of Waves: Superposition . . . . . . . . . . . . . . . . 439
10.3.2: Arbitrary Waveforms Propagating to the Left or Right . . . . . . . . . . . . . 439
10.3.3: Harmonic Waveforms Propagating to the Left or Right . . . . . . . . . . . . 440
10.3.4: Stationary Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441
10.4: Reflection of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442
10.5: Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443
Homework for Week 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448
Week 11: Sound

459

Sound Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459
11.1: Sound Waves in a Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461
11.2: Sound Wave Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462

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CONTENTS

11.3: Sound Wave Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462
11.3.1: Sound Displacement and Intensity In Terms of Pressure . . . . . . . . . . . . 463
11.3.2: Sound Pressure and Decibels . . . . . . . . . . . . . . . . . . . . . . . . . . . 465
11.4: Doppler Shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467
11.4.1: Moving Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467
11.4.2: Moving Receiver . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468
11.4.3: Moving Source and Moving Receiver . . . . . . . . . . . . . . . . . . . . . . . 469
11.5: Standing Waves in Pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469
11.5.1: Pipe Closed at Both Ends . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469
11.5.2: Pipe Closed at One End . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470
11.5.3: Pipe Open at Both Ends . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471
11.6: Beats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472
11.7: Interference and Sound Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472
11.8: The Ear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474
Homework for Week 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477
Week 12: Gravity

485

Gravity Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485
12.1: Cosmological Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489
12.2: Kepler’s Laws

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493

12.2.1: Ellipses and Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494
12.3: Newton’s Law of Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496
12.4: The Gravitational Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502
12.4.1: Spheres, Shells, General Mass Distributions . . . . . . . . . . . . . . . . . . . 503
12.5: Gravitational Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504
12.6: Energy Diagrams and Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505
12.7: Escape Velocity, Escape Energy

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506

Example 12.7.1: How to Cause an Extinction Event . . . . . . . . . . . . . . . . . . 507
12.8: Bridging the Gap: Coulomb’s Law and Electrostatics . . . . . . . . . . . . . . . . . 508
Homework for Week 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509

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Preface
This introductory mechanics text is intended to be used in the first semester of a two-semester series of courses teaching introductory physics at the college level, followed by a second semester course in introductory electricity and magnetism, and optics. The text is intended to support teaching the material at a rapid, but advanced level – it was developed to support teaching introductory calculus-based physics to potential physics majors, engineers, and other natural science majors at Duke University over a period of more than thirty years.
Students who hope to succeed in learning physics from this text will need, as a minimum prerequisite, a solid grasp of basic mathematics. It is strongly recommended that all students have mastered mathematics at least through single-variable differential calculus (typified by the AB advanced placement test or a first-semester college calculus course). Students should also be taking
(or have completed) single variable integral calculus (typified by the BC advanced placement test or a second-semester college calculus course). In the text it is presumed that students are competent in geometry, trigonometry, algebra, and single variable calculus; more advanced multivariate calculus is used in a number of places but it is taught in context as it is needed and is always “separable” into two or three independent one-dimensional integrals.
Many students are, unfortunately weak in their mastery of mathematics at the time they take physics. This enormously complicates the process of learning for them, especially if they are years removed from when they took their algebra, trig, and calculus classes (as is frequently the case for pre-medical students taking the course in their junior year of college). For that reason, a separate supplementary text intended specifically to help students of introductory physics quickly and efficiently review the required math is being prepared as a companion volume to all semesters of introductory physics. Indeed, it should really be quite useful for any course being taught with any textbook series and not just this one.
This book is located here: http://www.phy.duke.edu/∼rgb/Class/math for intro physics.php and I strongly suggest that all students who are reading these words preparing to begin studying physics pause for a moment, visit this site, and either download the pdf or bookmark the site.
Note that Week 0: How to Learn Physics is not part of the course per se, but I usually do a quick review of this material (as well as the course structure, grading scheme, and so on) in my first lecture of any given semester, the one where students are still finding the room, dropping and adding courses, and one cannot present real content in good conscience unless you plan to do it again in the second lecture as well. Students greatly benefit from guidance on how to study, as most enter physics thinking that they can master it with nothing but the memorization and rote learning skills that have served them so well for their many other fact-based classes. Of course this is completely false – physics is reason based and conceptual and it requires a very different pattern of study than simply staring at and trying to memorize lists of formulae or examples.
Students, however, should not count on their instructor doing this – they need to be self-actualized in their study from the beginning. It is therefore strongly suggested that all students read this xi xii

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preliminary chapter right away as their first “assignment” whether or not it is covered in the first lecture or assigned. In fact, (if you’re just such a student reading these words) you can always decide to read it right now (as soon as you finish this Preface). It won’t take you an hour, and might make as much as a full letter difference (to the good) in your final grade. What do you have to lose?
Even if you think that you are an excellent student and learn things totally effortlessly, I strongly suggest reading it. It describes a new perspective on the teaching and learning process supported by very recent research in neuroscience and psychology, and makes very specific suggestions as to the best way to proceed to learn physics.
Finally, the Introduction is a rapid summary of the entire course! If you read it and look at the pictures before beginning the course proper you can get a good conceptual overview of everything you’re going to learn. If you begin by learning in a quick pass the broad strokes for the whole course, when you go through each chapter in all of its detail, all those facts and ideas have a place to live in your mind.
That’s the primary idea behind this textbook – in order to be easy to remember, ideas need a house, a place to live. Most courses try to build you that house by giving you one nail and piece of wood at a time, and force you to build it in complete detail from the ground up.
Real houses aren’t built that way at all! First a foundation is established, then the frame of the whole house is erected, and then, slowly but surely, the frame is wired and plumbed and drywalled and finished with all of those picky little details. It works better that way. So it is with learning.

Textbook Layout and Design
This textbook has a design that is just about perfectly backwards compared to most textbooks that currently cover the subject. Here are its primary design features:
• All mathematics required by the student is reviewed in a standalone, cross-referenced (free) work at the beginning of the book rather than in an appendix that many students never find.
• There are only twelve chapters. The book is organized so that it can be sanely taught in a single college semester with at most a chapter a week.
• It begins each chapter with an “abstract” and chapter summary. Detail, especially lecture-note style mathematical detail, follows the summary rather than the other way around.
• This text does not spend page after page trying to explain in English how physics works
(prose which to my experience nobody reads anyway). Instead, a terse “lecture note” style presentation outlines the main points and presents considerable mathematical detail to support solving problems.
• Verbal and conceptual understanding is, of course, very important. It is expected to come from verbal instruction and discussion in the classroom and recitation and lab. This textbook relies on having a committed and competent instructor and a sensible learning process.
• Each chapter ends with a short (by modern standards) selection of challenging homework problems. A good student might well get through all of the problems in the book, rather than at most 10% of them as is the general rule for other texts.
• The problems are weakly sorted out by level, as this text is intended to support non-physics science and pre-health profession students, engineers, and physics majors all three. The material covered is of course the same for all three, but the level of detail and difficulty of the math used and required is a bit different.

CONTENTS

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• The textbook is entirely algebraic in its presentation and problem solving requirements – with very few exceptions no calculators should be required to solve problems. The author assumes that any student taking physics is capable of punching numbers into a calculator, but it is algebra that ultimately determines the formula that they should be computing. Numbers are used in problems only to illustrate what “reasonable” numbers might be for a given realworld physical situation or where the problems cannot reasonably be solved algebraically (e.g. resistance networks).
This layout provides considerable benefits to both instructor and student. This textbook supports a top-down style of learning, where one learns each distinct chapter topic by quickly getting the main points onboard via the summary, then derives them or explores them in detail, then applies them to example problems. Finally one uses what one has started to learn working in groups and with direct mentoring and support from the instructors, to solve highly challenging problems that cannot be solved without acquiring the deeper level of understanding that is, or should be, the goal one is striving for.
It’s without doubt a lot of work. Nobody said learning physics would be easy, and this book certainly doesn’t claim to make it so. However, this approach will (for most students) work.
The reward, in the end, is the ability to see the entire world around you through new eyes, understanding much of the “magic” of the causal chain of physical forces that makes all things unfold in time. Natural Law is a strange, beautiful sort of magic; one that is utterly impersonal and mechanical and yet filled with structure and mathematics and light. It makes sense, both in and of itself and of the physical world you observe.
Enjoy.

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I: Getting Ready to Learn Physics

1

Preliminaries
See, Do, Teach
If you are reading this, I assume that you are either taking a course in physics or wish to learn physics on your own. If this is the case, I want to begin by teaching you the importance of your personal engagement in the learning process. If it comes right down to it, how well you learn physics, how good a grade you get, and how much fun you have all depend on how enthusiastically you tackle the learning process. If you remain disengaged, detatched from the learning process, you almost certainly will do poorly and be miserable while doing it. If you can find any degree of engagement
– or open enthusiasm – with the learning process you will very likely do well, or at least as well as possible. Note that I use the term learning, not teaching – this is to emphasize from the beginning that learning is a choice and that you are in control. Learning is active; being taught is passive. It is up to you to seize control of your own educational process and fully participate, not sit back and wait for knowledge to be forcibly injected into your brain.
You may find yourself stuck in a course that is taught in a traditional way, by an instructor that lectures, assigns some readings, and maybe on a good day puts on a little dog-and-pony show in the classroom with some audiovisual aids or some demonstrations. The standard expectation in this class is to sit in your chair and watch, passive, taking notes. No real engagement is “required” by the instructor, and lacking activities or a structure that encourages it, you lapse into becoming a lecture transcription machine, recording all kinds of things that make no immediate sense to you and telling yourself that you’ll sort it all out later.
You may find yourself floundering in such a class – for good reason. The instructor presents an ocean of material in each lecture, and you’re going to actually retain at most a few cupfuls of it functioning as a scribe and passively copying his pictures and symbols without first extracting their sense. And the lecture makes little sense, at least at first, and reading (if you do any reading at all) does little to help. Demonstrations can sometimes make one or two ideas come clear, but only at the expense of twenty other things that the instructor now has no time to cover and expects you to get from the readings alone. You continually postpone going over the lectures and readings to understand the material any more than is strictly required to do the homework, until one day a big test draws nigh and you realize that you really don’t understand anything and have forgotten most of what you did, briefly, understand. Doom and destruction loom.
Sound familiar?
On the other hand, you may be in a course where the instructor has structured the course with a balanced mix of open lecture (held as a freeform discussion where questions aren’t just encouraged but required) and group interactive learning situations such as a carefully structured recitation and lab where discussion and doing blend together, where students teach each other and use what they have learned in many ways and contexts. If so, you’re lucky, but luck only goes so far.
3

4

Preliminaries

Even in a course like this you may still be floundering because you may not understand why it is important for you to participate with your whole spirit in the quest to learn anything you ever choose to study. In a word, you simply may not give a rodent’s furry behind about learning the material so that studying is always a fight with yourself to “make” yourself do it – so that no matter what happens, you lose. This too may sound very familiar to some.
The importance of engagement and participation in “active learning” (as opposed to passively being taught) is not really a new idea. Medical schools were four year programs in the year 1900.
They are four year programs today, where the amount of information that a physician must now master in those four years is probably ten times greater today than it was back then. Medical students are necessarily among the most efficient learners on earth, or they simply cannot survive.
In medical schools, the optimal learning strategy is compressed to a three-step adage: See one, do one, teach one.
See a procedure (done by a trained expert).
Do the procedure yourself, with the direct supervision and guidance of a trained expert.
Teach a student to do the procedure.
See, do, teach. Now you are a trained expert (of sorts), or at least so we devoutly hope, because that’s all the training you are likely to get until you start doing the procedure over and over again with real humans and with limited oversight from an attending physician with too many other things to do. So you practice and study on your own until you achieve real mastery, because a mistake can kill somebody.
This recipe is quite general, and can be used to increase your own learning in almost any class.
In fact, lifelong success in learning with or without the guidance of a good teacher is a matter of discovering the importance of active engagement and participation that this recipe (non-uniquely) encodes. Let us rank learning methodologies in terms of “probable degree of active engagement of the student”. By probable I mean the degree of active engagement that I as an instructor have observed in students over many years and which is significantly reinforced by research in teaching methodology, especially in physics and mathematics.
Listening to a lecture as a transcription machine with your brain in “copy machine” mode is almost entirely passive and is for most students probably a nearly complete waste of time. That’s not to say that “lecture” in the form of an organized presentation and review of the material to be learned isn’t important or is completely useless! It serves one very important purpose in the grand scheme of learning, but by being passive during lecture you cause it to fail in its purpose. Its purpose is not to give you a complete, line by line transcription of the words of your instructor to ponder later and alone. It is to convey, for a brief shining moment, the sense of the concepts so that you understand them.
It is difficult to sufficiently emphasize this point. If lecture doesn’t make sense to you when the instructor presents it, you will have to work much harder to achieve the sense of the material “later”, if later ever comes at all. If you fail to identify the important concepts during the presentation and see the lecture as a string of disconnected facts, you will have to remember each fact as if it were an abstract string of symbols, placing impossible demands on your memory even if you are extraordinarily bright. If you fail to achieve some degree of understanding (or synthesis of the material, if you prefer) in lecture by asking questions and getting expert explanations on the spot, you will have to build it later out of your notes on a set of abstract symbols that made no sense to you at the time. You might as well be trying to translate Egyptian Hieroglyphs without a Rosetta
Stone, and the best of luck to you with that.
Reading is a bit more active – at the very least your brain is more likely to be somewhat engaged if you aren’t “just” transcribing the book onto a piece of paper or letting the words and symbols happen in your mind – but is still pretty passive. Even watching nifty movies or cool-ee-oh demonstrations

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is basically sedentary – you’re still just sitting there while somebody or something else makes it all happen in your brain while you aren’t doing much of anything. At best it grabs your attention a bit better (on average) than lecture, but you are mentally passive.
In all of these forms of learning, the single active thing you are likely to be doing is taking notes or moving an eye muscle from time to time. For better or worse, the human brain isn’t designed to learn well in passive mode. Parts of your brain are likely to take charge and pull your eyes irresistably to the window to look outside where active things are going on, things that might not be so damn boring!
With your active engagement, with your taking charge of and participating in the learning process, things change dramatically. Instead of passively listening in lecture, you can at least try to ask questions and initiate discussions whenever an idea is presented that makes no intial sense to you.
Discussion is an active process even if you aren’t the one talking at the time. You participate! Even a tiny bit of participation in a classroom setting where students are constantly asking questions, where the instructor is constantly answering them and asking the students questions in turn makes a huge difference. Humans being social creatures, it also makes the class a lot more fun!
In summary, sitting on your ass1 and writing meaningless (to you, so far) things down as somebody says them in the hopes of being able to “study” them and discover their meaning on your own later is boring and for most students, later never comes because you are busy with many classes, because you haven’t discovered anything beautiful or exciting (which is the reward for figuring it all out – if you ever get there) and then there is partying and hanging out with friends and having fun. Even if you do find the time and really want to succeed, in a complicated subject like physics you are less likely to be able to discover the meaning on your own (unless you are so bright that learning methodology is irrelevant and you learn in a single pass no matter what). Most introductory students are swamped by the details, and have small chance of discovering the patterns within those details that constitute “making sense” and make the detailed information much, much easier to learn by enabling a compression of the detail into a much smaller set of connected ideas.
Articulation of ideas, whether it is to yourself or to others in a discussion setting, requires you to create tentative patterns that might describe and organize all the details you are being presented with. Using those patterns and applying them to the details as they are presented, you naturally encounter places where your tentative patterns are wrong, or don’t quite work, where something
“doesn’t make sense”. In an “active” lecture students participate in the process, and can ask questions and kick ideas around until they do make sense. Participation is also fun and helps you pay far more attention to what’s going on than when you are in passive mode. It may be that this increased attention, this consideration of many alternatives and rejecting some while retaining others with social reinforcement, is what makes all the difference. To learn optimally, even “seeing” must be an active process, one where you are not a vessel waiting to be filled through your eyes but rather part of a team studying a puzzle and looking for the patterns together that will help you eventually solve it.
Learning is increased still further by doing, the very essence of activity and engagement. “Doing” varies from course to course, depending on just what there is for you to do, but it always is the application of what you are learning to some sort of activity, exercise, problem. It is not just a recapitulation of symbols: “looking over your notes” or “(re)reading the text”. The symbols for any given course of study (in a physics class, they very likely will be algebraic symbols for real although
I’m speaking more generally here) do not, initially, mean a lot to you. If I write F = q(v × B) on the board, it means a great deal to me, but if you are taking this course for the first time it probably means zilch to you, and yet I pop it up there, draw some pictures, make some noises that hopefully make sense to you at the time, and blow on by. Later you read it in your notes to try to recreate that sense, but you’ve forgotten most of it. Am I describing the income I expect to make selling B
1I

mean, of course, your donkey. What did you think I meant?

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tons of barley with a market value of v and a profit margin of q?
To learn this expression (for yes, this is a force law of nature and one that we very much must learn this semester) we have to learn what the symbols stand for – q is the charge of a point-like object in motion at velocity v in a magnetic field B, and F is the resulting force acting on the particle. We have to learn that the × symbol is the cross product of evil (to most students at any rate, at least at first). In order to get a gut feeling for what this equation represents, for the directions associated with the cross product, for the trajectories it implies for charged particles moving in a magnetic field in a variety of contexts one has to use this expression to solve problems, see this expression in action in laboratory experiments that let you prove to yourself that it isn’t bullshit and that the world really does have cross product force laws in it. You have to do your homework that involves this law, and be fully engaged.
The learning process isn’t exactly linear, so if you participate fully in the discussion and the doing while going to even the most traditional of lectures, you have an excellent chance of getting to the point where you can score anywhere from a 75% to an 85% in the course. In most schools, say a C+ to B+ performance. Not bad, but not really excellent. A few students will still get A’s – they either work extra hard, or really like the subject, or they have some sort of secret, some way of getting over that barrier at the 90’s that is only crossed by those that really do understand the material quite well.
Here is the secret for getting yourself over that 90% hump, even in a physics class (arguably one of the most difficult courses you can take in college), even if you’re not a super-genius (or have never managed in the past to learn like one, a glance and you’re done): Work in groups!
That’s it. Nothing really complex or horrible, just get together with your friends who are also taking the course and do your homework together. In a well designed physics course (and many courses in mathematics, economics, and other subjects these days) you’ll have some aspects of the class, such as a recitation or lab, where you are required to work in groups, and the groups and group activities may be highly structured or freeform. “Studio” or “Team Based Learning” methods for teaching physics have even wrapped the lecture itself into a group-structured setting, so everything is done in groups/teams, and (probably by making it nearly impossible to be disengaged and sit passively in class waiting for learning to “happen”) this approach yields measureable improvements
(all things being equal) on at least some objective instruments for measurement of learning.
If you take charge of your own learning, though, you will quickly see that in any course, however taught, you can study in a group! This is true even in a course where “the homework” is to be done alone by fiat of the (unfortunately ignorant and misguided) instructor. Just study “around” the actual assignment – assign yourselves problems “like” the actual assignment – most textbooks have plenty of extra problems and then there is the Internet and other textbooks – and do them in a group, then (afterwards!) break up and do your actual assignment alone. Note that if you use a completely different textbook to pick your group problems from and do them together before looking at your assignment in your textbook, you can’t even be blamed if some of the ones you pick turn out to be ones your instructor happened to assign.
Oh, and not-so-subtly – give the instructor a PDF copy of this book (it’s free for instructors, after all, and a click away on the Internet) and point to this page and paragraph containing the following little message from me to them:

Yo! Teacher! Let’s wake up and smell the coffee! Don’t prevent your students from doing homework in groups – require it! Make the homework correspondingly more difficult!
Give them quite a lot of course credit for doing it well! Construct a recitation or review session where students – in groups – who still cannot get the most difficult problems can get socratic tutorial help after working hard on the problems on their own! Integrate discussion and deliberately teach to increase active engagement (instead of passive

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wandering attention) in lecture2 . Then watch as student performance and engagement spirals into the stratosphere compared to what it was before...
Then pray. Some instructors have their egos tied up in things to the point where they cannot learn, and then what can you do? If an instructor lets ego or politics obstruct their search for functional methodology, you’re screwed anyway, and you might as well just tackle the material on your own. Or heck, maybe their expertise and teaching experience vastly exceeds my own so that their naked words are sufficiently golden that any student should be able to learn by just hearing them and doing homework all alone in isolation from any peer-interaction process that might be of use to help them make sense of it all – all data to the contrary.
My own words and lecture – in spite of my 31 years of experience in the classroom, in spite of the fact that it has been well over twenty years since I actually used lecture notes to teach the course, in spite of the fact I never, ever prepare for recitation because solving the homework problems with the students “cold” as a peer member of their groups is useful where copying my privately worked out solutions onto a blackboard for them to passively copy on their papers in turn is useless, in spite of the fact that I wrote this book similarly without the use of any outside resource – my words and lecture are not. On the other hand, students who work effectively in groups and learn to use this book (and other resources) and do all of the homework “to perfection” might well learn physics quite well without my involvement at all!
Let’s understand why working in groups has such a dramatic effect on learning. What happens in a group? Well, a lot of discussion happens, because humans working on a common problem like to talk. There is plenty of doing going on, presuming that the group has a common task list to work through, like a small mountain of really difficult problems that nobody can possibly solve working on their own and are barely within their abilities working as a group backed up by the course instructor!
Finally, in a group everybody has the opportunity to teach!
The importance of teaching – not only seeing the lecture presentation with your whole brain actively engaged and participating in an ongoing discussion so that it makes sense at the time, not only doing lots of homework problems and exercises that apply the material in some way, but articulating what you have discovered in this process and answering questions that force you to consider and reject alternative solutions or pathways (or not) cannot be overemphasized. Teaching each other in a peer setting (ideally with mentorship and oversight to keep you from teaching each other mistakes) is essential!
This problem you “get”, and teach others (and actually learn it better from teaching it than they do from your presentation – never begrudge the effort required to teach your group peers even if some of them are very slow to understand). The next problem you don’t get but some other group member does – they get to teach you. In the end you all learn far more about every problem as a consequence of the struggle, the exploration of false paths, the discovery and articulation of the correct path, the process of discussion, resolution and agreement in teaching whereby everybody in the group reaches full understanding.
I would assert that it is all but impossible for someone to become a (halfway decent) teacher of anything without learning along the way that the absolute best way to learn any set of material deeply is to teach it – it is the very foundation of Academe and has been for two or three thousand
2 Perhaps by using Team Based Learning methods to structure and balance student groups and “flipping” classrooms to foist the lecture off onto videos of somebody else lecturing to increase the time spent in the class working in groups, but I’ve found that in mid-sized classes and smaller (less than around fifty students) one can get very good results from traditional lecture without a specially designed classroom by the Chocolate Method – I lecture without notes and offer a piece of chocolate or cheap toy or nifty pencil to any student who catches me making a mistake on the board before I catch it myself, who asks a particularly good question, who looks like they are nodding off to sleep (seriously, chocolate works wonders here, especially when ceremoniously offered). Anything that keeps students focussed during lecture by making it into a game, by allowing/encouraging them to speak out without raising their hands, by praising them and rewarding them for engagement makes a huge difference.

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years. It is, as we have noted, built right into the intensive learning process of medical school and graduate school in general. For some reason, however, we don’t incorporate a teaching component in most undergraduate classes, which is a shame, and it is basically nonexistent in nearly all K-12 schools, which is an open tragedy.
As an engaged student you don’t have to live with that! Put it there yourself, by incorporating group study and mutual teaching into your learning process with or without the help or permission of your teachers! A really smart and effective group soon learns to iterate the teaching – I teach you, and to make sure you got it you immediately use the material I taught you and try to articulate it back to me. Eventually everybody in the group understands, everybody in the group benefits, everybody in the group gets the best possible grade on the material. This process will actually make you (quite literally) more intelligent. You may or may not become smart enough to lock down an
A, but you will get the best grade you are capable of getting, for your given investment of effort.
This is close to the ultimate in engagement – highly active learning, with all cylinders of your brain firing away on the process. You can see why learning is enhanced. It is simply a bonus, a sign of a just and caring God, that it is also a lot more fun to work in a group, especially in a relaxed context with food and drink present. Yes, I’m encouraging you to have “physics study parties” (or history study parties, or psychology study parties). Hold contests. Give silly prizes. See. Do. Teach.

Other Conditions for Learning
Learning isn’t only dependent on the engagement pattern implicit in the See, Do, Teach rule. Let’s absorb a few more True Facts about learning, in particular let’s come up with a handful of things that can act as “switches” and turn your ability to learn on and off quite independent of how your instructor structures your courses. Most of these things aren’t binary switches – they are more like dimmer switches that can be slid up between dim (but not off) and bright (but not fully on). Some of these switches, or environmental parameters, act together more powerfully than they act alone.
We’ll start with the most important pair, a pair that research has shown work together to potentiate or block learning.
Instead of just telling you what they are, arguing that they are important for a paragraph or six, and moving on, I’m going to give you an early opportunity to practice active learning in the context of reading a chapter on active learning. That is, I want you to participate in a tiny mini-experiment.
It works a little bit better if it is done verbally in a one-on-one meeting, but it should still work well enough even if it is done in this text that you are reading.
I’m going to give you a string of ten or so digits and ask you to glance at it one time for a count of three and then look away. No fair peeking once your three seconds are up! Then I want you to do something else for at least a minute – anything else that uses your whole attention and interrupts your ability to rehearse the numbers in your mind in the way that you’ve doubtless learned permits you to learn other strings of digits, such as holding your mind blank, thinking of the phone numbers of friends or your social security number. Even rereading this paragraph will do.
At the end of the minute, try to recall the number I gave you and write down what you remember.
Then turn back to right here and compare what you wrote down with the actual number.
Ready? (No peeking yet...) Set? Go!
Ok, here it is, in a footnote at the bottom of the page to keep your eye from naturally reading ahead to catch a glimpse of it while reading the instructions above3 .
How did you do?
If you are like most people, this string of numbers is a bit too long to get into your immediate
3 1357986420

(one, two, three, quit and do something else for one minute...)

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memory or visual memory in only three seconds. There was very little time for rehearsal, and then you went and did something else for a bit right away that was supposed to keep you from rehearsing whatever of the string you did manage to verbalize in three seconds. Most people will get anywhere from the first three to as many as seven or eight of the digits right, but probably not in the correct order, unless...
...they are particularly smart or lucky and in that brief three second glance have time to notice that the number consists of all the digits used exactly once! Folks that happened to “see” this at a glance probably did better than average, getting all of the correct digits but maybe in not quite the correct order.
People who are downright brilliant (and equally lucky) realized in only three seconds (without cheating an extra second or three, you know who you are) that it consisted of the string of odd digits in ascending order followed by the even digits in descending order. Those people probably got it all perfectly right even without time to rehearse and “memorize” the string! Look again at the string, see the pattern now?
The moral of this little mini-demonstration is that it is easy to overwhelm the mind’s capacity for processing and remembering “meaningless” or “random” information. A string of ten measely
(apparently) random digits is too much to remember for one lousy minute, especially if you aren’t given time to do rehearsal and all of the other things we have to make ourselves do to “memorize” meaningless information.
Of course things changed radically the instant I pointed out the pattern! At this point you could very likely go away and come back to this point in the text tomorrow or even a year from now and have an excellent chance of remembering this particular digit string, because it makes sense of a sort, and there are plenty of cues in the text to trigger recall of the particular pattern that “compresses and encodes” the actual string. You don’t have to remember ten random things at all – only two and a half – odd ascending digits followed by the opposite (of both). Patterns rock!
This example has obvious connections to lecture and class time, and is one reason retention from lecture is so lousy. For most students, lecture in any nontrivial college-level course is a long-running litany of stuff they don’t know yet. Since it is all new to them, it might as well be random digits as far as their cognitive abilities are concerned, at least at first. Sure, there is pattern there, but you have to discover the pattern, which requires time and a certain amount of meditation on all of the information. Basically, you have to have a chance for the pattern to jump out of the stream of information and punch the switch of the damn light bulb we all carry around inside our heads, the one that is endlessly portrayed in cartoons. That light bulb is real – it actually exists, in more than just a metaphorical sense – and if you study long enough and hard enough to obtain a sudden, epiphinaic realization in any topic you are studying, however trivial or complex (like the pattern exposed above) it is quite likely to be accompanied by a purely mental flash of “light”. You’ll know it when it happens to you, in other words, and it feels great.
Unfortunately, the instructor doesn’t usually give students a chance to experience this in lecture.
No sooner is one seemingly random factoid laid out on the table than along comes a new, apparently disconnected one that pushes it out of place long before we can either memorize it the hard way or make sense out of it so we can remember it with a lot less work. This isn’t really anybody’s fault, of course; the light bulb is quite unlikely to go off in lecture just from lecture no matter what you or the lecturer do – it is something that happens to the prepared mind at the end of a process, not something that just fires away every time you hear a new idea.
The humble and unsurprising conclusion I want you to draw from this silly little mini-experiment is that things are easier to learn when they make sense! A lot easier. In fact, things that don’t make sense to you are never “learned” – they are at best memorized. Information can almost always be compressed when you discover the patterns that run through it, especially when the patterns all fit together into the marvelously complex and beautiful and mysterious process we call “deep

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understanding” of some subject.
There is one more example I like to use to illustrate how important this information compression is to memory and intelligence. I play chess, badly. That is, I know the legal moves of the game, and have no idea at all how to use them effectively to improve my position and eventually win. Ten moves into a typical chess game I can’t recall how I got myself into the mess I’m typically in, and at the end of the game I probably can’t remember any of what went on except that I got trounced, again. A chess master, on the other hand, can play umpty games at once, blindfolded, against pitiful fools like myself and when they’ve finished winning them all they can go back and recontruct each one move by move, criticizing each move as they go. Often they can remember the games in their entirety days or even years later.
This isn’t just because they are smarter – they might be completely unable to derive the Lorentz group from first principles, and I can, and this doesn’t automatically make me smarter than them either. It is because chess makes sense to them – they’ve achieved a deep understanding of the game, as it were – and they’ve built a complex meta-structure memory in their brains into which they can poke chess moves so that they can be retrieved extremely efficiently. This gives them the attendant capability of searching vast portions of the game tree at a glance, where I have to tediously work through each branch, one step at a time, usually omitting some really important possibility because
I don’t realize that that knight on the far side of the board can affect things on this side where we are both moving pieces.
This sort of “deep” (synthetic) understanding of physics is very much the goal of this course (the one in the textbook you are reading, since I use this intro in many textbooks), and to achieve it you must not memorize things as if they are random factoids, you must work to abstract the beautiful intertwining of patterns that compress all of those apparently random factoids into things that you can easily remember offhand, that you can easily reconstruct from the pattern even if you forget the details, and that you can search through at a glance. But the process I describe can be applied to learning pretty much anything, as patterns and structure exist in abundance in all subjects of interest. There are even sensible rules that govern or describe the anti-pattern of pure randomness!
There’s one more important thing you can learn from thinking over the digit experiment. Some of you reading this very likely didn’t do what I asked, you didn’t play along with the game. Perhaps it was too much of a bother – you didn’t want to waste a whole minute learning something by actually doing it, just wanted to read the damn chapter and get it over with so you could do, well, whatever the hell else it is you were planning to do today that’s more important to you than physics or learning in other courses.
If you’re one of these people, you probably don’t remember any of the digit string at this point from actually seeing it – you never even tried to memorize it. A very few of you may actually be so terribly jaded that you don’t even remember the little mnemonic formula I gave above for the digit string (although frankly, people that are that disengaged are probably not about to do things like actually read a textbook in the first place, so possibly not). After all, either way the string is pretty damn meaningless, pattern or not.
Pattern and meaning aren’t exactly the same thing. There are all sorts of patterns one can find in random number strings, they just aren’t “real” (where we could wax poetic at this point about information entropy and randomness and monkeys typing Shakespeare if this were a different course). So why bother wasting brain energy on even the easy way to remember this string when doing so is utterly unimportant to you in the grand scheme of all things?
From this we can learn the second humble and unsurprising conclusion I want you to draw from this one elementary thought experiment. Things are easier to learn when you care about learning them! In fact, they are damn near impossible to learn if you really don’t care about learning them.

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Let’s put the two observations together and plot them as a graph, just for fun (and because graphs help one learn for reasons we will explore just a bit in a minute). If you care about learning what you are studying, and the information you are trying to learn makes sense (if only for a moment, perhaps during lecture), the chances of your learning it are quite good. This alone isn’t enough to guarantee that you’ll learn it, but it they are basically both necessary conditions, and one of them is directly connected to degree of engagement.

Figure 1: Relation between sense, care and learning
On the other hand, if you care but the information you want to learn makes no sense, or if it makes sense but you hate the subject, the instructor, your school, your life and just don’t care, your chances of learning it aren’t so good, probably a bit better in the first case than in the second as if you care you have a chance of finding someone or some way that will help you make sense of whatever it is you wish to learn, where the person who doesn’t cares, well, they don’t care. Why should they remember it?
If you don’t give a rat’s ass about the material and it makes no sense to you, go home. Leave school. Do something else. You basically have almost no chance of learning the material unless you are gifted with a transcendent intelligence (wasted on a dilettante who lives in a state of perpetual ennui) and are miraculously gifted with the ability learn things effortlessly even when they make no sense to you and you don’t really care about them. All the learning tricks and study patterns in the world won’t help a student who doesn’t try, doesn’t care, and for whom the material never makes sense. If we worked at it, we could probably find other “logistic” controlling parameters to associate with learning – things that increase your probability of learning monotonically as they vary. Some of

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them are already apparent from the discussion above. Let’s list a few more of them with explanations just so that you can see how easy it is to sit down to study and try to learn and have “something wrong” that decreases your ability to learn in that particular place and time.
Learning is actual work and involves a fair bit of biological stress, just like working out. Your brain needs food – it burns a whopping 20-30% of your daily calorie intake all by itself just living day to day, even more when you are really using it or are somewhat sedentary in your physical habits. Note that your brain runs on pure, energy-rich glucose, so when your blood sugar drops your brain activity drops right along with it. This can happen (paradoxically) because you just ate a carbohydrate rich meal. A balanced diet containing foods with a lower glycemic index4 tends to be harder to digest and provides a longer period of sustained energy for your brain. A daily multivitamin (and various antioxidant supplements such as alpha lipoic acid) can also help maintain your body’s energy release mechanisms at the cellular level.
Blood sugar is typically lowest first thing in the morning, so this is a lousy time to actively study. On the other hand, a good hearty breakfast, eaten at least an hour before plunging in to your studies, is a great idea and is a far better habit to develop for a lifetime than eating no breakfast and instead eating a huge meal right before bed.
Learning requires adequate sleep. Sure this is tough to manage at college – there are no parents to tell you to go to bed, lots of things to do, and of course you’re in class during the day and then you study, so late night is when you have fun. Unfortunately, learning is clearly correlated with engagement, activity, and mental alertness, and all of these tend to shut down when you’re tired.
Furthermore, the formation of long term memory of any kind from a day’s experiences has been shown in both animal and human studies to depend on the brain undergoing at least a few natural sleep cycles of deep sleep alternating with REM (Rapid Eye Movement) sleep, dreaming sleep. Rats taught a maze and then deprived of REM sleep cannot run the maze well the next day; rats that are taught the same maze but that get a good night’s of rat sleep with plenty of rat dreaming can run the maze well the next day. People conked on the head who remain unconscious for hours and are thereby deprived of normal sleep often have permanent amnesia of the previous day – it never gets turned into long term memory.
This is hardly surprising. Pure common sense and experience tell you that your brain won’t work too well if it is hungry and tired. Common sense (and yes, experience) will rapidly convince you that learning generally works better if you’re not stoned or drunk when you study. Learning works much better when you have time to learn and haven’t put everything off to the last minute. In fact, all of Maslow’s hierarchy of needs5 are important parameters that contribute to the probability of success in learning.
There is one more set of very important variables that strongly affect our ability to learn, and they are in some ways the least well understood. These are variables that describe you as an individual, that describe your particular brain and how it works. Pretty much everybody will learn better if they are self-actualized and fully and actively engaged, if the material they are trying to learn is available in a form that makes sense and clearly communicates the implicit patterns that enable efficient information compression and storage, and above all if they care about what they are studying and learning, if it has value to them.
But everybody is not the same, and the optimal learning strategy for one person is not going to be what works well, or even at all, for another. This is one of the things that confounds “simple” empirical research that attempts to find benefit in one teaching/learning methodology over another.
4 Wikipedia:

http://www.wikipedia.org/wiki/glycemic index. http://www.wikipedia.org/wiki/Maslow’s hierarchy of needs. In a nutshell, in order to become selfactualized and realize your full potential in activities such as learning you need to have your physiological needs met, you need to be safe, you need to be loved and secure in the world, you need to have good self-esteem and the esteem of others. Only then is it particularly likely that you can become self-actualized and become a great learner and problem solver.
5 Wikipedia:

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Some students do improve, even dramatically improve – when this or that teaching/learning methodology is introduced. In others there is no change. Still others actually do worse. In the end, the beneficial effect to a selected subgroup of the students may be lost in the statistical noise of the study and the fact that no attempt is made to identify commonalities among students that succeed or fail.
The point is that finding an optimal teaching and learning strategy is technically an optimization problem on a high dimensional space. We’ve discussed some of the important dimensions above, isolating a few that appear to have a monotonic effect on the desired outcome in at least some range
(relying on common sense to cut off that range or suggest trade-offs – one cannot learn better by simply discussing one idea for weeks at the expense of participating in lecture or discussing many other ideas of equal and coordinated importance; sleeping for twenty hours a day leaves little time for experience to fix into long term memory with all of that sleep). We’ve omitted one that is crucial, however. That is your brain!

Your Brain and Learning
Your brain is more than just a unique instrument. In some sense it is you. You could imagine having your brain removed from your body and being hooked up to machinary that provided it with sight, sound, and touch in such a way that “you” remain6 . It is difficult to imagine that you still exist in any meaningful sense if your brain is taken out of your body and destroyed while your body is artificially kept alive.
Your brain, however, is an instrument. It has internal structure. It uses energy. It does “work”.
It is, in fact, a biological machine of sublime complexity and subtlety, one of the true wonders of the world! Note that this statement can be made quite independent of whether “you” are your brain per se or a spiritual being who happens to be using it (a debate that need not concern us at this time, however much fun it might be to get into it) – either way the brain itself is quite marvelous.
For all of that, few indeed are the people who bother to learn to actually use their brain effectively as an instrument. It just works, after all, whether or not we do this. Which is fine. If you want to get the most mileage out of it, however, it helps to read the manual.
So here’s at least one user manual for your brain. It is by no means complete or authoritative, but it should be enough to get you started, to help you discover that you are actually a lot smarter than you think, or that you’ve been in the past, once you realize that you can change the way you think and learn and experience life and gradually improve it.
In the spirit of the learning methodology that we eventually hope to adopt, let’s simply itemize in no particular order the various features of the brain7 that bear on the process of learning. Bear in mind that such a minimal presentation is more of a metaphor than anything else because simple
(and extremely common) generalizations such as “creativity is a right-brain function” are not strictly true as the brain is far more complex than that.
• The brain is bicameral: it has two cerebral hemispheres8 , right and left, with brain functions asymmetrically split up between them.
• The brain’s hemispheres are connected by a networked membrane called the corpus callosum that is how the two halves talk to each other.
• The human brain consists of layers with a structure that recapitulates evolutionary phylogeny; that is, the core structures are found in very primitive animals and common to nearly all
6 Imagine

very easily if you’ve ever seen The Matrix movie trilogy... http://www.wikipedia.org/wiki/brain. 8 Wikipedia: http://www.wikipedia.org/wiki/cerebral hemisphere.
7 Wikipedia:

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Preliminaries vertebrate animals, with new layers (apparently) added by evolution on top of this core as the various phyla differentiated, fish, amphibian, reptile, mammal, primate, human. The outermost layer where most actual thinking occurs (in animals that think) is known as the cerebral cortex.
• The cerebral cortex9 – especially the outermost layer of it called the neocortex – is where
“higher thought” activities associated with learning and problem solving take place, although the brain is a very complex instrument with functions spread out over many regions.
• An important brain model is a neural network10 . Computer simulated neural networks provide us with insight into how the brain can remember past events and process new information.
• The fundamental operational units of the brain’s information processing functionality are called neurons11 . Neurons receive electrochemical signals from other neurons that are transmitted through long fibers called axons12 Neurotransmitters13 are the actual chemicals responsible for the triggered functioning of neurons and hence the neural network in the cortex that spans the halves of the brain.
• Parts of the cortex are devoted to the senses. These parts often contain a map of sorts of the world as seen by the associated sense mechanism. For example, there exists a topographic map in the brain that roughly corresponds to points in the retina, which in turn are stimulated by an image of the outside world that is projected onto the retina by your eye’s lens in a way we will learn about later in this course! There is thus a representation of your visual field laid out inside your brain!
• Similar maps exist for the other senses, although sensations from the right side of your body are generally processed in a laterally inverted way by the opposite hemisphere of the brain.
What your right eye sees, what your right hand touches, is ultimately transmitted to a sensory area in your left brain hemisphere and vice versa, and volitional muscle control flows from these brain halves the other way.
• Neurotransmitters require biological resources to produce and consume bioenergy (provided as glucose) in their operation. You can exhaust the resources, and saturate the receptors for the various neurotransmitters on the neurons by overstimulation.
• You can also block neurotransmitters by chemical means, put neurotransmitter analogues into your system, and alter the chemical trigger potentials of your neurons by taking various drugs, poisons, or hormones. The biochemistry of your brain is extremely important to its function, and (unfortunately) is not infrequently a bit “out of whack” for many individuals, resulting in e.g. attention deficit or mood disorders that can greatly affect one’s ability to easily learn while leaving one otherwise highly functional.
• Intelligence14 , learning ability, and problem solving capabilities are not fixed; they can vary
(often improving) over your whole lifetime! Your brain is highly plastic and can sometimes even reprogram itself to full functionality when it is e.g. damaged by a stroke or accident.
On the other hand neither is it infinitely plastic – any given brain has a range of accessible capabilities and can be improved only to a certain point. However, for people of supposedly
“normal” intelligence and above, it is by no means clear what that point is! Note well that intelligence is an extremely controversial subject and you should not take things like your own measured “IQ” too seriously.
9 Wikipedia:

10 Wikipedia:
11 Wikipedia:
12 Wikipedia:
13 Wikipedia:
14 Wikipedia:

http://www.wikipedia.org/wiki/Cerebral cortex. http://www.wikipedia.org/wiki/Neural network. http://www.wikipedia.org/wiki/Neurons. http://www.wikipedia.org/wiki/axon. . http://www.wikipedia.org/wiki/neurotransmitters. http://www.wikipedia.org/wiki/intelligence.

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• Intelligence is not even fixed within a population over time. A phenomenon known as “the
Flynn effect”15 (after its discoverer) suggests that IQ tests have increased almost six points a decade, on average, over a timescale of tens of years, with most of the increases coming from the lower half of the distribution of intelligence. This is an active area of research (as one might well imagine) and some of that research has demonstrated fairly conclusively that individual intelligences can be improved by five to ten points (a significant amount) by environmentally correlated factors such as nutrition, education, complexity of environment.
• The best time for the brain to learn is right before sleep. The process of sleep appears to
“fix” long term memories in the brain and things one studies right before going to bed are retained much better than things studied first thing in the morning. Note that this conflicts directly with the party/entertainment schedule of many students, who tend to study early in the evening and then amuse themselves until bedtime. It works much better the other way around. • Sensory memory16 corresponds to the roughly 0.5 second (for most people) that a sensory impression remains in the brain’s “active sensory register”, the sensory cortex. It can typically hold less than 12 “objects” that can be retrieved. It quickly decays and cannot be improved by rehearsal, although there is some evidence that its object capacity can be improved over a longer term by practice.
• Short term memory is where some of the information that comes into sensory memory is transferred. Just which information is transferred depends on where one’s “attention” is, and the mechanics of the attention process are not well understood and are an area of active research. Attention acts like a filtering process, as there is a wealth of parallel information in our sensory memory at any given instant in time but the thread of our awareness and experience of time is serial. We tend to “pay attention” to one thing at a time. Short term memory lasts from a few seconds to as long as a minute without rehearsal, and for nearly all people it holds
4 − 5 objects17 . However, its capacity can be increased by a process called “chunking” that is basically the information compression mechanism demonstrated in the earlier example with numbers – grouping of the data to be recalled into “objects” that permit a larger set to still fit in short term memory.
• Studies of chunking show that the ideal size for data chunking is three. That is, if you try to remember the string of letters:
FBINSACIAIBMATTMSN
with the usual three second look you’ll almost certainly find it impossible. If, however, I insert the following spaces:
FBI NSA CIA IBM ATT MSN
It is suddenly much easier to get at least the first four. If I parenthesize:
(FBI NSA CIA) (IBM ATT MSN) so that you can recognize the first three are all government agencies in the general category of
“intelligence and law enforcement” and the last three are all market symbols for information technology mega-corporations, you can once again recall the information a day later with only the most cursory of rehearsals. You’ve taken eighteen ”random” objects that were meaningless and could hence be recalled only through the most arduous of rehearsal processes, converted them to six “chunks” of three that can be easily tagged by the brain’s existing long term memory (note that you are not learning the string FBI, you are building an association to the
15 Wikipedia:

http://www.wikipedia.org/wiki/flynn effect. http://www.wikipedia.org/wiki/memory. Several items in a row are connected to this page.
17 From this you can see why I used ten digits, gave you only a few seconds to look, and blocked rehearsal in our earlier exercise.
16 Wikipedia:

16

Preliminaries already existing memory of what the string FBI means, which is much easier for the brain to do), and chunking the chunks into two objects.
Eighteen objects without meaning – difficult indeed! Those same eighteen objects with meaning
– umm, looks pretty easy, doesn’t it...
Short term memory is still that – short term. It typically decays on a time scale that ranges from minutes for nearly everything to order of a day for a few things unless the information can be transferred to long term memory. Long term memory is the big payoff – learning is associated with formation of long term memory.
• Now we get to the really good stuff. Long term is memory that you form that lasts a long time in human terms. A “long time” can be days, weeks, months, years, or a lifetime. Long term memory is encoded completely differently from short term or sensory/immediate memory
– it appears to be encoded semantically18 , that is to say, associatively in terms of its meaning.
There is considerable evidence for this, and it is one reason we focus so much on the importance of meaning in the previous sections.
To miraculously transform things we try to remember from “difficult” to learn random factoids that have to be brute-force stuffed into disconnected semantic storage units created as it were one at a time for the task at hand into “easy” to learn factoids, all we have to do is discover meaning associations with things we already know, or create a strong memory of the global meaning or conceptualization of a subject that serves as an associative home for all those little factoids. A characteristic of this as a successful process is that when one works systematically to learn by means of the latter process, learning gets easier as time goes on. Every factoid you add to the semantic structure of the global conceptualization strengthens it, and makes it even easier to add new factoids. In fact, the mind’s extraordinary rational capacity permits it to interpolate and extrapolate, to fill in parts of the structure on its own without effort and in many cases without even being exposed to the information that needs to be “learned”!
• One area where this extrapolation is particularly evident and powerful is in mathematics. Any time we can learn, or discover from experience a formula for some phenomenon, a mathematical pattern, we don’t have to actually see something to be able to “remember” it. Once again, it is easy to find examples. If I give you data from sales figures over a year such as January
= $1000, October = $10,000, December = $12,000, March=$3000, May = $5000, February
= $2000, September = $9000, June = $6000, November = $11,000, July = $7000, August =
$8000, April = $4000, at first glance they look quite difficult to remember. If you organize them temporally by month and look at them for a moment, you recognize that sales increased linearly by month, starting at $1000 in January, and suddenly you can reduce the whole series to a simple mental formula (straight line) and a couple pieces of initial data (slope and starting point). One amazing thing about this is that if I asked you to “remember” something that you have not seen, such as sales in February in the next year, you could make a very plausible guess that they will be $14,000!
Note that this isn’t a memory, it is a guess. Guessing is what the mind is designed to do, as it is part of the process by which it “predicts the future” even in the most mundane of ways.
When I put ten dollars in my pocket and reach in my pocket for it later, I’m basically guessing, on the basis of my memory and experience, that I’ll find ten dollars there. Maybe my guess is wrong – my pocket could have been picked19 , maybe it fell out through a hole. My concept of object permanence plus my memory of an initial state permit me to make a predictive guess about the Universe!

18 Wikipedia:
19 With

http://www.wikipedia.org/wiki/semantics. three sons constantly looking for funds to attend movies and the like, it isn’t as unlikely as you might think!

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This is, in fact, physics! This is what physics is all about – coming up with a set of rules (like conservation of matter) that encode observations of object permanence, more rules (equations of motion) that dictate how objects move around, and allow me to conclude that “I put a ten dollar bill, at rest, into my pocket, and objects at rest remain at rest. The matter the bill is made of cannot be created or destroyed and is bound together in a way that is unlikely to come apart over a period of days. Therefore the ten dollar bill is still there!” Nearly anything that you do or that happens in your everyday life can be formulated as a predictive physics problem. • The hippocampus20 appears to be partly responsible for both forming spatial maps or visualizations of your environment and also for forming the cognitive map that organizes what you know and transforms short term memory into long term memory, and it appears to do its job
(as noted above) in your sleep. Sleep deprivation prevents the formation of long term memory.
Being rendered unconscious for a long period often produces short term amnesia as the brain loses short term memory before it gets put into long term memory. The hippocampus shows evidence of plasticity – taxi drivers who have to learn to navigate large cities actually have larger than normal hippocampi, with a size proportional to the length of time they’ve been driving. This suggests (once again) that it is possible to deliberately increase the capacity of your own hippocampus through the exercise of its functions, and consequently increase your ability to store and retrieve information, which is an important component (although not the only component) of intelligence!
• Memory is improved by increasing the supply of oxygen to the brain, which is best accomplished by exercise. Unsurprisingly. Indeed, as noted above, having good general health, good nutrition, good oxygenation and perfusion – having all the biomechanism in tip-top running order – is perfectly reasonably linked to being able to perform at your best in anything, mental activity included.
• Finally, the amygdala21 is a brain organ in our limbic system (part of our “old”, reptile brain).
The amygdala is an important part of our emotional system. It is associated with primitive survival responses, with sexual response, and appears to play a key role in modulating (filtering) the process of turning short term memory into long term memory. Basically, any short term memory associated with a powerful emotion is much more likely to make it into long term memory. There are clear evolutionary advantages to this. If you narrowly escape being killed by a saber-toothed tiger at a particular pool in the forest, and then forget that this happened by the next day and return again to drink there, chances are decent that the saber-tooth is still there and you’ll get eaten. On the other hand, if you come upon a particular fruit tree in that same forest and get a free meal of high quality food and forget about the tree a day later, you might starve.
We see that both negative and positive emotional experiences are strongly correlated with learning! Powerful experiences, especially, are correlated with learning. This translates into learning strategies in two ways, one for the instructor and one for the student. For the instructor, there are two general strategies open to helping students learn. One is to create an atmosphere of fear, hatred, disgust, anger – powerful negative emotions. The other is to create an atmosphere of love, security, humor, joy – powerful positive emotions. In between there is a great wasteland of bo-ring, bo-ring, bo-ring where students plod along, struggling to form memories because there is nothing “exciting” about the course in either a positive or negative way and so their amygdala degrades the memory formation process in favor of other more
“interesting” experiences.
20 Wikipedia:
21 Wikipedia:

http://www.wikipedia.org/wiki/hippocampus. http://www.wikipedia.org/wiki/amygdala. 18

Preliminaries
Now, in my opinion, negative experiences in the classroom do indeed promote the formation of long term memories, but they aren’t the memories the instructor intended. The student is likely to remember, and loath, the instructor for the rest of their life but is not more likely to remember the material except sporadically in association with particularly traumatic episodes.
They may well be less likely, as we naturally avoid negative experiences and will study less and work less hard on things we can’t stand doing.
For the instructor, then, positive is the way to go. Creating a warm, nurturing classroom environment, ensuring that the students know that you care about their learning and about them as individuals helps to promote learning. Making your lectures and teaching processes fun – and funny – helps as well. Many successful lecturers make a powerful positive impression on the students, creating an atmosphere of amazement or surprise. A classroom experience should really be a joy in order to optimize learning in so many ways.
For the student, be aware that your attitude matters! As noted in previous sections, caring is an essential component of successful learning because you have to attach value to the process in order to get your amygdala to do its job. However, you can do much more. You can see how many aspects of learning can be enhanced through the simple expedient of making it a positive experience! Working in groups, working with a team of peers, is fun, and you learn more when you’re having fun (or quavering in abject fear, or in an interesting mix of the two).
Attending an interesting lecture is fun, and you’ll retain more than average. Participation is fun, especially if you are “rewarded” in some way that makes a moment or two special to you, and you’ll remember more of what goes on.
Chicken or egg? We see a fellow student who is relaxed and appears to be having fun because they are doing really well in the course where we are constantly stressed out and struggling, and we write their happiness off as being due to their success and our misery as being caused by our failure. It is possible, however, that we have this backwards! Perhaps they are doing really well in the course because they are relaxed and having fun, perhaps we are doing not so well because for us, every minute in the classroom is a torture!
In any event, you’ve probably tried misery in the classroom in at least one class already. How’d that work out for you? Perhaps it is worth trying joy, instead!

From all of these little factoids (presented in a way that I’m hoping helps you to build at least the beginnings of a working conceptual model of your own brain) I’m hoping that you are coming to realize that all of this is at least partially under your control! Even if your instructor is scary or boring, the material at first glance seems dry and meaningless, and so on – all the negative-neutral things that make learning difficult, you can decide to make it fun and exciting, you can ferret out the meaning, you can adopt study strategies that focus on the formation of cognitive maps and organizing structures first and then on applications, rehearsal, factoids, and so on, you can learn to study right before bed, get enough sleep, become aware of your brain’s learning biorhythms.
Finally, you can learn to increase your functional learning capabilities by a significant amount.
Solving puzzles, playing mental games, doing crossword puzzles or sudoku, working homework problems, writing papers, arguing and discussing, just plain thinking about difficult subjects and problems even when you don’t have to all increase your active intelligence in initially small but cumulative ways. You too can increase the size of your hippocampus by navigating a new subject instead of a city, you too can learn to engage your amygdala by choosing in a self-actualized way what you value and learning to discipline your emotions accordingly, you too can create more conceptual maps within your brain that can be shared as components across the various things you wish to learn.
The more you know about anything, the easier it is to learn everything – this is the pure biology underlying the value of the liberal arts education.
Use your whole brain, exercise it often, don’t think that you “just” need math and not spatial relations, visualization, verbal skills, a knowledge of history, a memory of performing experiments

Preliminaries

19

with your hands or mind or both – you need it all! Remember, just as is the case with physical exercise (which you should get plenty of), mental exercise gradually makes you mentally stronger, so that you can eventually do easily things that at first appear insurmountably difficult. You can learn to learn three to ten times as fast as you did in high school, to have more fun while doing it, and to gain tremendous reasoning capabilities along the way just by trying to learn to learn more efficiently instead of continuing to use learning strategies that worked (possibly indifferently) back in elementary and high school.
The next section, at long last, will make a very specific set of suggestions for one very good way to study physics (or nearly anything else) in a way that maximally takes advantage of your own volitional biology to make learning as efficient and pleasant as it is possible to be.

How to Do Your Homework Effectively
By now in your academic career (and given the information above) it should be very apparent just where homework exists in the grand scheme of (learning) things. Ideally, you attend a class where a warm and attentive professor clearly explains some abstruse concept and a whole raft of facts in some moderately interactive way that encourages engagement and “being earnest”. Alas, there are too many facts to fit in short term/immediate memory and too little time to move most of them through into long term/working memory before finishing with one and moving on to the next one.
The material may appear to be boring and random so that it is difficult to pay full attention to the patterns being communicated and remain emotionally enthusiastic all the while to help the process along. As a consequence, by the end of lecture you’ve already forgotten many if not most of the facts, but if you were paying attention, asked questions as needed, and really cared about learning the material you would remember a handful of the most important ones, the ones that made your brief understanding of the material hang (for a brief shining moment) together.
This conceptual overview, however initially tenuous, is the skeleton you will eventually clothe with facts and experiences to transform it into an entire system of associative memory and reasoning where you can work intellectually at a high level with little effort and usually with a great deal of pleasure associated with the very act of thinking. But you aren’t there yet.
You now know that you are not terribly likely to retain a lot of what you are shown in lecture without engagement. In order to actually learn it, you must stop being a passive recipient of facts.
You must actively develop your understanding, by means of discussing the material and kicking it around with others, by using the material in some way, by teaching the material to peers as you come to understand it.
To help facilitate this process, associated with lecture your professor almost certainly gave you an assignment. Amazingly enough, its purpose is not to torment you or to be the basis of your grade
(although it may well do both). It is to give you some concrete stuff to do while thinking about the material to be learned, while discussing the material to be learned, while using the material to be learned to accomplish specific goals, while teaching some of what you figure out to others who are sharing this whole experience while being taught by them in turn. The assignment is much more important than lecture, as it is entirely participatory, where real learning is far more likely to occur.
You could, once you learn the trick of it, blow off lecture and do fine in a course in all other respects.
If you fail to do the assignments with your entire spirit engaged, you are doomed.
In other words, to learn you must do your homework, ideally at least partly in a group setting.
The only question is: how should you do it to both finish learning all that stuff you sort-of-got in lecture and to re-attain the moment(s) of clarity that you then experienced, until eventually it becomes a permanent characteristic of your awareness and you know and fully understand it all on your own?

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There are two general steps that need to be iterated to finish learning anything at all. They are a lot of work. In fact, they are far more work than (passively) attending lecture, and are more important than attending lecture. You can learn the material with these steps without ever attending lecture, as long as you have access to what you need to learn in some media or human form. You in all probability will never learn it, lecture or not, without making a few passes through these steps.
They are:
a) Review the whole (typically textbooks and/or notes)
b) Work on the parts (do homework, use it for something)
(iterate until you thoroughly understand whatever it is you are trying to learn).
Let’s examine these steps.
The first is pretty obvious. You didn’t “get it” from one lecture. There was too much material.
If you were lucky and well prepared and blessed with a good instructor, perhaps you grasped some of it for a moment (and if your instructor was poor or you were particularly poorly prepared you may not have managed even that) but what you did momentarily understand is fading, flitting further and further away with every moment that passes. You need to review the entire topic, as a whole, as well as all its parts. A set of good summary notes might contain all the relative factoids, but there are relations between those factoids – a temporal sequencing, mathematical derivations connecting them to other things you know, a topical association with other things that you know. They tell a story, or part of a story, and you need to know that story in broad terms, not try to memorize it word for word.
Reviewing the material should be done in layers, skimming the textbook and your notes, creating a new set of notes out of the text in combination with your lecture notes, maybe reading in more detail to understand some particular point that puzzles you, reworking a few of the examples presented.
Lots of increasingly deep passes through it (starting with the merest skim-reading or reading a summary of the whole thing) are much better than trying to work through the whole text one line at a time and not moving on until you understand it. Many things you might want to understand will only come clear from things you are exposed to later, as it is not the case that all knowledge is ordinal, hierarchical, and derivatory.
You especially do not have to work on memorizing the content. In fact, it is not desireable to try to memorize content at this point – you want the big picture first so that facts have a place to live in your brain. If you build them a house, they’ll move right in without a fuss, where if you try to grasp them one at a time with no place to put them, they’ll (metaphorically) slip away again as fast as you try to take up the next one. Let’s understand this a bit.
As we’ve seen, your brain is fabulously efficient at storing information in a compressed associative form. It also tends to remember things that are important – whatever that means – and forget things that aren’t important to make room for more important stuff, as your brain structures work together in understandable ways on the process. Building the cognitive map, the “house”, is what it’s all about. But as it turns out, building this house takes time.
This is the goal of your iterated review process. At first you are memorizing things the hard way, trying to connect what you learn to very simple hierarchical concepts such as this step comes before that step. As you do this over and over again, though, you find that absorbing new information takes you less and less time, and you remember it much more easily and for a longer time without additional rehearsal. Sometimes your brain even outruns the learning process and “discovers” a missing part of the structure before you even read about it! By reviewing the whole, well-organized structure over and over again, you gradually build a greatly compressed representation of it in your brain and tremendously reduce the amount of work required to flesh out that structure with increasing levels of detail and remember them and be able to work with them for a long, long time.

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21

Now let’s understand the second part of doing homework – working problems. As you can probably guess on your own at this point, there are good ways and bad ways to do homework problems. The worst way to do homework (aside from not doing it at all, which is far too common a practice and a bad idea if you have any intention of learning the material) is to do it all in one sitting, right before it is due, and to never again look at it.
Doing your homework in a single sitting, working on it just one time fails to repeat and rehearse the material (essential for turning short term memory into long term in nearly all cases). It exhausts the neurons in your brain (quite literally – there is metabolic energy consumed in thinking) as one often ends up working on a problem far too long in one sitting just to get done. It fails to incrementally build up in your brain’s long term memory the structures upon which the more complex solutions are based, so you have to constantly go back to the book to get them into short term memory long enough to get through a problem. Even this simple bit of repetition does initiate a learning process. Unfortunately, by not repeating them after this one sitting they soon fade, often without a discernable trace in long term memory.
Just as was the case in our experiment with memorizing the number above, the problems almost invariably are not going to be a matter of random noise. They have certain key facts and ideas that are the basis of their solution, and those ideas are used over and over again. There is plenty of pattern and meaning there for your brain to exploit in information compression, and it may well be very cool stuff to know and hence important to you once learned, but it takes time and repetition and a certain amount of meditation for the “gestalt” of it to spring into your awareness and burn itself into your conceptual memory as “high order understanding”.
You have to give it this time, and perform the repetitions, while maintaining an optimistic, philosophical attitude towards the process. You have to do your best to have fun with it. You don’t get strong by lifting light weights a single time. You get strong lifting weights repeatedly, starting with light weights to be sure, but then working up to the heaviest weights you can manage. When you do build up to where you’re lifting hundreds of pounds, the fifty pounds you started with seems light as a feather to you.
As with the body, so with the brain. Repeat broad strokes for the big picture with increasingly deep and “heavy” excursions into the material to explore it in detail as the overall picture emerges.
Intersperse this with sessions where you work on problems and try to use the material you’ve figured out so far. Be sure to discuss it and teach it to others as you go as much as possible, as articulating what you’ve figured out to others both uses a different part of your brain than taking it in (and hence solidifies the memory) and it helps you articulate the ideas to yourself ! This process will help you learn more, better, faster than you ever have before, and to have fun doing it!
Your brain is more complicated than you think. You are very likely used to working hard to try to make it figure things out, but you’ve probably observed that this doesn’t work very well.
A lot of times you simply cannot “figure things out” because your brain doesn’t yet know the key things required to do this, or doesn’t “see” how those parts you do know fit together. Learning and discovery is not, alas, “intentional” – it is more like trying to get a bird to light on your hand that flits away the moment you try to grasp it.
People who do really hard crossword puzzles (one form of great brain exercise) have learned the following. After making a pass through the puzzle and filling in all the words they can “get”, and maybe making a couple of extra passes through thinking hard about ones they can’t get right away, looking for patterns, trying partial guesses, they arrive at an impasse. If they continue working hard on it, they are unlikely to make further progress, no matter how long they stare at it.
On the other hand, if they put the puzzle down and do something else for a while – especially if the something else is go to bed and sleep – when they come back to the puzzle they often can immediately see a dozen or more words that the day before were absolutely invisible to them. Sometimes one of the long theme answers (perhaps 25 characters long) where they have no more than two letters just

22

Preliminaries

“gives up” – they can simply “see” what the answer must be.
Where do these answers come from? The person has not “figured them out”, they have “recognized” them. They come all at once, and they don’t come about as the result of a logical sequential process. Often they come from the person’s right brain22 . The left brain tries to use logic and simple memory when it works on crosswork puzzles. This is usually good for some words, but for many of the words there are many possible answers and without any insight one can’t even recall one of the possibilities. The clues don’t suffice to connect you up to a word. Even as letters get filled in this continues to be the case, not because you don’t know the word (although in really hard puzzles this can sometimes be the case) but because you don’t know how to recognize the word “all at once” from a cleverly nonlinear clue and a few letters in this context.
The right brain is (to some extent) responsible for insight and non-linear thinking. It sees patterns, and wholes, not sequential relations between the parts. It isn’t intentional – we can’t “make” our right brains figure something out, it is often the other way around! Working hard on a problem, then “sleeping on it” (to get that all important hippocampal involvement going) is actually a great way to develop “insight” that lets you solve it without really working terribly hard after a few tries.
It also utilizes more of your brain – left and right brain, sequential reasoning and insight, and if you articulate it, or use it, or make something with your hands, then it exercieses these parts of your brain as well, strengthening the memory and your understanding still more. The learning that is associated with this process, and the problem solving power of the method, is much greater than just working on a problem linearly the night before it is due until you hack your way through it using information assembled a part at a time from the book.
The following “Method of Three Passes” is a specific strategy that implements many of the tricks discussed above. It is known to be effective for learning by means of doing homework (or in a generalized way, learning anything at all). It is ideal for “problem oriented homework”, and will pay off big in learning dividends should you adopt it, especially when supported by a group oriented recitation with strong tutorial support and many opportunities for peer discussion and teaching.

The Method of Three Passes
Pass 1 Three or more nights before recitation (or when the homework is due), make a fast pass through all problems. Plan to spend 1-1.5 hours on this pass. With roughly 10-12 problems, this gives you around 6-8 minutes per problem. Spend no more than this much time per problem and if you can solve them in this much time fine, otherwise move on to the next. Try to do this the last thing before bed at night (seriously) and then go to sleep.
Pass 2 After at least one night’s sleep, make a medium speed pass through all problems. Plan to spend 1-1.5 hours on this pass as well. Some of the problems will already be solved from the first pass or nearly so. Quickly review their solution and then move on to concentrate on the still unsolved problems. If you solved 1/4 to 1/3 of the problems in the first pass, you should be able to spend 10 minutes or so per problem in the second pass. Again, do this right before bed if possible and then go immediately to sleep.
Pass 3 After at least one night’s sleep, make a final pass through all the problems. Begin as before by quickly reviewing all the problems you solved in the previous two passes. Then spend fifteen minutes or more (as needed) to solve the remaining unsolved problems. Leave any “impossible” problems for recitation – there should be no more than three from any given assignment, as a general rule. Go immediately to bed.
22 Note that this description is at least partly metaphor, for while there is some hemispherical specialization of some of these functions, it isn’t always sharp. I’m retaining them here (oh you brain specialists who might be reading this) because they are a valuable metaphor.

Preliminaries

23

This is an extremely powerful prescription for deeply learning nearly anything. Here is the motivation. Memory is formed by repetition, and this obviously contains a lot of that. Permanent (long term) memory is actually formed in your sleep, and studies have shown that whatever you study right before sleep is most likely to be retained. Physics is actually a “whole brain” subject – it requires a synthesis of both right brain visualization and conceptualization and left brain verbal/analytical processing – both geometry and algebra, if you like, and you’ll often find that problems that stumped you the night before just solve themselves “like magic” on the second or third pass if you work hard on them for a short, intense, session and then sleep on it. This is your right (nonverbal) brain participating as it develops intuition to guide your left brain algebraic engine.
Other suggestions to improve learning include working in a study group for that third pass (the first one or two are best done alone to “prepare” for the third pass). Teaching is one of the best ways to learn, and by working in a group you’ll have opportunities to both teach and learn more deeply than you would otherwise as you have to articulate your solutions.
Make the learning fun – the right brain is the key to forming long term memory and it is the seat of your emotions. If you are happy studying and make it a positive experience, you will increase retention, it is that simple. Order pizza, play music, make it a “physics homework party night”.
Use your whole brain on the problems – draw lots of pictures and figures (right brain) to go with the algebra (left brain). Listen to quiet music (right brain) while thinking through the sequences of events in the problem (left brain). Build little “demos” of problems where possible – even using your hands in this way helps strengthen memory.
Avoid memorization. You will learn physics far better if you learn to solve problems and understand the concepts rather than attempt to memorize the umpty-zillion formulas, factoids, and specific problems or examples covered at one time or another in the class. That isn’t to say that you shouldn’t learn the important formulas, Laws of Nature, and all of that – it’s just that the learning should generally not consist of putting them on a big sheet of paper all jumbled together and then trying to memorize them as abstract collections of symbols out of context.
Be sure to review the problems one last time when you get your graded homework back. Learn from your mistakes or you will, as they say, be doomed to repeat them.
If you follow this prescription, you will have seen every assigned homework problem a minimum of five or six times – three original passes, recitation itself, a final write up pass after recitation, and a review pass when you get it back. At least three of these should occur after you have solved all of the problems correctly, since recitation is devoted to ensuring this. When the time comes to study for exams, it should really be (for once) a review process, not a cram. Every problem will be like an old friend, and a very brief review will form a seventh pass or eighth pass through the assigned homework. With this methodology (enhanced as required by the physics resource rooms, tutors, and help from your instructors) there is no reason for you do poorly in the course and every reason to expect that you will do well, perhaps very well indeed! And you’ll still be spending only the 3 to 6 hours per week on homework that is expected of you in any college course of this level of difficulty!
This ends our discussion of course preliminaries (for nearly any serious course you might take, not just physics courses) and it is time to get on with the actual material for this course.

Mathematics
Physics, as was noted in the preface, requires a solid knowledge of all mathematics through calculus.
That’s right, the whole nine yards: number theory, algebra, geometry, trigonometry, vectors, differential calculus, integral calculus, even a smattering of differential equations. Somebody may have

24

Preliminaries

told you that you can go ahead and take physics having gotten C’s in introductory calculus, perhaps in a remedial course that you took because you had such a hard time with precalc or because you failed straight up calculus when you took it.
They lied.
Sorry to be blunt, but that’s the simple truth. Here’s a list of a few of the kinds of things you’ll have to be able to do during the next two semesters of physics. Don’t worry just yet about what they mean – that is part of what you will learn along the way. The question is, can you (perhaps with a short review of things you’ve learned and knew at one time but have not forgotten) evaluate these mathematical expressions or solve for the algebraic unknowns? You don’t necessarily have to be able to do all of these things right this instant, but you should at the very least recognize most of them and be able to do them with just a very short review:
• What are the two values of α that solve: α2 +
• What is:

• What is:

R
1
α+
= 0?
L
LC

Q(r) =

ρ0

R

r

r′3 dr′ ?
0

d cos(ωt + δ)
?
dt

y

A θ ?

? x • What are the x and y components of a vector of length A that makes an angle of θ with the positive x axis (proceeding, as usual, counterclockwise for positive θ)?
• What is the sum of the two vectors A = Ax x + Ay y and B = Bx y + By y ?
ˆ
ˆ
ˆ
ˆ
• What is the inner/dot product of the two vectors A = Ax x + Ay y and B = Bx y + By y ?
ˆ
ˆ
ˆ
ˆ
• What is the cross product of the two vectors r = rx x and F = Fy y (magnitude and direction)?
ˆ
ˆ
If all of these items are unfamiliar – you don’t remember the quadratic formula (needed to solve the first one), can’t integrate xn dx (needed to solve the second one), don’t recall how to differentiate a sine or cosine function, don’t recall your basic trigonometry so that you can’t find the components of a vector from its length and angle or vice versa, and don’t recall what the dot or cross product of two vectors are, then you are going to have to add to the burden of learning physics per se the burden of learning, or re-learning, all of the basic mathematics that would have permitted you to answer these questions easily.
Here are the answers, see if this jogs your memory:

25

Preliminaries
• Here are the two roots, found with the quadratic formula: α± =

Q(r) =


R 2
L

−R ±
L

2

ρ0

R



4
LC

=−

r

r′3 dr′ =
0

1
R2

2
4L
LC

R
±
2L

ρ0 r′4 4π
R
4

r

=
0

ρ0 πr4
R

d cos(ωt + δ)
= −ω sin(ωt + δ) dt •

Ax = A cos(θ)



Ay = A sin(θ)

A + B = (Ax + Bx )ˆ + (Ay + By )ˆ x y



A · B = Ax Bx + Ay By



r × F = rx x × Fy y = rx Fy (ˆ × y ) = rx Fy z
ˆ
ˆ x ˆ
ˆ

My strong advice to you, if you are now feeling the cold icy grip of panic because in fact you are signed up for physics using this book and you couldn’t answer any of these questions and don’t even recognize the answers once you see them, is to seek out the course instructor and review your math skills with him or her to see if, in fact, it is advisable for you to take physics at this time or rather should wait and strengthen your math skills first. You can, and will, learn a lot of math while taking physics and that is actually part of the point of taking it! If you are too weak going into it, though, it will cost you some misery and hard work and some of the grade you might have gotten with better preparation ahead of time.
So, what if you could do at least some of these short problems and can remember once learning/knowing the tools, like the Quadratic Formula, that you were supposed to use to solve them?
Suppose you are pretty sure that – given a chance and resource to help you out – you can do some review and they’ll all be fresh once again in time to keep up with the physics and still do well in the course? What if you have no choice but to take physics now, and are just going to have to do your best and relearn the math as required along the way? What if you did in fact understand math pretty well once upon a time and are sure it won’t be much of an obstacle, but you really would like a review, a summary, a listing of the things you need to know someplace handy so you can instantly look them up as you struggle with the problems that uses the math it contains? What if you are
(or were) really good at math, but want to be able to look at derivations or reread explanations to bring stuff you learned right back to your fingertips once again?
Hmmm, that set of questions spans the set of student math abilities from the near-tyro to the near-expert. In my experience, everybody but the most mathematically gifted students can probably benefit from having a math review handy while taking this course. For all of you, then, I provide the following free book online:
Mathematics for Introductory Physics
It is located here: http://www.phy.duke.edu/∼rgb/Class/math for intro physics.php

26

Preliminaries

It is a work in progress, and is quite possibly still somewhat incomplete, but it should help you with a lot of what you are missing or need to review, and if you let me know what you are missing that you didn’t find there, I can work to add it!
I would strongly advise all students of introductory physics (any semester) to visit this site right now and bookmark it or download the PDF, and to visit the site from time to time to see if I’ve posted an update. It is on my back burner, so to speak, until I finish the actual physics texts themselves that I’m working on, but I will still add things to them as motivated by my own needs teaching courses using this series of books.

Summary
That’s enough preliminary stuff. At this point, if you’ve read all of this “week”’s material and vowed to adopt the method of three passes in all of your homework efforts, if you’ve bookmarked the math help or downloaded it to your personal ebook viewer or computer, if you’ve realized that your brain is actually something that you can help and enhance in various ways as you try to learn things, then my purpose is well-served and you are as well-prepared as you can be to tackle physics.

Preliminaries

27

Homework for Week 0

Problem 1.
Skim read this entire section (Week 0: How to Learn Physics), then read it like a novel, front to back. Think about the connection between engagement and learning and how important it is to try to have fun in a physics course. Write a short essay (say, three paragraphs) describing at least one time in the past where you were extremely engaged in a course you were taking, had lots of fun in the class, and had a really great learning experience.

Problem 2.
Skim-read the entire content of Mathematics for Introductory Physics (linked above). Identify things that it covers that you don’t remember or don’t understand. Pick one and learn it.

Problem 3.
Apply the Method of Three Passes to this homework assignment. You can either write three short essays or revise your one essay three times, trying to improve it and enhance it each time for the first problem, and review both the original topic and any additional topics you don’t remember in the math review problem. On the last pass, write a short (two paragraph) essay on whether or not you found multiple passes to be effective in helping you remember the content.
Note well: You may well have found the content boring on the third pass because it was so familiar to you, but that’s not a bad thing. If you learn physics so thoroughly that its laws become boring, not because they confuse you and you’d rather play World of Warcraft but because you know them so well that reviewing them isn’t adding anything to your understanding, well damn you’ll do well on the exams testing the concept, won’t you?

28

Preliminaries

II: Elementary Mechanics

29

Preliminaries

31

OK, so now you are ready to learn physics. Your math skills are buffed and honed, you’ve practiced the method of three passes, you understand that success depends on your full engagement and a certain amount of hard work. In case you missed the previous section (or are unused to actually reading a math-y textbook instead of minimally skimming it to extract just enough “stuff” to be able to do the homework) I usually review its content on the first day of class at the same time I review the syllabus and set down the class rules and grading scheme that I will use.
It’s time to embark upon the actual week by week, day by day progress through the course material. For maximal ease of use for you the student and (one hopes) your instructor whether or not that instructor is me, the course is designed to cover one chapter per week-equivalent, whether or not the chapter is broken up into a day and a half of lecture (summer school), an hour a day (MWF), or an hour and a half a day (TTh) in a semester based scheme. To emphasize this preferred rhythm, each chapter will be referred to by the week it would normally be covered in my own semester-long course. A week’s work in all cases covers just about exactly one “topic” in the course. A very few are spread out over two weeks; one or two compress two related topics into one week, but in all cases the homework is assigned on a weekly rhythm to give you ample opportunity to use the method of three passes described in the first part of the book, culminating in an expected 2-3 hour recitation where you should go over the assigned homework in a group of three to six students, with a mentor handy to help you where you get stuck, with a goal of getting all of the homework perfectly correct by the end of recitation.
That is, at the end of a week plus its recitation, you should be able to do all of the week’s homework, perfectly, and without looking or outside help. You will usually need all three passes, the last one working in a group, plus the mentored recitation to achieve this degree of competence! But without it, surely the entire process is a waste of time. Just finishing the homework is not enough, the whole point of the homework is to help you learn the material and it is the latter that is the real goal of the activity not the mere completion of a task.
However, if you do this – attempt to really master the material – you are almost certain to do well on a quiz that terminates the recitation period, and you will be very likely to retain the material and not have to “cram” it in again for the hour exams and/or final exam later in the course. Once you achieve understanding and reinforce it with a fair bit of repetition and practice, most students will naturally transform this experience into remarkably deep and permanent learning.
Note well that each week is organized for maximal ease of learning with the week/chapter review first. Try to always look at this review before lecture even if you skip reading the chapter itself until later, when you start your homework. Skimming the whole week/chapter guided by this summary before lecture is, of course, better still. It is a “first pass” that can often make lecture much easier to follow and help free you from the tyranny of note-taking as you only need to note differences in the presentation from this text and perhaps the answers to questions that helped you understand something during the discussion. Then read or skim it again right before each homework pass.

32

Week 1: Newton’s Laws

Week 1: Newton’s Laws
Summary
• Physics is a language – in particular the language of a certain kind of worldview. For philosophically inclined students who wish to read more deeply on this, I include links to terms that provide background for this point of view.
– Wikipedia: http://www.wikipedia.org/wiki/Worldview
– Wikipedia: http://www.wikipedia.org/wiki/Semantics
– Wikipedia: http://www.wikipedia.org/wiki/Ontology
Mathematics is the natural language and logical language of physics, not for any particularly deep reason but because it works. The components of the semantic language of physics are thus generally expressed as mathematical or logical coordinates, and the semantic expressions themselves are generally mathematical/algebraic laws.
• Coordinates are the fundamental adjectival modifiers corresponding to the differentiating properties of “things” (nouns) in the real Universe, where the term fundamental can also be thought of as meaning irreducible – adjectival properties that cannot be readily be expressed in terms of or derived from other adjectival properties of a given object/thing. See:
– Wikipedia: http://www.wikipedia.org/wiki/Coordinate System
• Units. Physical coordinates are basically mathematical numbers with units (or can be so considered even when they are discrete non-ordinal sets). In this class we will consistently and universally use Standard International (SI) units unless otherwise noted. Initially, the irreducible units we will need are:
a) meters – the SI units of length
b) seconds – the SI units of time
c) kilograms – the SI units of mass
All other units for at least a short while will be expressed in terms of these three, for example units of velocity will be meters per second, units of force will be kilogram-meters per second squared. We will often give names to some of these combinations, such as the SI units of force:
1 Newton =

kg-m sec2 Later you will learn of other irreducible coordinates that describe elementary particles or extended macroscopic objects in physics such as electrical charge, as well as additional derivative quantities such as energy, momentum, angular momentum, electrical current, and more.
33

34

Week 1: Newton’s Laws
• Laws of Nature are essentially mathematical postulates that permit us to understand natural phenomena and make predictions concerning the time evolution or static relations between the coordinates associated with objects in nature that are consistent mathematical theorems of the postulates. These predictions can then be compared to experimental observation and, if they are consistent (uniformly successful) we increase our degree of belief in them. If they are inconsistent with observations, we decrease our degree of belief in them, and seek alternative or modified postulates that work better23 .
The entire body of human scientific knowledge is the more or less successful outcome of this process. This body is not fixed – indeed it is constantly changing because it is an adaptive process, one that self-corrects whenever observation and prediction fail to correspond or when new evidence spurs insight (sometimes revolutionary insight!)
Newton’s Laws built on top of the analytic geometry of Descartes (as the basis for at least the abstract spatial coordinates of things) are the dynamical principle that proved successful at predicting the outcome of many, many everyday experiences and experiments as well as cosmological observations in the late 1600’s and early 1700’s all the way up to the mid-19th century24 . When combined with associated empirical force laws they form the basis of the physics you will learn in this course.
• Newton’s Laws:
a) Law of Inertia: Objects at rest or in uniform motion (at a constant velocity) in an inertial reference frame remain so unless acted upon by an unbalanced (net) force. We can write this algebraically25 as
F i = 0 = ma = m i dv dt ⇒

v = constant vector

(1)

b) Law of Dynamics: The net force applied to an object is directly proportional to its acceleration in an inertial reference frame. The constant of proportionality is called the mass of the object. We write this algebraically as:
F =

F i = ma = i d(mv) dp = dt dt

(2)

where we introduce the momentum of a particle, p = mv, in the last way of writing it.
c) Law of Reaction: If object A exerts a force F AB on object B along a line connecting the two objects, then object B exerts an equal and opposite reaction force of F BA = −F AB on object A. We write this algebraically as:
F ij

(3)

F ij

(or)

= −F ji
=

(4)

0

i,j

(the latter form means that the sum of all internal forces between particles in any closed system of particles cancel).
23 Students of philosophy or science who really want to read something cool and learn about the fundamental basis of our knowledge of reality are encouraged to read e.g. Richard Cox’s The Algebra of Probable Reason or E. T. Jaynes’ book Probability Theory: The Logic of Science. These two related works quantify how science is not (as some might think) absolute truth or certain knowledge, but rather the best set of things to believe based on our overall experience of the world, that is to say, “the evidence”.
24 Although they failed in the late 19th and early 20th centuries, to be superceded by relativistic quantum mechanics.
Basically, everything we learn in this course is wrong, but it nevertheless works damn well to describe the world of macroscopic, slowly moving objects of our everyday experience.
P
25 For students who are still feeling very shaky about their algebra and notation, let me remind you that i Fi
P
stands for “The sum over i of all force F i ”, or F 1 + F 2 + F 3 + .... We will often use as shorthand for summing over a list of similar objects or components or parts of a whole.

35

Week 1: Newton’s Laws

• An inertial reference frame is a spatial coordinate system (or frame) that is either at rest or moving at a constant speed, a non-accelerating set of coordinates that can be used to describe the locations of real objects. In physics one has considerable leeway when it comes to choosing the (inertial) coordinate frame to be used to solve a problem – some lead to much simpler solutions than others!
• Forces of Nature (weakest to strongest):
a) Gravity
b) Weak Nuclear
c) Electromagnetic
d) Strong Nuclear
It is possible that there are more forces of nature waiting to be discovered. Because physics is not a dogma, this presents no real problem. If they are, we’ll simply give the discoverer a
Nobel Prize and add their name to the “pantheon” of great physicists, add the force to the list above, and move on. Science, as noted, is self-correcting.
• Force is a vector. For each force rule below we therefore need both a formula or rule for the magnitude of the force (which we may have to compute in the solution to a problem – in the case of forces of constraint such as the normal force (see below) we will usually have to do so) and a way of determining or specifying the direction of the force. Often this direction will be obvious and in corresponence with experience and mere common sense – strings pull, solid surfaces push, gravity points down and not up. Other times it will be more complicated or geometric and (along with the magnitude) may vary with position and time.
• Force Rules The following set of force rules will be used both in this chapter and throughout this course. All of these rules can be derived or understood (with some effort) from the forces of nature, that is to say from “elementary” natural laws.
a) Gravity (near the surface of the earth):
Fg = mg

(5)

The direction of this force is down, so one could write this in vector form as F g = −mg y
ˆ
in a coordinate system such that up is the +y direction. This rule follows from Newton’s
Law of Gravitation, the elementary law of nature in the list above, evaluated “near” the surface of the earth where it is approximately constant.
b) The Spring (Hooke’s Law) in one dimension:
Fx = −k∆x

(6)

This force is directed back to the equilibrium point of unstretched spring, in the opposite direction to ∆x, the displacement of the mass on the spring from equilibrium. This rule arises from the primarily electrostatic forces holding the atoms or molecules of the spring material together, which tend to linearly oppose small forces that pull them apart or push them together (for reasons we will understand in some detail later).
c) The Normal Force:
F⊥ = N

(7)

This points perpendicular and away from solid surface, magnitude sufficient to oppose the force of contact whatever it might be! This is an example of a force of constraint – a force whose magnitude is determined by the constraint that one solid object cannot generally interpenetrate another solid object, so that the solid surfaces exert whatever force is needed to prevent it (up to the point where the “solid” property itself fails). The

36

Week 1: Newton’s Laws physical basis is once again the electrostatic molecular forces holding the solid object together, and microscopically the surface deforms, however slightly, more or less like a spring. d) Tension in an Acme (massless, unstretchable, unbreakable) string:
Fs = T

(8)

This force simply transmits an attractive force between two objects on opposite ends of the string, in the directions of the taut string at the points of contact. It is another constraint force with no fixed value. Physically, the string is like a spring once again – it microscopically is made of bound atoms or molecules that pull ever so slightly apart when the string is stretched until the restoring force balances the applied force.
e) Static Friction fs ≤ µs N

(9)

(directed opposite towards net force parallel to surface to contact). This is another force of constraint, as large as it needs to be to keep the object in question travelling at the same speed as the surface it is in contact with, up to the maximum value static friction can exert before the object starts to slide. This force arises from mechanical interlocking at the microscopic level plus the electrostatic molecular forces that hold the surfaces themselves together.
f) Kinetic Friction fk = µk N

(10)

(opposite to direction of relative sliding motion of surfaces and parallel to surface of contact). This force does have a fixed value when the right conditions (sliding) hold.
This force arises from the forming and breaking of microscopic adhesive bonds between atoms on the surfaces plus some mechanical linkage between the small irregularities on the surfaces.
g) Fluid Forces, Pressure: A fluid in contact with a solid surface (or anything else) in general exerts a force on that surface that is related to the pressure of the fluid:
FP = P A

(11)

which you should read as “the force exerted by the fluid on the surface is the pressure in the fluid times the area of the surface”. If the pressure varies or the surface is curved one may have to use calculus to add up a total force. In general the direction of the force exerted is perpendicular to the surface. An object at rest in a fluid often has balanced forces due to pressure. The force arises from the molecules in the fluid literally bouncing off of the surface of the object, transferring momentum (and exerting an average force) as they do so. We will study this in some detail and will even derive a kinetic model for a gas that is in good agreement with real gases.
h) Drag Forces:
Fd = −bv n

(12)

(directed opposite to relative velocity of motion through fluid, n usually between 1 (low velocity) and 2 (high velocity). This force also has a determined value, although one that depends in detail on the motion of the object. It arises first because the surface of an object moving through a fluid is literally bouncing fluid particles off in the leading direction while moving away from particles in the trailing direction, so that there is a differential pressure on the two surfaces.

Week 1: Newton’s Laws

37

The first week summary would not be complete without some sort of reference to methodologies of problem solving using Newton’s Laws and the force laws or rules above. The following rubric should be of great use to you as you go about solving any of the problems you will encounter in this course, although we will modify it slightly when we treat e.g. energy instead of force, torque instead of force, and so on.

Dynamical Recipe for Newton’s Second Law
a) Draw a good picture of what is going on. In general you should probably do this even if one has been provided for you – visualization is key to success in physics.
b) On your drawing (or on a second one) decorate the objects with all of the forces that act on them, creating a free body diagram for the forces on each object.
c) Write Newton’s Second Law for each object (summing the forces and setting the result to mi ai for each – ith – object) and algebraically rearrange it into (vector) differential equations
2
x of motion (practically speaking, this means solving for or isolating the acceleration ai = ddt2 i of the particles in the equations of motion).
d) Decompose the 1, 2 or 3 dimensional equations of motion for each object into a set of independent 1 dimensional equations of motion for each of the orthogonal coordinates by choosing a suitable coordinate system (which may not be cartesian, for some problems) and using trig/geometry. Note that a “coordinate” here may even wrap around a corner following a string, for example – or we can use a different coordinate system for each particle, as long as we have a known relation between the coordinate systems.
e) Solve the independent 1 dimensional systems for each of the independent orthogonal coordinates chosen, plus any coordinate system constraints or relations. In many problems the constraints will eliminate one or more degrees of freedom from consideration. Note that in most nontrivial cases, these solutions will have to be simultaneous solutions, obtained by e.g. algebraic substitution or elimination.
f) Reconstruct the multidimensional trajectory by adding the vectors components thus obtained back up (for a common independent variable, time).
g) Answer algebraically any questions requested concerning the resultant trajectory.
Some parts of this rubric will require experience and common sense to implement correctly for any given kind of problem. That is why homework is so critically important! We want to make solving the problems (and the conceptual understanding of the underlying physics) easy, and they will only get to be easy with practice followed by a certain amount of meditation and reflection, practice followed by reflection, iterate until the light dawns...

38

Week 1: Newton’s Laws

1.1: Introduction: A Bit of History and Philosophy
It has been remarked by at least one of my colleagues that one reason we have such a hard time teaching Newtonian physics to college students is that we have to first unteach them their already prevailing “natural” worldview of physics, which dates all the way back to Aristotle.
In a nutshell and in very general terms (skipping all the “nature is a source or cause of being moved and of being at rest” as primary attributes, see Aristotle’s Physica, book II) Aristotle considered motion or the lack thereof of an object to be innate properties of materials, according to their proportion of the big four basic elements: Earth, Air, Fire and Water. He extended the idea of the moving and the immovable to cosmology and to his Metaphysics as well.
In this primitive view of things, the observation that (most) physical objects (being “Earth”) set in motion slow down is translated into the notion that their natural state is to be at rest, and that one has to add something from one of the other essences to produce a state of uniform motion. This was not an unreasonable hypothesis; a great deal of a person’s everyday experience (then and now) is consistent with this. When we pick something up and throw it, it moves for a time and bounces, rolls, slides to a halt. We need to press down on the accelerator of a car to keep it going, adding something from the “fire” burning in the motor to the “earth” of the body of the car. Even our selves seem to run on “something” that goes away when we die.
Unfortunately, it is completely and totally wrong. Indeed, it is almost precisely Newton’s first law stated backwards. It is very likely that the reason Newton wrote down his first law (which is otherwise a trivial consequence of his second law) is to directly confront the error of Aristotle, to force the philosophers of his day to confront the fact that his (Newton’s) theory of physics was irreconcilable with that of Aristotle, and that (since his actually worked to make precise predictions of nearly any kind of classical motion that were in good agreement with observation and experiments designed to test it) Aristotle’s physics was almost certainly wrong.
Newton’s discoveries were a core component of the Enlightment, a period of a few hundred years in which Europe went from a state of almost slavish, church-endorsed belief in the infallibility and correctness of the Aristotelian worldview to a state where humans, for the first time in history, let nature speak for itself by using a consistent framework to listen to what nature had to say26 .
Aristotle lost, but his ideas are slow to die because they closely parallel everyday experience. The first chore for any serious student of physics is thus to unlearn this Aristotelian view of things27 .
This is not, unfortunately, an abstract problem. It is very concrete and very current. Because I have an online physics textbook, and because physics is in some very fundamental sense the “magic” that makes the world work, I not infrequently am contacted by individuals who do not understand the material covered in this textbook, who do not want to do the very hard work required to master it, but who still want to be “magicians”. So they invent their own version of the magic, usually altering the mathematically precise meanings of things like “force”, “work”, “energy” to be something else altogether that they think that they understand but that, when examined closely, usually are dimensionally or conceptually inconsistent and mean nothing at all.
Usually their “new” physics is in fact a reversion to the physics of Aristotle. They recreate the magic of earth and air, fire and water, a magic where things slow down unless fire (energy) is added to sustain their motion or where it can be extracted from invisible an inexhaustible resources, a world
26 Students who like to read historical fiction will doubtless enjoy Neal Stephenson’s Baroque Cycle, a set of novels
– filled with sex and violence and humor, a good read – that spans the Enlightenment and in which Newton, Liebnitz,
Hooke and other luminaries to whom we owe our modern conceptualization of physics all play active parts.
27 This is not the last chore, by the way. Physicists have long since turned time into a coordinate just like space so that how long things take depends on one’s point of view, eliminated the assumption that one can measure any set of measureable quantities to arbitrary precision in an arbitrary order, replaced the determinism of mathematically precise trajectories with a funny kind of stochastic quasi-determinism, made (some) forces into just an aspect of geometry, and demonstrated a degree of mathematical structure (still incomplete, we’re working on it) beyond the wildest dreams of Aristotle or his mathematical-mystic buddies, the Pythagoreans.

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39

where the mathematical relations between work and energy and force and acceleration do not hold.
A world, in short, that violates a huge, vast, truly stupdendous body of accumulated experimental evidence including the very evidence that you yourselves will very likely observe in person in the physics labs associated with this course. A world in which things like perpetual motion machines are possible, where free lunches abound, and where humble dilettantes can be crowned “the next
Einstein” without having a solid understanding of algebra, geometry, advanced calculus, or the physics that everybody else seems to understand perfectly.
This world does not exist. Seriously, it is a fantasy, and a very dangerous one, one that threatens modern civilization itself. One of the most important reasons you are taking this course, whatever your long term dreams and aspirations are professionally, is to come to fully and deeply understand this. You will come to understand the magic of science, at the same time you learn to reject the notion that science is magic or vice versa.
There is nothing wrong with this. I personally find it very comforting that the individuals that take care of my body (physicians) and who design things like jet airplanes and automobiles
(engineers) share a common and consistent Newtonian28 view of just how things work, and would find it very disturbing if any of them believed in magic, in gods, in fairies, in earth, air, fire and water as constituent elements, in “crystal energies”, in the power of a drawn pentagram or ritually chanted words in any context whatsoever. These all represent a sort of willful wishful thinking on the part of the believer, a desire for things to not follow the rigorous mathematical rules that they appear to follow as they evolve in time, for there to be a “man behind the curtain” making things work out as they appear to do. Or sometimes an entire pantheon.
Let me be therefore be precise. In the physics we will study week by week below, the natural state of “things” (e.g. objects made of matter) will be to move uniformly. We will learn non-Aristotelian physics, Newtonian physics. It is only when things are acted on from outside by unbalanced forces that the motion becomes non-uniform; they will speed up or slow down. By the time we are done, you will understand how this can still lead to the damping of motion observed in everyday life, why things do generally slow down. In the meantime, be aware of the problem and resist applying the
Aristotelian view to real physics problems, and consider, based on the evidence and your experiences taking this course rejecting “magic” as an acceptable component of your personal worldview unless and until it too has some sort of objective empirical support. Which would, of course, just make it part of physics!

1.2: Dynamics
Physics is the study of dynamics. Dynamics is the description of the actual forces of nature that, we believe, underlie the causal structure of the Universe and are responsible for its evolution in time.
We are about to embark upon the intensive study of a simple description of nature that introduces the concept of a force, due to Isaac Newton. A force is considered to be the causal agent that produces the effect of acceleration in any massive object, altering its dynamic state of motion.
Newton was not the first person to attempt to describe the underlying nature of causality. Many, many others, including my favorite ‘dumb philosopher’, Aristotle, had attempted this. The major difference between Newton’s attempt and previous ones is that Newton did not frame his as a philosophical postulate per se. Instead he formulated it as a mathematical theory and proposed a set of laws that (he hoped) precisely described the regularities of motion in nature.
In physics a law is the equivalent of a postulated axiom in mathematics. That is, a physical law is, like an axiom, an assumption about how nature operates that not formally provable by any
28 Newtonian or better, that is. Of course actual modern physics is non-Newtonian quantum mechanics, but this is just as non-magical and concrete and it reduces to Newtonian physics in the macroscopic world of our quotidian experience. 40

Week 1: Newton’s Laws

means, including experience, within the theory. A physical law is thus not considered “correct” – rather we ascribe to it a “degree of belief” based on how well and consistently it describes nature in experiments designed to verify and falsify its correspondence.
It is important to do both. Again, interested students are are encouraged to look up Karl
Popper’s “Falsifiability”29 and the older Postivism30 . A hypothesis must successfully withstand the test of repeated, reproducible experiments that both seek to disprove it and to verify that it has predictive value in order to survive and become plausible. And even then, it could be wrong!
If a set of laws survive all the experimental tests we can think up and subject it to, we consider it likely that it is a good approximation to the true laws of nature; if it passes many tests but then fails others (often failing consistently at some length or time scale) then we may continue to call the postulates laws (applicable within the appropriate milieu) but recognize that they are only approximately true and that they are superceded by some more fundamental laws that are closer
(at least) to being the “true laws of nature”.
Newton’s Laws, as it happens, are in this latter category – early postulates of physics that worked remarkably well up to a point (in a certain “classical” regime) and then failed. They are “exact”
(for all practical purposes) for massive, large objects moving slowly compared to the speed of light31 for long times such as those we encounter in the everyday world of human experience (as described by SI scale units). They fail badly (as a basis for prediction) for microscopic phenomena involving short distances, small times and masses, for very strong forces, and for the laboratory description of phenomena occurring at relativistic velocities. Nevertheless, even here they survive in a distorted but still recognizable form, and the constructs they introduce to help us study dynamics still survive.
Interestingly, Newton’s laws lead us to second order differential equations, and even quantum mechanics appears to be based on differential equations of second order or less. Third order and higher systems of differential equations seem to have potential problems with temporal causality
(where effects always follow, or are at worst simultaneous with, their causes in time); it is part of the genius of Newton’s description that it precisely and sufficiently allows for a full description of causal phenomena, even where the details of that causality turn out to be incorrect.
Incidentally, one of the other interesting features of Newton’s Laws is that Newton invented calculus to enable him to solve the problems they described. Now you know why calculus is so essential to physics: physics was the original motivation behind the invention of calculus itself.
Calculus was also (more or less simultaneously) invented in the more useful and recognizable form that we still use today by other mathematical-philosophers such as Leibnitz, and further developed by many, many people such as Gauss, Poincare, Poisson, Laplace and others. In the overwhelming majority of cases, especially in the early days, solving one or more problems in the physics that was still being invented was the motivation behind the most significant developments in calculus and differential equation theory. This trend continues today, with physics providing an underlying structure and motivation for the development of much of the most advanced mathematics.

29 Wikipedia: http://www.wikipedia.org/wiki/Falsifiability. Popper considered the ability to in principle disprove a hypothesis as an essential criterion for it to have objective meaning. Students might want to purchase and read
Nassim Nicholas Taleb’s book The Black Swan to learn of the dangers and seductions of worldview-building gone awry due to insufficient skepticism or a failure to allow for the disproportionate impact of the unexpected but true anyway – such as an experiment that falsifies a conclusion that was formerly accepted as being verified.
30 Wikipedia: http://www.wikipedia.org/wiki/Positivism. This is the correct name for “verifiability”, or the ability to verify a theory as the essential criterion for it to have objective meaning. The correct modern approach in physics is to do both, following the procedure laid out by Richard Cox and E. T. Jaynes wherein propositions are never proven or disproven per se, but rather are shown to be more or less “plausible”. A hypothesis in this approach can have meaning as a very implausible notion quite independent of whether or not it can be verified or falsified – yet.
31 c = 3 × 108 meters/second

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1.3: Coordinates
Think about any thing, any entity that objectively exists in the real, visible, Universe. What defines the object and differentiates it from all of the other things that make up the Universe? Before we can talk about how the Universe and its contents change in time, we have to talk about how to describe its contents (and time itself) at all.
As I type this I’m looking over at just such a thing across the room from me, an object that I truly believe exists in the real Universe. To help you understand this object, I have to use language.
I might tell you how large it is, what its weight is, what it looks like, where it is, how long it has been there, what it is for, and – of course – I have to use words to do this, not just nouns but a few adjectival modifiers, and speak of an “empty beer glass sitting on a table in my den just to my side”, where now I have only to tell you just where my den is, where the table is in the den, and perhaps differentiate this particular beer glass from other beer glasses you might have seen. Eventually, if I use enough words, construct a detailed map, make careful measurements, perhaps include a picture,
I can convey to you a very precise mental picture of the beer glass, one sufficiently accurate that you can visualize just where, when and what it is (or was).
Of course this prose description is not the glass itself ! If you like, the map is not the territory32 !
That is, it is an informational representation of the glass, a collection of symbols with an agreed upon meaning (where meaning in this context is a correspondence between the symbols and the presumed general sensory experience of the glass that one would have if one looked at the glass from my current point of view.
Constructing just such a map is the whole point of physics, only the map is not just of mundane objects such as a glass; it is the map of the whole world, the whole Universe. To the extent that this worldview is successful, especially in a predictive sense and not just hindsight, the physical map in your mind works well to predict the Universe you perceive through your sensory apparatus. A perfect understanding of physics (and a knowledge of certain data) is equivalent to a possessing a perfect map, one that precisely locates every thing within the Universe for all possible times.
Maps require several things. It is convenient, but not necessary, to have a set of single term descriptors, symbols for various “things” in the world the map is supposed to describe. So this symbol might stand for a house, that one for a bridge, still another one for a street or railroad crossing. Another absolutely essential part of a map is the actual coordinates of things that it is describing. The coordinate representation of objects drawn in on the map is supposed to exist in an accurate one-to-one but abstract correspondence with the concrete territory in the real world where the things represented by the symbols actually exist and move about33 .
Of course the symbols such as the term “beer glass” can themselves be abstractly modeled as points in some sort of space; Complex or composite objects with “simple” coordinates can be represented as a collection of far more coordinates for the smaller objects that make up the composite object. Ultimately one arrives at elementary objects, objects that are not (as far as we know or can tell so far ) made up of other objects at all. The various kinds of elementary objects, the list of their variable properties, and their spatial and temporal coordinates are in some deep sense all coordinates, and every object in the universe can be thought of as a point in or volume of this enormous and highly complex coordinate space!
In this sense “coordinates” are the fundamental adjectival modifiers corresponding to the differentiating properties of “named things” (nouns) in the real Universe, where the term fundamental
32 This is an adage of a field called General Semantics and is something to keep carefully in mind while studying physics. Not even my direct perception of the glass is the glass itself; it is just a more complex and dynamical kind of map.
33 Of course in the old days most actual maps were stationary, and one had to work hard to see “time” on them, but nowadays nearly everybody has or at least has seen GPS maps and video games, where things or the map coordinates themselves move.

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can also be thought of as meaning elementary or irreducible – adjectival properties that cannot be readily be expressed in terms of or derived from other adjectival properties of a given object/thing.
Physical coordinates are, then, basically mathematical numbers with units (and can be so considered even when they are discrete non-ordinal sets). At first we will omit most of the details from the objects we study to keep things simple. In fact, we will spend most of the first part of the course interested in only three quantities that set the scale for the coordinate system we need to describe the classical physics of a rather generic “particle”: space (where it is), time (when it is there), and mass (an intrinsic property).
This is our first idealization – the treatment of an extended (composite) object as if it were itself an elementary object. This is called the particle approximation, and later we will justify this approximation ex post facto (after the fact) by showing that there really is a special point in a collective system of particles that behaves like a particle as far as Newton’s Laws are concerned.
For the time being, then, objects such as porpoises, puppies, and ponies are all idealized and will be treated as particles 34 . We’ll talk more about particles in a page or two.
We need units to describe intervals or values in all three coordinates so that we can talk or think about those particles (idealized objects) in ways that don’t depend on the listener. In this class we will consistently and universally use Standard International (SI) units unless otherwise noted.
Initially, the irreducible units we will need are:
a) meters – the SI units of length
b) seconds – the SI units of time
c) kilograms – the SI units of mass
All other units for at least a short while will be expressed in terms of these three.
For example units of velocity will be meters per second, units of force will be kilogram-meters per second squared. We will often give names to some of these combinations, such as the SI units of force:
1 Newton =

kg-m sec2 Later you may learn of other irreducible coordinates that describe elementary particles or extended macroscopic objects in physics such as electrical charge, as well as additional derivative quantities such as energy, momentum, angular momentum, electrical current, and more.
As for what the quantities that these units represent are – well, that’s a tough question to answer.
I know how to measure distances between points in space and times between events that occur in space, using things like meter sticks and stopwatches, but as to just what the space and time in which these events are embedded really is I’m as clueless as a cave-man. It’s probably best to just define distance as that which we might measure with a meter stick or other “standard” of length, time as that which we might measure with a clock or other “standard” of time, and mass that which we might measure compared to some “standard” of mass using methods we’ll have to figure out below. Existential properties cannot really be defined, only observed, quantified, and understood in the context of a complete, consistent system, the physical worldview, the map we construct that works to establish a useful semantic representation of that which we observe.
Sorry if that’s difficult to grasp, but there it is. It’s just as difficult for me (after most of a lifetime studying physics) as it is for you right now. Dictionaries are, after all, written in words that are in
34 I teach physics in the summers at the Duke Marine Lab, where there are porpoises and wild ponies visible from the windows of our classroom. Puppies I threw in for free because they are cute and also begin with “p”. However, you can think of a particle as a baseball or bullet or ball bearing if you prefer less cute things that begin with the letter “b”, which is a symmetry transformed “p”, sort of.

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the dictionaries and hence are self-referential and in some deep sense should be abstract, arbitrary, meaningless – yet somehow we learn to speak and understand them. So it is, so it will be for you, with physics, and the process takes some time. y m

y(t)

∆x

x(t) x(t+ ∆t)

x(t)

x

Figure 2: Coordinatized visualization of the motion of a particle of mass m along a trajectory x(t).
Note that in a short time ∆t the particle’s position changes from x(t) to x(t + ∆t).
Coordinates are enormously powerful ideas, the very essence of mapmaking and knowledge itself.
To assist us in working with them, we introduce the notion of coordinate frame – a system of all of the relevant coordinates that describe at least the position of a particle (in one, two or three dimensions, usually). In figure 2 is a picture of a simple single particle with mass m (that might represent my car) on a set of coordinates that describes at least part of the actual space where my car is sitting. The solid line on this figure represents the trajectory of my car as it moves about.
Armed with a watch, an apparatus for measuring mass, meter sticks and some imagination, one can imagine a virtual car rolling up or down along its virtual trajectory and compare its motion in our conceptual map35 with the correspondent happenings in the world outside of our minds where the real car moves along a real track.
Note well that we idealize my car by treating the whole thing as a single object with a single position (located somewhere “in the middle”) when we know that it is really made up steering wheels and bucket seats that are “objects” in their own right that are further assembled into a “car” All of these wheels and panels, nuts and bolts are made up of still smaller objects – molecules – and molecules are made up of atoms, and atoms are made of protons and neutrons and electrons, and protons and neutrons are made up of quarks, and we don’t really know for certain if electrons and quarks are truly elementary particles or are themselves composite objects36 . Later in this semester we will formally justify our ability to do this, and we will improve on our description of things like cars and wheels and solid composite objects by learning how they can move, rotate, and even explode into little bits of car and still have some parts of their collective coordinate motion that behaves as though the ex-car is still a “single point-like object”.
In the meantime, we will simply begin with this idealization and treat discrete solid objects as particles – masses that are at a single point in space at a single time. So we will treat objects such as planets, porpoises, puppies, people, baseballs and blocks, cars and cannonballs and much more as if they have a single mass and a single spatial location at any single instant in time – as a particle. One
35 This map need not be paper, in other words – I can sit here and visualize the entire drive from my house to the grocery store, over time. Pictures of trajectories on paper are just ways we help our brains manage this sort of understanding. 36 Although the currently accepted belief is that they are. However, it would take only one good, reproducible experiment to make this belief less plausible, more probably false. Evidence talks, belief walks.

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advantage of this is that the mathematical expressions for all of these quantities become functions of time37 and possibly other coordinates.
In physical dynamics, we will be concerned with finding the trajectory of particles or systems – the position of each particle in the system expressed as a function of time. We can write the trajectory as a vector function on a spatial coordinate system (e.g. cartesian coordinates in 2 dimensions): x(t) = x(t)ˆ + y(t)ˆ x y

(13)

Note that x(t) stands for a vector from the origin to the particle, where x(t) by itself (no boldface or vector sign) stands for the x-component of this vector. An example trajectory is visualized in figure 2 (where as noted, it might stand for the trajectory of my car, treated as a particle). In all of the problems we work on throughout the semester, visualization will be a key component of the solution. The human brain doesn’t, actually, excel at being able to keep all of these details onboard in your “mind’s eye”, the virtual visual field of your imagination. Consequently, you must always draw figures, usually with coordinates, forces, and other “decorations”, when you solve a physics problem. The paper (or blackboard or whiteboard) then becomes an extension of your brain – a kind of “scratch space” that augments your visualization capabilities and your sequential symbolic reasoning capabilities. To put it bluntly, you are more intelligent when you reason with a piece of paper and pen than you are when you are forced to rely on your brain alone. To succeed in physics, you need all of the intelligence you can get, and you need to synthesize solutions using both halves of your brain, visualization and sequential reason. Paper and pen facilitate this process and one of the most important lessons of this course is how to use them to attain the full benefit of the added intelligence they enable not just in physics problems, but everywhere else as well.
If we know the trajectory function of a particle, we know a lot of other things too. Since we know where it is at any given time, we can easily tell how fast it is going in what direction. This combination of the speed of the particle and its direction forms a vector called the velocity of the particle. Speed, we all hopefully know from our experience in real life doing things like driving cars, is a measure of how far you go in a certain amount of time, expressed in units of distance
(length) divided by time. Miles per hour. Furlongs per fortnight. Or, in a physics course, meters per second 38 .
The average velocity of the particle is by definition the vector change in its position ∆x in some time ∆t divided by that time:
∆x
v av =
(14)
∆t
Sometimes average velocity is useful, but often, even usually, it is not. It can be a rather poor measure for how fast a particle is actually moving at any given time, especially if averaged over times that are long enough for interesting changes in the motion to occur.
For example, I might get in my car and drive around a racetrack at speed of 50 meters per second
– really booking it, tires squealing on the turns, smoke coming out of my engine (at least if I tried this in my car, as it would likely explode if I tried to go 112 mph for any extended time), and screech to a halt right back where I began. My average velocity is then zero – I’m back where I started!
That zero is a poor explanation for the heat waves pulsing out from under the hood of the car and the wear on its tires.
More often we will be interested in the instantaneous velocity of the particle. This is basically the average velocity, averaged over as small a time interval as possible – one so short that it is just
37 Recall that a function is a quantity that depends on a set of argument(s) that is single-valued, that is, has a single value for each unique value of its argument(s).
38 A good rule of thumb for people who have a practical experience of speeds in miles per hour trying to visualize meters per second is that 1 meter per second is approximately equal to 2 1 miles per hour, hence four meters per
4
second is nine miles per hour. A cruder but still quite useful approximation is (meters per second) equals (miles per hour/2). 45

Week 1: Newton’s Laws

long enough for the car to move at all. Calculus permits us to take this limit, and indeed uses just this limit as the definition of the derivative. We thus define the instantaneous velocity vector as the time-derivative of the position vector: v(t) = lim

∆t→0

x(t + ∆t) − x(t)
∆x
dx
= lim
=
∆t→0 ∆t
∆t
dt

(15)

Sometimes we will care about “how fast” a car is going but not so much about the direction.
Speed is defined to be the magnitude of the velocity vector: v(t) = |v(t)|

(16)

We could say more about it, but I’m guessing that you already have a pretty good intuitive feel for speed if you drive a car and pay attention to how your speedometer reading corresponds to the way things zip by or crawl by outside of your window.
The reason that average velocity is a poor measure is that (of course) our cars speed up and slow down and change direction, often. Otherwise they tend to run into things, because it is usually not possible to travel in perfectly straight lines at only one speed and drive to the grocery store. To see how the velocity changes in time, we will need to consider the acceleration of a particle, or the rate at which the velocity changes. As before, we can easily define an average acceleration over a possibly long time interval ∆t as: v(t + ∆t) − v(t)
∆v
aav =
=
(17)
∆t
∆t
Also as before, this average is usually a poor measure of the actual acceleration a particle (or car) experiences. If I am on a straight track at rest and stamp on the accelerator, burning rubber until I reach 50 meters per second (112 miles per hour) and then stamp on the brakes to quickly skid to a halt, tires smoking and leaving black streaks on the pavement, my average acceleration is once again zero, but there is only one brief interval (between taking my foot off of the accelerator and before I pushed it down on the brake pedal) during the trip where my actual acceleration was anything close to zero. Yet, my average acceleration is zero.
Things are just as bad if I go around a circular track at a constant speed! As we will shortly see, in that case I am always accelerating towards the center of the circle, but my average acceleration is systematically approaching zero as I go around the track more and more times.
From this we conclude that the acceleration that really matters is (again) the limit of the average over very short times; the time derivative of the velocity. This limit is thus defined to be the instantaneous accleration:
∆v
dv d2 x a(t) = lim
=
= 2,
(18)
∆t→0 ∆t dt dt the acceleration of the particle right now.
Obviously we could continue this process and define the time derivative of the acceleration39 and still higher order derivatives of the trajectory. However, we generally do not have to in physics
– we will not need to continue this process. As it turns out, the dynamic principle that appears sufficient to describe pretty much all classical motion will involve force and acceleration, and pretty much all of the math we do (at least at first) will involve solving backwards from a knowledge of the acceleration to a knowledge of the velocity and position vectors via integration or more generally
(later) solving the differential equation of motion.
We are now prepared to formulate this dynamical principle – Newton’s Second Law. While we’re at it, let’s study his First and Third Laws too – might as well collect the complete set...

39 A

quantity that actually does have a name – it is called the jerk – but we won’t need it.

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1.4: Newton’s Laws
The following are Newton’s Laws as you will need to know them to both solve problems and answer conceptual questions in this course. Note well that they are framed in terms of the spatial coordinates defined in the previous section plus mass and time.
a) Law of Inertia: Objects at rest or in uniform motion (at a constant velocity) in an inertial reference frame remain so unless acted upon by an unbalanced (net, total) force. We can write this algebraically as:
F =

F i = 0 = ma = m i dv dt ⇒

v = constant vector

(19)

b) Law of Dynamics: The total force applied to an object is directly proportional to its acceleration in an inertial reference frame. The constant of proportionality is called the mass of the object. We write this algebraically as:
F =

F i = ma = i d(mv) dp = dt dt

(20)

where we introduce the momentum of a particle, p = mv, in the last way of writing it.
c) Law of Reaction: If object A exerts a force F AB on object B along a line connecting the two objects, then object B exerts an equal and opposite reaction force of F BA = −F AB on object A. We write this algebraically as:
F ij

(21)

F ij

(or)

= −F ji
=

(22)

0

i,j

where i and j are arbitrary particle labels. The latter form will be useful to us later; it means that the sum of all internal forces between particles in any closed system of particles cancels!.
Note that these laws are not all independent as mathematics goes. The first law is a clear and direct consequence of the second. The third is not – it is an independent statement. The first law historically, however, had an important purpose. It rejected the dynamics of Aristotle, introducing the new idea of intertia where an object in motion continues in that motion unless acted upon by some external agency. This is directly opposed to the Aristotelian view that things only moved when acted on by an external agency and that they naturally came to rest when that agency was removed.
The second law is our basic dynamical principle. It tells one how to go from a problem description
(in words) plus a knowledge of the force laws of nature to an “equation of motion” (typically a statement of Newton’s second law). The equation of motion, generally solved for the acceleration, becomes a kinematical equation from which we can develop a full knowledge of the solution using mathematics guided by our experience and physical intuition.
The third law leads (as we shall see) to the surprising result that systems of particles behave collectively like a particle! This is indeed fortunate! We know that something like a baseball is really made up of a lot of teensy particles itself, and yet it obeys Newton’s Second law as if it is a particle.
We will use the third law to derive this and the closely related Law of Conservation of Momentum in a later week of the course.
An inertial reference frame is a coordinate system (or frame) that is either at rest or moving at a constant speed, a non-accelerating frame of reference. For example, the ground, or lab frame, is a coordinate system at rest relative to the approximately non-accelerating ground or lab and is considered to be an inertial frame to a good approximation. A (coordinate system inside a) car

Week 1: Newton’s Laws

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travelling at a constant speed relative to the ground, a spaceship coasting in a region free from fields, a railroad car rolling on straight tracks at constant speed are also inertial frames. A (coordinate system inside a) car that is accelerating (say by going around a curve), a spaceship that is accelerating, a freight car that is speeding up or slowing down – these are all examples of non-inertial frames. All of Newton’s laws suppose an inertial reference frame (yes, the third law too) and are generally false for accelerations evaluated in an accelerating frame as we will prove and discuss next week.
In the meantime, please be sure to learn the statements of the laws including the condition “in an inertial reference frame”, in spite of the fact that you don’t yet really understand what this means and why we include it. Eventually, it will be the other important meaning and use of Newton’s First
Law – it is the law that defines an inertial reference frame as any frame where an object remains in a state of uniform motion if no forces act on it!
You’ll be glad that you did.

1.5: Forces
Classical dynamics at this level, in a nutshell, is very simple. Find the total force on an object.
Use Newton’s second law to obtain its acceleration (as a differential equation of motion). Solve the equation of motion by direct integration or otherwise for the position and velocity.
That’s it!
Well, except for answering those pesky questions that we typically ask in a physics problem, but we’ll get to that later. For the moment, the next most important problem is: how do we evaluate the total force?
To answer it, we need a knowledge of the forces at our disposal, the force laws and rules that we are likely to encounter in our everyday experience of the world. Some of these forces are fundamental forces – elementary forces that we call “laws of nature” because the forces themselves aren’t caused by some other force, they are themselves the actual causes of dynamical action in the visible
Universe. Other force laws aren’t quite so fundamental – they are more like “approximate rules” and aren’t exactly correct. They are also usually derivable from (or at least understandable from) the elementary natural laws, although it may be quite a lot of work to do so.
We quickly review the forces we will be working with in the first part of the course, both the forces of nature and the force rules that apply to our everyday existence in approximate form.

1.5.1: The Forces of Nature
At this point in your life, you almost certainly know that all normal matter of your everyday experience is made up of atoms. Most of you also know that an atom itself is made up of a positively charged atomic nucleus that is very tiny indeed surrounded by a cloud of negatively charged electrons that are much lighter. Atoms in turn bond together to make molecules, atoms or molecules in turn bind together (or not) to form gases, liquids, solids – “things”, including those macroscopic things that we are so far treating as particles.
The actual elementary particles from which they are made are much tinier than atoms. It is worth providing a greatly simplified table of the “stuff” from which normal atoms (and hence molecules, and hence we ourselves) are made:
In this table, up and down quarks and electrons are so-called elementary particles – things that are not made up of something else but are fundamental components of nature. Quarks bond together three at a time to form nucleons – a proton is made up of “up-up-down” quarks and has a charge of +e, where e is the elementary electric charge. A neutron is made up of “up-down-down” and has

48

Week 1: Newton’s Laws
Particle
Up or Down Quark
Proton
Neutron
Nucleus
Electron
Atom
Molecule

Location
Nucleon (Proton or Neutron)
Nucleus
Nucleus
Atom
Atom
Molecules or Objects
Objects

Size pointlike 10−15 meters
10−15 meters
10−15 meters pointlike ∼ 10−10 meters
> 10−10 meters

Table 1: Basic building blocks of normal matter as of 2011, subject to change as we discover and understand more about the Universe, ignoring pesky things like neutrinos, photons, gluons, heavy vector bosons, heavier leptons that physics majors (at least) will have to learn about later... no charge.
Neutrons and protons, in turn, bond together to make an atomic nucleus. The simplest atomic nucleus is the hydrogen nucleus, which can be as small as a single proton, or can be a proton bound to one neutron (deuterium) or two neutrons (tritium). No matter how many protons and neutrons are bound together, the atomic nucleus is small – order of 10−15 meters in diameter40 . The quarks, protons and neutrons are bound together by means of a force of nature called the strong nuclear interaction, which is the strongest force we know of relative to the mass of the interacting particles.
The positive nucleus combines with electrons (which are negatively charged and around 2000 times lighter than a proton) to form an atom. The force responsible for this binding is the electromagnetic force, another force of nature (although in truth nearly all of the interaction is electrostatic in nature, just one part of the overall electromagnetic interaction).
The light electrons repel one another electrostatically almost as strongly as they are attracted to the nucleus that anchors them. They also obey the Pauli exclusion principle which causes them to avoid one another’s company. These things together cause atoms to be much larger than a nucleus, and to have interesting “structure” that gives rise to chemistry and molecular bonding and
(eventually) life.
Inside the nucleus (and its nucleons) there is another force that acts at very short range. This force can cause e.g. neutrons to give off an electron and turn into a proton or other strange things like that. This kind of event changes the atomic number of the atom in question and is usually accompanied by nuclear radiation. We call this force the weak nuclear force. The two nuclear forces thus both exist only at very short length scales, basically in the quantum regime inside an atomic nucleus, where we cannot easily see them using the kinds of things we’ll talk about this semester.
For our purposes it is enough that they exist and bind stable nuclei together so that those nuclei in turn can form atoms, molecules, objects, us.
Our picture of normal matter, then, is that it is made up of atoms that may or may not be bonded together into molecules, with three forces all significantly contributing to what goes on inside a nucleus and only one that is predominantly relevant to the electronic structure of the atoms themselves. There is, however, a fourth force (that we know of – there may be more still, but four is all that we’ve been able to positively identify and understand). That force is gravity. Gravity is a bit
“odd”. It is a very long range, but very weak force – by far the weakest force of the four forces of nature. It only is signficant when one builds planet or star sized objects, where it can do anything from simply bind an atmosphere to a planet and cause moons and satellites to go around it in nice orbits to bring about the catastrophic collapse of a dying star. The physical law for gravitation will
40 ...with

the possible exception of neutrons bound together by gravity to form neutron stars. Those can be thought of, very crudely, as very large nuclei.

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Week 1: Newton’s Laws

be studied over an entire week of work – later in the course. I put it down now just for completeness, but initially we’ll focus on the force rules in the following section.
F 21 = −

Gm1 m2 r 12
ˆ
2 r12 (23)

Don’t worry too much about what all of these symbols mean and what the value of G is – we’ll get to all of that but not now.
Since we live on the surface of a planet, to us gravity will be an important force, but the forces we experience every day and we ourselves are primarily electromagnetic phenomena, with a bit of help from quantum mechanics to give all that electromagnetic stuff just the right structure.
Let’s summarize this in a short table of forces of nature, strongest to weakest:
a) Strong Nuclear
b) Electromagnetic
c) Weak Nuclear
d) Gravity
Note well: It is possible that there are more forces of nature waiting to be discovered. Because physics is not a dogma, this presents no real problem. If a new force of nature (or radically different way to view the ones we’ve got) emerges as being consistent with observation and predictive, and hence possibly/plausibly true and correct, we’ll simply give the discoverer a Nobel Prize, add their name to the “pantheon of great physicists”, add the force itself to the list above, and move on.
Science, as noted above, is a self-correcting system of reasoning, at least when it is done right.

1.5.2: Force Rules
The following set of force rules will be used both in this chapter and throughout this course. All of these rules can be derived or understood (with some effort) from the forces of nature, that is to say from “elementary” natural laws, but are not quite laws themselves.
a) Gravity (near the surface of the earth):
Fg = mg

(24)

The direction of this force is down, so one could write this in vector form as F g = −mg y in
ˆ
a coordinate system such that up is the +y direction. This rule follows from Newton’s Law of Gravitation, the elementary law of nature in the list above, evaluated “near” the surface of the earth where it is varies only very slowly with height above the surface (and hence is
“constant”) as long as that height is small compared to the radius of the Earth.
The measured value of g (the gravitational “constant” or gravitational field close to the Earth’s surface) thus isn’t really constant – it actually varies weakly with latitude and height and the local density of the earth immediately under your feet and is pretty complicated41 . Some
“constant”, eh?
Most physics books (and the wikipedia page I just linked) give g’s value as something like: g ≈ 9.81

meters second2 (25)

41 Wikipedia: http://www.wikipedia.org/wiki/Gravity of Earth. There is a very cool “rotating earth” graphic on this page that shows the field variation in a color map. This page goes into much more detail than I will about the causes of variation of “apparent gravity”.

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Week 1: Newton’s Laws
(which is sort of an average of the variation) but in this class to the extent that we do arithmetic with it we’ll just use meters g ≈ 10
(26)
second2 because hey, so it makes a 2% error. That’s not very big, really – you will be lucky to measure g in your labs to within 2%, and it is so much easier to multiply or divide by 10 than 9.80665.
b) The Spring (Hooke’s Law) in one dimension:
Fx = −k∆x

(27)

This force is directed back to the equilibrium point (the end of the unstretched spring where the mass is attached) in the opposite direction to ∆x, the displacement of the mass on the spring away from this equilibrium position. This rule arises from the primarily electrostatic forces holding the atoms or molecules of the spring material together, which tend to linearly oppose small forces that pull them apart or push them together (for reasons we will understand in some detail later).
c) The Normal Force:
F⊥ = N

(28)

This points perpendicular and away from solid surface, magnitude sufficient to oppose the force of contact whatever it might be! This is an example of a force of constraint – a force whose magnitude is determined by the constraint that one solid object cannot generally interpenetrate another solid object, so that the solid surfaces exert whatever force is needed to prevent it
(up to the point where the “solid” property itself fails). The physical basis is once again the electrostatic molecular forces holding the solid object together, and microscopically the surface deforms, however slightly, more or less like a spring to create the force.
d) Tension in an Acme (massless, unstretchable, unbreakable) string:
Fs = T

(29)

This force simply transmits an attractive force between two objects on opposite ends of the string, in the directions of the taut string at the points of contact. It is another constraint force with no fixed value. Physically, the string is like a spring once again – it microscopically is made of bound atoms or molecules that pull ever so slightly apart when the string is stretched until the restoring force balances the applied force.
e) Static Friction fs ≤ µs N

(30)

(directed opposite towards net force parallel to surface to contact). This is another force of constraint, as large as it needs to be to keep the object in question travelling at the same speed as the surface it is in contact with, up to the maximum value static friction can exert before the object starts to slide. This force arises from mechanical interlocking at the microscopic level plus the electrostatic molecular forces that hold the surfaces themselves together.
f) Kinetic Friction fk = µk N

(31)

(opposite to direction of relative sliding motion of surfaces and parallel to surface of contact).
This force does have a fixed value when the right conditions (sliding) hold. This force arises from the forming and breaking of microscopic adhesive bonds between atoms on the surfaces plus some mechanical linkage between the small irregularities on the surfaces.

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Week 1: Newton’s Laws

g) Fluid Forces, Pressure: A fluid in contact with a solid surface (or anything else) in general exerts a force on that surface that is related to the pressure of the fluid:
FP = P A

(32)

which you should read as “the force exerted by the fluid on the surface is the pressure in the fluid times the area of the surface”. If the pressure varies or the surface is curved one may have to use calculus to add up a total force. In general the direction of the force exerted is perpendicular to the surface. An object at rest in a fluid often has balanced forces due to pressure. The force arises from the molecules in the fluid literally bouncing off of the surface of the object, transferring momentum (and exerting an average force) as they do so. We will study this in some detail and will even derive a kinetic model for a gas that is in good agreement with real gases.
h) Drag Forces:
Fd = −bv n

(33)

(directed opposite to relative velocity of motion through fluid, n usually between 1 (low velocity) and 2 (high velocity). It arises in part because the surface of an object moving through a fluid is literally bouncing fluid particles off in the leading direction while moving away from particles in the trailing direction, so that there is a differential pressure on the two surfaces, in part from “kinetic friction” that exerts a force component parallel to a surface in relative motion to the fluid. It is really pretty complicated – so complicated that we can only write down a specific, computable expression for it for very simple geometries and situations. Still, it is a very important and ubiquitous force and we’ll try to gain some appreciation for it along the way.

1.6: Force Balance – Static Equilibrium
Before we start using dynamics at all, let us consider what happens when all of the forces acting on an object balance. That is, there are several non-zero (vector) forces acting on an object, but those forces sum up to zero force. In this case, Newton’s First Law becomes very useful. It tells us that the object in question will remain at rest if it is initially at rest. We call this situation where the forces are all balanced static force equilibrium:
F tot =

F i = ma = 0

(34)

i

This works both ways; if an object is at rest and stays that way, we can be certain that the forces acting on it balance!
We will spend some time later studying static equilibrium in general once we have learned about both forces and torques, but for the moment we will just consider a single example of what is after all a pretty simple idea. This will also serve as a short introduction to one of the forces listed above,
Hooke’s Law for the force exerted by a spring on an attached mass.

Example 1.6.1: Spring and Mass in Static Force Equilibrium
Suppose we have a mass m hanging on a spring with spring constant k such that the spring is stretched out some distance ∆x from its unstretched length. This situation is pictured in figure 3.
We will learn how to really solve this as a dynamics problem later – indeed, we’ll spend an entire week on it! Right now we will just write down Newton’s laws for this problem so we can find a. Let the x direction be up. Then (using Hooke’s Law from the list above):

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Week 1: Newton’s Laws

m

∆x

Figure 3: A mass m hangs on a spring with spring constant k. We would like to compute the amount
∆x by which the string is stretched when the mass is at rest in static force equilibrium.

Fx = −k(x − x0 ) − mg = max

(35)

or (with ∆x = x − x0 , so that ∆x is negative as shown) ax = −

k
∆x − g m (36)

Note that this result doesn’t depend on where the origin of the x-axis is, because x and x0 both change by the same amount as we move it around. In most cases, we will find the equilibrium position of a mass on a spring to be the most convenient place to put the origin, because then x and
∆x are the same!
In static equilibrium, ax = 0 (and hence, Fx = 0) and we can solve for ∆x: ax = −

k
∆x − g = 0 m k
∆x = g m mg
∆x = k (37)

You will see this result appear in several problems and examples later on, so bear it in mind.

1.7: Simple Motion in One Dimension
Finally! All of that preliminary stuff is done with. If you actually read and studied the chapter up to this point (many of you will not have done so, and you’ll be SORRReeee...) you should:
a) Know Newton’s Laws well enough to recite them on a quiz – yes, I usually just put a question like “What are Newton’s Laws” on quizzes just to see who can recite them perfectly, a really useful thing to be able to do given that we’re going to use them hundreds of times in the next
12 weeks of class, next semester, and beyond; and
b) Have at least started to commit the various force rules we’ll use this semester to memory.

Week 1: Newton’s Laws

53

I don’t generally encourage rote memorization in this class, but for a few things, usually very fundamental things, it can help. So if you haven’t done this, go spend a few minutes working on this before starting the next section.
All done? Well all rightie then, let’s see if we can actually use Newton’s Laws (usually Newton’s
Second Law, our dynamical principle) and force rules to solve problems. We will start out very gently, trying to understand motion in one dimension (where we will not at first need multiple coordinate dimensions or systems or trig or much of the other stuff that will complicate life later) and then, well, we’ll complicate life later and try to understand what happens in 2+ dimensions.
Here’s the basic structure of a physics problem. You are given a physical description of the problem. A mass m at rest is dropped from a height H above the ground at time t = 0; what happens to the mass as a function of time? From this description you must visualize what’s going on (sometimes but not always aided by a figure that has been drawn for you representing it in some way). You must select a coordinate system to use to describe what happens. You must write
Newton’s Second Law in the coordinate system for all masses, being sure to include all forces or force rules that contribute to its motion. You must solve Newton’s Second Law to find the accelerations of all the masses (equations called the equations of motion of the system). You must solve the equations of motion to find the trajectories of the masses, their positions as a function of time, as well as their velocities as a function of time if desired. Finally, armed with these trajectories, you must answer all the questions the problem poses using algebra and reason and – rarely in this class
– arithmetic!
Simple enough.
Let’s put this simple solution methodology to the test by solving the following one dimensional, single mass example problem, and then see what we’ve learned.

Example 1.7.1: A Mass Falling from Height H
Let’s solve the problem we posed above, and as we do so develop a solution rubric – a recipe for solving all problems involving dynamics42 ! The problem, recall, was to drop a mass m from rest from a height H, algebraically find the trajectory (the position function that solves the equations of motion) and velocity (the time derivative of the trajectory), and then answer any questions that might be asked using a mix of algebra, intuition, experience and common sense. For this first problem we’ll postpone actually asking any question until we have these solutions so that we can see what kinds of questions one might reasonably ask and be able to answer.
The first step in solving this or any physics problem is to visualize what’s going on! Mass m?
Height H? Drop? Start at rest? Fall? All of these things are input data that mean something when translated into algebraic ”physicsese”, the language of physics, but in the end we have to coordinatize the problem (choose a coordinate system in which to do the algebra and solve our equations for an answer) and to choose a good one we need to draw a representation of the problem.
Physics problems that you work and hand in that have no figure, no picture, not even additional hand-drawn decorations on a provided figure will rather soon lose points in the grading scheme!
At first we (the course faculty) might just remind you and not take points off, but by your second assignment you’d better be adding some relevant artwork to every solution43 . Figure 4 is what an
42 At

least for the next couple of weeks... but seriously, this rubric is useful all the way up to graduate physics. has two benefits – one is that it actually is a critical step in solving the problem, the other is that drawing engages the right hemisphere of your brain (the left hemisphere is the one that does the algebra). The right hemisphere is the one that controls formation of long term memory, and it can literally get bored, independently of the left hemisphere and interrupt your ability to work. If you’ve ever worked for a very long time on writing something very dry (left hemisphere) or doing lots of algebraic problems (left hemisphere) and found your eyes being almost irresistably drawn up to look out the window at the grass and trees and ponies and bright sun, then know that it is your “right brain” that has taken over your body because it is dying in there, bored out of its (your!) gourd.
43 This

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Week 1: Newton’s Laws

Figure 4: A picture of a ball being dropped from a height H, with a suitable one-dimensional coordinate system added. Note that the figure clearly indicates that it is the force of gravity that makes it fall. The pictures of Satchmo (my border collie) and the tree and sun and birds aren’t strictly necessary and might even be distracting, but my right brain was bored when I drew this picture and they do orient the drawing and make it more fun! actual figure you might draw to accompany a problem might look like.
Note a couple of things about this figure. First of all, it is large – it took up 1/4+ of the unlined/white page I drew it on. This is actually good practice – do not draw postage-stamp sized figures! Draw them large enough that you can decorate them, not with Satchmo but with things like coordinates, forces, components of forces, initial data reminders. This is your brain we’re talking about here, because the paper is functioning as an extension of your brain when you use it to help solve the problem. Is your brain postage-stamp sized? Don’t worry about wasting paper – paper is cheap, physics educations are expensive. Use a whole page (or more) per problem solution at this point, not three problems per page with figures that require a magnifying glass to make out.
When I (or your instructor) solve problems with you, this is the kind of thing you’ll see us draw, over and over again, on the board, on paper at a table, wherever. In time, physicists become pretty good schematic artists and so should you. However, in a textbook we want things to be clearer and prettier, so I’ll redraw this in figure 5, this time with a computer drawing tool (xfig) that I’ll use for drawing most of the figures included in the textbook. Alas, it won’t have Satchmo, but it does have all of the important stuff that should be on your hand-drawn figures.
Note that I drew two alternative ways of adding coordinates to the problem. The one on the left is appropriate if you visualize the problem from the ground, looking up like Satchmo, where the ground is at zero height. This might be e.g. dropping a ball off of the top of Duke Chapel, for example, with you on the ground watching it fall.
The one or the right works if you visualize the problem as something like dropping the same ball
To keep the right brain happy while you do left brained stuff, give it something to do – listen to music, draw pictures or visualize a lot, take five minute right-brain-breaks and deliberately look at something visually pleasing. If your right brain is happy, you can work longer and better. If your right brain is engaged in solving the problem you will remember what you are working on much better, it will make more sense, and your attention won’t wander as much.
Physics is a whole brain subject, and the more pathways you use while working on it, the easier it is to understand and remember!

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Week 1: Newton’s Laws

+y

v=0@t=0

+y

v=0@t=0

m

m

mg

H

mg

x

H

x
Figure 5: The same figure and coordinate system, drawn “perfectly” with xfig, plus a second (alternative) coordinatization. into a well, where the ground is still at “zero height” but now it falls down to a negative height H from zero instead of starting at H and falling to height zero. Or, you dropping the ball from the top of the Duke Chapel and counting “y = 0 as the height where you are up there (and the initial position in y of the ball), with the ground at y = −H below the final position of the ball after it falls. Now pay attention, because this is important: Physics doesn’t care which coordinate system you use! Both of these coordinatizations of the problem are inertial reference frames. If you think about it, you will be able to see how to transform the answers obtained in one coordinate system into the corresponding answers in the other (basically subtract a constant H from the values of y in the left hand figure and you get y ′ in the right hand figure, right?). Newton’s Laws will work perfectly in either inertial reference frame44 , and truthfully there are an infinite number of coordinate frames you could choose that would all describe the same problem in the end. You can therefore choose the frame that makes the problem easiest to solve!
Personally, from experience I prefer the left hand frame – it makes the algebra a tiny bit prettier
– but the one on the right is really almost as good. I reject without thinking about it all of the frames where the mass m e.g. starts at the initial position yi = H/2 and falls down to the final position yf = −H/2. I do sometimes consider a frame like the one on the right with y positive pointing down, but it often bothers students to have “down” be positive (even though it is very natural to orient our coordinates so that F points in the positive direction of one of them) so we’ll work into that gently. Finally, I did draw the x (horizontal) coordinate and ignored altogether for now the z coordinate that in principle is pointing out of the page in a right-handed coordinate frame. We don’t really need either of these because no aspect of the motion will change x or z (there are no forces acting in those directions) so that the problem is effectively one-dimensional.
Next, we have to put in the physics, which at this point means: Draw in all of the forces that act on the mass as proportionate vector arrows in the direction of the force. The
“proportionate” part will be difficult at first until you get a feeling for how large the forces are likely to be relative to one another but in this case there is only one force, gravity that acts, so we can write on our page (and on our diagram) the vector relation:
F = −mg y
ˆ

(38)

or if you prefer, you can write the dimension-labelled scalar equation for the magnitude of the force in the y-direction:
Fy = −mg
(39)
44 For

the moment you can take my word for this, but we will prove it in the next week/chapter when we learn how to systematically change between coordinate frames!

56

Week 1: Newton’s Laws

Note well! Either of these is acceptable vector notation because the force is a vector (magnitude and direction). So is the decoration on the figure – an arrow for direction labelled mg.
What is not quite right (to the tune of minus a point or two at the discretion of the grader) is to just write F = mg on your paper without indicating its direction somehow. Yes, this is the magnitude of the force, but in what direction does it point in the particular coordinate system you drew into your figure? After all, you could have made +x point down as easily as −y! Practice connecting your visualization of the problem in the coordinates you selected to a correct algebraic/symbolic description of the vectors involved.
In context, we don’t really need to write Fx = Fz = 0 because they are so clearly irrelevant.
However, in many other problems we will need to include either or both of these. You’ll quickly get a feel for when you do or don’t need to worry about them.
Now comes the key step – setting up all of the algebra that leads to the solution. We write
Newton’s Second Law for the mass m, and algebraically solve for the acceleration! Since there is only one relevant component of the force in this one-dimensional problem, we only need to do this one time for the scalar equation for that component.:
Fy = −mg may ay d2 y dt2 = may
= −mg

= −g
=

dvy
= −g dt (40)

where g = 10 m/second2 is the constant (within 2%, close to the Earth’s surface, remember).
We are all but done at this point. The last line (the algebraic expression for the acceleration) is called the equation of motion for the system, and one of our chores will be to learn how to solve several common kinds of equation of motion. This one is a constant acceleration problem. Let’s do it.
Here is the algebra involved. Learn it. Practice doing this until it is second nature when solving simple problems like this. I do not recommend memorizing the solution you obtain at the end, even though when you have solved the problem enough times you will probably remember it anyway for the rest of your share of eternity. Start with the equation of motion for a constant acceleration: dvy dt dvy dvy vy (t)

= −g

Next, multiply both sides by dt to get:

= −g dt
= −

g dt

Then integrate both sides: doing the indefinite integrals to get:

= −gt + C

(41)

The final C is the constant of integration of the indefinite integrals. We have to evaluate it using the given (usually initial) conditions. In this case we know that: vy (0) = −g · 0 + C = C = 0

(42)

(Recall that we even drew this into our figure to help remind us – it is the bit about being “dropped from rest” at time t = 0.) Thus: vy (t) = −gt

(43)

We now know the velocity of the dropped ball as a function of time! This is good, we are likely to need it. However, the solution to the dynamical problem is the trajectory function, y(t). To find

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Week 1: Newton’s Laws it, we repeat the same process, but now use the definition for vy in terms of y: dy dt dy dy

= vy (t) = −gt
= −gt dt
= −

Multiply both sides by dt to get:

Next, integrate both sides:

gt dt

to get:

1 y(t) = − gt2 + D
2

(44)

The final D is again the constant of integration of the indefinite integrals. We again have to evaluate it using the given (initial) conditions in the problem. In this case we know that:
1
y(0) = − g · 02 + D = D = H
2

(45)

because we dropped it from an initial height y(0) = H. Thus:
1
y(t) = − gt2 + H
2

(46)

and we know everything there is to know about the motion! We know in particular exactly where it is at all times (until it hits the ground) as well as how fast it is going and in what direction. Sure, later we’ll learn how to evaluate other quantities that depend on these two, but with the solutions in hand evaluating those quantities will be (we hope) trivial.
Finally, we have to answer any questions that the problem might ask! Note well that the problem may not have told you to evaluate y(t) and vy (t), but in many cases you’ll need them anyway to answer the questions they do ask. Here are a couple of common questions you can now answer using the solutions you just obtained:
a) How long will it take for the ball to reach the ground?
b) How fast is it going when it reaches the ground?
To answer the first one, we use a bit of algebra. “The ground” is (recall) y = 0 and it will reach there at some specific time (the time we want to solve for) tg . We write the condition that it is at the ground at time tg :
1
y(tg ) = − gt2 + H = 0
(47)
2 g
If we rearrange this and solve for tg we get: tg = ±

2H g (48)

Hmmm, there seem to be two times at which y(tg ) equals zero, one in the past and one in the future. The right answer, of course, must be the one in the future: tg = + 2H/g, but you should think about what the one in the past means, and how the algebraic solution we’ve just developed is ignorant of things like your hand holding the ball before t = 0 and just what value of y corresponds to “the ground”...
That was pretty easy. To find the speed at which it hits the ground, one can just take our correct
(future) time and plug it into vy ! That is: vg = vy (tg ) = −gtg = −g

2H
= − 2gH g (49)

Note well that it is going down (in the negative y direction) when it hits the ground. This is a good hint for the previous puzzle. What direction would it have been going at the negative time? What

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Week 1: Newton’s Laws

kind of motion does the overall solution describe, on the interval from t = (−∞, ∞)? Do we need to use a certain amount of common sense to avoid using the algebraic solution for times or values of y for which they make no sense, such as y < 0 or t < 0 (in the ground or before we let go of the ball, respectively)?
The last thing we might look at I’m going to let you do on your own (don’t worry, it’s easy enough to do in your head). Assuming that this algebraic solution is valid for any reasonable H, how fast does the ball hit the ground after falling (say) 5 meters? How about 20 = 4 ∗ 5 meters?
How about 80 = 16 ∗ 5 meters? How long does it take for the ball to fall 5 meters, 20 meters, 80 meters, etc? In this course we won’t do a lot of arithmetic, but whenever we learn a new idea with parameters like g in it, it is useful to do a little arithmetical exploration to see what a “reasonable” answer looks like. Especially note how the answers scale with the height – if one drops it from 4x the height, how much does that increase the time it falls and speed with which it hits?
One of these heights causes it to hit the ground in one second, and all of the other answers scale with it like the square root. If you happen to remember this height, you can actually estimate how long it takes for a ball to fall almost any height in your head with a division and a square root, and if you multiply the time answer by ten, well, there is the speed with which it hits! We’ll do some conceptual problems that help you understand this scaling idea for homework.
This (a falling object) is nearly a perfect problem archetype or example for one dimensional motion. Sure, we can make it more complicated, but usually we’ll do that by having more than one thing move in one dimension and then have to figure out how to solve the two problems simultaneously and answer questions given the results.
Let’s take a short break to formally solve the equation of motion we get for a constant force in one dimension, as the general solution exhibits two constants of integration that we need to be able to identify and evaluate from initial conditions. Note well that the next problem is almost identical to the former one. It just differs in that you are given the force F itself, not a knowledge that the force is e.g. “gravity”.

Example 1.7.2: A Constant Force in One Dimension
This time we’ll imagine a different problem. A car of mass m is travelling at a constant speed v0 as it enters a long, nearly straight merge lane. A distance d from the entrance, the driver presses the accelerator and the engine exerts a constant force of magnitude F on the car.
a) How long does it take the car to reach a final velocity vf > v0 ?
b) How far (from the entrance) does it travel in that time?
As before, we need to start with a good picture of what is going on. Hence a car:

y

t=0 m t=tf v0 v0

vf

F

x

d
D

Figure 6: One possible way to portray the motion of the car and coordinatize it.

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Week 1: Newton’s Laws

In figure 6 we see what we can imagine are three “slices” of the car’s position as a function of time at the moments described in the problem. On the far left we see it “entering a long, nearly straight merge lane”. The second position corresponds to the time the car is a distance d from the entrance, which is also the time the car starts to accelerate because of the force F . I chose to start the clock then, so that I can integrate to find the position as a function of time while the force is being applied. The final position corresponds to when the car has had the force applied for a time tf and has acquired a velocity vf . I labelled the distance of the car from the entrance D at that time. The mass of the car is indicated as well.
This figure completely captures the important features of the problem! Well, almost. There are two forces I ignored altogether. One of them is gravity, which is pulling the car down. The other is the so-called normal force exerted by the road on the car – this force pushes the car up. I ignored them because my experience and common sense tell me that under ordinary circumstances the road doesn’t push on the car so that it jumps into the air, nor does gravity pull the car down into the road
– the two forces will balance and the car will not move or accelerate in the vertical direction. Next week we’ll take these forces into explicit account too, but here I’m just going to use my intuition that they will cancel and hence that the y-direction can be ignored, all of the motion is going to be in the x-direction as I’ve defined it with my coordinate axes.
It’s time to follow our ritual. We will write Newton’s Second Law and solve for the acceleration
(obtaining an equation of motion). Then we will integrate twice to find first vx (t) and then x(t). We will have to be extra careful with the constants of integration this time, and in fact will get a very general solution, one that can be applied to all constant acceleration problems, although I do not recommend that you memorize this solution and try to use it every time you see Newton’s Second
Law! For one thing, we’ll have quite a few problems this year where the force, and acceleration, are not constant and in those problems the solution we will derive is wrong. Alas, to my own extensive and direct experience, students that memorize kinematic solutions to the constant acceleration problem instead of learning to solve it with actual integration done every time almost invariably try applying the solution to e.g. the harmonic oscillator problem later, and I hate that. So don’t memorize the answer; learn how to derive it and practice the derivation until (sure) you know the result, and also know when you can use it.
Thus:
F ax dvx dt = max
F
=
= a0 m (a constant)

= a0

(50)

Next, multiply through by dt and integrate both sides: vx (t) =

dvx =

a0 dt = a0 t + V =

F t+V m

(51)

Either of the last two are valid answers, provided that we define a0 = F/m somewhere in the solution and also provided that the problem doesn’t explicitly ask for an answer to be given in terms of F and m. V is a constant of integration that we will evaluate below.
Note that if a0 = F/m was not a constant (say that F(t) is a function of time) then we would have to do the integral :
1
F (t)
F (t) dt =???
(52)
dt = vx (t) = m m
At the very least, we would have to know the explicit functional form of F (t) to proceed, and the answer would not be linear in time.

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Week 1: Newton’s Laws
At time t = 0, the velocity of the car in the x-direction is v0 , so (check for yourself) V = v0 and: vx (t) = a0 t + v0 =

dx dt (53)

We multiply this equation by dt on both sides, integrate, and get: x(t) =

dx =

(a0 t + v0 ) dt =

1 2 a0 t + v0 t + x0
2

(54)

where x0 is the constant of integration. We note that at time t = 0, x(0) = d, so x0 = d. Thus: x(t) =

1 2 a0 t + v0 t + d
2

(55)

It is worth collecting the two basic solutions in one place. It should be obvious that for any one-dimensional (say, in the x-direction) constant acceleration ax = a0 problem we will always find that: vx (t) = a0 t + v0
1 2 x(t) = a0 t + v0 t + x0
2

(56)
(57)

where x0 is the x-position at time t = 0 and v0 is the x-velocity at time t = 0. You can see why it is so very tempting to just memorize this result and pretend that you know a piece of physics, but don’t! The algebra that led to this answer is basically ordinary math with units. As we’ve seen, “math with units” has a special name all its own – kinematics – and the pair of equations 56 and 57 are called the kinematic solutions to the constant acceleration problem. Kinematics should be contrasted with dynamics, the physics of forces and laws of nature that lead us to equations of motion. One way of viewing our solution strategy is that – after drawing and decorating our figure, of course – we solve first the dynamics problem of writing our dynamical principle (Newton’s Second Law with the appropriate vector total force), turning it into a differential equation of motion, then solving the resulting kinematics problem represented by the equation of motion with calculus. Don’t be tempted to skip the calculus and try to memorize the kinematic solutions – it is just as important to understand and be able to do the kinematic calculus quickly and painlessly as it is to be able to set up the dynamical part of the solution.
Now, of course, we have to actually answer the questions given above. To do this requires as before logic, common sense, intuition, experience, and math. First, at what time tf does the car have speed vf ? When: vx (tf ) = vf = a0 tf + v0

(58)

of course. You can easily solve this for tf . Note that I just transformed the English statement “At tf , the car must have speed vf ” into an algebraic equation that means the exact same thing!
Second, what is D? Well in English, the distance D from the entrance is where the car is at time tf , when it is also travelling at speed vf . If we turn this sentence into an equation we get: x(tf ) = D =

1 2 a0 t + v0 tf + d
2 f

(59)

Again, having solved the previous equation algebraically, you can substitute the result for tf into this equation and get D in terms of the originally given quantities! The problem is solved, the questions are answered, we’re finished.
Or rather, you will be finished, after you fill in these last couple of steps on your own!

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Week 1: Newton’s Laws

1.7.1: Solving Problems with More Than One Object
One of the keys to answering the questions in both of these examples has been turning easy-enough statements in English into equations, and then solving the equations to obtain an answer to a question also framed in English. So far, we have solved only single equations, but we will often be working with more than one thing at a time, or combining two or more principles, so that we have to solve several simultaneous equations.
The only change we might make to our existing solution strategy is to construct and solve the equations of motion for each object or independent aspect (such as dimension) of the problem. In a moment, we’ll consider problems of the latter sort, where this strategy will work when the force in one coordinate direction is independent of the force in another coordinate direction! . First, though, let’s do a couple of very simple one-dimensional problems with two objects with some sort of constraint connecting the motion of one to the motion of the other.

Example 1.7.3: Atwood’s Machine

T

T

m1 m2 m 1g m 2g
Figure 7: Atwood’s Machines consists of a pair of masses (usually of different mass) connected by a string that runs over a pulley as shown. Initially we idealize by considering the pulley to be massless and frictionless, the string to be massless and unstretchable, and ignore drag forces.
A mass m1 and a second mass m2 are hung at both ends of a massless, unstretchable string that runs over a frictionless, massless pulley as shown in figure 7. Gravity near the Earth’s surface pulls both down. Assuming that the masses are released from rest at time t = 0, find:
a) The acceleration of both masses;
b) The tension T in the string;
c) The speed of the masses after they have moved through a distance H in the direction of the more massive one.
The trick of this problem is to note that if mass m2 goes down by a distance (say) x, mass m1 goes up by the same distance x and vice versa. The magnitude of the displacement of one is the same as that of the other, as they are connected by a taut unstretchable string. This also means that the speed of one rising equals the speed of the other falling, the magnitude of the acceleration of one up equals the magnitude of the acceleration of the other down. So even though it at first looks like you need two coordinate systems for this problem, x1 (measured from m1 ’s initial position, up or down) will equal x2 (measured from m2 ’s initial position, down or up) be the same. We therefore

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Week 1: Newton’s Laws

can just use x to describe this displacement (the displacement of m1 up and m2 down from its starting position), with vx and ax being the same for both masses with the same convention.
This, then, is a wraparound one-dimensional coordinate system, one that “curves around the pulley”. In these coordinates, Newton’s Second Law for the two masses becomes the two equations:
F1
F2

= T − m1 g = m1 ax

= m2 g − T = m2 ax

(60)
(61)

This is a set of two equations and two unknowns (T and ax ). It is easiest to solve by elimination.
If we add the two equations we eliminate T and get: m2 g − m1 g = (m2 − m1 )g = m1 ax + m2 ax = (m1 + m2 )ax

(62)

or

m2 − m1 g (63) m1 + m2
In the figure above, if m2 > m1 (as the figure suggests) then both mass m2 will accelerate down and m1 will accelerate up with this constant acceleration. ax =

We can find T by substituting this value for ax into either force equation:

T − m1 g
T − m1 g
T
T
T

= m1 ax m2 − m 1 m1 g
=
m1 + m 2 m2 − m 1
=
m 1 g + m1 g m1 + m 2 m2 + m1 m2 − m 1 m1 g + m1 g
=
m1 + m 2 m1 + m2
2m2 m1
=
g m1 + m 2

(64)

ax is constant, so we can evaluate vx (t) and x(t) exactly as we did for a falling ball: dvx dt

=

dvx

=

dvx

=

vx

=

vx (t)

=

ax =

m2 − m1 g m1 + m2 m2 − m1 g dt m1 + m2 m2 − m1 g dt m1 + m2 m2 − m1 gt + C m1 + m2 m2 − m1 gt m1 + m2

(65)

and then: dx dt

=

dx

=

dx

vx (t) =

=

x = x(t) =

m2 − m1 gt m1 + m2 m2 − m1 gt dt m1 + m2 m2 − m1 gt dt m1 + m2
1 m2 − m1 2 gt + C ′
2 m1 + m2
1 m2 − m1 2 gt 2 m1 + m2

(66)

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Week 1: Newton’s Laws

(where C and C ′ are set from our knowledge of the initial conditions, x(0) = 0 and v(0) = 0 in the coordinates we chose).
Now suppose that the blocks “fall” a height H (only m2 actually falls, m1 goes up). Then we can, as before, find out how long it takes for x(th ) = H, then substitute this into vx (th ) to find the speed. I leave it as an exercise to show that this answer is: m2 − m1 m1 + m2

vx (th ) =

2gH

(67)

Example 1.7.4: Braking for Bikes, or Just Breaking Bikes?
A car of mass M is overtaking a bicyclist. Initially, the car is travelling at speed v0c and the bicyclist is travelling at v0b < v0c in the same direction. At a time that the bicyclist is D meters away, the driver of the car suddenly sees that he is on a collision course and applies the brakes, exerting a force
−F on his car (where the minus sign just means that he is slowing down, diminishing his velocity.
Assuming that the bicyclist doesn’t speed up or slow down, does he hit the bike?
At this point you should have a pretty good idea how to proceed for each object. First, we’ll draw a figure with both objects and formulate the equations of motion for each object separately.
Second, we’ll solve the equations of motion for reach object. Third, we’ll write an equation that captures the condition that the car hits the bike, and see if that equation has any solutions. If so, then it is likely that the car will be breaking, not braking (in time)!

y t=0 F

M

m v0c v0b x D

Figure 8: The initial picture of the car overtaking the bike at the instant it starts to brake. Again we will ignore the forces in the y-direction as we know that the car doesn’t jump over the bike and we’ll pretend that the biker can’t just turn and get out of the way as well.
Here’s the solution, without most of the details. You should work through this example, filling in the missing details and making the solution all pretty. The magnitude of the acceleration of the car is ac = F/M , and we’ll go ahead and use this constant acceleration ac to formulate the answer – we can always do the arithmetic and substitute at the end, given some particular values for F and
M.
Integrating this (and using xc (0) = 0, vc (0) = v0c ) you will get: vc (t) xc (t)

= −ac t + v0c
1
= − ac t2 + v0c t
2

(68)
(69)

The acceleration of the bike is ab = 0. This means that: vb (t) = ab t + v0b = v0b

(70)

The velocity of the bike is constant because there is no (net) force acting on it and hence it has no acceleration. Integrating this one gets (using xb (0) = D): xb (t) = v0b t + D

(71)

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Week 1: Newton’s Laws

Now the big question: Does the car hit the bike? If it does, it does so at some real time, call it th . “Hitting” means that there is no distance between them – they are at the same place at the same time, in particular at this time th . Turning this sentence into an equation, the condition for a collision is algebraically:
1
xb (th ) = v0b th + D = − ac t2 + v0c th = xc (th )
2 h

(72)

Rearranged, this is a quadratic equation:
1 2 ac t − (v0c − v0b ) th + D = 0
2 h

(73)

and therefore has two roots. If we write down the quadratic formula: th =

(v0c − v0b ) ±

(v0c − v0b )2 − 2ac D ac (74)

we can see that there will only be a real (as opposed to imaginary) time th that solves the collision condition if the argument of the square root is non-negative. That is:
(v0c − v0b )2 ≥ 2ac D

(75)

If this is true, there will be a collision. If it is false, the car will never reach the bike.
There is actually a second way to arrive at this result. One can find the time ts that the car is travelling at the same speed as the bike. That’s really pretty easy: v0b = vc (ts ) = −ac ts + v0c or ts =

(v0c − v0b ) ac (76)

(77)

Now we locate the car relative to the bike. If the collision hasn’t happened by ts it never will, as afterwards the car will be slower than the bike and the bike will pull away. If the position of the car is behind (or barely equal to) the position of the bike at ts , all is well, no collision occurs. That is:
1
xc (ts ) = − ac t2 + v0c ts ≤ v0b ts + D
2 s

(78)

if no collision occurs. It’s left as an exercise to show that this leads to the same condition that the quadratic gives you.
Next, let’s see what happens when we have only one object but motion in two dimensions.

1.8: Motion in Two Dimensions
The idea of motion in two or more dimensions is very simple. Force is a vector, and so is acceleration.
Newton’s Second Law is a recipe for taking the total force and converting it into a differential equation of motion: d2 r
F tot a= =
(79)
2 dt m
In the most general case, this can be quite difficult to solve. For example, consider the forces that act upon you throughout the day – every step you take, riding in a car, gravity, friction, even the wind exert forces subtle or profound on your mass and accelerate you first this way, then that as you move around. The total force acting on you varies wildly with time and place, so even though your trajectory is a solution to just such an equation of motion, computing it algebraically is out of

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Week 1: Newton’s Laws

the question. Computing it with a computer would be straightforward if the forces were all known, but of course they vary according to your volition and the circumstances of the moment and are hardly knowable ahead of time.
However, much of what happens in the world around you can actually be at least approximated by relatively simple (if somewhat idealized) models and explicitly solved. These simple models generally arise when the forces acting are due to the “well-known” forces of nature or force rules listed above and hence point in specific directions (so that their vector description can be analyzed) and are either constant in time or vary in some known way so that the calculus of the solution is tractable45 .
We will now consider only these latter sorts of forces: forces that act in a well-defined direction with a computable value (initially, with a computable constant value, or a value that varies in some simple way with position or time). If we write the equation of motion out in components: ax =

ay

=

az

=

d2 x
Ftot,x
= dt2 m
Ftot,y
d2 y
=
dt2 m Ftot,z d2 z
=
dt2 m (80)
(81)
(82)

we will often reduce the complexity of the problem from a “three dimensional problem” to three
“one dimensional problems” of the sort we just learned to solve in the section above.
Of course, there’s a trick to it. The trick is this:
Select a coordinate system in which one of the coordinate axes is aligned with the total force.
We won’t always be able to do this, but when it can it will get us off to a very good start, and trying it will help us understand what to do when we hit problems where this alone won’t quite work or help us solve the problem.
The reason this step (when possible) simplifies the problem is simple enough to understand: In this particular coordinate frame (with the total force pointing in a single direction along one of the coordinate axes), the total force in the other directions adds up to zero! That means that all acceleration occurs only along the selected coordinate direction. Solving the equations of motion in the other directions is then trivial – it is motion with a constant velocity (which may be zero, as in the case of dropping a ball vertically down from the top of a tower in the problems above). Solving the equation of motion in the direction of the total force itself is then “the problem”, and you will need lots of practice and a few good examples to show you how to go about it.
To make life even simpler, we will now further restrict ourselves to the class of problems where the acceleration and velocity in one of the three dimensions is zero. In that case the value of that coordinate is constant, and may as well be taken to be zero. The motion (if any) then occurs in the remaining two dimensional plane that contains the origin. In the problems below, we will find it useful to use one of two possible two-dimensional coordinate systems to solve for the motion:
Cartesian coordinates (which we’ve already begun to use, at least in a trivial way) and Plane Polar coordinates, which we will review in context below.
As you will see, solving problems in two or three dimensions with a constant force direction simply introduces a few extra steps into the solution process:
45 “Tractable” here means that it can either be solved algebraically, true for many of the force laws or rules, or at least solved numerically. In this course you may or may not be required or expected to explore numerical solutions to the differential equations with e.g. matlab, octave, or mathematica.

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Week 1: Newton’s Laws
• Decomposing the known forces into a coordinate system where one of the coordinate axes lines up with the (expected) total force...
• Solving the individual one-dimensional motion problems (where one or two of the resulting solutions will usually be “trivial”, e.g. constant motion)...
• Finally, reconstructing the overall (vector) solution from the individual solutions for the independent vector coordinate directions...

and answering any questions as usual.

1.8.1: Free Flight Trajectories – Projectile Motion
Perhaps the simplest example of this process adds just one small change to our first example. Instead of dropping a particle straight down let us imagine throwing the ball off of a tower, or firing a cannon, or driving a golf ball off of a tee or shooting a basketball. All of these are examples of projectile motion
– motion under the primary action of gravity where the initial velocity in some horizontal direction is not zero.
Note well that we will necessarily idealize our treatment by (initially) neglecting some of the many things that might affect the trajectory of all of these objects in the real world – drag forces which both slow down e.g. a golf ball and exert “lift” on it that can cause it to hook or slice, the fact that the earth is not really an inertial reference frame and is rotating out underneath the free flight trajectory of a cannonball, creating an apparent deflection of actual projectiles fired by e.g. naval cannons. That is, only gravity near the earth’s surface will act on our ideal particles for now.
The easiest way to teach you how to handle problems of this sort is just to do a few examples
– there are really only three distinct cases one can treat – two rather special ones and the general solution. Let’s start with the simplest of the special ones.

Example 1.8.1: Trajectory of a Cannonball y m v0 mg θ R

x

Figure 9: An idealized cannon, neglecting the drag force of the air. Let x be the horizontal direction and y be the vertical direction, as shown. Note well that F g = −mg y points along one of the
ˆ
coordinate directions while Fx = (Fz =)0 in this coordinate frame.
A cannon fires a cannonball of mass m at an initial speed v0 at an angle θ with respect to the ground as shown in figure 9. Find:
a) The time the cannonball is in the air.
b) The range of the cannonball.

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Week 1: Newton’s Laws

We’ve already done the first step of a good solution – drawing a good figure, selecting and sketching in a coordinate system with one axis aligned with the total force, and drawing and labelling all of the forces (in this case, only one). We therefore proceed to write Newton’s Second Law for both coordinate directions.
Fx

= max = 0

(83)

Fy

d2 y
= may = m 2 = −mg dt (84)

We divide each of these equations by m to obtain two equations of motion, one for x and the other for y: ax =

0

(85)

ay

= −g

(86)

ax =

dvx
=0
dt

(87)

We solve them independently. In x:

The derivative of any constant is zero, so the x-component of the velocity does not change in time.
We find the initial (and hence constant) component using trigonometry: vx (t) = v0x = v0 cos(θ)

(88)

We then write this in terms of derivatives and solve it: dx dt dx = v0 cos(θ) dt

dx

vx =

= v0 cos(θ)

x(t)

= v0 cos(θ)

dt

= v0 cos(θ)t + C

We evaluate C (the constant of integration) from our knowledge that in the coordinate system we selected, x(0) = 0 so that C = 0. Thus: x(t) = v0 cos(θ)t

(89)

The solution in y is more or less identical to the solution that we obtained above dropping a ball, except the constants of integration are different: dvy dt dvy = −g

dvy

ay =

= −

= −g dt g dt

vy (t) = −gt + C ′

(90)

For this problem, we know from trigonometry that: vy (0) = v0 sin(θ)

(91)

vy (t) = −gt + v0 sin(θ)

(92)

so that C ′ = v0 sin(θ) and:

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Week 1: Newton’s Laws
We write vy in terms of the time derivative of y and integrate: dy dt dy = vy (t) = −gt + v0 sin(θ)

dy

=

=

(−gt + v0 sin(θ)) dt
(−gt + v0 sin(θ)) dt

1 y(t) = − gt2 + v0 sin(θ)t + D
2

(93)

Again we use y(0) = 0 in the coordinate system we selected to set D = 0 and get:
1
y(t) = − gt2 + v0 sin(θ)t
2

(94)

Collecting the results from above, our overall solution is thus: x(t) = v0 cos(θ)t
1
y(t) = − gt2 + v0 sin(θ)t
2
vx (t) = v0x = v0 cos(θ)

(95)
(96)

vy (t) = −gt + v0 sin(θ)

(98)

(97)

We know exactly where the cannonball is at all times, and we know exactly what its velocity is as well. Now let’s see how we can answer the equations.
To find out how long the cannonball is in the air, we need to write an algebraic expression that we can use to identify when it hits the ground. As before (dropping a ball) “hitting the ground” in algebra-speak is y(tg ) = 0, so finding tg such that this is true should do the trick:
1
y(tg ) = − gt2 + v0 sin(θ)tg
2 g
1
− gtg + v0 sin(θ) tg
2

=

0

=

0

or tg,1 =

tg,2

=

0
2v0 sin(θ) g (99)
(100)

are the two roots of this (factorizable) quadratic. The first root obviously describes when the ball was fired, so it is the second one we want. The ball hits the ground after being in the air for a time tg,2 =

2v0 sin(θ) g (101)

Now it is easy to find the range of the cannonball, R. R is just the value of x(t) at the time that the cannonball hits!
2v 2 sin(θ) cos(θ)
R = x(tg,2 ) = 0
(102)
g
Using a trig identity one can also write this as:
R=

2 v0 sin(2θ) g (103)

The only reason to do this is so that one can see that the range of this projectile is symmetric: It is the same for θ = π/4 ± φ for any φ ∈ [0, π/4].

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Week 1: Newton’s Laws

For your homework you will do a more general case of this, one where the cannonball (or golf ball, or arrow, or whatever) is fired off of the top of a cliff of height H. The solution will proceed identically except that the initial and final conditions may be different. In general, to find the time and range in this case one will have to solve a quadratic equation using the quadratic formula (instead of simple factorization) so if you haven’t reviewed or remembered the quadratic formula before now in the course, please do so right away.

1.8.2: The Inclined Plane
The inclined plane is another archetypical problem for motion in two dimensions. It has many variants. We’ll start with the simplest one, one that illustrates a new force, the normal force. Recall from above that the normal force is whatever magnitude it needs to be to prevent an object from moving in to a solid surface, and is always perpendicular (normal) to that surface in direction.
In addition, this problem beautifully illustrates the reason one selects coordinates aligned with the total force when that direction is consistent throughout a problem, if at all possible.

Example 1.8.2: The Inclined Plane

y θ N m mg
H

L

θ x Figure 10: This is the naive/wrong coordinate system to use for the inclined plane problem. The problem can be solved in this coordinate frame, but the solution (as you can see) would be quite difficult. A block m rests on a plane inclined at an angle of θ with respect to the horizontal. There is no friction (yet), but the plane exerts a normal force on the block that keeps it from falling straight down. At time t = 0 it is released (at a height H = L sin(θ) above the ground), and we might then be asked any of the “usual” questions – how long does it take to reach the ground, how fast is it going when it gets there and so on.
The motion we expect is for the block to slide down the incline, and for us to be able to solve the problem easily we have to use our intuition and ability to visualize this motion to select the best coordinate frame.
Let’s start by doing the problem foolishly. Note well that in principle we actually can solve the problem set up this way, so it isn’t really wrong, but in practice while I can solve it in this frame
(having taught this course for 30 years and being pretty good at things like trig and calculus) it is somewhat less likely that you will have much luck if you haven’t even used trig or taken a derivative

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Week 1: Newton’s Laws

for three or four years. Kids, Don’t Try This at Home46 ...
In figure 10, I’ve drawn a coordinate frame that is lined up with gravity. However, gravity is not the only force acting any more. We expect the block to slide down the incline, not move straight down. We expect that the normal force will exert any force needed such that this is so. Let’s see what happens when we try to decompose these forces in terms of our coordinate system.
We start by finding the components of N , the vector normal force, in our coordinate frame:
Nx

= N sin(θ)

(104)

Ny

= N cos(θ)

(105)

where N = |N | is the (unknown) magnitude of the normal force.
We then add up the total forces in each direction and write Newton’s Second Law for each direction’s total force :
Fx

= N sin(θ) = max

(106)

Fy

= N cos(θ) − mg = may

(107)

Finally, we write our equations of motion for each direction: ax =

ay

=

N sin(θ) m N cos(θ) − mg m (108)
(109)

Unfortunately, we cannot solve these two equations as written yet. That is because we do not know the value of N ; it is in fact something we need to solve for! To solve them we need to add a condition on the solution, expressed as an equation. The condition we need to add is that the motion is down the incline, that is, at all times: y(t) = tan(θ)
L cos(θ) − x(t)

(110)

must be true as a constraint47 . That means that: y(t) dy(t) dt d2 y(t) dt2 ay

=

(L cos(θ) − x(t)) tan(θ) dx(t) = − tan(θ) dt d2 x(t)
= − tan(θ) dt2
= −ax tan(θ)

(111)

where we used the fact that the time derivative of L cos(θ) is zero! We can use this relation to eliminate (say) ay from the equations above, solve for ax , then backsubstitute to find ay . Both are constant acceleration problems and hence we can easily enough solve them. But yuk! The solutions we get will be so very complicated (at least compared to choosing a better frame), with both x and y varying nontrivially with time.
Now let’s see what happens when we choose the right (or at least a “good”) coordinate frame according to the prescription given. Such a frame is drawn in 11:
As before, we can decompose the forces in this coordinate system, but now we need to find the components of the gravitational force as N = N y is easy! Furthermore, we know that ay = 0 and
ˆ
46 Or

rather, by all means give it a try, especially after reviewing my solution. that the tangent involves the horizontal distance of the block from the lower apex of the inclined plane, x′ = L cos(θ) − x where x is measured, of course, from the origin.
47 Note

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Week 1: Newton’s Laws

y
N
m θ mg

L

H

θ x Figure 11: A good choice of coordinate frame has (say) the x-coordinate lined up with the total force and hence direction of motion. hence Fy = 0.
Fx = mg sin(θ) = max

(112)

Fy = N − mg cos(θ) = may = 0

(113)

We can immediately solve the y equation for:
N = mg cos(θ)

(114)

and write the equation of motion for the x-direction: ax = g sin(θ)

(115)

which is a constant.
From this point on the solution should be familiar – since vy (0) = 0 and y(0) = 0, y(t) = 0 and we can ignore y altogether and the problem is now one dimensional! See if you can find how long it takes for the block to reach bottom, and how fast it is going when it gets there. You should find

that vbottom = 2gH, a familiar result (see the very first example of the dropped ball) that suggests that there is more to learn, that gravity is somehow “special” if a ball can be dropped or slide down from a height H and reach the bottom going at the same speed either way!

1.9: Circular Motion
So far, we’ve solved only two dimensional problems that involved a constant acceleration in some specific direction. Another very general (and important!) class of motion is circular motion. This could be: a ball being whirled around on a string, a car rounding a circular curve, a roller coaster looping-the-loop, a bicycle wheel going round and round, almost anything rotating about a fixed axis has all of the little chunks of mass that make it up going in circles!
Circular motion, as we shall see, is “special” because the acceleration of a particle moving in a circle towards the center of the circle has a value that is completely determined by the geometry of this motion. The form of centripetal acceleration we are about to develop is thus a kinematic relation – not dynamical. It doesn’t matter which force(s) or force rule(s) off of the list above make something actually move around in a circle, the relation is true for all of them. Let’s try to understand this.

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Week 1: Newton’s Laws

v v ∆θ

∆ s = r ∆θ r Figure 12: A way to visualize the motion of a particle, e.g. a small ball, moving in a circle of radius r.
We are looking down from above the circle of motion at a particle moving counterclockwise around the circle. At the moment, at least, the particle is moving at a constant speed v (so that its velocity is always tangent to the circle.

1.9.1: Tangential Velocity
First, we have to visualize the motion clearly. Figure 12 allows us to see and think about the motion of a particle moving in a circle of radius r (at a constant speed, although later we can relax this to instantaneous speed) by visualizing its position at two successive times. The first position (where the particle is solid/shaded) we can imagine as occurring at time t. The second position (empty/dashed) might be the position of the particle a short time later at (say) t + ∆t.
During this time, the particle travels a short distance around the arc of the circle. Because the length of a circular arc is the radius times the angle subtended by the arc we can see that:
∆s = r∆θ

(116)

Note Well! In this and all similar equations θ must be measured in radians, never degrees. In fact, angles measured in degrees are fundamentally meaningless, as degrees are an arbitrary partitioning of the circle. Also note that radians (or degrees, for that matter) are dimensionless – they are the ratio between the length of an arc and the radius of the arc (think 2π is the ratio of the circumference of a circle to its radius, for example).
The average speed v of the particle is thus this distance divided by the time it took to move it: vavg =

∆s
∆θ
=r
∆t
∆t

(117)

Of course, we really don’t want to use average speed (at least for very long) because the speed might be varying, so we take the limit that ∆t → 0 and turn everything into derivatives, but it is much easier to draw the pictures and visualize what is going on for a small, finite ∆t: v = lim r
∆t→0


∆θ
=r
∆t
dt

(118)

This speed is directed tangent to the circle of motion (as one can see in the figure) and we will often refer to it as the tangential velocity. Sometimes I’ll even put a little “t” subscript on it to emphasize the point, as in: dθ vt = r
(119)
dt but since the velocity is always tangent to the trajectory (which just happens to be circular in this case) we don’t really need it.

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Week 1: Newton’s Laws

In this equation, we see that the speed of the particle at any instant is the radius times the rate that the angle is being swept out by by the particle per unit time. This latter quantity is a very, very useful one for describing circular motion, or rotating systems in general. We define it to be the angular velocity : dθ (120) ω= dt
Thus:
v = rω

(121)

v r (122)

or ω= are both extremely useful expressions describing the kinematics of circular motion.

1.9.2: Centripetal Acceleration
∆v = v∆θ

∆θ

v

Figure 13: The velocity of the particle at t and t + ∆t. Note that over a very short time ∆t the speed of the particle is at least approximately constant, but its direction varies because it always has to be perpendicular to r, the vector from the center of the circle to the particle. The velocity swings through the same angle ∆θ that the particle itself swings through in this (short) time.
Next, we need to think about the velocity of the particle (not just its speed, note well, we have to think about direction). In figure 13 you can see the velocities from figure 12 at time t and t + ∆t placed so that they begin at a common origin (remember, you can move a vector anywhere you like as long as the magnitude and direction are preserved).
The velocity is perpendicular to the vector r from the origin to the particle at any instant of time. As the particle rotates through an angle ∆θ, the velocity of the particle also must rotate through the angle ∆θ while its magnitude remains (approximately) the same.
In time ∆t, then, the magnitude of the change in the velocity is:
∆v = v∆θ

(123)

Consequently, the average magnitude of the acceleration is: aavg =

∆v
∆θ
=v
∆t
∆t

(124)

As before, we are interested in the instantaneous value of the acceleration, and we’d also like to determine its direction as it is a vector quantity. We therefore take the limit ∆t → 0 and inspect the figure above to note that the direction in that limit is to the left, that is to say in the negative r direction! (You’ll need to look at both figures, the one representing position and the

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Week 1: Newton’s Laws

other representing the velocity, in order to be able to see and understand this.) The instantaneous magnitude of the acceleration is thus: a = lim v
∆t→0

dθ v2 ∆θ
=v
= vω =
= rω 2
∆t
dt r (125)

where we have substituted equation 122 for ω (with a bit of algebra) to get the last couple of equivalent forms. The direction of this vector is towards the center of the circle.
The word “centripetal” means “towards the center”, so we call this kinematic acceleration the centripetal acceleration of a particle moving in a circle and will often label it: ac = vω =

v2
= rω 2 r (126)

A second way you might see this written or referred to is as the r-component of a vector in plane polar coordinates. In that case “towards the center” is in the −ˆ direction and we could write: r ar = −vω = −

v2
= −rω 2 r (127)

In most actual problems, though, it is easiest to just compute the magnitude ac and then assign the direction in terms of the particular coordinate frame you have chosen for the problem, which might well make “towards the center” be the positive x direction or something else entirely in your figure at the instant drawn.
This is an enormously useful result. Note well that it is a kinematic result – math with units
– not a dynamic result. That is, I’ve made no reference whatsoever to forces in the derivations above; the result is a pure mathematical consequence of motion in a 2 dimensional plane circle, quite independent of the particular forces that cause that motion. The way to think of it is as follows: If a particle is moving in a circle at instantaneous speed v, then its acceleration towards the center of that circle is v 2 /r (or rω 2 if that is easier to use in a given problem).
This specifies the acceleration in the component of Newton’s Second Law that points towards the center of the circle of motion! No matter what forces act on the particle, if it moves in a circle the component of the total force acting on it towards the center of the circle must be mac = mv 2 /r.
If the particle is moving in a circle, then the centripetal component of the total force must have this value, but this quantity isn’t itself a force law or rule! There is no such thing as a “centripetal force”, although there are many forces that can cause a centripetal acceleration in a particle moving in circular trajectory.
Let me say it again, with emphasis: A common mistake made by students is to confuse mv 2 /r with a “force rule” or “law of nature”. It is nothing of the sort. No special/new force “appears” because of circular motion, the circular motion is caused by the usual forces we list above in some combination that add up to mac = mv 2 /r in the appropriate direction. Don’t make this mistake one a homework problem, quiz or exam! Think about this a bit and discuss it with your instructor if it isn’t completely clear.

Example 1.9.1: Ball on a String
At the bottom of the trajectory, the tension T in the string points straight up and the force mg points straight down. No other forces act, so we should choose coordinates such that one axis lines up with these two forces. Let’s use +y vertically up, aligned with the string. Then:
Fy = T − mg = may = m

v2
L

(128)

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Week 1: Newton’s Laws

L
T
m mg v

Figure 14: A ball of mass m swings down in a circular arc of radius L suspended by a string, arriving at the bottom with speed v. What is the tension in the string? or T = mg + m

v2
L

(129)

Wow, that was easy! Easy or not, this simple example is a very useful one as it will form part of the solution to many of the problems you will solve in the next few weeks, so be sure that you understand it. The net force towards the center of the circle must be algebraically equal to mv 2 /r, where I’ve cleverly given you L as the radius of the circle instead of r just to see if you’re paying attention48 .

Example 1.9.2: Tether Ball/Conic Pendulum

θ

L

r

m vt Figure 15: Ball on a rope (a tether ball or conical pendulum). The ball sweeps out a right circular cone at an angle θ with the vertical when launched appropriately.
Suppose you hit a tether ball (a ball on a string or rope, also called a conic pendulum as the rope sweeps out a right circular cone) so that it moves in a plane circle at an angle θ at the end of a string of length L. Find T (the tension in the string) and v, the speed of the ball such that this is true.
48 There is actually an important lesson here as well: Read the problem! I can’t tell you how often students miss points because they don’t solve the problem given, they solve a problem like the problem given that perhaps was a class example or on their homework. This is easily avoided by reading the problem carefully and using the variables and quantities it defines. Read the problem!

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Week 1: Newton’s Laws

We note that if the ball is moving in a circle of radius r = L sin θ, its centripetal acceleration
2
must be ar = − vr . Since the ball is not moving up and down, the vertical forces must cancel. This suggests that we should use a coordinate system with +y vertically up and x in towards the center of the circle of motion, but we should bear in mind that we will also be thinking of the motion in plane polar coordinates in the plane and that the angle θ is specified relative to the vertical! Oooo, head aching, must remain calm and visualize, visualize.
Visualization is aided by a good figure, like the one (without coordinates, you can add them) in figure 15. Note well in this figure that the only “real” forces acting on the ball are gravity and the tension T in the string. Thus in the y-direction we have:
Fy = T cos θ − mg = 0

(130)

and in the x-direction (the minus r-direction, as drawn) we have:
Fx = T sin θ = mar =
Thus

mv 2
.
r

(131)

mg
,
cos θ

(132)

T r sin θ m (133)

gL sin θ tan θ

(134)

T = v2 = or v=

Nobody said all of the answers will be pretty...

1.9.3: Tangential Acceleration
Sometimes we will want to solve problems where a particle speeds up or slows down while moving in a circle. Obviously, this means that there is a nonzero tangential acceleration changing the magnitude of the tangential velocity.
Let’s write F (total) acting on a particle moving in a circle in a coordinate system that rotates
ˆ
along with the particle – plane polar coordinates. The tangential direction is the θ direction, so we will get:
ˆ
(135)
F = Fr r + Ft θ
ˆ
From this we will get two equations of motion (connecting this, at long last, to the dynamics of two dimensional motion): v2 r

Fr

= −m

Ft

= mat = m

(136) dv dt

(137)

The acceleration on the right hand side of the first equation is determined from m, v, and r, but v(t) itself is determined from the second equation. You will use these two equations together to solve the
“bead sliding on a wire” problem in the next week’s homework assignment, so keep this in mind.
That’s about it for the first week. We have more to do, but to do it we’ll need more forces.
Next week we move on to learn some more forces from our list, especially friction and drag forces.
We’ll wrap the week’s work up with a restatement of our solution rubric for “standard” dynamics problems. I would recommend literally ticking off the steps in your mind (and maybe on the paper!) as you work this week’s homework. It will really help you later on!

Week 1: Newton’s Laws

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1.10: Conclusion: Rubric for Newton’s Second Law Problems
a) Draw a good picture of what is going on. In general you should probably do this even if one has been provided for you – visualization is key to success in physics.
b) On your drawing (or on a second one) decorate the objects with all of the forces that act on them, creating a free body diagram for the forces on each object.
c) Write Newton’s Second Law for each object (summing the forces and setting the result to mi ai for each – ith – object) and algebraically rearrange it into (vector) differential equations
2
x of motion (practically speaking, this means solving for or isolating the acceleration ai = ddt2 i of the particles in the equations of motion).
d) Decompose the 1, 2 or 3 dimensional equations of motion for each object into a set of independent 1 dimensional equations of motion for each of the orthogonal coordinates by choosing a suitable coordinate system (which may not be cartesian, for some problems) and using trig/geometry. Note that a “coordinate” here may even wrap around a corner following a string, for example – or we can use a different coordinate system for each particle, as long as we have a known relation between the coordinate systems.
e) Solve the independent 1 dimensional systems for each of the independent orthogonal coordinates chosen, plus any coordinate system constraints or relations. In many problems the constraints will eliminate one or more degrees of freedom from consideration. Note that in most nontrivial cases, these solutions will have to be simultaneous solutions, obtained by e.g. algebraic substitution or elimination.
f) Reconstruct the multidimensional trajectory by adding the vectors components thus obtained back up (for a common independent variable, time).
g) Answer algebraically any questions requested concerning the resultant trajectory.

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Week 1: Newton’s Laws

Homework for Week 1

Problem 1.

Physics Concepts
In order to solve the following physics problems for homework, you will need to have the following physics and math concepts first at hand, then in your long term memory, ready to bring to bear whenever they are needed. Every week (or day, in a summer course) there will be new ones.
To get them there efficiently, you will need to carefully organize what you learn as you go along.
This organized summary will be a standard, graded part of every homework assignment!
Your homework will be graded in two equal parts. Ten points will be given for a complete crossreferenced summary of the physics concepts used in each of the assigned problems. One problem will be selected for grading in detail – usually one that well-exemplifies the material covered that week – for ten more points.
Points will be taken off for egregiously missing concepts or omitted problems in the concept summary. Don’t just name the concepts; if there is an equation and/or diagram associated with the concept, put that down too. Indicate (by number) all of the homework problems where a concept was used.
This concept summary will eventually help you prioritize your study and review for exams! To help you understand what I have in mind, I’m building you a list of the concepts for this week, and indicating the problems that (will) need them as a sort of template, or example. However, Note
Well! You must write up, and hand in, your own version this week as well as all of the other weeks to get full credit.
In the end, if you put your homework assignments including the summaries for each week into a three-ring binder as you get them back, you will have a nearly perfect study guide to go over before all of the exams and the final. You might want to throw the quizzes and hour exams in as well, as you get them back. Remember the immortal words of Edmund Burke: ”Those who don’t know history are destined to repeat it” – know your own “history”, by carefully saving, and going over, your own work throughout this course!
• Writing a vector in cartesian coordinates. For example:
A = Ax x + Ay y + Az z
ˆ
ˆ
ˆ
Used in problems 2,3,4,5,6,7,8,9,10,12
• Decomposition of a vector at some angle into components in a (2D) coordinate system. Given a vector A with length A at angle θ with respect to the x-axis:
Ax = A cos(θ)
Ay = A sin(θ)
Used in problem 5,6,9,10,11,12
• Definition of trajectory, velocity and acceleration of a particle:
The trajectory is the vector x(t), the vector position of the particle as a function of the time.

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Week 1: Newton’s Laws

The velocity of the particle is the (vector) rate at which its position changes as a function of time, or the time derivative of the trajectory:

v=

∆x dx =
∆t
dt

The acceleration is the (vector) rate at which its velocity changes as a function of time, or the time derivative of the velocity:

a=

∆v dv =
∆t
dt

Used in all problems.
• Inertial reference frame
A set of coordinates in which (if you like) the laws of physics that describe the trajectory of particles take their simplest form. In particular a frame in which Newton’s Laws (given below) hold in a consistent manner. A set of coordinates that is not itself accelerating with respect to all of the other non-accelerating coordinate frames in which Newton’s Laws hold.
Used in all problems (when I choose a coordinate system that is an inertial reference frame).
• Newton’s First Law
In an inertial reference frame, an object in motion will remain in motion, and an object at rest will remain at rest, unless acted on by a net force.
If F = 0, then v is a constant vector.
A consequence, as one can see, of Newton’s Second Law. Not used much yet.
• Newton’s Second Law
In an inertial reference frame, the net vector force on an object equals its mass times its acceleration. F = ma
Used in every problem! Very important! Key! Five stars! *****
• Newton’s Third Law
If one object exerts a force on a second object (along the line connecting the two objects), the second object exerts an equal and opposite force on the first.
F ij = −F ji
Not used much yet.
• Differentiating xn dxn = nxn−1 dx Not used much yet.

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Week 1: Newton’s Laws
• Integrating xn dx xn dx =

xn+1 n+1 Used in every problem where we implicitly use kinematic solutions to constant acceleration to find a trajectory.
Problems 2
• The force exerted by gravity near the Earth’s surface
F = −mg y
ˆ
(down).
Used in problems 2,3,4,5,6,8,9,10,11,12
Problems 2
• Centripetal acceleration. ar = −

v2 r Used in problems 11,13
This isn’t a perfect example – if I were doing this by hand I would have drawn pictures to accompany, for example, Newton’s second and third law, the circular motion acceleration, and so on. I also included more concepts than are strictly needed by the problems – don’t hesitate to add important concepts to your list even if none of the problems seem to need them! Some concepts
(like that of inertial reference frames) are ideas and underlie problems even when they aren’t actually/obviously used in an algebraic way in the solution!

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Week 1: Newton’s Laws
Problem 2.

t=0 m, v 0 = 0

mg

H

A ball of mass m is dropped at time t = 0 from the top of the Duke Chapel (which has height
H) to fall freely under the influence of gravity.
a) How long does it take for the ball to reach the ground?
b) How fast is it going when it reaches the ground?
To solve this first problem, be sure that you use the following ritual:
• Draw a good figure – in this case a chapel tower, the ground, the ball falling. Label the distance
H in the figure, indicate the force on the mass with a vector arrow labelled mg pointing down.
This is called a force diagram (or sometimes a free body diagram). Note well! Solutions without a figure will lose points!
• Choose coordinates! In this case you could (for example) put an origin at the bottom of the tower with a y-axis going up so that the height of the object is y(t).
• Write Newton’s second law for the mass.
• Transform it into a (differential) equation of motion. This is the math problem that must be solved. • In this case, you will want to integrate the constant dy dt = vy (t)) to get y(t).

dvy dt = ay = −g to get vy (t), then integrate

• Express the algebraic condition that is true when the mass reaches the ground, and solve for the time it does so, answering the first question.
• Use the answer to the first question (plus your solutions) to answer the second.
The first four steps in this solution will nearly always be the same for Newton’s Law problems.
Once one has the equation of motion, solving the rest of the problem depends on the force law(s) in question, and answering the questions requires a bit of insight that only comes from practice. So practice! 82

Week 1: Newton’s Laws

Problem 3.

t=0 m, v 01 = 0

H v02 m

A baseball of mass m is dropped at time t = 0 from rest (v01 = 0) from the top of the Duke
Chapel (which has height H) to fall freely under the influence of gravity. At the same instant, a second baseball of mass m is thrown up from the ground directly beneath at a speed v02 (so that if the two balls travel far enough, fast enough, they will collide). Neglect drag.
a) Draw a free body diagram for and compute the net force acting on each mass separately.
b) From the equation of motion for each mass, determine their one dimensional trajectory functions, y1 (t) and y2 (t).
c) Sketch a qualitatively correct graph of y1 (t) and y2 (t) on the same set of axes in the case where the two collide before they hit the ground, and draw a second graph of y1 (t) and y2 (t) on a new set of axes in the case where they do not. From your two pictures, determine a criterion for whether or not the two balls will actually collide before they hit the ground. Express this criterion as an algebraic expression (inequality) involving H, g, and v02 .
d) The Duke Chapel is roughly 100 meters high. What (also roughly, you may estimate and don’t need a calculator) is the minimum velocity v0 2 a the second mass must be thrown up in order for the two to collide? Note that you should give an actual numerical answer here. What is the
(again approximate, no calculators) answer in miles per hour, assuming that 1 meter/second
≈ 9/4 miles per hour? Do you think you can throw a baseball that fast?

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Week 1: Newton’s Laws
Problem 4.

F m mg
A model rocket of mass m blasts off vertically from rest at time t = 0 being pushed by an engine that produces a constant thrust force F (up). The engine blasts away for tb seconds and then stops.
Assume that the mass of the rocket remains more or less unchanged during this time, and that the only forces acting are the thrust and gravity near the earth’s surface.
a) Find the height yb and vertical velocity vb that the rocket has reached by the end of the blast at time tb (neglect any drag forces from the air).
b) Find the maximum height ym that the rocket reaches. You may want to reset your clock to be zero at tb , solving for v(t′ ) and y(t′ ) in terms of the reset clock t′ . Your answer may be expressed in terms of the symbols vb and yb (which are now initial data for the second part of the motion after the rocket engine goes off).
c) Find the speed of the rocket as it hits the ground, vg (note that this is a magnitude and won’t need the minus sign). You may find it easiest to express this answer in terms of ym .
d) Sketch v(t) and y(t) for the entire time the rocket is in the air. Indicate and label (on both graphs) tb , tm (the time the rocket reaches its maximum height) and tg (the time it reaches the ground again).
e) Evaluate the numerical value of your algebraic answers to a-c if m = 0.1 kg, F = 5 N, and tb = 3 seconds. You may use g = 10 m/sec2 (now and for the rest of the course) for simplicity.
Note that you will probably want to evaluate the numbers piecewise – find yb and vb , then put these and the other numbers into your algebraic answer for ym , put that answer into your algebraic answer for vg .

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Week 1: Newton’s Laws

Problem 5.

m v0 H θ R

A cannon sits on a horizontal plain. It fires a cannonball of mass m at speed v0 at an angle θ relative to the ground. Find:
a) The maximum height H of the cannonball’s trajectory.
b) The time ta the cannonball is in the air.
c) The range R of the cannonball.
Questions to discuss in recitation: How does the time the cannonball remains in the air depend on its maximum height? If the cannon is fired at different angles and initial speeds, does the cannonballs with the greatest range always remain in the air the longest? Use the trigonometric identity:
2 sin(θ) cos(θ) = sin(2θ) to express your result for the range. For a fixed v0 , how many angles (usually) can you set the cannon to that will have the same range?

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Week 1: Newton’s Laws
Problem 6.

v0

m

θ

ymax
H

R

A cannon sits on at the top of a rampart of height (to the mouth of the cannon) H. It fires a cannonball of mass m at speed v0 at an angle θ relative to the ground. Find:
a) The maximum height ymax of the cannonball’s trajectory.
b) The time the cannonball is in the air.
c) The range of the cannonball.
Discussion: In your solution to b) above you should have found two times, one of them negative.
What does the negative time correspond to? (Does our mathematical solution “know” about the actual prior history of the cannonball?
You might find the quadratic formula useful in solving this problem. We will be using this a lot in this course, and on a quiz or exam you won’t be given it, so be sure that you really learn it now in case you don’t know or have forgotten it. The roots of a quadratic: ax2 + bx + c = 0 are √

b2 − 4ac
2a
You can actually derive this for yourself if you like (it helps you remember it). Just divide the whole equation by a and complete the square by adding and subtracting the right algebraic quantities, then factor. x= −b ±

86

Week 1: Newton’s Laws

Problem 7.

Fx = −kx

k

m

eq

x

A mass m on a frictionless table is connected to a spring with spring constant k (so that the force on it is Fx = −kx where x is the distance of the mass from its equilibrium postion. It is then pulled so that the spring is stretched by a distance x from its equilibrium position and at t = 0 is released. Write Newton’s Second Law and solve for the acceleration. Solve for the acceleration and write the result as a second order, homogeneous differential equation of motion for this system.
Discussion in your recitation group: Based on your experience and intuition with masses on springs, how do you expect the mass to move in time? Since x(t) is not constant, and a is proportional to x(t), a is a function of time! Do you expect the solution to resemble the kinds of solutions you derived in constant acceleration problems above at all ?
The moral of this story is that not everything moves under the influence of a constant force! If the force/acceleration vary in time, we cannot use e.g. the constant acceleration solution x(t) = 1 at2 !
2
Yet this is a very common mistake made by intro physics students, often as late as the final exam. Try to make sure that you are not one of them!

87

Week 1: Newton’s Laws
Problem 8.

m1

m2

A mass m1 is attached to a second mass m2 by an Acme (massless, unstretchable) string. m1 sits on a frictionless table; m2 is hanging over the ends of a table, suspended by the taut string from an Acme (frictionless, massless) pulley. At time t = 0 both masses are released.
a) Draw the force/free body diagram for this problem.
b) Find the acceleration of the two masses.
c) The tension T in the string.
d) How fast are the two blocks moving when mass m2 has fallen a height H (assuming that m1 hasn’t yet hit the pulley)?
Discussion: Your answer should look something like: The total unopposed force acting on the system accelerates both masses. The string just transfers force from one mass to the other so that they accelerate together! This is a common feature to many problems involving multiple masses and internal forces, as we’ll see and eventually formalize.
Also, by this point you should be really internalizing the ritual for finding the speed of something when it has moved some distance while acclerating as in d) above: find the time it takes to move the distance, backsubstitute to find the speed/velocity. We could actually do this once and for all algebraically for constant accelerations and derive a formula that saves these steps:
2
2 v1 − v0 = 2a∆x

However, very soon we will formally eliminate time as a variable altogether from Newton’s Second
Law, and the resulting work-energy theorem is a better version of this same result that will work even for non-constant forces and accelerations (and is the basis of a fundamental law of nature!), so we won’t do this yet.

88

Week 1: Newton’s Laws

Problem 9.

m1

m2

θ

A mass m1 is attached to a second mass m2 > m1 by an Acme (massless, unstretchable) string. m1 sits on a frictionless inclined plane at an angle θ with the horizontal; m2 is hanging over the high end of the plane, suspended by the taut string from an Acme (frictionless, massless) pulley. At time t = 0 both masses are released from rest.
a) Draw the force/free body diagram for this problem.
b) Find the acceleration of the two masses.
c) Find the tension T in the string.
d) How fast are the two blocks moving when mass m2 has fallen a height H (assuming that m1 hasn’t yet hit the pulley)?

89

Week 1: Newton’s Laws
Problem 10.

θ m F
M

A block m is sitting on a frictionless inclined block with mass M at an angle θ0 as shown. With what force F should you push on the large block in order that the small block will remain motionless with respect to the large block and neither slide up nor slide down?
BTW, I made the angle θ0 sit in the upper corner just to annoy you and make you actually think about sines and cosines of angles. This is good for you – don’t just memorize the trig for an inclined plane, understand it! Talk about it in your groups until you do!

90

Week 1: Newton’s Laws

Problem 11.

θ

v

A tether ball of mass m is suspended by a rope of length L from the top of a pole. A youngster gives it a whack so that it moves with some speed v in a circle of radius r = L sin(θ) < L around the pole.
a) Find an expression for the tension T in the rope as a function of m, g, and θ.
b) Find an expression for the speed v of the ball as a function of θ.
Discussion: Why don’t you need to use L or v in order to find the tension T ? Once the tension
T is known, how does it constrain the rest of your solution?
By now you should have covered, and understood, the derivation of the True Fact that if a particle is moving in a circle of radius r, it must have a total acceleration towards the center of the circle of: v2 ac = r This acceleration (or rather, the acceleration times the mass, mac ) is not a force!. The force that produces this acceleration has to come from the many real forces of nature pushing and pulling on the object (in this case tension in the string and/or gravity).

91

Week 1: Newton’s Laws
Problem 12.

v0 θ A researcher aims her tranquiler gun directly at a monkey in a distant tree. Just as she fires, the monkey lets go and drops in free fall towards the ground.
Show that the sleeping dart hits the monkey.
Discussion: There are some unspoken assumptions in this problem. For example, if the gun shoots the dart too slowly (v0 too small), what will really happen? Also, real guns fire a bullet so fast that the trajectory is quite flat. We must neglect drag forces (discussed next chapter) or the problem is absurdly difficult and we could not possibly answer it here. Finally and most importantly, real hunters allow for the drop in their dart/bullet and would aim the gun at a point above the monkey to hit it if it did not drop (the default assumption).
Be at peace. No monkeys, real or virtual, were harmed in this problem.

92

Week 1: Newton’s Laws

Problem 13.

A train engine of mass m is chugging its way around a circular curve of radius R at a constant speed v. Draw a free body/force diagram for the train engine showing all of the forces acting on it.
Evaluate the total vector force acting on the engine as a function of its speed in a plane perpendicular to its velocity v.
You may find the picture above of a train’s wheels useful. Note that they are notched so that they fit onto the rails – the thin rim of metal that rides on the inside of each rail is essential to the train being able to go around a curve and stay on a track!
Draw a schematic picture of the wheel and rail in cross-section and draw in the forces using the force rules we have learned so far that illustrate how a rail can exert both components of the force needed to hold a train up and curve its trajectory around in a circle.
Discussion: What is the mechanical origin of the force responsible for making the train go in a curve without coming off of the track (and for that matter, keeping it on the track in the first place, even when it is going “straight”)? What would happen if there were no rim on the train’s wheels?

Week 2: Newton’s Laws: Continued

93

Optional Problems
The following problems are not required or to be handed in, but are provided to give you some extra things to work on or test yourself with after mastering the required problems and concepts above and to prepare for quizzes and exams.
No optional problems this week.

94

Week 2: Newton’s Laws: Continued

Week 2: Newton’s Laws:
Continued
Summary
We now continue our discussion of dynamics and Newton’s Laws, adding a few more very important force rules to our repertoire. So far our idealizations have carefully excluded forces that bring things to rest as they move, forces that always seem to act to slow things down unless we constantly push on them. The dissipative forces are, of course, ubiquitous and we cannot afford to ignore them for long. We’d also like to return to the issue of inertial reference frames and briefly discuss the topic of pseudoforces introduced in the “weight in an elevator” example above. Naturally, we will also see many examples of the use of these ideas, and will have to do even more problems for homework to make them intelligible.
The ideas we will cover include:
• Static Friction 49 is the force exerted by one surface on another that acts parallel to the surfaces to prevent the two surfaces from sliding.
a) Static friction is as large as it needs to be to prevent any sliding motion, up to a maximum value, at which point the surfaces begin to slide.
b) The maximum force static friction can exert is proportional to both the pressure between the surfaces and the area in contact. This makes it proportional to the product of the pressure and the area, which equals the normal force. We write this as: max fs ≤ fs
= µs N

(138)

where µs is the coefficient of static friction, a dimensionless constant characteristic of the two surfaces in contact, and N is the normal force.
c) The direction of static friction is parallel to the surfaces in contact and opposes the component of the difference between the total force acting on the object in in that plane – it therefore acts to hold the object stationary until the applied force exceeds the maximum max fs . Note that in general it does not matter which direction the applied force points in the plane of contact – static friction usually acts symmetrically to the right or left, backwards or forwards and required to hold an object stationary.
• Kinetic Friction is the force exerted by one surface on another that is sliding across it.
It, also, acts parallel to the surfaces and opposes the direction of relative motion of the two surfaces. That is:
49 Wikipedia: http://www.wikipedia.org/wiki/Friction. This article describes some aspects of friction in more detail than my brief introduction below. The standard model of friction I present is at best an approximate, idealized one.
Wikipedia: http://www.wikipedia.org/wiki/Tribology describes the science of friction and lubrication in more detail.

95

96

Week 2: Newton’s Laws: Continued
a) The force of kinetic friction is proportional to both the pressure between the surfaces and the area in contact. This makes it proportional to the product of the pressure and the area, which equals the normal force. Thus again fk = µk N

(139)

where µk is the coefficient of kinetic friction, a dimensionless constant characteristic of the two surfaces in contact, and N is the normal force.
Note well that kinetic friction equals µk N in magnitude, where static friction is whatever it needs to be to hold the surfaces static up to a maximum of µs N . This is often a point of confusion for students when they first start to solve problems.
b) The direction of kinetic friction is parallel to the surfaces in contact and opposes the relative direction of the sliding surfaces. That is, if the bottom surface has a velocity
(in any frame) of v b and the top frame has a velocity of v t = v b , the direction of kinetic friction on the top object is the same as the direction of the vector −(v t − v b ) = v b − v t .
The bottom surface “drags” the top one in the (relative) direction it slides, as it were
(and vice versa).
Note well that often the circumstances where you will solve problems involving kinetic friction will involve a stationary lower surface, e.g. the ground, a fixed inclined plane, a roadway – all cases where kinetic friction simply opposes the direction of motion of the upper object – but you will be given enough problems where the lower surface is moving and “dragging” the upper one that you should be able to learn to manage them as well.
• Drag Force50 is the “frictional” force exerted by a fluid (liquid or gas) on an object that moves through it. Like kinetic friction, it always opposes the direction of relative motion of the object and the medium: “drag force” equally well describes the force exerted on a car by the still air it moves through and the force exerted on a stationary car in a wind tunnel.
Drag is an extremely complicated force. It depends on a vast array of things including but not limited to:
– The size of the object.
– The shape of the object.
– The relative velocity of the object through the fluid.
– The state of the fluid (e.g. its internal turbulence).
– The density of the fluid.
– The viscosity of the fluid (we will learn what this is later).
– The properties and chemistry of the surface of the object (smooth versus rough, strong or weak chemical interaction with the fluid at the molecular level).
– The orientation of the object as it moves through the fluid, which may be fixed in time or varying in time (as e.g. an object tumbles).
The long and the short of this is that actually computing drag forces on actual objects moving through actual fluids is a serious job of work for fluid engineers and physicists. To obtain mastery in this, one must first study for years, although then one can make a lot of money
(and have a lot of fun, I think) working on cars, jets, turbine blades, boats, and many other things that involve the utilization or minimization of drag forces in important parts of our society. To simplify drag forces to where we learn to understand in general how they work, we will use following idealizations:
50 Wikipedia:

http://www.wikipedia.org/wiki/Drag (physics). This article explains a lot of the things we skim over below, at least in the various links you can follow if you are particularly interested.

97

Week 2: Newton’s Laws: Continued

a) We will only consider smooth, uniform, nonreactive surfaces of convex, bluff objects
(like spheres) or streamlined objects (like rockets or arrows) moving through uniform, stationary fluids where we can ignore or treat separately e.g. bouyant forces.
b) We will wrap up all of our ignorance of the shape and cross-sectional area of the object, the density and viscosity of the fluid, and so on in a single number, b. This dimensioned number will only be actually computable for certain particularly “nice” shapes and so on (see the Wikipedia article on drag linked above) but allows us to treat drag relatively simply. We will treat drag in two limits:
c) Low velocity, non-turbulent (streamlined, laminar) motion leads to Stokes’ drag, described by: F d = −bv
(140)
This is the simplest sort of drag – a drag force directly proportional to the velocity of
(relative) motion of the object through the fluid and oppositely directed.
d) High velocity, turbulent (high Reynolds number) drag that is described by a quadratic dependence on the relative velocity:
F d = −b|v|v

(141)

It is still directed opposite to the relative velocity of the object and the fluid but now is proportional to that velocity squared.
e) In between, drag is a bit of a mess – changing over from one from to the other. We will ignore this transitional region where turbulence is appearing and so on, except to note that it is there and you should be aware of it.
• Pseudoforces in an accelerating frame are gravity-like “imaginary” forces we must add to the real forces of nature to get an accurate Newtonian description of motion in a noninertial reference frame. In all cases it is possible to solve Newton’s Laws without recourse to pseudoforces (and this is the general approach we promote in this textbook) but it is useful in a few cases to see how to proceed to solve or formulate a problem using pseudoforces such as
“centrifugal force” or “coriolis force” (both arising in a rotating frame) or pseudogravity in a linearly accelerating frame. In all cases if one tries to solve force equations in an accelerating frame, one must modify the actual force being exerted on a mass m in an inertial frame by:
F accelerating = F intertial − maframe

(142)

where −maframe is the pseudoforce.
This sort of force is easily exemplified – indeed, we’ve already seen such an example in our treatment of apparent weight in an elevator in the first week/chapter.

2.1: Friction
So far, our picture of natural forces as being the cause of the acceleration of mass seems fairly successful. In time it will become second nature to you; you will intuitively connect forces to all changing velocities. However, our description thus far is fairly simplistic – we have massless strings, frictionless tables, drag-free air. That is, we are neglecting certain well-known and important facts or forces that appear in real-world problems in order to concentrate on “ideal” problems that illustrate the methods simply.
It is time to restore some of the complexity to the problems we solve. The first thing we will add is friction.
Experimentally

98

Week 2: Newton’s Laws: Continued

Applied Force F
Normal Force
Frictional Contact Force

Figure 16: magnified. the degree features in

A cartoon picture representing two “smooth” surfaces in contact when they are highly
Note the two things that contribute to friction – area in actual contact, which regulates of chemical bonding between the surfaces, and a certain amount of “keyholing” where one surface fit into and are physically locked by features in the other.

a) fs ≤ µs |N |. The force exerted by static friction is less than or equal to the coefficient of static friction mus times the magnitude of the normal force exerted on the entire (homogeneous) surface of contact. We will sometimes refer to this maximum possible value of static friction max as fs
= µs |N |. It opposes the component of any (otherwise net) applied force in the plane of the surface to make the total force component parallel to the surface zero as long as it is able to do so (up to this maximum).
b) fk = µk |N |. The force exerted by kinetic friction (produced by two surfaces rubbing against or sliding across each other in motion) is equal to the coefficient of kinetic friction times the magnitude of the normal force exerted on the entire (homogeneous) surface of contact. It opposes the direction of the relative motion of the two surfaces.
c) µk < µs
d) µk is really a function of the speed v (see discussion on drag forces), but for “slow” speeds µk ∼ constant and we will idealize it as a constant throughout this book.
e) µs and µk depend on the materials in “smooth” contact, but are independent of contact area.
We can understand this last observation by noting that the frictional force should depend on the pressure (the normal force/area ≡ N/m2 ) and the area in contact. But then fk = µk P ∗ A = µk

N
∗ A = µk N
A

(143)

and we see that the frictional force will depend only on the total force, not the area or pressure separately. The idealized force rules themselves, we see, are pretty simple: fs ≤ µs N and fk = µk N . Let’s see how to apply them in the context of actual problems.

Example 2.1.1: Inclined Plane of Length L with Friction
In figure 17 the problem of a block of mass m released from rest at time t = 0 on a plane of length L inclined at an angle θ relative to horizontal is once again given, this time more realistically, including the effects of friction. The inclusion of friction enables new questions to be asked that require the

99

Week 2: Newton’s Laws: Continued

y f s,k

N m θ mg H

L

θ x Figure 17: Block on inclined plane with both static and dynamic friction. Note that we still use the coordinate system selected in the version of the problem without friction, with the x-axis aligned with the inclined plane. use of your knowledge of both the properties and the formulas that make up the friction force rules to answer, such as:
a) At what angle θc does the block barely overcome the force of static friction and slide down the incline.? b) Started at rest from an angle θ > θc (so it definitely slides), how fast will the block be going when it reaches the bottom?
To answer the first question, we note that static friction exerts as much force as necessary to max = µs N . We therefore decompose the keep the block at rest up to the maximum it can exert, fs known force rules into x and y components, sum them componentwise, write Newton’s Second Law for both vector components and finally use our prior knowledge that the system remains in static force equilibrium to set ax = ay = 0. We get:
Fx = mg sin(θ) − fs = 0
(for θ ≤ θc and v(0) = 0) and

(144)

Fy = N − mg cos(θ) = 0

(145)

So far, fs is precisely what it needs to be to prevent motion: fs = mg sin(θ)

(146)

N = mg cos(θ)

(147)

while is true at any angle, moving or not moving, from the Fy equation51 .
You can see that as one gradually and gently increases the angle θ, the force that must be exerted by static friction to keep the block in static force equilibrium increases as well. At the same time, the
51 Here

again is an appeal to experience and intuition – we know that masses placed on inclines under the influence of gravity generally do not “jump up” off of the incline or “sink into” the (solid) incline, so their acceleration in the perpendicular direction is, from sheer common sense, zero. Proving this in terms of microscopic interactions would be absurdly difficult (although in principle possible) but as long as we keep our wits about ourselves we don’t have to! 100

Week 2: Newton’s Laws: Continued

normal force exerted by the plane decreases (and hence the maximum force static friction can exert decreases as well. The critical angle is the angle where these two meet; where fs is as large as it can be such that the block barely doesn’t slide (or barely starts to slide, as you wish – at the boundary the slightest fluctuation in the total force suffices to trigger sliding). To find it, we can substitute max fs
= µs Nc where Nc = mg cos(θc ) into both equations, so that the first equation becomes:
Fx = mg sin(θc ) − µs mg cos(θc ) = 0

(148)

at θc . Solving for θc , we get: θc = tan−1 (µs )

(149)

Once it is moving (either at an angle θ > θc or at a smaller angle than this but with the initial condition vx (0) > 0, giving it an initial “push” down the incline) then the block will (probably) accelerate and Newton’s Second Law becomes:
Fx = mg sin(θ) − µk mg cos(θ) = max

(150)

which we can solve for the constant acceleration of the block down the incline: ax = g sin(θ) − µk g cos(θ) = g(sin(θ) − µk cos(θ))

(151)

Given ax , it is now straightforward to answer the second question above (or any of a number of others) using the methods exemplified in the first week/chapter. For example, we can integrate twice and find vx (t) and x(t), use the latter to find the time it takes to reach the bottom, and substitute that time into the former to find the speed at the bottom of the incline. Try this on your own, and get help if it isn’t (by now) pretty easy.
Other things you might think about: Suppose that you started the block at the top of an incline at an angle less than θc but at an initial speed vx (0) = v0 . In that case, it might well be the case that fk > mg sin(θ) and the block would slide down the incline slowing down. An interesting question might then be: Given the angle, µk , L and v0 , does the block come to rest before it reaches the bottom of the incline? Does the answer depend on m or g? Think about how you might formulate and answer this question in terms of the givens.

Example 2.1.2: Block Hanging off of a Table
+y
N

f s,k

m1

+x
T

m 1g

T

m2

+y

m 2g
+x

Figure 18: Atwood’s machine, sort of, with one block resting on a table with friction and the other dangling over the side being pulled down by gravity near the Earth’s surface. Note that we should use an “around the corner” coordinate system as shown, since a1 = a2 = a if the string is unstretchable.

101

Week 2: Newton’s Laws: Continued

Suppose a block of mass m1 sits on a table. The coefficients of static and kinetic friction between the block and the table are µs > µk and µk respectively. This block is attached by an “ideal” massless unstretchable string running over an “ideal” massless frictionless pulley to a block of mass m2 hanging off of the table as shown in figure 18. The blocks are released from rest at time t = 0.
Possible questions include:
a) What is the largest that m2 can be before the system starts to move, in terms of the givens and knowns (m1 , g, µk , µs ...)?
b) Find this largest m2 if m1 = 10 kg and µs = 0.4.
c) Describe the subsequent motion (find a, v(t), the displacement of either block x(t) from its starting position). What is the tension T in the string while they are stationary?
d) Suppose that m2 = 5 kg and µk = 0.3. How fast are the masses moving after m2 has fallen one meter? What is the tension T in the string while they are moving?
Note that this is the first example you have been given with actual numbers. They are there to tempt you to use your calculators to solve the problem. Do not do this! Solve both of these problems algebraically and only at the very end, with the full algebraic answers obtained and dimensionally checked, consider substituting in the numbers where they are given to get a numerical answer. In most of the rare cases you are given a problem with actual numbers in this book, they will be simple enough that you shouldn’t need a calculator to answer them! Note well that the right number answer is worth very little in this course – I assume that all of you can, if your lives (or the lives of others for those of you who plan to go on to be physicians or aerospace engineers) depend on it, can punch numbers into a calculator correctly. This course is intended to teach you how to correctly obtain the algebraic expression that you need to numerically evaluate, not “drill” you in calculator skills52 .
We start by noting that, like Atwood’s Machine and one of the homework problems from the first week, this system is effectively “one dimensional”, where the string and pulley serve to “bend” the contact force between the blocks around the corner without loss of magnitude. I crudely draw such a coordinate frame into the figure, but bear in mind that it is really lined up with the string.
The important thing is that the displacement of both blocks from their initial position is the same, and neither block moves perpendicular to “x” in their (local) “y” direction.
At this point the ritual should be quite familiar. For the first (static force equilibrium) problem we write Newton’s Second Law with ax = ay = 0 for both masses and use static friction to describe the frictional force on m1 :
Fx1

= T − fs = 0

Fy1

= N − m1 g = 0

Fx2

= m2 g − T = 0

Fy2

=

0

(152)

From the second equation, N = m1 g. At the point where m2 is the largest it can be (given m1 and max so on) fs = fs
= µs N = µs m1 g. If we substitute this in and add the two x equations, the T cancels and we get: mmax g − µs m1 g = 0
(153)
2
Thus
mmax = µs m1
2

(154)

52 Indeed, numbers are used as rarely as they are to break you of the bad habit of thinking that a calculator, or computer, is capable of doing your intuitive and formal algebraic reasoning for you, and are only included from time to time to give you a “feel” for what reasonable numbers are for describing everyday things.

102

Week 2: Newton’s Laws: Continued

which (if you think about it) makes both dimensional and physical sense. In terms of the given numbers, m2 > mus m1 = 4 kg is enough so that the weight of the second mass will make the whole system move. Note that the tension T = m2 g = 40 Newtons, from Fx2 (now that we know m2 ).
Similarly, in the second pair of questions m2 is larger than this minimum, so m1 will slide to the right as m2 falls. We will have to solve Newton’s Second Law for both masses in order to obtain the non-zero acceleration to the right and down, respectively:
Fx1

= T − fk = m1 a

Fy1

= N − m1 g = 0

Fx2

= m2 g − T = m2 a

Fy2

=

0

(155)

If we substitute the fixed value for fk = µk N = µk m1 g and then add the two x equations once again (using the fact that both masses have the same acceleration because the string is unstretchable as noted in our original construction of round-the-corner coordinates), the tension T cancels and we get: m2 g − µk m1 g = (m1 + m2 )a
(156)
or a= m2 − µk m1 g m1 + m2

(157)

is the constant acceleration.
This makes sense! The string forms an “internal force” not unlike the molecular forces that glue the many tiny components of each block together. As long as the two move together, these internal forces do not contribute to the collective motion of the system any more than you can pick yourself up by your own shoestrings! The net force “along x” is just the weight of m2 pulling one way, and the force of kinetic friction pulling the other. The sum of these two forces equals the total mass times the acceleration!
Solving for v(t) and x(t) (for either block) should now be easy and familiar. So should finding the time it takes for the blocks to move one meter, and substituting this time into v(t) to find out how fast they are moving at this time. Finally, one can substitute a into either of the two equations of motion involving T and solve for T . In general you should find that T is less than the weight of the second mass, so that the net force on this mass is not zero and accelerates it downward. The tension T can never be negative (as drawn) because strings can never push an object, only pull.
Basically, we are done. We know (or can easily compute) anything that can be known about this system. Example 2.1.3: Find The Minimum No-Skid Braking Distance for a Car
One of the most important everyday applications of our knowledge of static versus kinetic friction is in anti-lock brake systems (ABS)53 ABS brakes are implemented in every car sold in the European
Union (since 2007) and are standard equipment in almost every car sold in the United States, where for reasons known only to congress it has yet to be formally mandated. This is in spite of the fact that road tests show that on average, stopping distances for ABS-equipped cars are some 18 to 35% shorter than non-ABS equipped cars, for all but the most skilled drivers (who still find it difficult to actually beat ABS stopping distances but who can equal them).
One small part of the reason may be that ABS braking “feels strange” as the car pumps the brakes for you 10-16 times per second, making it “pulse” as it stops. This causes drivers unprepared
53 Wikipedia:

http://www.wikipedia.org/wiki/Anti-lock Braking System.

103

Week 2: Newton’s Laws: Continued

a)
N
fs

v0 mg Ds

b)
N
fk

v0 mg Dk

Figure 19: Stopping a car with and without locking the brakes and skidding. The coordinate system
(not drawn) is x parallel to the ground, y perpendicular to the ground, and the origin in both cases is at the point where the car begins braking. In panel a), the anti-lock brakes do not lock and the car is stopped with the maximum force of static friction. In panel b) the brakes lock and the car skids to a stop, slowed by kinetic/sliding friction.

for the feeling to back off of the brake pedal and not take full advantage of the ABS feature, but of course the simpler and better solution is for drivers to educate themselves on the feel of anti-lock brakes in action under safe and controlled conditions and then trust them.
This problem is designed to help you understand why ABS-equipped cars are “better” (safer) than non-ABS-equipped cars, and why you should rely on them to help you stop a car in the minimum possible distance. We achieve this by answer the following questions:
Find the minimum braking distance of a car travelling at speed v0 30 m/sec running on tires with µs = 0.5 and µk = 0.3:
a) equipped with ABS such that the tires do not skid, but rather roll (so that they exert the maximum static friction only);
b) the same car, but without ABS and with the wheels locked in a skid (kinetic friction only)
c) Evaluate these distances for v0 = 30 meters/second (∼ 67 mph), and both for µs = 0.8, µk = 0.7 (reasonable values, actually, for good tires on dry pavement) and for µs = 0.7, µk = 0.3 (not unreasonable values for wet pavement). The latter are, however, highly variable, depending on the kind and conditions of the treads on your car (which provide channels for water to be displaced as a thin film of water beneath the treads lubricates the point of contact between the tire and the road. With luck they will teach you why you should slow down and allow the distance between your vehicle and the next one to stretch out when driving in wet, snowy, or icy conditions.
To answer all of these questions, it suffices to evalute the acceleration of the car given either max = µs N = µs mg (for a car being stopped by peak static friction via ABS) and fk = µk N = fs 104

Week 2: Newton’s Laws: Continued

µk mg. In both cases we use Newton’s Law in the x-direction to find ax :
Fx

= −µ(s,k) N = max

(158)

Fy

= N − mg = may = 0

(159)

x

y

(where µs is for static friction and µk is for kinetic friction), or: max = −µ(s,k) mg

(160)

ax = −µ(s,k) g

(161)

so which is a constant.
We can then easily determine how long a distance D is required to make the car come to rest.
We do this by finding the stopping time ts from: vx (ts ) = 0 = v0 − µ(s,k) gts or: ts =

(162)

v0 µ(s,k) g

(163)

and using it to evaluate:

1
D(s,k) = x(ts ) = − µ(s,k) gt2 + v0 ts
(164)
s
2
I will leave the actual completion of the problem up to you, because doing these last few steps four times will provide you with a valuable lesson that we will exploit shortly to motivate learing about energy, which will permit us to answer questions like this without always having to find times as intermediate algebraic steps.
Note well! The answers you obtain for D (if correctly computed) are reasonable! That is, yes, it can easily take you order of 100 meters to stop your car with an initial speed of 30 meters per second, and this doesn’t even allow for e.g. reaction time. Anything that shortens this distance makes it easier to survive an emergency situation, such as avoiding a deer that “appears” in the middle of the road in front of you at night.

Example 2.1.4: Car Rounding a Banked Curve with Friction
+y
θ

θ

N fs θ

mg

R

+x

Figure 20: Friction and the normal force conspire to accelerate car towards the center of the circle in which it moves, together with the best coordinate system to use – one with one axis pointing in the direction of actual acceleration. Be sure to choose the right coordinates for this problem!
A car of mass m is rounding a circular curve of radius R banked at an angle θ relative to the horizontal. The car is travelling at speed v (say, into the page in figure 20 above). The coefficient of static friction between the car’s tires and the road is µs . Find:

105

Week 2: Newton’s Laws: Continued
a) The normal force exerted by the road on the car.
b) The force of static friction exerted by the road on the tires.

c) The range of speeds for which the car can round the curve successfully (without sliding up or down the incline).
Note that we don’t know fs , but we are certain that it must be less than or equal to µs N in order for the car to successfully round the curve (the third question). To be able to formulate the range problem, though, we have to find the normal force (in terms of the other/given quantities and the force exerted by static friction (in terms of the other quantities), so we start with that.
As always, the only thing we really know is our dynamical principle – Newton’s Second Law – plus our knowledge of the force rules involved plus our experience and intuition, which turn out to be crucial in setting up this problem.
For example, what direction should fs point? Imagine that the inclined roadway is coated with frictionless ice and the car is sitting on it (almost) at rest (for a finite but tiny v → 0). What will happen (if µs = µk = 0)? Well, obviously it will slide down the hill which doesn’t qualify as
‘rounding the curve’ at a constant height on the incline. Now imagine that the car is travelling at an enormous v; what will happen? The car will skid off of the road to the outside, of course. We know (and fear!) that from our own experience rounding curves too fast.
We now have two different limiting behaviors – in the first case, to round the curve friction has to keep the car from sliding down at low speeds and hence must point up the incline; in the second case, to round the curve friction has to point down to keep the car from skidding up and off of the road. We have little choice but to pick one of these two possibilities, solve the problem for that possibility, and then solve it again for the other (which should be as simple as changing the sign of fs in the algebra. I therefore arbitrarily picked fs pointing down (and parallel to, remember) the incline, which will eventually give us the upper limit on the speed v with which we can round the curve.
As always we use coordinates lined up with the eventual direction of F tot and the actual acceleration of the car: +x parallel to the ground (and the plane of the circle of movement with radius R).
We write Newton’s second law: mv 2
R

Fx

= N sin θ + fs cos θ = max =

Fy

= N cos θ − mg − fs sin θ = may = 0

x

y

(165)
(166)

(where so far fs is not its maximum value, it is merely whatever it needs to be to make the car round the curve for a v presumed to be in range) and solve the y equation for N :
N=

mg + fs sin θ cos θ

(167)

substitute into the x equation:
(mg + fs sin θ) tan θ + fs cos θ = and finally solve for fs : fs =
From this we see that if

mv 2
R

− mg tan θ sin θ tan θ + cos θ

mv 2
> mg tan θ
R

mv 2
R

(168)

(169)

(170)

106

Week 2: Newton’s Laws: Continued

or

v2
> tan θ
Rg

(171)
2

v then fs is positive (down the incline), otherwise it is negative (up the incline). When Rg = tan θ, fs = 0 and the car would round the curve even on ice (as you determined in a previous homework problem). max
See if you can use your knowledge of the algebraic form for fs to determine the range of v given µs that will permit the car to round the curve. It’s a bit tricky! You may have to go back max max a couple of steps and find N max (the N associated with fs ) and fs in terms of that N at the same time, because both N and fs depend, in the end, on v...

2.2: Drag Forces viscous friction

turbulence

Pressure increase
Fd
v

Pressure decrease
Figure 21: A “cartoon” illustrating the differential force on an object moving through a fluid. The drag force is associated with a differential pressure where the pressure on the side facing into the
‘wind’ of its passage is higher than the pressure of the trailing/lee side, plus a “dynamic frictional” force that comes from the fluid rubbing on the sides of the object as it passes. In very crude terms, the former is proportional to the cross-sectional area; the latter is proportional to the surface area exposed to the flow. However, the details of even this simple model, alas, are enormously complex.
As we will discuss later in more detail in the week that we cover fluids, when an object is sitting at rest in a fluid at rest with a uniform temperature, pressure and density, the fluid around it presses on it, on average, equally on all sides54 .
Basically, the molecules of the fluid on one side of the object hit it, on average, with as much force per unit area area as molecules on the other side and the total cross-sectional area of the object seen from any given direction or the opposite of that direction is the same. By the time one works out all of the vector components and integrates the force component along any line over the whole surface area of the object, the force cancels. This “makes sense” – the whole system is in average static force equilibrium and we don’t expect a tree to bend in the wind when there is no wind!
When the same object is moving with respect to the fluid (or the fluid is moving with respect to the object, i.e. – there is a wind in the case of air) then we empirically observe that a friction-like force is exerted on the object (and back on the fluid) called drag 55 .
We can make up at least an heuristic description of this force that permits us to intuitively reason about it. As an object moves through a fluid, one expects that the molecules of the fluid will hit
54 We are ignoring variations with bulk fluid density and pressure in e.g. a gravitational field in this idealized statement; later we will see how the field gradient gives rise to buoyancy through Archimedes’ Principle. However, lateral forces perpendicular to the gravitational field and pressure gradient still cancel even then.
55 Wikipedia: http://www.wikipedia.org/wiki/Drag (physics). This is a nice summary and well worth at least glancing at to take note of the figure at the top illustrating the progression from laminar flow and skin friction to highly turbulent flow and pure form drag.

Week 2: Newton’s Laws: Continued

107

on the side facing the direction of motion harder, on the average, then molecules on the other side.
Even though we will delay our formal treatment of fluid pressure until later, we should all be able to understand that these stronger collisions correspond (on average) to a greater pressure on the side of the object moving against the fluid or vice versa, and a lower pressure in the turbulent flow on the far side, where the object is moving away from the “chasing” and disarranged molecules of fluid. This pressure-linked drag force we might expect to be proportional to the cross-sectional area of the object perpendicular to its direction of relative motion through the fluid and is called form drag to indicate its strong dependence on the shape of the object.
However, the fluid that flows over the sides of the object also tends to “stick” to the surface of the object because of molecular interactions that occur during the instant of the molecular collision between the fluid and the surface. These collisions exert transverse “frictional” forces that tend to speed up the recoiling air molecules in the direction of motion of the object and slow the object down.
The interactions can be strong enough to actually “freeze” a thin layer called the boundary layer of the fluid right up next to the object so that the frictional forces are transmitted through successive layers of fluid flowing and different speeds relative to the object. This sort of flow in layers is often called laminar (layered) flow and the frictional force exerted on the object transmitted through the rubbing of the layers on the sides of the object as it passes through the fluid is called skin friction or laminar drag.
Note well: When an object is enlongated and passes through a fluid parallel to its long axis with a comparatively small forward-facing cross section compared to its total area, we say that it is a streamlined object as the fluid tends to pass over it in laminar flow. A streamlined object will often have its total drag dominated by skin friction. A bluff object, in contrast has a comparatively large cross-sectional surface facing forward and will usually have the total drag dominated by form drag. Note that a single object, such as an arrow or piece of paper, can often be streamlined moving through the fluid one way and bluff another way or be crumpled into a different shape with any mix in between. A sphere is considered to be a bluff body, dominated by form drag.
Unfortunately, this is only the beginning of an heuristic description of drag. Drag is a very complicated force, especially when the object isn’t smooth or convex but is rather rough and irregularly shaped, or when the fluid through which it moves is not in an “ideal” state to begin with, when the object itself tumbles as it moves through the fluid causing the drag force to constantly change form and magnitude. Flow over different parts of a single object can be laminar here, or turbulent there (with portions of the fluid left spinning in whirlpool-like eddies in the wake of the object after it passes).
The full Newtonian description of a moving fluid is given by the Navier-Stokes equation56 which is too hard for us to even look at.
We will therefore need to idealize; learn a few nearly universal heuristic rules that we can use to conceptually understand fluid flow for at least simple, smooth, convex geometries.
It would be nice, perhaps, to be able to skip all of this but we can’t, not even for future physicians as opposed to future engineers, physicists or mathematicians. As it happens, the body contains at least two major systems of fluid flow – the vasculature and the lymphatic system – as well as numerous minor ones (the renal system, various sexual systems, even much of the digestive system is at least partly a fluid transport problem). Drag forces play a critical role in understanding blood pressure, heart disease, and lots of other stuff. Sorry, my beloved students, you gotta learn it at least well enough to qualitatively and semi-quantitatively understand it.

56 Wikipedia:

http://www.wikipedia.org/wiki/Navier-Stokes Equation. A partial differential way, way beyond the scope of this course. To give you an idea of how difficult the Navier-Stokes equation is to solve (in all but a few relatively simple geometries) simply demonstrating that solutions to it always exist and are smooth is one of the seven most important questions in mathematics and you could win a million dollar prize if you were to demonstrate it (or offer a proven counterexample).

108

Week 2: Newton’s Laws: Continued

Besides, this section is the key to understanding how to at least in principle fall out of an airplane without a parachute and survive. Drag forces significantly modify the idealized trajectory functions we derived in week 1, so much so that anyone relying on them to aim a cannon would almost certainly consistently miss any target they aimed at using the idealized no-drag trajectories.
Drag is an extremely complicated force. It depends on a vast array of things including but not limited to:
• The size of the object.
• The shape of the object.
• The relative velocity of the object through the fluid.
• The state of the fluid (e.g. its velocity field including any internal turbulence).
• The density of the fluid.
• The viscosity of the fluid (we will learn what this is later).
• The properties and chemistry of the surface of the object (smooth versus rough, strong or weak chemical interaction with the fluid at the molecular level).
• The orientation of the object as it moves through the fluid, which may be fixed in time
(streamlined versus bluff motion) or varying in time (as, for example, an irregularly shaped object tumbles).
To eliminate most of this complexity and end up with “force rules” that will often be quantitatively predictive we will use a number of idealizations. We will only consider smooth, uniform, nonreactive surfaces of convex bluff objects (like spheres) or streamlined objects (like rockets or arrows) moving through uniform, stationary fluids where we can ignore or treat separately the other non-drag (e.g. buoyant) forces acting on the object.
There are two dominant contributions to drag for objects of this sort.
The first, as noted above, is form drag – the difference in pressure times projective area between the front of an object and the rear of an object. It is strongly dependent on both the shape and orientation of an object and requires at least some turbulence in the trailing wake in order to occur.
The second is skin friction, the friction-like force resulting from the fluid rubbing across the skin at right angles in laminar flow.
In this course, we will wrap up all of our ignorance of the shape and cross-sectional area of the object, the density and viscosity of the fluid, and so on into a single number: b. This (dimensioned) number will only be actually computable for certain particularly “nice” shapes, but it allows us to understand drag qualitatively and treat drag semi-quantitatively relatively simply in two important limits. 2.2.1: Stokes, or Laminar Drag
The first is when the object is moving through the fluid relatively slowly and/or is arrow-shaped or rocket-ship-shaped so that streamlined laminar drag (skin friction) is dominant. In this case there is relatively little form drag, and in particular, there is little or no turbulence – eddies of fluid spinning around an axis – in the wake of the object as the presence of turbulence (which we will discuss in more detail later when we consider fluid dynamics) breaks up laminar flow.
This “low-velocity, streamlined” skin friction drag is technically named Stokes’ drag (as Stokes was the first to derive it as a particular limit of the Navier-Stokes equation for a sphere moving

109

Week 2: Newton’s Laws: Continued through a fluid) or laminar drag and has the idealized force rule:
F d = −bl v

(172)

This is the simplest sort of drag – a drag force directly proportional to the velocity of relative motion of the object through the fluid and oppositely directed.
Stokes derived the following relation for the dimensioned number bl (the laminar drag coefficient) that appears in this equation for a sphere of radius R: bl = −6πµR

(173)

where µ is the dynamical viscosity. Different objects will have different laminar drag coefficients bl , and in general it will be used as a simple given parameter in any problem involving Stokes drag.
Sadly – sadly because Stokes drag is remarkably mathematically tractable compared to e.g. turbulent drag below – spheres experience pure Stokes drag only when they are very small or moving very slowly through the fluid. To given you an idea of how slowly – a sphere moving at
1 meter per second through water would have to be on the order of one micron (a millionth of a meter) in size in order to experience predominantly laminar/Stokes drag. Equivalently, a sphere a meter in diameter would need to be moving at a micron per second. This is a force that is relevant for bacteria or red blood cells moving in water, but not too relevant to baseballs.
It becomes more relevant for streamlined objects – objects whose length along the direction of motion greatly exceeds the characteristic length of the cross-sectional area perpendicular to this direction. We will therefore still find it useful to solve a few problems involving Stokes drag as it will be highly relevant to our eventual studies of harmonic oscillation and is not irrelevant to the flow of blood in blood vessels.

2.2.2: Rayleigh, or Turbulent Drag
On the other hand, if one moves an object through a fluid too fast – where the actual speed depends in detail on the actual size and shape of the object, how bluff or streamlined it is – pressure builds up on the leading surface and turbulence 57 appears in its trailing wake in the fluid (as illustrated in figure 21 above) when the Reynolds number Re of the relative motion (which is a function of the relative velocity, the kinetic viscosity, and the characteristic length of the object) exceeds a critical threshold. Again, we will learn more about this (and perhaps define the Reynolds number) later – for the moment it suffices to know that most macroscopic objects moving through water or air at reasonable velocities experience turbulent drag, not Stokes drag.
This high velocity, turbulent drag exerts a force described by a quadratic dependence on the relative velocity due to Lord Rayleigh:
1
F d = − ρCd A|v|v = −bt |v|v
2

(174)

It is still directed opposite to the relative velocity of the object and the fluid but now is proportional to that velocity squared. In this formula ρ is the density of the fluid through which the object moves (so denser fluids exert more drag as one would expect) and A is the cross-sectional area of the object perpendicular to the direction of motion, also known as the orthographic projection of the object on any plane perpendicular to the motion. For example, for a sphere of radius R, the orthographic projection is a circle of radius R and the area A = πR2 .
57 Wikipedia:

http://www.wikipedia.org/wiki/Turbulence. Turbulence – eddies spun out in the fluid as it moves off of the surface passing throughout it – is arguably the single most complex phenomenon physics attempts to describe, dwarfing even things like quantum field theory in its difficulty. We can “see” a great deal of structure in it, but that structure is fundamentally chaotic and hence subject to things like the butterfly effect. In the end it is very difficult to compute except in certain limiting and idealized cases.

110

Week 2: Newton’s Laws: Continued

The number Cd is called the drag coefficient and is a dimensionless number that depends on relative speed, flow direction, object position, object size, fluid viscosity and fluid density. In other words, the expression above is only valid in certain domains of all of these properties where Cd is slowly varying and can be thought of as a “constant”! Hence we can say that for a sphere moving through still air at speeds where turbulent drag is dominant it is around 0.47 ≈ 0.5, or:
1
ρπR2
(175)
4 which one can compare to bl = 6µπR for the Stokes drag of the same sphere, moving much slower. bt ≈

To get a feel for non-spherical objects, bluff convex objects like potatoes or cars or people have drag coefficients close to but a bit more or less than 0.5, while highly bluff objects might have a drag coefficient over 1.0 and truly streamlined objects might have a drag coefficient as low as 0.04.
As one can see, the functional complexity of the actual non-constant drag coefficient Cd even for such a simple object as a sphere has to manage the entire transition from laminar drag force for low velocities/Reynold’s numbers to turbulent drag for high velocites/Reynold’s numbers, so that at speeds in between the drag force is at best a function of a non-integer power of v in between 1 and
2 or some arcane mixture of form drag and skin friction. We will pretty much ignore this transition.
It is just too damned difficult for us to mess with, although you should certainly be aware that it is there. You can see that in our actual expression for the drag force above, as promised, we have simplified things even more and express all of this dependence – ρ, µ, size and shape and more – wrapped up in the turbulent dimensioned constant bt , which one can think of as an overall turbulent drag coefficient that plays the same general role as the laminar drag coefficient bl we similarly defined above. However, it is impossible for the heuristic descriptors bl and bt to be the same for Stokes’ and turbulent drag – they don’t even have the same units – and for most objects most of the time the total drag is some sort of mixture of these limiting forces, with one or the other (probably) dominant. As you can see, drag forces are complicated! In the end, they turn out to be most useful (to us) as heuristic rules with drag coefficients bl or bt given so that we can see what we can reasonably compute or estimate in these limits.

2.2.3: Terminal velocity

F d v mg Figure 22: A simple object falling through a fluid experiences a drag force that increasingly cancels the force of gravity as the object accelerates until a terminal velocity vt is asymptotically reached.
For bluff objects such as spheres, the Fd = −bv 2 force rule is usually appropriate.
One immediate consequence of this is that objects dropped in a gravitational field in fluids such as air or water do not just keep speeding up ad infinitum. When they are dropped from rest, at first

111

Week 2: Newton’s Laws: Continued

their speed is very low and drag forces may well be negligible58 . The gravitational force accelerates them downward and their speed gradually increases.
As it increases, however, the drag force in all cases increases as well. For many objects the drag forces will quickly transition over to turbulent drag, with a drag force magnitude of bt v 2 . For others, the drag force may remain Stokes’ drag, with a drag force magnitude of bl v (in both cases opposing the directioin of motion through the fluid). Eventually, the drag force will balance the gravitational force and the object will no longer accelerate. It will fall instead at a constant speed. This speed is called the terminal velocity.
It is extremely easy to compute the terminal velocity for a falling object, given the form of its drag force rule. It is the velocity where the net force on the object vanishes. If we choose a coordinate system with “down” being e.g. x positive (so gravity and the velocity are both positive pointing down) we can write either: mg − bl vt

vt

= max = 0 mg = bl or

= max = 0 mg = bt and

(176)

(for Stokes’ drag) or
2
mg − bt vt

vt

(177)

for turbulent drag.
We expect vx (t) to asymptotically approach vt with time. Rather than draw a generic asymptotic curve (which is easy enough, just start with the slope of v being g and bend the curve over to smoothly approach vt ), we will go ahead and see how to solve the equations of motion for at least the two limiting (and common) cases of Stokes’ and turbulent drag.
The entire complicated set of drag formulas above can be reduced to the following “rule of thumb” that applies to objects of water-like density that have sizes such that turbulent drag determines their terminal velocity – raindrops, hail, live animals (including humans) falling in air just above sea level near the surface of the Earth. In this case terminal velocity is roughly equal to

vt = 90 d

(178)

where d is the characteristic size of the object in meters.
For a human body d ≈ 0.6 so vt ≈ 70 meters per second or 156 miles per hour. However, if one falls in a bluff position, one can reduce this to anywhere from 40 to 55 meters per second, say 90 to
120 miles per hour.
Note that the characteristic size of a small animal such as a squirrel or a cat might be 0.05
(squirrel) to 0.1 (cat). Terminal velocity for a cat is around 28 meters per second, lower if the cat falls in a bluff position (say, 50 to 60 mph) and for a squirrel in a bluff position it might be as low as 10 to 20 mph. Smaller animals – especially ones with large bushy tails or skin webs like those observed in the flying squirrel 59 – have a much lower terminal velocity than (say) humans and hence have a much better chance of survival. One rather imagines that this provided a direct evolutionary path to actual flight for small animals that lived relatively high above the ground in arboreal niches.
58 In air and other low viscosity, low density compressible gases they probably are; in water or other viscous, dense, incompressible liquids they may not be.
59 Wikipedia: http://www.wikipedia.org/wiki/Flying Squirrel. A flying squirrel doesn’t really fly – rather it skydives in a highly bluff position so that it can glide long transverse distances and land with a very low terminal velocity.

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Week 2: Newton’s Laws: Continued

Example 2.2.1: Falling From a Plane and Surviving
As noted above, the terminal velocity for humans in free fall near the Earth’s surface is (give or take, depending on whether you are falling in a streamlined swan dive or falling in a bluff skydiver’s belly flop position) anywhere from 40 to 70 meters per second (90-155 miles per hour). Amazingly, humans can survive60 collisions at this speed.
The trick is to fall into something soft and springy that gradually slows you from high speed to zero without ever causing the deceleration force to exceed 100 times your weight, applied as uniformly as possible to parts of your body you can live without such as your legs (where your odds go up the smalller this multiplier is, of course). It is pretty simple to figure out what kinds of things might do.
Suppose you fall from a large height (long enough to reach terminal velocity) to hit a haystack of height H that exerts a nice, uniform force to slow you down all the way to the ground, smoothly compressing under you as you fall. In that case, your initial velocity at the top is vt , down. In order to stop you before y = 0 (the ground) you have to have a net acceleration −a such that: v(tg )

=

y(tg )

=

0 = vt − atg

(179)

1
0 = H − vt tg − at2
2 g

(180)

If we solve the first equation for tg (something we have done many times now) and substitute it into the second and solve for the magnitude of a, we will get:
2
− vt

= −2aH
2
vt a =
2H

or
(181)

We know also that
Fhaystack − mg = ma or Fhaystack = ma + mg = m(a + g) = mg ′ = m

(182)
2
vt
+g
2H

(183)

Let’s suppose the haystack was H = 1.25 meter high and, because you cleverly landed on it in a
“bluff” position to keep vt as small as possible, you start at the top moving at only vt = 50 meters per second. Then g ′ = a + g is approximately 1009.8 meters/second2 , 103 ‘gees’, and the force the haystack must exert on you is 103 times your normal weight. You actually have a small chance of surviving this stopping force, but it isn’t a very large one.
To have a better chance of surviving, one needs to keep the g-force under 100, ideally well under
100, although a very few people are known to have survived 100 g accelerations in e.g. race car crashes. Since the “haystack” portion of the acceleration needed is inversely proportional to H we can see that a 2.5 meter haystack would lead to 51 gees, a 5 meter haystack would lead to 26 gees, and a 10 meter haystack would lead to a mere 13.5 gees, nothing worse than some serious bruising.
If you want to get up and walk to your press conference, you need a haystack or palette at the mattress factory or thick pine forest that will uniformly slow you over something like 10 or more meters. I myself would prefer a stack of pillows at least 40 meters high... but then I have been known to crack a rib just falling a meter or so playing basketball.
The amazing thing is that a number of people have been reliably documented61 to have survived just such a fall, often with a stopping distance of only a very few meters if that, from falls as high
60 Wikipedia:

http://www.wikipedia.org/wiki/Free fall#Surviving falls. ...and have survived...
This website contains ongoing and constantly updated links to contemporary survivor stories as well as historical ones. It’s a fun read.
61 http://www.greenharbor.com/fffolder/ffresearch.html

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Week 2: Newton’s Laws: Continued

as 18,000 feet. Sure, they usually survive with horrible injuries, but in a very few cases, e.g. falling into a deep bank of snow at a grazing angle on a hillside, or landing while strapped into an airline seat that crashed down through a thick forest canopy they haven’t been particularly badly hurt...
Kids, don’t try this at home! But if you ever do happen to fall out of an airplane at a few thousand feet, isn’t it nice that your physics class helps you have the best possible chance at surviving?

Example 2.2.2: Solution to Equations of Motion for Stokes’ Drag
We don’t have to work very hard to actually find and solve the equations of motion for a streamlined object that falls subject to a Stokes’ drag force.
We begin by writing the total force equation for an object falling down subject to near-Earth gravity and Stokes’ drag, with down being positive: mg − bv = m

dv dt (184)

(where we’ve selected the down direction to be positive in this one-dimensional problem).
We rearrange this to put the velocity derivative by itself, factor out the coefficient of v on the right, divide through the v-term from the right, multiply through by dt, integrate both sides, exponentiate both sides, and set the constant of integration. Of course...
Was that too fast for you62 ? Like this: dv dt dv dt dv v − mg b mg ln v − b mg v− b

= g−

b v m

b mg v− m b b = − dt m b
= − t+C m = −

b

b

= e− m t eC = v0 e− m t

v(t) = v(t) =

b mg + v0 e− m t b b mg 1 − e− m t b (185)

or b v(t) = vt 1 − e− m t

(186)

(where we used the fact that v(0) = 0 to set the constant of integration v0 , which just happened to be vt , the terminal velocity!
Objects falling through a medium under the action of Stokes’ drag experience an exponential approach to a constant (terminal) velocity. This is an enormously useful piece of calculus to master; we will have a number of further opportunities to solve equations of motion this and next semester that are first order, linear, inhomogeneous ordinary differential equations such as this one.
Given v(t) it isn’t too difficult to integrate again and find x(t), if we care to, but in this class we will usually stop here as x(t) has pieces that are both linear and exponential in t and isn’t as
“pretty” as v(t) is.
62 Just

kidding! I know you (probably) have no idea how to do this. That’s why you’re taking this course!

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Week 2: Newton’s Laws: Continued
60

50

v(t)

40

30

20

10

0
0

5

10

15

20

25

30

t
Figure 23: A simple object falling through a fluid experiences a drag force of Fd = −bl v. In the figure above m = 100 kg, g = 9.8 m/sec2 , and bl = 19.6, so that terminal velocity is 50 m/sec.
Compare this figure to figure 25 below and note that it takes a relatively longer time to reach the same terminal velocity for an object of the same mass. Note also that the bl that permits the terminal velocities to be the same is much larger than bt !

2.2.4: Advanced: Solution to Equations of Motion for Turbulent Drag
Turbulent drag is set up exactly the same way that Stokes’ drag is We suppose an object is dropped from rest and almost immediately converts to a turbulent drag force. This can easily happen because it has a bluff shape or an irregular surface together with a large coupling between that surface and the surrounding fluid (such as one might see in the following example, with a furry, fluffy ram).
The one “catch” is that the integral you have to do is a bit difficult for most physics students to do, unless they were really good at calculus. We will use a special method to solve this integral in the example below, one that I commend to all students when confronted by problems of this sort.

Example 2.2.3: Dropping the Ram
The UNC ram, a wooly beast of mass Mram is carried by some naughty (but intellectually curious)
Duke students up in a helicopter to a height H and is thrown out. On the ground below a student armed with a radar gun measures and records the velocity of the ram as it plummets toward the vat of dark blue paint below63 . Assume that the fluffy, cute little ram experiences a turbulent drag force on the way down of −bt v 2 in the direction shown.
In terms of these quantities (and things like g):
a) Describe qualitatively what you expect to see in the measurements recorded by the student
(v(t)).
b) What is the actual algebraic solution v(t) in terms of the givens.
c) Approximately how fast is the fat, furry creature going when it splashes into the paint, more or less permanently dying it Duke Blue, if it has a mass of 100 kg and is dropped from a height
63 Note

well: No real sheep are harmed in this physics problem – this actual experiment is only conducted with soft, cuddly, stuffed sheep...

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Week 2: Newton’s Laws: Continued

Fd
M ram v mg

H

!

hh

hh

aa
Ba

Figure 24: The kidnapped UNC Ram is dropped a height H from a helicopter into a vat of Duke
Blue paint! of 1000 meters, given bt = 0.392 Newton-second2 /meter2 ?
I’ll get you started, at least. We know that:
Fx = mg − bv 2 = ma = m

dv dt (187)

or

mg b b dv v2 −
(188)
= g − v2 = − dt m m b much as before. Also as before, we divide all of the stuff with a v in it to the left, multiply the dt to the right, integrate, solve for v(t), set the constant of integration, and answer the questions. a= I’ll do the first few steps in this for you, getting you set up with a definite integral: v2 vf
0
vf
0

dv
− mg b dv v 2 − mg b dv v 2 − mg b = −

b dt m

tf b dt m 0 b = − tf m = −

(189)

Unfortunately, the remaining integral is one you aren’t likely to remember. I’m not either!
Does this mean that we are done? Not at all! We use the look it up in an integral table method of solving it, also known as the famous mathematician method! Once upon a time famous

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Week 2: Newton’s Laws: Continued

mathematicians (and perhaps some not so famous ones) worked all of this sort of thing out. Once upon a time you and I probably worked out how to solve this in a calculus class. But we forgot (at least I did – I took integral calculus in the spring of 1973, almost forty years ago as I write this). So what the heck, look it up!
We discover that:

√ tanh−1 (x/ a) dx √
=−
(190) x2 − a a Now you know what those rarely used buttons on your calculator are for. We substitute x− > v, a → mg/b, multiply out the mg/b and then take the hyperbolic tangent of both sides and then multiply by mg/b again to get the following result for the speed of descent as a function of time: v(t) =

mg tanh b

gb t m

(191)

This solution is plotted for you as a function of time in figure 25 below.
60

50

v(t)

40

30

20

10

0
0

5

10

15

20

25

30

t
Figure 25: A simple object falling through a fluid experiences a drag force of Fd = −bt v 2 . In the figure above (generated using the numbers given in the ram example), m = 100 kg, g = 9.8 m/sec2 , and bt = 0.392, so that terminal velocity is 50 m/sec. Note that the initial acceleration is g, but that after falling around 14 seconds the object is travelling at a speed very close to terminal velocity.

Since even without drag forces it takes 2H/g ≈ 200 ≈ 14 seconds to fall 1000 meters, it is almost certain that the ram will be travelling at the terminal velocity of 50 m/sec as it hits the paint!
Clearly this is a lot of algebra, but that’s realistic (or more so than Stokes’ drag for most problems). It’s just the way nature really is, tough luck and all that. If we want any consolation, at least we didn’t have to try to integrate over the transition between Stokes’ drag and full-blown turbulent drag for the specific shape of a furry ram being dropped from underneath a helicopter
(that no doubt has made the air it falls through initially both turbulent and beset by a substantial downdraft). Week 2: Newton’s Laws: Continued

117

Real physics is often not terribly easy to compute, but the good thing is that it is still easy enough to understand. Even if we have a hard time answering question b) above, we should all be able to understand and draw a qualitative picture for a) and we should really even be able to guess that the ram is moving at or near terminal velocity by the time it has fallen 1000 meters.

2.3: Inertial Reference Frames – the Galilean Transformation
We have already spoken about coordinate systems, or “frames”, that we need to imagine when we create the mental map between a physics problem in the abstract and the supposed reality that it describes. One immediate problem we face is that there are many frames we might choose to solve a problem in, but that our choice of frames isn’t completely arbitrary. We need to reason out how much freedom we have, so that we can use that freedom to make a “good choice” and select a frame that makes the problem relatively simple.
Students that go on in physics will learn that there is more to this process than meets the eye – the symmetries of frames that preserve certain quantities actually leads us to an understanding of conserved quantities and restricts acceptable physical theories in certain key ways. But even students with no particular interest in relativity theory or quantum theory or advanced classical mechanics
(where all of this is developed) have to understand the ideas developed in this section, simply to be able to solve problems efficiently.

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Week 2: Newton’s Laws: Continued

2.3.1: Time
Let us start by thinking about time. Suppose that you wish to time a race (as physicists). The first thing one has to do is understand what the conditions are for the start of the race and the end of the race. The start of the race is the instant in time that the gun goes off and the racers (as particles) start accelerating towards the finish line. This time is concrete, an actual event that you can “instantly” observe64 . Similarly, the end of the race is the instant in time that the racers (as particles) cross the finish line.
Consider three observers timing the same racer. One uses a “perfect” stop watch, one that is triggered by the gun and stopped by the racer crossing the finish line. The race starts at time t = 0 on the stop watch, and stops at time tf , the time it took the racer to complete the race.
The second doesn’t have a stop watch – she has to use their watch set to local time. When the gun goes off she records t0 , the time her watch reads at the start of the race. When the racer crosses the finish line, she records t1 , the finish time in local time coordinates. To find the time of the race, she converts her watch’s time to seconds and subtracts to obtain tf = t1 − t0 , which must non-relativistically65 agree with the first observer.
The third has just arrived from India, and hasn’t had time to reset his watch. He records t′ for
0
the start, t′ for the finish, and subtracts to once again obtain tf = t′ − t′ .
0
1
1
All three of these times must agree because clearly the time required for the racer to cross the finish line has nothing to do with the observers. We could use any clock we wished, set to any time zone or started so that “t = 0” occurs at any time you like to time the race as long as it records times in seconds accurately. In physics we express this invariance by stating that we can change clocks at will when considering a particular problem by means of the transformation: t′ = t − t0

(192)

where t0 is the time in our old time-coordinate frame that we wish to be zero in our new, primed frame. This is basically a linear change of variables, a so-called “u-substitution” in calculus, but because we shift the “zero” of our clock in all cases by a constant, it is true that: dt′ = dt

(193)

so differentiation by t′ is identical to differentiation by t and:
F =m

d2 x d2 x
= m ′2
2
dt dt (194)

That is, Newton’s second law is invariant under uniform translations of time, so we can start our clocks whenever we wish and still accurately describe all motion relative to that time.

2.3.2: Space
We can reason the same way about space. If we want to measure the distance between two points on a line, we can do so by putting the zero on our meter stick at the first and reading off the distance of the second, or we can put the first at an arbitrary point, record the position of the second, and subtract to get the same distance. In fact, we can place the origin of our coordinate system anywhere we like and measure all of our locations relative to this origin, just as we can choose to start our clock at any time and measure all times relative to that time.
64 For the purpose of this example we will ignore things like the speed of sound or the speed of light and assume that our observation of the gun going off is instantaneous.
65 Students not going on in physics should just ignore this adverb. Students going on in physics should be aware that the real, relativistic Universe those times might not agree.

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Week 2: Newton’s Laws: Continued
Displacing the origin is described by: x′ = x − x0

(195)

dx′ = dx

(196)

and as above, so differentiating by x is the same as differentiating by a displaced x′ .
However, there is another freedom we have in coordinate transformations. Suppose you are driving a car at a uniform speed down the highway. Inside the car is a fly, flying from the back of the car to the front of the car. To you, the fly is moving slowly – in fact, if you place a coordinate frame inside the car, you can describe the fly’s position and velocity relative to that coordinate frame very easily. To an observer on the ground, however, the fly is flying by at the speed of the car plus the speed of the car. The observer on the ground can use a coordinate frame on the ground and can also describe the position of the fly perfectly well.
S’

S

x’ x vt

Figure 26: The frame S can be thought of as a coordinate system describing positions relative to the ground, or laboratory, “at rest”. The frame S ′ can be thought of as the coordinate system inside
(say) the car moving at a constant velocity v relative to the coordinate system on the ground. The position of a fly in the ground coordinate frame is the position of the car in the ground frame plus the position of the fly in the coordinate frame inside the car. The position of the fly in the car’s frame is the position of the fly in the ground frame minus the position of the car in the ground frame. This is an easy mental model to use to understand frame transformations.
In figure 26 one can consider the frame S to be the “ground” frame. x is the position of the fly relative to the ground. The frame S ′ is the car, moving at a constant velocity v relative to the ground, and x′ is the position of the fly relative to the car. Repeat the following ritual expression
(and meditate) until it makes sense forwards and backwards:
The position of the fly in the coordinate frame of the ground is the position of the fly in the coordinate frame of the car, plus the position of the car in the coordinate frame of the ground.
In this way we can relate the position of the fly in time in either one of the two frames to its position in the other, as (looking at the triangle of vectors in figure 26): x(t) ′

x (t)

= x′ (t) + v frame t
= x(t) − v frame t

or
(197)

We call the transformation of coordinates in equations 197 from one (inertial) reference frame to another moving at a constant velocity relative to the first the Galilean transformation. Note that we use the fact that the displacement of the origin of the two frames is vt, the velocity of the

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Week 2: Newton’s Laws: Continued

moving frame times time. In a bit we’ll show that this is formally correct, but you probably already understand this pretty well based on your experiences driving cars and the like.
So much for description; what about dynamics? If we differentiate this equation twice, we get: dx dt d2 x dt2 =
=

dx′
+ v frame dt d2 x′ dt2 (198)
(199)

(where we use the fact that the velocity v frame is a constant so that it disappears from the second derivative) so that if we multiply both sides by m we prove:
F =m

d2 x d2 x′
=m 2 dt2 dt

(200)

or Newton’s second law is invariant under the Galilean transformation represented by equation 197 – the force acting on the mass is the same in both frames, the acceleration is the same in both frames, the mass itself is the same in both frames, and so the motion is the same except that the translation of the S ′ frame itself has to be added to the trajectory in the S frame to get the trajectory in the
S ′ frame. It makes sense!
Any coordinate frame travelling at a constant velocity (in which Newton’s first law will thus apparently hold66 ) is called an inertial reference frame, and since our law of dynamics is invariant with respect to changes of inertial frame (as long as the force law itself is), we have complete freedom to choose the one that is the most convenient.
The physics of the fly relative to (expressed in) the coordinate frame in the car are identical to the physics of the fly relative to (expressed in) the coordinate frame on the ground when we account for all of the physical forces (in either frame) that act on the fly.
Equation 197, differentiated with respect to time, can be written as: v ′ = v − v frame

(201)

which you can think of as the velocity relative to the ground is the velocity in the frame plus the velocity of the frame. This is the conceptual rule for velocity transformations: The fly may be moving only at 1 meter per second in the car, but if the car is travelling at 19 meters per second relative to the ground in the same direction, the fly is travelling at 20 meters per second relative to the ground!
The Galilean transformation isn’t the only possible way to relate frames, and in fact it doesn’t correctly describe nature. A different transformation called the Lorentz transformation from the theory of relativity works much better, where both length intervals and time intervals change when changing inertial reference frames. However, describing and deriving relativistic transformations
(and the postulates that lead us to consider them in the first place) is beyond the scope of this course, and they are not terribly important in the classical regime where things move at speeds much less than that of light.
66 This is a rather subtle point, as my colleague Ronen Plesser pointed out to me. If velocity itself is always defined relative to and measured within some frame, then “constant velocity” relative to what frame? The Universe doesn’t come with a neatly labelled Universal inertial reference frame – or perhaps it does, the frame where the blackbody background radiation leftover from the big bang is isotropic – but even if it does the answer is “relative to another inertial reference frame” which begs the question, a very bad thing to do when constructing a consistent physical theory. To avoid this, an inertial reference frame may be defined to be “any frame where Newton’s First Law is true, that is, a frame where objects at rest remain at rest and objects in motion remain in uniform motion unless acted on by an actual external force.

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Week 2: Newton’s Laws: Continued

2.4: Non-Inertial Reference Frames – Pseudoforces
Note that if the frame S ′ is not travelling at a constant velocity and we differentiate equation 201, one more time with respect to time then: dx dt d2 x dt2 =
=

dx′
+ v(t) dt d2 x ′
+ aframe dt2 (202)

or a = a′ + aframe

(203)

Note that the velocity transformation is unchanged from that in an inertial frame – the velocity of the fly relative to the ground is always the velocity of the fly in the car plus the velocity of the car, even if the car is accelerating.
However, the acceleration transformation is now different – to find the acceleration of an object
(e.g. a fly in a car) in the lab/ground frame S we have to add the acceleration of the accelerating frame (the car) in S to the acceleration of in the accelerating frame S’.
Newton’s second law is then not invariant. If S is an inertial frame where Newton’s Second Law is true, then:
F = ma = ma′ + maframe
(204)
We would like to be able to write something that looks like Newton’s Second Law in this frame that can also be solved like Newton’s Second Law in the (accelerating) frame coordinates. That is, we would like to write:

F = ma′
(205)
If we compare these last two equations, we see that this is possible if and only if:


F = F − maframe = F − F p

(206)

where F p is a pseudoforce – a force that does not exist as a force or force rule of nature – that arises within the accelerated frame from the acceleration of the frame.
In the case of uniform frame accelerations, this pseudoforce is proportional to the mass times a the constant acceleration of the frame and behaves a lot like the only force rule we have so far which produces uniform forces proportional to the mass – gravity near Earth’s surface! Indeed, it feels to our senses like gravity has been modified if we ride along in an accelerating frame – made weaker, stronger, changing its direction. However, our algebra above shows that a pseudoforce behaves consistently like that – we can actually solve equations of motion in the accelerating frame using the additional “force rule” of the pseudoforce and we’ll get the right answers within the frame and, when we add the coordinates in the frame to the ground/inertial frame coordinates of the frame, in those coordinates as well.
Pseudoforces are forces which aren’t really there. Why, then, you might well ask, do we deal with them? From the previous paragraph you should be able to see the answer: because it is psychologically and occasionally computationally useful to do so. Psychologically because they describe what we experience in such a frame; computationally because we live in a non-inertial frame (the surface of the rotating earth) and for certain problems it is the solution in the natural coordinates of this non-inertial frame that matters.
We have encountered a few pseudoforces already, either in the course or in our life experience.
We will encounter more in the weeks to come. Here is a short list of places where one experiences pseudoforces, or might find the concept itself useful in the mathematical description of motion in an accelerating frame:

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Week 2: Newton’s Laws: Continued

a) The force added or subtracted to a real force (i.e. – mg, or a normal force) in a frame accelerating uniformly. The elevator and boxcar examples below illustrate this nearly ubiquitous experience. This is the “force” that pushes you back in your seat when riding in a jet as it takes off, or a car that is speeding up. Note that it isn’t a force at all – all that is really happening is that the seat of the car is exerting a normal force on you so that you accelerate at the same rate as the car, but this feels like gravity has changed to you, with a new component added to mg straight down.
b) Rotating frames account for lots of pseudoforces, in part because we live on one (the Earth).
The “centrifugal” force mv 2 /r that apparently acts on an object in a rotating frame is a pseudoforce. Note that this is just minus the real centripetal force that pushes the object toward the center. The centrifugal force is the normal force that a scale might read as it provides the centripetal push. It is not uniform, however – its value depends on your distance r from the axis of rotation!
c) Because of this r dependence, there are slightly different pseudoforces acting on objects falling towards or away from a rotating sphere (such as the earth). These forces describe the apparent deflection of a particle as its straight-line trajectory falls ahead of or behind the rotating frame
(in which the ”rest” velocity is a function of ω and r).
d) Finally, objects moving north or south along the surface of a rotating sphere also experience a similar deflection, for similar reasons. As a particle moves towards the equator, it is suddenly travelling too slowly for its new radius (and constant ω) and is apparently “deflected” west.
As it travels away from the equator it is suddenly traveling too fast for its new radius and is deflected east. These effects combine to produce clockwise rotation of large air masses in the northern hemisphere and anticlockwise rotations in the southern hemisphere.
Note Well: Hurricanes rotate counterclockwise in the northern hemisphere because the counterclockwise winds meet to circulate the other way around a defect at the center. This defect is called the “eye”. Winds flowing into a center have to go somewhere. At the defect they must go up or down. In a hurricane the ocean warms air that rushes toward the center and rises. This warm wet air dumps (warm) moisture and cools. The cool air circulates far out and gets pulled back along the ocean surface, warming as it comes in. A hurricane is a heat engine! There is an optional section on hurricanes down below “just for fun”. I live in North Carolina and teach physics in the summers at the Duke Marine Lab at Beaufort, NC, which is more or less one end of the “bowling alley” where hurricanes spawned off of the coast of Africa eventually come to shore. For me, then, hurricanes are a bit personal – every now and then they come roaring overhead and do a few billion dollars worth of damage and kill people. It’s interesting to understand at least a bit about them and how the rotation of the earth is key to their formation and structure.
The two forces just mentioned (pseudoforces in a rotating spherical frame) are commonly called coriolis forces and are a major driving factor in the time evolution of weather patterns in general, not just hurricanes. They also complicate naval artillery trajectories, missile launches, and other long range ballistic trajectories in the rotating frame, as the coriolis forces combine with drag forces to produce very real and somewhat unpredictable deflections compared to firing right at a target in a presumed cartesian inertial frame. One day, pseudoforces will one day make pouring a drink in a space station that is being rotated to produce a kind of ‘pseudogravity’ an interesting process (hold the cup just a bit antispinward, as things will not – apparently – fall in a straight line!).

2.4.1: Identifying Inertial Frames
We are now finally prepared to tackle a very difficult concept. All of our dynamics so far is based on the notion that we can formulate it in an inertial frame. It’s right there in Newton’s Laws – valid

Week 2: Newton’s Laws: Continued

123

only in inertial frames, and we can now clearly see that if we are not in such a frame we have to account for pseudoforces before we can solve Newtonian problems in that frame.
This is not a trivial question. The Universe doesn’t come with a frame attached – frames are something we imagine, a part of the conceptual map we are trying to build in our minds in an accurate correspondence with our experience of that Universe. When we look out of our window, the world appears flat so we invent a Cartesian flat Earth. Later, further experience on longer length scales reveals that the world is really a curved, approximately spherical object that is only locally flat – a manifold 67 in fact.
Similarly we do simple experiments – suspending masses from strings, observing blocks sliding down inclined planes, firing simple projectiles and observing their trajectories – under the assumption that our experiential coordinates associated with the Earth’s surface form an inertial frame, and
Newton’s Laws appear to work pretty well in them at first. But then comes the day when we fire a naval cannon at a target twenty kilometers to the north of us and all of our shots consistently miss to the east of it because of that pesky coriolis force – the pseudoforce in the rotating frame of the earth and we learn of our mistake.
We learn to be cautious in our system of beliefs. We are always doing our experiments and making our observations in a sort of “box”, a box of limited range and resolution. We have to accept the fact that any set of coordinates we might choose might or might not be inertial, might or might not be “flat”, that at best they might be locally flat and inertial within the box we can reach so far.
In this latter, highly conservative point of view, how do we determine that the coordinates we are using are truly inertial? To put it another way, is there a rest frame for the Universe, a Universal inertial frame S from which we can transform to all other frames S ′ , inertial or not?
The results of the last section provide us with one possible way. If we systematically determine the force laws of nature, Newton tells us that all of those laws involve two objects (at least). I cannot be pushed unless I push back on something. In more appropriate language (although not so conceptually profound) all of the force laws are laws of interaction. I interact with the Earth by means of gravity, and with a knowledge of the force law I can compute the force I exert on the Earth and the force the Earth exerts on me given only a knowledge of our mutual relative coordinates in any coordinate system. Later we will learn that the same is more or less true for the electromagnetic interaction – it is a lot more complicated, in the end, than gravity, but it is still true that a knowledge of the trajectories of two charged objects suffices to determine their electromagnetic interaction, and there is a famous paper by Wheeler and Feynman that suggests that even “radiation reaction”
(something that locally appears as a one-body self-force) is really a consequence of interaction with a Universe of charge pairs.
The point is that in the end, the operational definition of an inertial frame is that it is a frame where Newton’s Laws are true for a closed, finite set of (force) Law of Nature that all involve welldefined interaction in the coordinates of the inertial frame. In that case we can add up all of the actual forces acting on any mass. If the observed movement of that mass is in correspondence with
Newton’s Laws given that total force, the frame must be inertial! Otherwise, there must be at least one “force” that we cannot identify in terms of any interaction pair, and examined closely, it will have a structure that suggests some sort of acceleration of the frame rather than interaction per se with a (perhaps still undiscovered) interaction law.
There is little more that we can do, and even this will not generally suffice to prove that any given frame is truly inertial. For example, consider the “rest frame” of the visible Universe, which can be thought of as a sphere some 13.7 billion Light Years68 in radius surrounding the Earth. To
67 Wikipedia: http://www.wikipedia.org/wiki/manifold. A word that students of physics or mathematics would do well to start learning...
68 Wikipedia: http://www.wikipedia.org/wiki/Light Year. The distance light travels in a year.

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Week 2: Newton’s Laws: Continued

the best of our ability to tell, there is no compelling asymmetry of velocity or relative acceleration within that sphere – all motion is reasonably well accounted for by means of the known forces plus an as yet unknown force, the force associated with “dark matter” and “dark energy”, that still appears to be a local interaction but one we do not yet understand.
How could we tell if the entire sphere were uniformly accelerating in some direction, however?
Note well that we can only observe near-Earth gravity by its opposition – in a freely falling box all motion in box coordinates is precisely what one would expect if the box were not falling! The pseudoforce associated with the motion only appears when relating the box coordinates back to the actual unknown inertial frame.
If all of this gives you a headache, well, it gives me a bit of a headache too. The point is once again that an inertial frame is practically speaking a frame where Newton’s Laws hold, and that while the coordinate frame of the visible Universe is probably the best that we can do for a Universal rest frame, we cannot be certain that it is truly inertial on a much larger length scale – we might be able to detect it if it wasn’t, but then, we might not.
Einstein extended these general meditations upon the invariance of frames to invent first special relativity (frame transformations that leave Maxwell’s Equations form invariant and hence preserve the speed of light in all inertial frames) and then general relativity, which is discussed a bit further below. Example 2.4.1: Weight in an Elevator

a

Figure 27: An elevator accelerates up with a net acceleration of a. The normal force exerted by the
(scale on the floor) of the elevator overcomes the force of gravity to provide this acceleration.
Let’s compute our apparent weight in an elevator that is accelerating up (or down, but say up) at some net rate a. If you are riding in the elevator, you must be accelerating up with the same acceleration. Therefore the net force on you must be
F = ma

(207)

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Week 2: Newton’s Laws: Continued

where the coordinate direction of these forces can be whatever you like, x, y or z, because the problem is really one dimensional and you can name the dimension whatever seems most natural to you. That net force is made up of two “real” forces: The force of gravity which pulls you “down” (in whatever coordinate frame you choose), and the (normal) force exerted by all the molecules in the scale upon the soles of your feet. This latter force is what the scale indicates as your ”weight”69 .
Thus:
F = ma = N − mg

(208)

or
N = W = mg + ma

(209)

where W is your weight.
Note well that we could also write this as:
W = m(g + a) = mg ′

(210)

The elevator is an example of a non-inertial reference frame and its acceleration causes you to experiences something that feels like an additional force of gravity, as if g → g ′ . Similarly, if the elevator accelerates down, gravity g ′ = g − a feels weaker and you feel lighter.

Example 2.4.2: Pendulum in a Boxcar
Tx
m

+y

Ty mg’ θ

+x

T

+y’ a θ

mg

−ma

+x’

Figure 28: A plumb bob or pendulum hangs “at rest” at an angle θ in the frame of a boxcar that is uniformly accelerating to the right.
In figure 28, we see a railroad boxcar that is (we imagine) being uniformly and continuously accelerated to the right at some constant acceleration a = ax x in the (ground, inertial) coordinate
ˆ
frame shown. A pendulum of mass m has been suspended “at rest” (in the accelerating frame of the boxcar) at a stationary angle θ relative to the inertial frame y axis as shown
We would like to be able to answer questions such as:
a) What is the tension T in the string suspending the mass m?
b) What is the angle θ in terms of the givens and knows?
We can solve this problem and answer these questions two ways (in two distinct frames). The first, and I would argue “right” way, is to solve the Newton’s Second Law dynamics problem in the inertial coordinate system corresponding to the ground. This solution (as we will see) is simple
69 Mechanically, a non-digital bathroom scale reads the net force applied to/by its top surface as that force e.g. compresses a spring, which in turn causes a little geared needle to spin around a dial. This will make more sense later, as we come study springs in more detail.

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Week 2: Newton’s Laws: Continued

enough to obtain, but it does make it relatively difficult to relate the answer in ground coordinates
(that isn’t going to be terribly simple) to the extremely simple solution in the primed coordinate system of the accelerating boxcar shown in figure 28. Alternatively, we can solve and answer it directly in the primed accelerating frame – the coordinates you would naturally use if you were riding in the boxcar – by means of a pseudoforce.
Let’s proceed the first way. In this approach, we as usual decompose the tension in the string in terms of the ground coordinate system:
Fx

= Tx = T sin(θ) = max

(211)

Fy

= Ty − mg = T cos(θ) − mg = may = 0

(212)

where we see that ay is 0 because the mass is “at rest” in y as the whole boxcar frame moves only in the x-direction and hence has no y velocity or net acceleration.
From the second equation we get:
T =

mg cos(θ) (213)

and if we substitute this for T into the first equation (eliminating T ) we get: mg tan(θ) = max or tan(θ) =

ax g (214)

(215)

We thus know that θ = tan−1 (ax /g) and we’ve answered the second question above. To answer the first, we look at the right triangle that makes up the vector force of the tension (also from Newton’s
Laws written componentwise above):
Tx

= max

(216)

Ty

= mg

(217)

and find:
T =

2
2
Tx + Ty = m

a2 + g 2 = mg ′ x (218)

where g ′ = a2 + g 2 is the effective gravitation that determines the tension in the string, an idea x that won’t be completely clear yet. At any rate, we’ve answered both questions.
To make it clear, let’s answer them both again, this time using a pseudoforce in the accelerating frame of the boxcar. In the boxcar, according to the work we did above, we expect to have a total effective force:

F = F − maframe
(219)
where F is the sum of the actual force laws and rules in the inertial/ground frame and −maframe is the pseudoforce associated with the acceleration of the frame of the boxcar. In this particular problem this becomes:
− mg ′ y ′ = −mg y − max x
ˆ
ˆ
ˆ
(220) or the magnitude of the effective gravity in the boxcar is mg ′ , and it points “down” in the boxcar frame in the y ′ direction. Finding g ′ from its components is now straightforward:
ˆ
g′ =

a2 + g 2 x (221)

as before and the direction of y ′ is now inclined at the angle
ˆ
θ = tan−1

ax g (222)

127

Week 2: Newton’s Laws: Continued

also as before. Now we get T directly from the one dimensional statics problem along the y ′ direction:
ˆ
T − mg ′ = may′ = 0

(223)

or
T = mg ′ = m

a2 + g 2 x (224)

as – naturally – before. We get the same answer either way, and there isn’t much difference in the work required. I personally prefer to think of the problem, and solve it, in the inertial ground frame, but what you experience riding along in the boxcar is much closer to what the second approach yields – gravity appears to have gotten stronger and to be pointing back at an angle as the boxcar accelerates, which is exactly what one feels standing up in a bus or train as it starts to move, in a car as it rounds a curve, in a jet as it accelerates down the runway during takeoff.
Sometimes (rarely, in my opinion) it is convenient to solve problems (or gain a bit of insight into behavior) using pseudoforces in an accelerating frame (and the latter is certainly in better agreement with our experience in those frames) but it will lead us to make silly and incorrect statements and get problems wrong if we do things carelessly, such as call mv 2 /r a force where it is really just mac , the right and side of Newton’s Second Law where the left hand side is made up of actual force rules.
In this kind of problem and many others it is better to just use the real forces in an inertial reference frame, and we will fairly religiously stick to this in this textbook. As the next discussion (intended only for more advanced or intellectually curious students who want to be guided on a nifty wikiromp of sorts) suggests, however, there is some advantage to thinking more globally about the apparent equivalence between gravity in particular and pseudoforces in accelerating frames.

2.4.2: Advanced: General Relativity and Accelerating Frames
As serious students of physics and mathematics will one day learn, Einstein’s Theory of Special
Relativity70 and the associated Lorentz Transformation71 will one day replace the theory of inertial
“relativity” and the Galilean transformation between inertial reference frames we deduced in week
1. Einstein’s result is based on more or less the same general idea – the laws of physics need to be invariant under inertial frame transformation. The problem is that Maxwell’s Equations (as you will learn in detail in part 2 of this course, if you continue) are the actual laws of nature that describe electromagnetism and hence need to be so invariant. Since Maxwell’s equations predict the speed of light, the speed of light has to be the same in all reference frames!
This has the consequence – which we will not cover in any sort of detail at this time – of causing space and time to become a system of four dimensional spacetime, not three space dimension plus time as an independent variable. Frame transformations nonlinearly mix space coordinates and time as a coordinate instead of just making simple linear tranformations of space coordinates according to “Galilean relativity”.
Spurred by his success, Einstein attempted to describe force itself in terms of curvature of spacetime, working especially on the ubiquitous force of gravity. The idea there is that the pseudoforce produced by the acceleration of a frame is indistinguishible from a gravitational force, and that a generalized frame transformation (describing acceleration in terms of curvature of spacetime) should be able to explain both.
This isn’t quite true, however. A uniformly accelerating frame can match the local magnitude of a gravitational force, but gravitational fields have (as we will learn) a global geometry that cannot be matched by a uniform acceleration – this hypothesis “works” only in small volumes of space where gravity is approximately uniform, for example in the elevator or train above. Nor can one match it
70 Wikipedia:
71 Wikipedia:

http://www.wikipedia.org/wiki/Special Relativity. http://www.wikipedia.org/wiki/Lorentz Transformation.

128

Week 2: Newton’s Laws: Continued

with a rotating frame as the geometric form of the coriolis force that arises in a rotating frame does not match the 1/r2 r gravitational force law.
ˆ
The consequence of this “problem” is that it is considerably more difficult to derive the theory of general relativity than it is the theory of special relativity – one has to work with manifolds 72 .
In a sufficiently small volume Einstein’s hypothesis is valid and gives excellent results that predict sometimes startling but experimentally verified deviations from classical expectations (such as the precession of the perihelion of Mercury)73
The one remaining problem with general relativity – also beyond the scope of this textbook – is its fundamental, deep incompatibility with quantum theory. Einstein wanted to view all forces of nature as being connected to spacetime curvature, but quantum mechanics provides a spectacularly different picture of the cause of interaction forces – the exchange of quantized particles that mediate the field and force, e.g. photons, gluons, heavy vector bosons, and by extension – gravitons74 . So far, nobody has found an entirely successful way of unifying these two rather distinct viewpoints, although there are a number of candidates75 .

72 Wikipedia:

http://www.wikipedia.org/wiki/Manifold. A manifold is a topological curved space that is locally
“flat” in a sufficiently small volume. For example, using a simple cartesian map to navigate on the surface of the
“flat” Earth is quite accurate up to distances of order 10 kilometers, but increasingly inaccurate for distances of order
100 kilometers, where the fact that the Earth’s surface is really a curved spherical surface and not a flat plane begins to matter. Calculus on curved spaces is typically defined in terms of a manifold that covers the space with locally
Euclidean patches. Suddenly the mathematics has departed from the relatively simple calculus and geometry we use in this book to something rather difficult...
73 Wikipedia: http://www.wikipedia.org/wiki/Tests of general relativity. This is one of several “famous” tests of the theory of general relativity, which is generally accepted as being almost correct, or rather, correct in context.
74 Wikipedia: http://www.wikipedia.org/wiki/gravitons. The quantum particle associated with the gravitational field. 75 Wikipedia: http://www.wikipedia.org/wiki/quantum gravity. Perhaps the best known of these is “string theory”, but as this article indicates, there are a number of others, and until dark matter and dark energy are better understood as apparent modifiers of gravitational force we may not be able to experimentally choose between them.

Week 2: Newton’s Laws: Continued

129

2.5: Just For Fun: Hurricanes

Figure 29: Satellite photo of Hurricane Ivan as of September 8, 2004. Note the roughly symmetric rain bands circulating in towards the center and the small but clearly defined “eye”.
Hurricanes are of great interest, at least in the Southeast United States where every fall several of them (on average) make landfall somewhere on the Atlantic or Gulf coast between Texas and
North Carolina. Since they not infrequently do billions of dollars worth of damage and kill dozens of people (usually drowned due to flooding) it is worth taking a second to look over their Coriolis dynamics. In the northern hemisphere, air circulates around high pressure centers in a generally clockwise direction as cool dry air “falls” out of them in all directions, deflecting west as it flows out south and east as it flow out north.
Air circulates around low pressure centers in a counterclockwise direction as air rushes to the center, warms, and lifts. Here the eastward deflection of north-travelling air meets the west deflection of south-travelling air and creates a whirlpool spinning opposite to the far curvature of the incoming air (often flowing in from a circulation pattern around a neighboring high pressure center).
If this circulation occurs over warm ocean water it picks up considerable water vapor and heat.
The warm, wet air cools as it lifts in the central pattern of the low and precipitation occurs, releasing the energy of fusion into the rapidly expanding air as wind flowing out of the low pressure center at high altitude in the usual clockwise direction (the “outflow” of the storm). If the low remains over warm ocean water and no “shear” winds blow at high altitude across the developing eye and interfere with the outflow, a stable pattern in the storm emerges that gradually amplifies into a hurricane with a well defined “eye” where the air has very low pressure and no wind at all.
Figure 32 shows a “snapshot” of the high and low pressure centers over much of North and South
America and the Atlantic on September 8, 2004. In it, two “extreme low” pressure centers are clearly visible that are either hurricanes or hurricane remnants. Note well the counterclockwise circulation around these lows. Two large high pressure regions are also clearly visible, with air circulating around them (irregularly) clockwise. This rotation smoothly transitions into the rotation around the lows across boundary regions.

130

Week 2: Newton’s Laws: Continued
(North−−Counter Clockwise)
Hurricane

N

Eye
Southward
trajectory
North to South deflects west

direction of rotation
Falling deflects east

S

Figure 30:
Figure 31: Coriolis dynamics associated with tropical storms. Air circulating clockwise (from surrounding higher pressure regions) meets at a center of low pressure and forms a counterclockwise
“eye”.
As you can see the dynamics of all of this are rather complicated – air cannot just “flow” on the surface of the Earth – it has to flow from one place to another, being replaced as it flows. As it flows north and south, east and west, up and down, pseudoforces associated with the Earth’s rotation join the real forces of gravitation, air pressure differences, buoyancy associated with differential heating and cooling due to insolation, radiation losses, conduction and convection, and moisture accumulation and release, and more. Atmospheric modelling is difficult and not terribly skilled
(predictive) beyond around a week or at most two, at which point small fluctuations in the initial conditions often grow to unexpectedly dominate global weather patterns, the so-called “butterfly effect”76 .
In the specific case of hurricanes (that do a lot of damage, providing a lot of political and economic incentive to improve the predictive models) the details of the dynamics and energy release are only gradually being understood by virtue of intense study, and at this point the hurricane models are quite good at predicting motion and consequence within reasonable error bars up to five or six days in advance. There is a wealth of information available on the Internet77 to any who wish to learn more. An article78 on the Atlantic Oceanographic and Meteorological Laboratory’s Hurricane
Frequently Asked Questions79 website contains a lovely description of the structure of the eye and the inflowing rain bands.
Atlantic hurricanes usually move from Southeast to Northwest in the Atlantic North of the
76 Wikipedia: http://www.wikipedia.org/wiki/Butterfly Effect. So named because “The flap of a butterfly’s wings in Brazil causes a hurricane in the U.S. some months later.” This latter sort of statement isn’t really correct, of course – many things conspire to cause the hurricane. It is intended to reflect the fact that weather systems exhibit deterministic chaotic dynamics – infinite sensitivity to initial conditions so that tiny differences in initial state lead to radically different states later on.
77 Wikipedia: http://www.wikipedia.org/wiki/Tropical Cyclone.
78 http://www.aoml.noaa.gov/hrd/tcfaq/A11.html
79 http://www.aoml.noaa.gov/hrd/weather sub/faq.html

Week 2: Newton’s Laws: Continued

131

Figure 32: Pressure/windfield of the Atlantic on September 8, 2004. Two tropical storms are visible
– the remnants of Hurricane Frances poised over the U.S. Southeast, and Hurricane Ivan just north of South America. Two more low pressure “tropical waves” are visible between South America and Africa – either or both could develop into tropical storms if shear and water temperature are favorable. The low pressure system in the middle of the Atlantic is extratropical and very unlikely to develop into a proper tropical storm. equator until they hook away to the North or Northeast. Often they sweep away into the North
Atlantic to die as mere extratropical storms without ever touching land. When they do come ashore, though, they can pack winds well over a hundred miles an hour. This is faster than the “terminal velocity” associated with atmospheric drag and thereby they are powerful enough to lift a human or a cow right off their feet, or a house right off its foundations. In addition, even mere “tropical storms” (which typically have winds in the range where wind per se does relatively little damage) can drop a foot of rain in a matter of hours across tens of thousands of square miles or spin down local tornadoes with high and damaging winds. Massive flooding, not wind, is the most common cause of loss of life in hurricanes and other tropical storms.
Hurricanes also can form in the Gulf of Mexico, the Carribean, or even the waters of the Pacific close to Mexico. Tropical cyclones in general occur in all of the world’s tropical oceans except for the Atlantic south of the equator, with the highest density of occurrence in the Western Pacific
(where they are usually called “typhoons” instead of “hurricanes”). All hurricanes tend to be highly unpredictable in their behavior as they bounce around between and around surrounding air pressure ridges and troughs like a pinball in a pinball machine, and even the best of computational models, updated regularly as the hurricane evolves, often err by over 100 kilometers over the course of just a day or two.

132

Week 2: Newton’s Laws: Continued

Homework for Week 2

Problem 1.

Physics Concepts: Make this week’s physics concepts summary as you work all of the problems in this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s) they were key to, and include concepts from previous weeks as necessary. Do the work carefully enough that you can (after it has been handed in and graded) punch it and add it to a three ring binder for review and study come finals!

Problem 2.

m

θ
L

A block of mass m sits at rest on a rough plank of length L that can be gradually tipped up until the block slides. The coefficient of static friction between the block and the plank is µs ; the coefficient of dynamic friction is µk and as usual, µk < µs .
a) Find the angle θ0 at which the block first starts to move.
b) Suppose that the plank is lifted to an angle θ > θ0 (where the mass will definitely slide) and the mass is released from rest at time t0 = 0. Find its acceleration a down the incline.
c) Finally, find the time tf that the mass reaches the lower end of the plank.

133

Week 2: Newton’s Laws: Continued
Problem 3.

r

m1 v m2
A hockey puck of mass m1 is tied to a string that passes through a hole in a frictionless table, where it is also attached to a mass m2 that hangs underneath. The mass is given a push so that it moves in a circle of radius r at constant speed v when mass m2 hangs free beneath the table. Find r as a function of m1 , m2 , v, and g.

Problem 4.

θ m F
M

A small square block m is sitting on a larger wedge-shaped block of mass M at an upper angle θ0 such that the little block will slide on the big block if both are started from rest and no other forces are present. The large block is sitting on a frictionless table. The coefficient of static friction between the large and small blocks is µs . With what range of force F can you push on the large block to the right such that the small block will remain motionless with respect to the large block and neither slide up nor slide down?

134

Week 2: Newton’s Laws: Continued

Problem 5.

m1

m2

A mass m1 is attached to a second mass m2 by an Acme (massless, unstretchable) string. m1 sits on a table with which it has coefficients of static and dynamic friction µs and µk respectively. m2 is hanging over the ends of a table, suspended by the taut string from an Acme (frictionless, massless) pulley. At time t = 0 both masses are released.
a) What is the minimum mass m2,min such that the two masses begin to move?
b) If m2 = 2m2,min , determine how fast the two blocks are moving when mass m2 has fallen a height H (assuming that m1 hasn’t yet hit the pulley)?

Problem 6.

m

v
R

(top view)

towards center θ (side view)

A car of mass m is rounding a banked curve that has radius of curvature R and banking angle
θ. Find the speed v of the car such that it succeeds in making it around the curve without skidding on an extremely icy day when µs ≈ 0.

135

Week 2: Newton’s Laws: Continued
Problem 7.

m

v
R

(top view)

towards center θ (side view)

A car of mass m is rounding a banked curve that has radius of curvature R and banking angle θ.
The coefficient of static friction between the car’s tires and the road is µs . Find the range of speeds v of the car such that it can succeed in making it around the curve without skidding.

Problem 8.

v m You and a friend are working inside a cylindrical new space station that is a hundred meters long and thirty meters in radius and filled with a thick air mixture. It is lunchtime and you have a bag of oranges. Your friend (working at the other end of the cylinder) wants one, so you throw one at him at speed v0 at t = 0. Assume Stokes drag, that is F d = −bv (this is probably a poor assumption depending on the initial speed, but it makes the algebra relatively easy and qualitatively describes the motion well enough).
a) Derive an algebraic expression for the velocity of the orange as a function of time.
b) How long does it take the orange to lose half of its initial velocity?

136

Week 2: Newton’s Laws: Continued

Problem 9.

R m v

A bead of mass m is threaded on a metal hoop of radius R. There is a coefficient of kinetic friction µk between the bead and the hoop. It is given a push to start it sliding around the hoop with initial speed v0 . The hoop is located on the space station, so you can ignore gravity.
a) Find the normal force exerted by the hoop on the bead as a function of its speed.
b) Find the dynamical frictional force exerted by the hoop on the bead as a function of its speed.
c) Find its speed as a function of time. This involves using the frictional force on the bead in
Newton’s second law, finding its tangential acceleration on the hoop (which is the time rate of change of its speed) and solving the equation of motion.
All answers should be given in terms of m, µk , R, v (where requested) and v0 .

137

Week 2: Newton’s Laws: Continued
Problem 10.

m2

m1

A block of mass m2 sits on a rough table. The coefficients of friction between the block and the table are µs and µk for static and kinetic friction respectively. A mass m1 is suspended from an massless, unstretchable, unbreakable rope that is looped around the two pulleys as shown and attached to the support of the rightmost pulley. At time t = 0 the system is released at rest.
a) Find an expression for the minimum mass m1,min such that the masses will begin to move.
b) Suppose m1 = 2m1,min (twice as large as necessary to start it moving). Solve for the accelerations of both masses. Hint: Is there a constraint between how far mass m2 moves when mass m1 moves down a short distance?
c) Find the speed of both masses after the small mass has fallen a distance H. Remember this answer and how hard you had to work to find it – next week we will find it much more easily.

138

Week 2: Newton’s Laws: Continued

Problem 11.

1 m 2 m θ
Two blocks, each with the same mass m but made of different materials, sit on a rough plane inclined at an angle θ such that they will slide (so that the component of their weight down the incline exceeds the maximum force exerted by static friction). The first (upper) block has a coefficient of kinetic friction of µk1 between block and inclined plane; the second (lower) block has coefficient of kinetic friction µk2 . The two blocks are connected by an Acme string.
Find the acceleration of the two blocks a1 and a2 down the incline:
a) when µk1 > µk2 ;
b) when µk2 > µk1 .

139

Week 2: Newton’s Laws: Continued
Advanced Problem 12.

ω

θ

R m A small frictionless bead is threaded on a semicircular wire hoop with radius R, which is then spun on its vertical axis as shown above at angular velocity ω.
a) Find an expression for θ in terms of R, g and ω.
b) What is the smallest angular frequency ωmin such that the bead will not sit at the bottom at θ = 0, for a given R.

140

Week 3: Work and Energy

Optional Problems
The following problems are not required or to be handed in, but are provided to give you some extra things to work on or test yourself with after mastering the required problems and concepts above and to prepare for quizzes and exams.
No optional problems (yet) this week.

Week 3: Work and Energy
Summary
• The Work-Kinetic Energy Theorem in words is “The work done by the total force acting on an object between two points equals the change in its kinetic energy.” As is frequently the case, though, this is more usefully learned in terms of its algebraic forms: x2 W (x1 → x2 ) =

Fx dx =

1
1
2
2
mv2 − mv1 = ∆K
2
2

(225)

F · dℓ =

1
1
2
2
mv2 − mv1 = ∆K
2
2

(226)

x1

in one dimension or x2 W (x1 → x2 ) =

x1

in two or more dimensions, where the integral in the latter is along some specific path between the two endpoints.
• A Conservative Force F c is one where the integral: x2 W (x1 → x2 ) =

x1

F c · dℓ

(227)

does not depend on the particular path taken between x1 and x2 . In that case going from x1 to x2 by one path and coming back by another forms a loop (a closed curve containing both points). We must do the same amount of positive work going one way as we do negative the other way and therefore we can write the condition as:

C

F c · dℓ = 0

(228)

for all closed curves C.
Note Well: If you have no idea what the dot-product in these equations is or how to evaluate it, if you don’t know what an integral along a curve is, it might be a good time to go over the former in the online math review and pay close attention to the pictures below that explain it in context. Don’t worry about this – it’s all part of what you need to learn in the course, and
I don’t expect that you have a particularly good grasp of it yet, but it is definitely something to work on!
• Potential Energy is the negative work done by a conservative force (only) moving between two points. The reason that we bother defining it is because for known, conservative force rules, we can do the work integral once and for all for the functional form of the force and obtain an answer that is (within a constant) the same for all problems! We can then simplify the WorkKinetic Energy Theorem for problems involving those conservative forces, changing them into energy conservation problems (see below). Algebraically:
U (x) = −
141

F c · dℓ + U0

(229)

142

Week 3: Work and Energy where the integral is the indefinite integral of the force and U0 is an arbitrary constant of integration (that may be set by some convention though it doesn’t really have to be, be wary) or else the change in the potential energy is: x1 ∆U (x0 → x1 ) = −

x0

F c · dℓ

(230)

(independent of the choice of path between the points).
• The Law of Conservation of Mechanical Energy states that if no non-conservative forces are acting, the sum of the potential and kinetic energies of an object are constant as the object moves around:
Ei = U0 + K0 = Uf + Kf = Ef
(231)
2 where U0 = U (x0 ), K0 = 1 mv0 etc.
2

• The Generalized Non-Conservative Work-Mechanical Energy Theorem states that if both conservative and non-conservative forces are acting on an object (particle), the work done by the non-conservative forces (e.g. friction, drag) equals the change in the total mechanical energy: x1
F nc · dℓ = ∆Emech = (Uf + Kf ) − (U0 + K0 )
(232)
Wnc = x0 In general, recall, the work done by non-conservative forces depends on the path taken, so the left hand side of this must be explicitly evaluated for a particular path while the right hand side depends only on the values of the functions at the end points of that path.
Note well: This is a theorem only if one considers the external forces acting on a particle.
When one considers systems of particles or objects with many “internal” degrees of freedom, things are not this simple because there can be non-conservative internal forces that (for example) can add or remove macroscopic mechanical energy to/from the system and turn it into microscopic mechanical energy, for example chemical energy or “heat”. Correctly treating energy at this level of detail requires us to formulate thermodynamics and is beyond the scope of the current course, although it requires a good understanding of its concepts to get started.
• Power is the work performed per unit time by a force:

dW dt In many mechanics problems, power is most easily evaluated by means of:
P =

P =

d dℓ F · dℓ = F ·
=F ·v dt dt

(233)

(234)

• An object is in force equilibrium when its potential energy function is at a minimum or maximum. This is because the other way to write the definition of potential energy is:
Fx = −

dU dx (235)

so that if

dU
=0
dx then Fx = 0, the condition for force equilibrium in one dimension.

(236)

For advanced students: In more than one dimension, the force is the negative gradient of the potential energy:
∂U
∂U
∂U
F = −∇U = − x− ˆ y− ˆ z ˆ
(237)
∂x
∂y
∂z

(where ∂x stands for the partial derivative with respect to x, the derivative of the function one takes pretending the other coordinates are constant.

143

Week 3: Work and Energy

• An equilibrium point xe is stable if U (xe ) is a minimum. A mass hanging at rest from a string is at a stable equilibrium at the bottom.
• An equilibrium point xe is unstable if U (xe ) is a maximum. A pencil balanced on its point
(if you can ever manage such a feat) is in unstable equilibrium – the slightest disturbance and it will fall.
• An equilibrium point xe is neutral if U (xe ) is flat to either side, neither ascending or descending. A disk placed on a perfectly level frictionless table is in neutral equilibrium – if it is place at rest, it will remain at rest no matter where you place it, but of course if it has the slightest nonzero velocity it will coast until it either reaches the edge of the table or some barrier that traps it. In the latter sense a perfect neutral equilibrium is often really unstable, as it is essentially impossible to place an object at rest, but friction or drag often conspire to
“stabilize” a neutral equilibrium so that yes, if you put a penny on a table it will be there the next day, unmoved, as far as physics is concerned...

3.1: Work and Kinetic Energy
If you’ve been doing all of the work assigned so far, you may have noticed something. In many of the problems, you were asked to find the speed of an object (or, if the direction was obvious, its velocity) after it moved from some initial position to a final position. The solution strategy you employed over and over again was to solve the equations of motion, solve for the time, substitute the time, find the speed or velocity. We used this in the very first example in the book and the first actual homework problem to show that a mass dropped from rest that falls a height H hits the ground at

speed v = 2gH, but later we discovered that a mass that slides down a frictionless inclined plane

starting from rest a height H above the ground arrives at the ground as a speed 2gH independent of the slope of the incline!
If you were mathematically inclined – or used a different textbook, one with a separate section on the kinematics of constant acceleration motion (a subject this textbook has assiduously avoided, instead requiring you to actually solve the equations of motion using calculus repeatedly and then use algebra as needed to answer the questions) you might have noted that you can actually do the algebra associated with this elimination of time once and for all for a constant acceleration problem in one dimension. It is simple.
If you look back at week 1, you can see if that if you integrate a constant acceleration of an object twice, you obtain: v(t) = at + v0
1 2 x(t) = at + v0 t + x0
2
as a completely general kinematic solution in one dimension, where v0 is the initial speed and x0 is the initial x position at time t = 0.
Now, suppose you want to find the speed v1 the object will have when it reaches position x1 .
One can algebraically, once and for all note that this must occur at some time t1 such that: v(t1 ) = at1 + v0 = v1
1 2 x(t1 ) = at + v0 t1 + x0 = x1
2 1
We can algebraically solve the first equation once and for all for t1 : t1 =

v1 − v0 a (238)

144

Week 3: Work and Energy

and substitute the result into the second equation, elminating time altogether from the solutions:
2

v1 − v 0 v 1 − v0
1
+ x0
+ v0 a 2 a a
2
1 v0 v1 − v0
2
2 v1 − 2v0 v1 + v0 +
2a
a
2
2
2
v1 − 2v0 v1 + v0 + 2v0 v1 − 2v0

= x1
= x1 − x0
=

2a(x1 − x0 )

or
2
2 v1 − v0 = 2a(x1 − x0 )

(239)

Many textbooks encourage students to memorize this equation as well as the two kinematic solutions for constant acceleration very early – often before one has even learned Newton’s Laws – so that students never have to actually learn why these solutions are important or where they come from, but at this point you’ve hopefully learned both of those things well and it is time to make solving problems of this kinds a little bit easier.
However, we will not do so using this constant acceleration kinematic equation even now! There is no need! As we will see below, it is quite simple to eliminate time from Newton’s Second Law itself once and for all, and obtain a powerful way of solving many, many physics problems – in particular, ones where the questions asked do not depend on specific times – without the tedium of integrating out the equations of motion. This “time independent” formulation of force laws and motion turns out, in the end, to be even more general and useful than Newton’s Laws themselves, surviving the transition to quantum theory where the concepts of force and acceleration do not.
One very good thing about waiting as we have done and not memorizing anything, let alone kinematic constant acceleration solutions, is that this new formulation in terms of work and energy works just fine for non-constant forces and accelerations, where the kinematic solutions above are
(as by now you should fully appreciate, having worked through e.g. the drag force and investigated the force exerted by springs, neither of which are constant in space or in time) completely useless and wrong.
Let us therefore begin now with this relatively meaningless kinematical result that arises from eliminating time for a constant acceleration in one dimension only – planning to use it only long enough to ensure that we never have to use it because we’ve found something even better that is far more meaningful :
2
2 v1 − v0 = 2a∆x
(240)
where ∆x is the displacement of the object x1 − x0 .
If we multiply by m (the mass of the object) and move the annoying 2 over to the other side, we can make the constant acceleration a into a constant force Fx = ma:
(ma)∆x

=

Fx ∆x =

1 mv 2 −
2 1
1
mv 2 −
2 1

1 mv 2
2 0
1
mv 2
2 0

(241)
(242)

We now define the work done by the constant force Fx on the mass m as it moves through the distance ∆x to be:
∆W = Fx ∆x.
(243)
The work can be positive or negative.
Of course, not all forces are constant. We have to wonder, then, if this result or concept is as fragile as the integral of a constant acceleration (which does not “work”, so to speak, for springs!) or if it can handle springs, pendulums, real gravity (not near the Earth’s surface) and so on. As you might guess, the answer is yes – we wouldn’t have bothered introducing and naming the concept if

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all we cared about was constant acceleration problems as we already had a satisfactory solution for them – but before we turn this initial result into a theorem that follows directly from the axiom of
Newton’s Second Law made independent of time, we should discuss units of work, energy, and all that. 3.1.1: Units of Work and Energy
Work is a form of energy. As always when we first use a new named quantity in physics, we need to define its units so we can e.g. check algebraic results for kinematic consistency, correctly identify work, and learn to quantitatively appreciated it when people refer to quantities in other sciences or circumstances (such as the energy yield of a chemical reaction, the power consumed by an electric light bulb, or the energy consumed and utilized by the human body in a day) in these units.
In general, the definition of SI units can most easily be remembered and understood from the basic equations that define the quantity of interest, and the units of energy are no exception. Since work is defined above to be a force times a distance, the SI units of energy must be the SI units of force (Newtons) times the SI units of length (meters). The units themselves are named (as many are) after a Famous Physicist, James Prescott Joule80 . Thus:
1 Joule = 1 Newton-meter = 1

kilogram-meter2 second2 (244)

3.1.2: Kinetic Energy
The latter, we also note, are the natural units of mass times speed squared. We observe that this is the quantity that changes when we do work on a mass, and that this energy appears to be a characteristic of the moving mass associated with the motion itself (dependent only on the speed
v). We therefore define the quantity changed by the work to be the kinetic energy 81 and will use the symbol K to represent it in this work:
K=

1 mv 2
2

(245)

Note that kinetic energy is a relative quantity – it depends upon the inertial frame in which it is measured. Suppose we consider the kinetic energy of a block of mass m sliding forward at a constant speed vb in a railroad car travelling at a constant speed vc . The frame of the car is an inertial reference frame and so Newton’s Laws must be valid there. In particular, our definition of kinetic energy that followed from a solution to Newton’s Laws ought to be valid there. It should be equally valid on the ground, but these two quantities are not equal.
Relative to the ground, the speed of the block is: vg = v b + v c

(246)

and the kinetic energy of the block is:
Kg =
80 Wikipedia:

1
1
1
2
2
2
mvg = mvb + mvc + mvb vc
2
2
2

(247)

http://www.wikipedia.org/wiki/James Prescott Joule. He worked with temperature and heat and was one of the first humans on Earth to formulate and initially experimentally verify the Law of Conservation of Energy, discussed below. He also discovered and quantified resistive electrical heating (Joule heating) and did highly precise experiments that showed that mechanical energy delivered into a closed system increased its temperature is the work converted into heat.
81 The work “kinetic” means “related to the motion of material bodies”, although we also apply it to e.g. hyperkinetic people... 146

Week 3: Work and Energy

or

1
2
Kg = Kb + mvc + mvb vc
2
where Kb is the kinetic energy of the block in the frame of the train.

(248)

Worse, the train is riding on the Earth, which is not exactly at rest relative to the sun, so we could describe the velocity of the block by adding the velocity of the Earth to that of the train and the block within the train. The kinetic energy in this case is so large that the difference in the energy of the block due to its relative motion in the train coordinates is almost invisible against the huge energy it has in an inertial frame in which the sun is approximately at rest. Finally, as we discussed last week, the sun itself is moving relative to the galactic center or the “rest frame of the visible Universe”.
What, then, is the actual kinetic energy of the block?
I don’t know that there is such a thing. But the kinetic energy of the block in the inertial
1
reference frame of any well-posed problem is 2 mv 2 , and that will have to be enough for us. As we will prove below, this definition makes the work done by the forces of nature consistent within the frame, so that our computations will give us answers consistent with experiment and experience in the frame coordinates.

3.2: The Work-Kinetic Energy Theorem
Let us now formally state the result we derived above using the new definitions of work and kinetic energy as the Work-Kinetic Energy Theorem (which I will often abbreviate, e.g. WKET) in one dimension in English:
The work done on a mass by the total force acting on it is equal to the change in its kinetic energy. and as an equation that is correct for constant one dimensional forces only:
∆W = Fx ∆x =

1
1
2
2
mvf − mvi = ∆K
2
2

(249)

You will note that in the English statement of the theorem, I didn’t say anything about needing the force to be constant or one dimensional. I did this because those conditions aren’t necessary –
I only used them to bootstrap and motivate a completely general result. Of course, now it is up to us to prove that the theorem is general and determine its correct and general algebraic form. We can, of course, guess that it is going to be the integral of this difference expression turned into a differential expression: dW = Fx dx = dK
(250)
but actually deriving this for an explicitly non-constant force has several important conceptual lessons buried in the derivation. So much so that I’ll derive it in two completely different.

3.2.1: Derivation I: Rectangle Approximation Summation
First, let us consider a force that varies with position so that it can be mathematically described as a function of x, Fx (x). To compute the work done going between (say) x0 and some position xf that will ultimately equal the total change in the kinetic energy, we can try to chop the interval xf −x0 up into lots of small pieces, each of width ∆x. ∆x needs to be small enough that Fx basically doesn’t change much across it, so that we are justified in saying that it is “constant” across each interval, even though the value of the constant isn’t exactly the same from interval to interval. The

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Week 3: Work and Energy

Fx

x 0 x1 x 2 x 3 x 4 x 5 x 6 x 7 x f

x

Figure 33: The work done by a variable force can be approximated arbitrarily accurately by summing
Fx ∆x using the average force as if it were a constant force across each of the “slices” of width ∆x one can divide the entire interval into. In the limit that the width ∆x → dx, this summation turns into the integral.

actual value we use as the constant in the interval isn’t terribly important – the easiest to use is the average value or value at the midpoint of the interval, but no matter what sensible value we use the error we make will vanish as we make ∆x smaller and smaller.
In figure 33, you can see a very crude sketch of what one might get chopping the total interval x0 → xf up into eight pieces such that e.g. x1 = x0 + ∆x, x2 = x1 + ∆x,... and computing the work done across each sub-interval using the approximately constant value it has in the middle of the sub-interval. If we let F1 = Fx (x0 + ∆x/2), then the work done in the first interval, for example, is
F1 ∆x, the shaded area in the first rectangle draw across the curve. Similarly we can find the work done for the second strip, where F2 = Fx (x1 + ∆x/2) and so on. In each case the work done equals the change in kinetic energy as the particle moves across each interval from x0 to xf .
We then sum the constant acceleration Work-Kinetic-Energy theorem for all of these intervals:
Wtot

F1 ∆x + F2 ∆x + . . .

= F1 (x1 − x0 ) + F2 (x2 − x1 ) + . . .
1
1
1
1
= ( mv 2 (x1 ) − mv 2 (x0 )) + ( mv 2 (x2 ) − mv 2 (x1 )) + . . .
2
2
2
2
1
1
2
2
=
mv − mv
2 f 2 i

8

Fi ∆x i=1 =

1
1
2
2
mvf − mvi
2
2

(251)

where the internal changes in kinetic energy at the end of each interval but the first and last cancel.
Finally, we let ∆x go to zero in the usual way, and replace summation by integration. Thus:
Wtot = lim

∆x→0

∞ i=1 xf

Fx dx = ∆K

Fx (x0 + i∆x)∆x =

(252)

x0

and we have generalized the theorem to include non-constant forces in one dimension82 .
This approach is good in that it makes it very clear that the work done is the area under the curve Fx (x), but it buries the key idea – the elimination of time in Newton’s Second Law – way back in the derivation and relies uncomfortably on constant force/acceleration results. It is much more elegant to directly derive this result using straight up calculus, and honestly it is a lot easier, too.
82 This is notationally a bit sloppy, as I’m not making it clear that as Deltax gets smaller, you have to systematically increase the number of pieces you divide xf − x0 into and so on, hoping that you all remember you intro calculus course and recognize this picture as being one of the first things you learned about integration...

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3.2.2: Derivation II: Calculus-y (Chain Rule) Derivation
To do that, we simply take Newton’s Second Law and eliminate dt using the chain rule. The algebra is: Fx

= max = m

Fx

= m

Fx
Fx dx

dvx dx dx dt dvx vx
= m dx = mvx dvx

(chain rule)
(definition of vx )
(rearrange)

v1

x1

Fx dx

vx dvx

= m

(integrate both sides)

v0

x0 x1 Fx dx

Wtot =

dvx dt =

x0

1
1
mv 2 − mv 2
2 1 2 0

(The WKE Theorem, QED!)

(253)

This is an elegant proof, one that makes it completely clear that time dependence is being eliminated in favor of the direct dependence of v on x. It is also clearly valid for very general one dimensional force functions; at no time did we assume anything about Fx other than its general integrability in the last step.
What happens if F is actually a vector force, not necessarily in acting only in one dimension?
Well, the first proof above is clearly valid for Fx (x), Fy (y) and Fz (z) independently, so:
F · dℓ =

Fx dx +

Fy dy +

Fz dz = ∆Kx + ∆Ky + ∆Kz = ∆K

(254)

However, this doesn’t make the meaning of the integral on the left very clear.

x(t)
F
θ

F||

dl

Figure 34: Consider the work done going along the particular trajectory x(t) where there is a force
F (x) acting on it that varies along the way. As the particle moves across the small/differential section dℓ, only the force component along dℓ does work. The other force component changes the direction of the velocity without changing its magnitude.
The best way to understand that is to examine a small piece of the path in two dimensions. In figure 34 a small part of the trajectory of a particle is drawn. A small chunk of that trajectory dℓ represents the vector displacement of the object over a very short time under the action of the force
F acting there.
The component of F perpendicular to dℓ doesn’t change the speed of the particle; it behaves like a centripetal force and alters the direction of the velocity without altering the speed. The component parallel to dℓ, however, does alter the speed, that is, does work. The magnitude of the component in this direction is (from the picture) F cos(θ) where θ is the angle between the direction of F and the direction of dℓ.

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Week 3: Work and Energy

That component acts (over this very short distance) like a one dimensional force in the direction of motion, so that
1
(255) dW = F cos(θ)dℓ = d( mv 2 ) = dK
2
The next little chunk of x(t) has a different force and direction but the form of the work done and change in kinetic energy as the particle moves over that chunk is the same. As before, we can integrate from one end of the path to the other doing only the one dimensional integral of the path element dℓ times F|| , the component of F parallel to the path at that (each) point.
The vector operation that multiplies a vector by the component of another vector in the same direction as the first is the dot (or scalar) product. The dot product between two vectors A and
B can be written more than one way (all equally valid):
A·B

= AB cos(θ)

(256)

= Ax Bx + Ay By + Az Bz

(257)

The second form is connected to what we got above just adding up the independent cartesian component statements of the Work-Kinetic Energy Theorem in one (each) dimension, but it doesn’t help us understand how to do the integral between specific starting and ending coordinates along some trajectory. The first form of the dot product, however, corresponds to our picture: dW = F cos(θ)dℓ = F · dℓ = dK

(258)

Now we can see what the integral means. We have to sum this up along some specific path between x0 and x1 to find the total work done moving the particle along that path by the force. For differential sized chunks, the “sum” becomes an integral and we integrate this along the path to get the correct statement of the Work-Kinetic Energy Theorem in 2 or 3 dimensions: x1 W (x0 → x1 ) =

x0

F · dℓ =

1
1
2
2
mv1 − mv0 = ∆K
2
2

(259)

Note well that this integral may well be different for different paths connecting points x0 to x1 !
In the most general case, one cannot find the work done without knowing the path taken, because there are many ways to go between any two points and the forces could be very different along them.
Also, Note well: Energy is a scalar – just a number with a magnitude and units but no direction
– and hence is considerably easier to treat than vector quantities like forces.
Note well: Normal forces (perpendicular to the direction of motion) do no work :
∆W = F ⊥ · ∆x = 0.

(260)

In fact, force components perpendicular to the trajectory bend the trajectory with local curvature
F⊥ = mv 2 /R but don’t speed the particle up or slow it down. This really simplifies problem solving, as we shall see.
We should think about using time-independent work and energy instead of time dependent
Newtonian dynamics whenever the answer requested for a given problem is independent of time. The reason for this should now be clear: we derived the work-energy theorem (and energy conservation) from the elimination of t from the dynamical equations.
Let’s look at a few examples to see how work and energy can make our problem solving lives much better.

Example 3.2.1: Pulling a Block
Suppose we have a block of mass m being pulled by a string at a constant tension T at an angle θ with the horizontal along a rough table with coefficients of friction µs > µk . Typical questions

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Week 3: Work and Energy

T θ M

Figure 35: A block is connected to an Acme (massless, unstretchable) string and is pulled so that it exerts a constant tension T on the block at the angle θ. might be: At what value of the tension does the block begin to move? If it is pulled with exactly that tension, how fast is it moving after it is pulled a horizontal distance L?
We see that in the y-direction, N + T sin(θ) − mg = 0, or N = mg − T sin(θ). In the x-direction,
Fx = T cos(θ) − µs N = 0 (at the point where block barely begins to move).
Therefore:
T cos(θ) − µs mg + T µs sin(θ) = 0 or T =

(261)

µs mg cos(θ) + µs sin(θ)

(262)

With this value of the tension T, the work energy theorem becomes:
W = Fx L = ∆K

(263)

where Fx = T cos(θ) − µk (mg − T sin(θ). That is:
(T cos(θ) − µk (mg − T sin(θ)) L =

1 mv 2 − 0
2 f

(since vi = 0)

(264)

or (after a bit of algebra, substituting in our value for T from the first part): vf =

2µs gL cos(θ)
2µk µs gL sin(θ)
− 2µk gL + cos(θ) + µs sin(θ) cos(θ) + µs sin(θ)

1
2

(265)

Although it is difficult to check exactly, we can see that if µk = µs , vf = 0 (or the mass doesn’t accelerate). This is consistent with our value of T – the value at which the mass will exactly not move against µs alone, but will still move if “tapped” to get it started so that static friction falls back to weaker dynamic friction.
This is an example of how we can combine Newton’s Laws or statics with work and energy for different parts of the same problem. So is the next example:

Example 3.2.2: Range of a Spring Gun
Suppose we have a spring gun with a bullet of mass m compressing a spring with force constant k a distance ∆x. When the trigger is pulled, the bullet is released from rest. It passes down a horizontal, frictionless barrel and comes out a distance H above the ground. What is the range of the gun?
If we knew the speed that the bullet had coming out of the barrel, we’d know exactly how to solve this as in fact we have solved it for homework (although you shouldn’t look – see if you can do this on your own or anticipate the answer below for the extra practice and review). To find that speed, we can use the Work-Kinetic Energy Theorem if we can compute the work done by the spring! 151

Week 3: Work and Energy

H

∆x
R?

Figure 36: A simple spring gun is fired horizontally a height H above the ground. Compute its range R.
So our first chore then is to compute the work done by the spring that is initially compressed a distance ∆x, and use that in turn to find the speed of the bullet leaving the barrel. x0 W

= x1 −k(x − x0 )dx

1
= − k(x − x0 )2 |x0 x1 2
1
1
2
k(∆x)2 = mvf − 0
=
2
2

(266)
(267)
(268)

or vf =

k
|∆x|
m

(269)

As you can see, this was pretty easy. It is also a result that we can get no other way, so far, because we don’t know how to solve the equations of motion for the mass on the spring to find x(t), solve for t, find v(t), substitute to find v and so on. If we hadn’t derived the WKE theorem for non-constant forces we’d be screwed!
The rest should be familiar. Given this speed (in the x-direction), find the range from Newton’s
Laws:
F = −mg y
ˆ
(270) or ax = 0, ay = −g, v0x = vf , v0y = 0, x0 = 0, y0 = H. Solving as usual, we find:
R

= vx0 t0
= vf
=

2H g 2kH
|∆x|
mg

(271)
(272)
(273)

where you can either fill in the details for yourself or look back at your homework. Or get help, of course. If you can’t do this second part on your own at this point, you probably should get help, seriously. 3.3: Conservative Forces: Potential Energy
We have now seen two kinds of forces in action. One kind is like gravity. The work done on a particle by gravity doesn’t depend on the path taken through the gravitational field – it only depends on the relative height of the two endpoints. The other kind is like friction. Friction not only depends on the path a particle takes, it is usually negative work; typically friction turns macroscopic mechanical energy into “heat”, which can crudely be thought of an internal microscopic mechanical energy that can no longer easily be turned back into macroscopic mechanical energy. A proper discussion of

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Week 3: Work and Energy

path 1

C

x2

x1

path 2

Figure 37: The work done going around an arbitrary loop by a conservative force is zero. This ensures that the work done going between two points is independent of the path taken, its defining characteristic. heat is beyond the scope of this course, but we will remark further on this below when we discuss non-conservative forces.
We define a conservative force to be one such that the work done by the force as you move a point mass from point x1 to point x2 is independent of the path used to move between the points: x2 x2
Wloop =

x1 (path 1)

F · dl =

x1 (path 2)

F · dl

(274)

In this case (only), the work done going around an arbitrary closed path (starting and ending on the same point) will be identically zero!
Wloop =
C

F · dl = 0

(275)

This is illustrated in figure 37. Note that the two paths from x1 to x2 combine to form a closed loop C, where the work done going forward along one path is undone coming back along the other.
Since the work done moving a mass m from an arbitrary starting point to any point in space is the same independent of the path, we can assign each point in space a numerical value: the work done by us on mass m, against the conservative force, to reach it. This is the negative of the work done by the force. We do it with this sign for reasons that will become clear in a moment. We call this function the potential energy of the mass m associated with the conservative forceF : x U (x) = −

x0

F · dx = −W

(276)

Note Well: that only one limit of integration depends on x; the other depends on where you choose to make the potential energy zero. This is a free choice. No physical result that can be measured or observed can uniquely depend on where you choose the potential energy to be zero.
Let’s understand this.

3.3.1: Force from Potential Energy
In one dimension, the x-component of −F · dℓ is: dU = −dW = −Fx dx
If we rearrange this, we get:
Fx = −

dU dx (277)

(278)

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Week 3: Work and Energy

U(x)

+x

Figure 38: A tiny subset of the infinite number of possible U (x) functions that might lead to the same physical force Fx (x). One of these is highlighted by means of a thick line, but the only thing that might make it “preferred” is whether or not it makes solving any given problem a bit easier.
That is, the force is the slope of the potential energy function. This is actually a rather profound result and relationship.
Consider the set of transformations that leave the slope of a function invariant. One of them is quite obvious – adding a positive or negative constant to U (x) as portrayed in figure 38 does not affect its slope with respect to x, it just moves the whole function up or down on the U -axis. That means that all of this infinite set of candidate potential energies that differ by only a constant overall energy lead to the same force!
That’s good, as force is something we can often measure, even “at a point” (without necessarily moving the object), but potential energy is not. To measure the work done by a conservative force on an object (and hence measure the change in the potential energy) we have to permit the force to move the object from one place to another and measure the change in its speed, hence its kinetic energy. We only measure a change, though – we cannot directly measure the absolute magnitude of the potential energy, any more than we can point to an object and say that the work of that object is zero Joules, or ten Joules, or whatever. We can talk about the amount of work done moving the object from here to there but objects do not possess “work” as an attribute, and potential energy is just a convenient renaming of the work, at least so far.
I cannot, then, tell you precisely what the near-Earth gravitational potential energy of a 1 kilogram mass sitting on a table is, not even if you tell me exactly where the table and the mass are in some sort of Universal coordinate system (where if the latter exists, as now seems dubious given our discussion of inertial frames and so on, we have yet to find it). There are literally an infinity of possible answers that will all equally well predict the outcome of any physical experiment involving near-Earth gravity acting on the mass, because they all lead to the same force acting on the object.
In more than one dimesion we have to use a bit of vector calculus to write this same result per component: ∆U dU = −

F · dℓ

= −F · dℓ

(279)
(280)

It’s a bit more work than we can do in this course to prove it, but the result one gets by “dividing through but dℓ” in this case is:
F = −∇U = −

∂U
∂U
∂U x− ˆ y− ˆ z ˆ
∂x
∂y
∂z

(281)

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Week 3: Work and Energy

which is basically the one dimensional result written above, per component. If you are a physics or math major (or have had or are in multivariate calculus) this last form should be studied until it makes sense, but everybody should know the first form (per component) and should be able to see that it should reasonably hold (subject to working out some more math than you may yet know) for all coordinate directions. Note that non-physics majors won’t (in my classes) be held responsible for knowing this vector calculus form, but everybody should understand the concept underlying it.
We’ll discuss this a bit further below, after we have learned about the total mechanical energy.
So much for the definition of a conservative force, its potential energy, and how to get the force back from the potential energy and our freedom to choose add a constant energy to the potential energy and still get the same answers to all physics problems83 we had a perfectly good theorem, the
Work-Kinetic Energy Theorem Why do we bother inventing all of this complication, conservative forces, potential energies? What was wrong with plain old work?
Well, for one thing, since the work done by conservative forces is independent of the path taken by definition, we can do the work integrals once and for all for the well-known conservative forces, stick a minus sign in front of them, and have a set of well-known potential energy functions that are generally even simpler and more useful. In fact, since one can easily differentiate the potential energy function to recover the force, one can in fact forget thinking in terms of the force altogether and formulate all of physics in terms of energies and potential energy functions!
In this class, we won’t go to this extreme – we will simply learn both the forces and the associated potential energy functions where appropriate (there aren’t that many, this isn’t like learning all of organic chemistry’s reaction pathways or the like), deriving the second from the first as we go, but in future courses taken by a physics major, a chemistry major, a math major it is quite likely that you will relearn even classical mechanics in terms of the Lagrangian 84 or Hamiltonian 85 formulation, both of which are fundamentally energy-based, and quantum physics is almost entirely derived and understood in terms of Hamiltonians.
For now let’s see how life is made a bit simpler by deriving general forms for the potential energy functions for near-Earth gravity and masses on springs, both of which will be very useful indeed to us in the weeks to come.

3.3.2: Potential Energy Function for Near-Earth Gravity
The potential energy of an object experiencing a near-Earth gravitational force is either: y Ug (y) = −

(−mg)dy ′ = mgy

(282)

0

where we have effectively set the zero of the potential energy to be “ground level”, at least if we put the y-coordinate origin at the ground. Of course, we don’t really need to do this – we might well want the zero to be at the top of a table over the ground, or the top of a cliff well above that, and we are free to do so. More generally, we can write the gravitational potential energy as the indefinite integral: Ug (y) = −

(−mg)dy = mgy + U0

(283)

where U0 is an arbitrary constant that sets the zero of gravitational potential energy. For example, suppose we did want the potential energy to be zero at the top of a cliff of height H, but for one
83 Wikipedia: http://www.wikipedia.org/wiki/Gauge Theory. For students intending to continue with more physics, this is perhaps your first example of an idea called Gauge freedom – the invariance of things like energy under certain sets of coordinate transformations and the implications (like invariance of a measured force) of the symmetry groups of those transformations – which turns out to be very important indeed in future courses. And if this sounds strangely like I’m speaking Martian to you or talking about your freedom to choose a 12 gauge shotgun instead of a 20 gauge shotgun – gauge freedom indeed – well, don’t worry about it...
84 Wikipedia: http://www.wikipedia.org/wiki/Lagrangian.
85 Wikipedia: http://www.wikipedia.org/wiki/Hamiltonian.

155

Week 3: Work and Energy reason or another selected a coordinate system with the y-origin at the bottom. Then we need:
Ug (y = H) = mgH + U0 = 0

(284)

U0 = −mgH

(285)

Ug (y) = mgy − mgH = mg(y − H) = mgy ′

(286)

or so that: where in the last step we changed variables (coordinate systems) to a new one y ′ = y − H with the origin at the top of the cliff!
From the latter, we see that our freedom to choose any location for the zero of our potential energy function is somehow tied to our freedom to choose an arbitrary origin for our coordinate frame. It is actually even more powerful (and more general) than that – we will see examples later where potential energy can be defined to be zero on entire planes or lines or “at infinity”, where of course it is difficult to put an origin at infinity and have local coordinates make any sense.
You will find it very helpful to choose a coordinate system and set the zero of potential energy in such as way as to make the problem as computationally simple as possible. Only experience and practice will ultimately be your best guide as to just what those are likely to be.

3.3.3: Springs
Springs also exert conservative forces in one dimension – the work you do compressing or stretching an ideal spring equals the work the spring does going back to its original position, whatever that position might be. We can therefore define a potential energy function for them.
In most cases, we will choose the zero of potential energy to be the equilibrium position of the spring – other choices are possible, though, and one in particular will be useful (a mass hanging from a spring in near-Earth gravity).
With the zero of both our one dimensional coordinate system and the potential energy at the equilibrium position of the unstretched spring (easiest) Hooke’s Law is just:
Fx = −kx

(287)

and we get: x Us (x)

= −
=

(−kx′ ) dx′
0

1 2 kx 2

(288)

This is the function you should learn – by deriving this result several times on your own, not by memorizing – as the potential energy of a spring.
More generally, if we do the indefinite integral in this coordinate frame instead we get:
U (x) = −

(−kx) dx =

1 2 kx + U0
2

(289)

To see how this is related to one’s choice of coordinate origin, suppose we choose the origin of coordinates to be at the end of the spring fixed to a wall, so that the equilibrium length of the unstretched, uncompressed spring is xeq . Hooke’s Law is written in these coordinates as:
Fx (x) = −k(x − xeq )

(290)

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Week 3: Work and Energy

Now we can choose the zero of potential energy to be at the position x = 0 by doing the definite integral: x
1
1
(−k(x′ − xeq )) dx′ = k(x − xeq )2 − kx2
(291)
Us (x) = −
2
2 eq
0
If we now change variables to, say, y = x − xeq , this is just:
Us (y) =

1 2 1 2
1
ky − kxeq = ky 2 + U0
2
2
2

(292)

which can be compared to the indefinite integral form above. Later, we’ll do a problem where a mass hangs from a spring and see that our freedom to add an arbitrary constant of integration allows us to change variables to an ”easier” origin of coordinates halfway through a problem.
Consider: our treatment of the spring gun (above) would have been simpler, would it not, if we could have simply started knowing the potential energy function for (and hence the work done by) a spring?
There is one more way that using potential energy instead of work per se will turn out to be useful to us, and it is the motivation for including the leading minus sign in its definition. Suppose that you have a mass m that is moving under the influence of a conservative force. Then the Work-Kinetic
Energy Theorem (259) looks like:
WC = ∆K
(293)
where WC is the ordinary work done by the conservative force. Subtracting WC over to the other side and substituting, one gets:
∆K − WC = ∆K + ∆U = 0

(294)

Since we can now assign U (x) a unique value (once we set the constant of integration or place(s)
U (x) is zero in its definition above) at each point in space, and since K is similarly a function of position in space when time is eliminated in favor of position and no other (non-conservative) forces are acting, we can define the total mechanical energy of the particle to be:
Emech = K + U

(295)

∆Emech = 0

(296)

in which case we just showed that

Wait, did we just prove that Emech is a constant any time a particle moves around under only the influence of conservative forces? We did...

3.4: Conservation of Mechanical Energy
OK, so maybe you missed that last little bit. Let’s make it a bit clearer and see how enormously useful and important this idea is.
First we will state the principle of the Conservation of Mechanical Energy :
The total mechanical energy (defined as the sum of its potential and kinetic energies) of a particle being acted on by only conservative forces is constant.
Or (in much more concise algebra), if only conservative forces act on an object and U is the potential energy function for the total conservative force, then
Emech = K + U = A scalar constant

(297)

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Week 3: Work and Energy
The proof of this statement is given above, but we can recapitulate it here.
Suppose
Emech = K + U

(298)

Because the change in potential energy of an object is just the path-independent negative work done by the conservative force,
∆K + ∆U = ∆K − WC = 0
(299)
is just a restatement of the WKE Theorem, which we derived and proved. So it must be true! But then ∆K + ∆U = ∆(K + U ) = ∆Emech = 0
(300)
and Emech must be constant as the conservative force moves the mass(es) around.

3.4.1: Force, Potential Energy, and Total Mechanical Energy
Now that we know what the total mechanical energy is, the following little litany might help you conceptually grasp the relationship between potential energy and force. We will return to this still again below, when we talk about potential energy curves and equilibrium, but repetition makes the ideas easier to understand and remember, so skim it here first, now.
The fact that the force is the negative derivative of (or gradient of) the potential energy of an object means that the force points in the direction the potential energy decreases in.
This makes sense. If the object has a constant total energy, and it moves in the direction of the force, it speeds up! Its kinetic energy increases, therefore its potential energy decreases. If it moves from lower potential energy to higher potential energy, its kinetic energy decreases, which means the force pointed the other way, slowing it down.
There is a simple metaphor for all of this – the slope of a hill. We all know that things roll slowly down a shallow hill, rapidly down a steep hill, and just fall right off of cliffs. The force that speeds them up is related to the slope of the hill, and so is the rate at which their gravitational potential energy increases as one goes down the slope! In fact, it isn’t actually just a metaphor, more like an example. Either way, “downhill” is where potential energy variations push objects – in the direction that the potential energy maximally decreases, with a force proportional to the rate at which it decreases.
The WKE Theorem itself and all of our results in this chapter, after all, are derived from Newton’s
Second Law – energy conservation is just Newton’s Second Law in a time-independent disguise.

Example 3.4.1: Falling Ball Reprise
To see how powerful this is, let us look back at a falling object of mass m (neglecting drag and friction). First, we have to determine the gravitational potential energy of the object a height y above the ground (where we will choose to set U (0) = 0): y U (y) = −

(−mg)dy = mgy

(301)

0

Wow, that was kind of – easy!
Now, suppose we have our ball of mass m at the height H and drop it from rest, yadda yadda.
How fast is it going when it hits the ground? This time we simply write the total energy of the ball at the top (where the potential is mgH and the kinetic is zero) and the bottom (where the potential is zero and kinetic is 1 mv 2 ) and set the two equal! Solve for v, done:
2
1
Ei = mgH + (0) = (0) + mv 2 = Ef
2

(302)

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Week 3: Work and Energy

or v= 2gH

(303)

Even better:

Example 3.4.2: Block Sliding Down Frictionless Incline Reprise
The block starts out a height H above the ground, with potential energy mgH and kinetic energy of 0. It slides to the ground (no non-conservative friction!) and arrives with no potential energy and
1
kinetic energy 2 mv 2 . Whoops, time to block-copy the previous solution:
1
Ei = mgH + (0) = (0) + mv 2 = Ef
2

(304)

or v= 2gH

(305)

Example 3.4.3: A Simple Pendulum
Here are two versions of a pendulum problem: Imagine a pendulum (ball of mass m suspended on a string of length L that we have pulled up so that the ball is a height H < L above its lowest point on the arc of its stretched string motion. We release it from rest. How fast is it going at the bottom?
Yep, you guessed it – block copy again:
1
Ei = mgH + (0) = (0) + mv 2 = Ef
2

(306)

or v= 2gH

(307)

It looks as though it does not matter what path a mass takes as it goes down a height H starting from rest – as long as no forces act to dissipate or add energy to the particle, it will arrive at the bottom travelling at the same speed.

θ

L

m

L

v0

H

Figure 39: Find the maximum angle through which the pendulum swings from the initial conditions.
Here’s the same problem, formulated a different way: A mass m is hanging by a massless thread of length L and is given an initial speed v0 to the right (at the bottom). It swings up and stops at some maximum height H at an angle θ as illustrated in figure 39 (which can be used “backwards” as the figure for the first part of this example, of course). Find θ.
Again we solve this by setting Ei = Ef (total energy is conserved).

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Initial:

1
1
2 mv 2 + mg(0) = mv0
2 0
2

(308)

1 m(0)2 + mgH = mgL(1 − cos(θ))
2

(309)

Ei =
Final:
Ef =
(Note well: H = L(1 − cos(θ))!)
Set them equal and solve:

cos(θ) = 1 − or θ = cos−1 (1 −

2 v0 2gL

(310)

2 v0 ).
2gL

(311)

Example 3.4.4: Looping the Loop m v
H
R

Figure 40:
Here is a lovely problem – so lovely that you will solve it five or six times, at least, in various forms throughout the semester, so be sure that you get to where you understand it – that requires you to use three different principles we’ve learned so far to solve:
What is the minimum height H such that a block of mass m loops-the-loop (stays on the frictionless track all the way around the circle) in figure 40 above?
Such a simple problem, such an involved answer. Here’s how you might proceed. First of all, let’s understand the condition that must be satisfied for the answer “stay on the track”. For a block to stay on the track, it has to touch the track, and touching something means “exerting a normal force on it” in physicsspeak. To barely stay on the track, then – the minimal condition – is for the normal force to barely go to zero.
Fine, so we need the block to precisely “kiss” the track at near-zero normal force at the point where we expect the normal force to be weakest. And where is that? Well, at the place it is moving the slowest, that is to say, the top of the loop. If it comes off of the loop, it is bound to come off at or before it reaches the top.
Why is that point key, and what is the normal force doing in this problem. Here we need two physical principles: Newton’s Second Law and the kinematics of circular motion since the mass is undoubtedly moving in a circle if it stays on the track. Here’s the way we reason: “If the

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Week 3: Work and Energy

block moves in a circle of radius R at speed v, then its acceleration towards the center must be ac = v 2 /R. Newton’s Second Law then tells us that the total force component in the direction of the center must be mv 2 /R. That force can only be made out of (a component of) gravity and the normal force, which points towards the center. So we can relate the normal force to the speed of the block on the circle at any point.”
At the top (where we expect v to be at its minimum value, assuming it stays on the circle) gravity points straight towards the center of the circle of motion, so we get: mg + N =

mv 2
R

(312)

and in the limit that N → 0 (“barely” looping the loop) we get the condition: mg =

2 mvt R

(313)

where vt is the (minimum) speed at the top of the track needed to loop the loop.
Now we need to relate the speed at the top of the circle to the original height H it began at.
This is where we need our third principle – Conservation of Mechanical Energy! Note that we cannot possible integrate Newton’s Second Law and solve an equation of motion for the block on the frictionless track – I haven’t given you any sort of equation for the track (because I don’t know it) and even a physics graduate student forced to integrate N2 to find the answer for some relatively
“simple” functional form for the track would suffer mightily finding the answer. With energy we don’t care about the shape of the track, only that the track do no work on the mass which (since it is frictionless and normal forces do no work) is in the bag.
Thus:

1
2
Ei = mgH = mg2R + mvt = Ef
2

(314)

2
If you put these two equations together (e.g. solve the first for mvt and substitute it into the second, then solve for H in terms of R) you should get Hmin = 5R/2. Give it a try. You’ll get even more practice in your homework, for some more complicated situations, for masses on strings or rods – they’re all the same problem, but sometimes the Newton’s Law condition will be quite different! Use your intuition and experience with the world to help guide you to the right solution in all of these causes.

So any time a mass moves down a distance H, its change in potential energy is mgH, and since total mechanical energy is conserved, its change in kinetic energy is also mgH the other way. As one increases, the other decreases, and vice versa!
This makes kinetic and potential energy bone simple to use. It also means that there is a lovely analogy between potential energy and your savings account, kinetic energy and your checking account, and cash transfers (conservative movement of money from checking to savings or vice versa where your total account remains constant.
Of course, it is almost too much to expect for life to really be like that. We know that we always have to pay banking fees, teller fees, taxes, somehow we never can move money around and end up with as much as we started with. And so it is with energy.
Well, it is and it isn’t. Actually conservation of energy is a very deep and fundamental principle of the entire Universe as best we can tell. Energy seems to be conserved everywhere, all of the time, in detail, to the best of our ability to experimentally check. However, useful energy tends to decrease over time because of “taxes”. The tax collectors, as it were, of nature are non-conservative forces!
What happens when we try to combine the work done by non-conservative forces (which we must tediously calculate per problem, per path) with the work done by conservative ones, expressed in terms of potential and total mechanical energy? You get the...

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Week 3: Work and Energy

3.5: Generalized Work-Mechanical Energy Theorem
So, as suggested above let’s generalize the WKE one further step by considering what happens if both conservative and nonconservative forces are acting on a particle. In that case the argument above becomes:
Wtot = WC + WN C = ∆K
(315)
or
WN C = ∆K − WC = ∆K + ∆U = ∆Emech

(316)

which we state as the Generalized Non-Conservative Work-Mechanical Energy Theorem:
The work done by all the non-conservative forces acting on a particle equals the change in its total mechanical energy.
Our example here is very simple.

Example 3.5.1: Block Sliding Down a Rough Incline
Suppose a block of mass m slides down an incline of length L at an incline θ with respect to the horizontal and with kinetic friction (coefficient µk ) acting against gravity. How fast is it going
(released from rest at an angle where static friction cannot hold it) when it reaches the ground?
Here we have to do a mixture of several things. First, let’s write Newton’s Second Law for just the (static) y direction:
N − mg cos(θ) = 0
(317)
or
N = mg cos(θ)

(318)

fk = µk N = µk mg cos(θ)

(319)

Next, evaluate:
(up the incline, opposite to the motion of the block).
We ignore dynamics in the direction down the plane. Instead, we note that the work done by friction is equal to the change in the mechanical energy of the block. Ei = mgH = mgL sin(θ).
Ef = 1 mv 2 . So:
2
1
− µk mgL cos(θ) = Ef − Ei = mv 2 − mgH
(320)
2 or 1 mv 2 = mgH − µk mgL cos(θ)
(321)
2 so that:
(322)
v = ± 2gH − µk 2gL cos(θ)
Here we really do have to be careful and choose the sign that means “going down the incline” at the bottom. As an extra bonus, our answer tells us the condition on (say) the angle such that the mass doesn’t or just barely makes it to the bottom. v = 0 means “barely” (gets there and stops) and if v is imaginary, it doesn’t make all the way to the bottom at all.
I don’t know about you, but this seems a lot easier than messing with integrating Newton’s Law, solving for v(t) and x(t), solving for t, back substituting, etc. It’s not that this is all that difficult, but work-energy is simple bookkeeping, anybody can do it if they just know stuff like the form of the potential energy, the magnitude of the force, some simple integrals.
Here’s another example.

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Week 3: Work and Energy

Example 3.5.2: A Spring and Rough Incline

H max

∆x k m

θ

Figure 41: A spring compressed an initial distance ∆x fires a mass m across a smooth (µk ≈ 0) floor to rise up a rough µk = 0) incline. How far up the incline does it travel before coming to rest?
In figure 41 a mass m is released from rest from a position on a spring with spring constant k compressed a distance ∆x from equilibrium. It slides down a frictionless horizontal surface and then slides up a rough plane inclined at an angle θ. What is the maximum height that it reaches on the incline? This is a problem that is basically impossible, so far, for us to do using Newton’s Laws alone.
This is because we are weeks away from being able to solve the equation of motion for the mass on the spring! Even if/when we can solve the equation of motion for the mass on the spring, though, this problem would still be quite painful to solve using Newton’s Laws and dynamical solutions directly. Using the GWME Theorem, though, it is pretty easy. As before, we have to express the initial total mechanical energy and the final total mechanical energy algebraically, and set their difference equal to the non-conservative work done by the force of kinetic friction sliding up the incline.
I’m not going to do every step for you, as this seems like it would be a good homework problem, but here are a few:
1
1
1
Ei = Ugi + Usi + Ki = mg(0) + k∆x2 + m(0)2 = k∆x2
2
2
2

(323)

1
Ef = Ugf + Kf = mgHmax + m(0)2 = mgHmax
2

(324)

Remaining for you to do: Find the force of friction down the incline (as it slides up). Find the work done by friction. Relate that work to Hmax algebraically, write the GWME Theorem algebraically, and solve for Hmax . Most of the steps involving friction and the inclined plane can be found in the previous example, if you get stuck, but try to do it without looking first!

3.5.1: Heat and Conservation of Energy
Note well that the theorem above only applies to forces acting on particles, or objects that we consider in the particle approximation (ignoring any internal structure and treating the object like a single mass). In fact, all of the rules above (so far) from Newton’s Laws on down strictly speaking only apply to particles in inertial reference frames, and we have some work to do in order to figure out how to apply them to systems of particles being pushed on both by internal forces between particles in the system as well as external forces between the particles of the system and particles that are not part of the system.

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Week 3: Work and Energy

What happens to the energy added to or removed from an object (that is really made up of many particles bound together by internal e.g. molecular forces) by things like my non-conservative hand as I give a block treated as a “particle” a push, or non-conservative kinetic friction and drag forces as they act on the block to slow it down as it slides along a table? This is not a trivial question.
To properly answer it we have to descend all the way into the conceptual abyss of treating every single particle that makes up the system we call “the block” and every single particle that makes up the system consisting of “everything else in the Universe but the block” and all of the internal forces between them – which happen, as far as we can tell, to be strictly conservative forces – and then somehow average over them to recover the ability to treat the block like a particle, the table like a fixed, immovable object it slides on, and friction like a comparatively simple force that does non-conservative work on the block.
It requires us to invent things like statistical mechanics to do the averaging, thermodynamics to describe certain kinds of averaged systems, and whole new sciences such as chemistry and biology that use averaged energy concepts with their own fairly stable rules that cannot easily be connected back to the microscopic interactions that bind quarks and electrons into atoms and atoms together into molecules. It’s easy to get lost in this, because it is both fascinating and really difficult.
I’m therefore going to give you a very important empirical law (that we can understand well enough from our treatment of particles so far) and a rather heuristic description of the connections between microscopic interactions and energy and the macroscopic mechanical energy of things like blocks, or cars, or human bodies.
The important empirical law is the Law of Conservation of Energy 86 . Whenever we examine a physical system and try very hard to keep track of all of the mechanical energy exchanges withing that system and between the system and its surroundings, we find that we can always account for them all without any gain or loss. In other words, we find that the total mechanical energy of an isolated system never changes, and if we add or remove mechanical energy to/from the system, it has to come from or go to somewhere outside of the system. This result, applied to well defined systems of particles, can be formulated as the First Law of Thermodynamics:
∆Qin = ∆Eof + Wby

(325)

In words, the heat energy flowing in to a system equals the change in the internal total mechanical energy of the system plus the external work (if any) done by the system on its surroundings. The total mechanical energy of the system itself is just the sum of the potential and kinetic energies of all of its internal parts and is simple enough to understand if not to compute. The work done by the system on its surroundings is similarly simple enough to understand if not to compute. The hard part of this law is the definition of heat energy, and sadly, I’m not going to give you more than the crudest idea of it right now and make some statements that aren’t strictly true because to treat heat correctly requires a major chunk of a whole new textbook on textbfThermodynamics. So take the following with a grain of salt, so to speak.
When a block slides down a rough table from some initial velocity to rest, kinetic friction turns the bulk organized kinetic energy of the collectively moving mass into disorganized microscopic energy – heat. As the rough microscopic surfaces bounce off of one another and form and break chemical bonds, it sets the actual molecules of the block bounding, increasing the internal microscopic mechanical energy of the block and warming it up. Some of it similarly increasing the internal microscopic mechanical energy of the table it slide across, warming it up. Some of it appears as light energy (electromagnetic radiation) or sound energy – initially organized energy forms that themselves become ever more disorganized. Eventually, the initial organized energy of the block becomes a tiny increase in the average internal mechanical energy of a very, very large number of objects both inside and outside of the original system that we call the block, a process we call being
“lost to heat”.
86 More

properly, mass-energy, but we really don’t want to get into that in an introductory course.

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Week 3: Work and Energy

We have the same sort of problem tracking energy that we add to the system when I give the block a push. Chemical energy in sugars causes muscle cells to change their shape, contracting muscles that do work on my arm, which exchanges energy with the block via the normal force between block and skin. The chemical energy itself originally came from thermonuclear fusion reactions in the sun, and the free energy released in those interactions can be tracked back to the Big Bang, with a lot of imagination and sloughing over of details. Energy, it turns out, has “always” been around (as far back in time as we can see, literally) but is constantly changing form and generally becoming more disorganized as it does so.
In this textbook, we will say a little more on this later, but this is enough for the moment. We will summarize this discussion by remarking that non-conservative forces, both external (e.g. friction acting on a block) and internal (e.g. friction or collision forces acting between two bodies that are part of “the system” being considered) will often do work that entirely or partially “turns into heat”
– disappears from the total mechanical energy we can easily track. That doesn’t mean that it is has truly disappeared, and more complex treatments or experiments can indeed track and/or measure it, but we just barely learned what mechanical energy is and are not yet ready to try to deal with what happens when it is shared among (say) Avogadro’s number of interacting gas molecules.

3.6: Power
The energy in a given system is not, of course, usually constant in time. Energy is added to a given mass, or taken away, at some rate. We accelerate a car (adding to its mechanical energy). We brake a car (turning its kinetic energy into heat). There are many times when we are given the rate at which energy is added or removed in time, and need to find the total energy added or removed. This rate is called the power.
Power: The rate at which work is done, or energy released into a system.
This latter form lets us express it conveniently for time-varying forces: dW = F · dx = F ·

dx dt dt

(326)

or
P =

dW
=F ·v dt (327)

so that
∆W = ∆Etot =

P dt

(328)

The units of power are clearly Joules/sec = Watts. Another common unit of power is “Horsepower”, 1 HP = 746 W. Note that the power of a car together with its drag coefficient determine how fast it can go. When energy is being added by the engine at the same rate at which it is being dissipated by drag and friction, the total mechanical energy of the car will remain constant in time.

Example 3.6.1: Rocket Power
A model rocket engine delivers a constant thrust F that pushes the rocket (of approximately constant mass m) up for a time tr before shutting off. Show that the total energy delivered by the rocket engine is equal to the change in mechanical energy the hard way – by solving Newton’s Second Law for the rocket to obtain v(t), using that to find the power P , and integrating the power from 0 to
1
tr to find the total work done by the rocket engine, and comparing this to mgy(tr ) + 2 mv(tr )2 , the total mechanical energy of the rocket at time tr .

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Week 3: Work and Energy
To outline the solution, following a previous homework problem, we write:
F − mg = ma or a=

F − mg m (329)

(330)

We integrate twice to obtain (starting at y(0) = 0 and v(0) = 0):
F − mg t m
1 F − mg 2
1 2 at = t 2
2 m

v(t) = at =

(331)

y(t) =

(332)
(333)

From this we can find:
1
= mgy(tr ) + mv(tr )2
2
2
1 F − mg 2 1
F − mg
= mg tr + m tr 2 m
2
m
F2
1
F g − mg 2 +
+ mg 2 − 2F g t2
=
r
2
m
1
=
F 2 − F mg t2 r 2m

Emech (tr )

(334)

Now for the power:
P

= F · v = F v(t)
F 2 − F mg
=
t m (335)

We integrate this from 0 to tr to find the total energy delivered by the rocket engine: tr P dt =

W =
0

1
F 2 − F mg t2 = Emech (tr ) r 2m

(336)

For what it is worth, this should also just be W = F × y(tr ), the force through the distance:
W =F ×

1
1 F − mg 2 tr =
F 2 − F mg t2 r 2 m
2m

(337)

and it is.
The main point of this example is to show that all of the definitions and calculus above are consistent. It doesn’t matter how you proceed – compute ∆Emech , find P (t) and integrate, or just straight up evaluate the work W = F ∆y, you will get the same answer.
Power is an extremely important quantity, especially for engines because (as you see) the faster you go at constant thrust, the larger the power delivery. Most engines have a limit on the amount of power they can generate. Consequently the forward directed force or thrust tends to fall off as the speed of the e.g. rocket or car increases. In the case of a car, the car must also overcome a (probably nonlinear!) drag force. One of your homework problems explores the economic consequences of this.

3.7: Equilibrium
Recall that the force is given by the negative gradient of the potential energy:
F = −∇U

(338)

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Week 3: Work and Energy

or (in each direction87 ):
Fx = −

dU
,
dx

Fy = −

dU
,
dy

Fz = −

dU dz (339)

or the force is the negative slope of the potential energy function in this direction. As discussed above, the meaning of this is that if a particle moves in the direction of the (conservative) force, it speeds up. If it speeds up, its kinetic energy increases. If its kinetic energy increases, its potential energy must decrease. The force (component) acting on a particle is thus the rate at which the potential energy decreases (the negative slope) in any given direction as shown.
In the discussion below, we will concentrate on one-dimensional potentials to avoid overstressing students’ calculus muscles while they are still under development, but the ideas all generalize beautifully to two or three (or in principle still more) dimensions.

U(x)

F
F

b

a

x

F = 0 Equilibrium
Figure 42: A one-dimensional potential energy curve U (x). This particular curve might well represent U (x) = 1 kx2 for a mass on a spring, but the features identified and classified below are
2
generic.
In one dimension, we can use this to rapidly evaluate the behavior of a system on a qualitative basis just by looking at a graph of the curve! Consider the potential energy curves in figure 42. At the point labelled a, the x-slope of U (x) is positive. The x (component of the) force, therefore, is in the negative x direction. At the point b, the x-slope is negative and the force is correspondingly positive. Note well that the force gets larger as the slope of U (x) gets larger (in magnitude).
The point in the middle, at x = 0, is special. Note that this is a minimum of U (x) and hence the x-slope is zero. Therefore the x-directed force F at that point is zero as well. A point at which the force on an object is zero is, as we previously noted, a point of static force equilibrium – a particle placed there at rest will remain there at rest.
In this particular figure, if one moves the particle a small distance to the right or the left of the equilibrium point, the force pushes the particle back towards equilibrium. Points where the force is zero and small displacements cause a restoring force in this way are called stable equilibrium points.
As you can see, the isolated minima of a potential energy curve (or surface, in higher dimensions) are all stable equilibria.
Figure 43 corresponds to a more useful “generic” atomic or molecular interaction potential energy.
It corresponds roughly to a Van der Waals Force 88 between two atoms or molecules, and exhibits a number of the features that such interactions often have.
87 Again, more advanced math or physics students will note that these should all be partial derivatives in correspondance with the force being the gradient of the potential energy surface U (x, y, z), but even then each component is the local slope along the selected direction.
88 Wikipedia: http://www.wikipedia.org/wiki/Van der Waals Force.

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Week 3: Work and Energy

U(r)

F

Unstable Equilibrium
Neutral Equilibrium r Stable Equilibrium
Figure 43: A fairly generic potential energy shape for microscopic (atomic or molecular) interactions, drawn to help exhibits features one might see in such a curve more than as a realistically scaled potential energy in some set of units. In particular, the curve exhibits stable, unstable, and neutral equilibria for a radial potential energy as a fuction of r, the distance between two e.g. atoms.
At very long ranges, the forces between neutral atoms are extremely small, effectively zero. This is illustrated as an extended region where the potential energy is flat for large r. Such a range is called neutral equilibrium because there are no forces that either restore or repel the two atoms.
Neutral equilibrium is not stable in the specific sense that a particle placed there with any nonzero velocity will move freely (according to Newton’s First Law). Since it is nearly impossible to prepare an atom at absolute rest relative to another particle, one basically “never” sees two unbound microscopic atoms with a large, perfectly constant spatial orientation.
As the two atoms near one another, their interaction becomes first weakly attractive due to e.g. quantum dipole-induced dipole interactions and then weakly repulsive as the two atoms start to
“touch” each other. There is a potential energy minimum in between where two atoms separated by a certain distance can be in stable equilibrium without being chemically bound.
Atoms that approach one another still more closely encounter a second potential energy well that is at first strongly attractive (corresponding, if you like, to an actual chemical interaction between them) followed by a hard core repulsion as the electron clouds are prevented from interpenetrating by e.g. the Pauli exclusion principle. This second potential energy well is often modelled by a
Lennard-Jones potential energy (or “6-12 potential energy”, corresponding to the inverse powers of r used in the model89 . It also has a point of stable equilibrium.
In between, there is a point where the growing attraction of the inner potential energy well and the growing repulsion of the outer potential energy well balance, so that the potential energy function has a maximum. At this maximum the slope is zero (so it is a position of force equilibrium) but because the force on either side of this point pushes the particle away from it, this is a point of unstable equilibrium. Unstable equilibria occur at isolated maxima in the potential energy function, just as stable equilibria occur at isolated minima.
Note for advanced students: In more than one dimension, a potential energy curve can have
“saddle points” that are maxima in one dimension and minima in another (so called because the potential energy surface resembles the surface of a saddle, curved up front-to-back to hold the
89 Wikipedia: http://www.wikipedia.org/wiki/Lennard-Jones Potential. We will learn the difference between a “potential energy” and a “potential” later in this course, but for the moment it is not important. The shapes of the two curves are effectively identical.

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rider in and curved down side to side to allow the legs to straddle the horse). Saddle points are unstable equilibria (because instability in any direction means unstable) and are of some conceptual importance in more advanced studies of physics or in mathematics when considering asymptotic convergence. 3.7.1: Energy Diagrams: Turning Points and Forbidden Regions
U(r)
Ea
Classically forbidden domains (shaded)

Eb

c a b

c b b

r

K < 0 forbidden
Ec
Quadratic region
K = E − U > 0 allowed
Figure 44: The same potential energy curve, this time used to illustrate turning points and classically allowed and forbidden regions. Understanding the role of the total energy on potential energy diagrams and how transitions from a higher energy state to a lower energy state can “bind” a system provide insight into chemistry, orbital dynamics, and more.
We now turn to another set of extremely useful information one can extract from potential energy curves in cases where one knows the total mechanical energy of the particle in addition to the potential energy curve. In figure 44 we again see the generic (Van der Waals) atomic interaction curve this time rather “decorated” with information. To understand this information and how to look at the diagram and gain insight, please read the following description very carefully while following along in the figure.
Consider a particle with total energy mechanical Ea . Since the total mechanical energy is a constant, we can draw the energy in on the potential energy axes as a straight line with zero slope
– the same value for all r. Now note carefully that:
K(r) = Emech − U (r)

(340)

which is the difference between the total energy curve and the potential energy curve. The kinetic energy of a particle is 1 mv 2 which is non-negative. This means that we can never observe a particle
2
with energy Ea to the left of the position marked a on the r-axis – only point where Ea ≥ U lead to K > 0. We refer to the point a as a turning point of the motion for any given energy – when r = a, Emech = Ea = U (a) and K(a) = 0.
We can interpret the motion associated with Ea very easily. An atom comes in at more or less a constant speed from large r, speeds and slows and speeds again as it reaches the support of the potential energy90 , “collides” with the central atom at r = 0 (strongly repelled by the hard core
90 The

“support” of a function is the set values of the argument for which the function is not zero, in this case a finite sphere around the atom out where the potential energy first becomes attractive.

Week 3: Work and Energy

169

interaction) and recoils, eventually receding from the central atom at more or less the same speed it initially came in with. Its distance of closest approach is r = a.
Now consider a particle coming in with energy Emech = Eb . Again, this is a constant straight line on the potential energy axes. Again K(r) = Eb − U (r) ≥ 0. The points on the r-axis labelled b are the turning points of the motion, where K(b) = 0. The shaded regions indicate classically forbidden regions where the kinetic energy would have to be negative for the particle (with the given total energy) to be found there. Since the kinetic energy can never be negative, the atom can never be found there.
Again we can visualize the motion, but now there are two possibilities. If the atom comes in from infinity as before, it will initially be weakly attracted ultimately be slowed and repelled not by the hard core, but by the much softer force outside of the unstable maximum in U (r). This sort of
“soft” collision is an example of an interaction barrier a chemical reaction that cannot occur at low temperatures (where the energy of approach is too low to overcome this initial repulsion and allow the atoms to get close enough to chemically interact.
However, a second possibility emerges. If the separation of the two atoms (with energy Eb , recall) is in the classically allowed region between the two inner turning points, then the atoms will oscillate between those two points, unable to separate to infinity without passing through the classically forbidden region that would require the kinetic energy to be negative. The atoms in this case are said to be bound in a classically stable configuration around the stable equilibrium point associated with this well.
In nature, this configuration is generally not stably bound with an energy Eb > 0 – quantum theory permits an atom outside with this energy to tunnel into the inner well and an atom in the inner well to tunnel back to the outside and thence be repelled to r → ∞. Atoms bound in this inner well are then said to be metastable (which means basically “slowly unstable”) – they are classically bound for a while but eventually escape to infinity.
However, in nature pairs of atoms in the metastable configuration have a chance of giving up some energy (by, for example, giving up a photon or phonon, where you shouldn’t worry too much about what these are just yet) and make a transistion to a still lower energy state such as that represented by Ec < 0.
When the atoms have total energy Ec as drawn in this figure, they have only two turning points
(labelled c in the figure). The classically permitted domain is now only the values of r in between these two points; everything less than the inner turning point or outside of the outer turning point corresponds to a kinetic energy that is less than 0 which is impossible. The classically forbidden regions for Ec are again shaded on the diagram. Atoms with this energy oscillate back and forth between these two turning points.
They oscillate back and forth very much like a mass on a spring! Note that this regions is labelled the quadratic region on the figure. This means that in this region, a quadratic function of r − re (where re is the stable equilibrium at the minimum of U (r) in this well) is a very good approximation to the actual potential energy. The potential energy of a mass on a spring aligned with r and with its equilibrium length moved so that it is re is just 1 k(r − re )2 + U0 , which can be
2
fit to U (r) in the quadratic region with a suitable choice of k and U0 .

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Homework for Week 3

Problem 1.

Physics Concepts: Make this week’s physics concepts summary as you work all of the problems in this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s) they were key to, and include concepts from previous weeks as necessary. Do the work carefully enough that you can (after it has been handed in and graded) punch it and add it to a three ring binder for review and study come finals!

Problem 2.
Derive the Work-Kinetic Energy (WKE) theorem in one dimension from Newton’s second law. You may use any approach used in class or given and discussed in this textbook (or any other), but do it yourself and without looking after studying.

Problem 3.

m

L µ k

at rest

θ
D?
A block of mass m slides down a smooth (frictionless) incline of length L that makes an angle θ with the horizontal as shown. It then reaches a rough surface with a coefficient of kinetic friction µk .
Use the concepts of work and/or mechanical energy to find the distance D the block slides across the rough surface before it comes to rest. You will find that using the generalized non-conservative work-mechanical energy theorem is easiest, but you can succeed using work and mechanical energy conservation for two separate parts of the problem as well.

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Week 3: Work and Energy
Problem 4.

d

H

m

A simple child’s toy is a jumping frog made up of an approximately massless spring of uncompressed length d and spring constant k that propels a molded plastic “frog” of mass m. The frog is pressed down onto a table (compressing the spring by d) and at t = 0 the spring is released so that the frog leaps high into the air.
Use work and/or mechanical energy to determine how high the frog leaps.

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Week 3: Work and Energy

Problem 5.

m2

m1

H

A block of mass m2 sits on a rough table. The coefficients of friction between the block and the table are µs and µk for static and kinetic friction respectively. A much larger mass m1 (easily heavy enough to overcome static friction) is suspended from a massless, unstretchable, unbreakable rope that is looped around the two pulleys as shown and attached to the support of the rightmost pulley.
At time t = 0 the system is released at rest.
Use work and/or mechanical energy (where the latter is very easy since the internal work done by the tension in the string cancels) to find the speed of both masses after the large mass m1 has fallen a distance H. Note that you will still need to use the constraint between the coordinates that describe the two masses. Remember how hard you had to “work” to get this answer last week?
When time isn’t important, energy is better!

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Week 3: Work and Energy
Problem 6.

D m F = F oe

−x/D

x

A simple schematic for a paintball gun with a barrel of length D is shown above; when the trigger is pulled carbon dioxide gas under pressure is released into the approximately frictionless barrel behind the paintball (which has mass m). As it enters, the expanding gas is cut off by a special valve so that it exerts a force on the ball of magnitude:
F = F0 e−x/D on the ball, pushing it to the right, where x is measured from the paintball’s initial position as shown, until the ball leaves the barrel.
a) Find the work done on the paintball by the force as the paintball is accelerated a total distance
D down the barrel.
b) Use the work-kinetic-energy theorem to compute the kinetic energy of the paintball after it has been accelerated.
c) Find the speed with which the paintball emerges from the barrel after the trigger is pulled.

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Week 3: Work and Energy

Problem 7.

m

v
H
R θ R

A block of mass M sits at rest at the top of a frictionless hill of height H leading to a circular frictionless loop-the-loop of radius R.
a) Find the minimum height Hmin for which the block barely goes around the loop staying on the track at the top. (Hint: What is the condition on the normal force when it “barely” stays in contact with the track? This condition can be thought of as “free fall” and will help us understand circular orbits later, so don’t forget it.).
Discuss within your recitation group why your answer is a scalar number times R and how this kind of result is usually a good sign that your answer is probably right.
b) If the block is started at height Hmin , what is the normal force exerted by the track at the bottom of the loop where it is greatest?
If you have ever ridden roller coasters with loops, use the fact that your apparent weight is the normal force exerted on you by your seat if you are looping the loop in a roller coaster and discuss with your recitation group whether or not the results you derive here are in accord with your experiences. If you haven’t, consider riding one aware of the forces that are acting on you and how they affect your perception of weight and change your direction on your next visit to e.g. Busch Gardens to be, in a bizarre kind of way, a physics assignment. (Now c’mon, how many classes have you ever taken that assign riding roller coasters, even as an optional activity?:-) 175

Week 3: Work and Energy
Problem 8.

vmin

R
T

m

vo
A ball of mass m is attached to a (massless, unstretchable) string and is suspended from a pivot.
It is moving in a vertical circle of radius R such that it has speed v0 at the bottom as shown. The ball is in a vacuum; neglect drag forces and friction in this problem. Near-Earth gravity acts down.
a) Find an expression for the force exerted on the ball by the rod at the top of the loop as a function of m, g, R, and vtop , assuming that the ball is still moving in a circle when it gets there. b) Find the minimum speed vmin that the ball must have at the top to barely loop the loop
(staying on the circular trajectory) with a precisely limp string with tension T = 0 at the top.
c) Determine the speed v0 the ball must have at the bottom to arrive at the top with this minimum speed. You may use either work or potential energy for this part of the problem.

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Week 3: Work and Energy

Problem 9.

vmin

R
T

m

vo
A ball of mass m is attached to a massless rod (note well) and is suspended from a frictionless pivot. It is moving in a vertical circle of radius R such that it has speed v0 at the bottom as shown.
The ball is in a vacuum; neglect drag forces and friction in this problem. Near-Earth gravity acts down. a) Find an expression for the force exerted on the ball by the rod at the top of the loop as a function of m, g, R, and vtop , assuming that the ball is still moving in a circle when it gets there. b) Find the minimum speed vmin that the ball must have at the top to barely loop the loop
(staying on the circular trajectory). Note that this is easy, once you think about how the rod is different from a string!
c) Determine the speed v0 the ball must have at the bottom to arrive at the top with this minimum speed. You may use either work or potential energy for this part of the problem.

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Week 3: Work and Energy
Problem 10.

m

H

θ R

v

A block of mass M sits at the top of a frictionless hill of height H. It slides down and around a loop-the-loop of radius R, so that its position on the circle can be identified with the angle θ with respect to the vertical as shown
a) Find the magnitude of the normal force as a function of the angle θ.
b) From this, deduce an expression for the angle θ0 at which the block will leave the track if the block is started at a height H = 2R.

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Week 3: Work and Energy

Advanced Problem 11.

2v0

D v0 In the figure above we see two cars, one moving at a speed v0 and an identical car moving at a speed 2v0 . The cars are moving at a constant speed, so their motors are pushing them forward with a force that precisely cancels the drag force exerted by the air. This drag force is quadratic in their speed: Fd = −bv 2
(in the opposite direction to their velocity) and we assume that this is the only force acting on the car in the direction of motion besides that provided by the motor itself, neglecting various other sources of friction or inefficiency.
a) Prove that the engine of the faster car has to be providing eight times as much power to maintain the higher constant speed than the slower car.
b) Prove that the faster car has to do four times as much work to travel a fixed distance D than the slower car.
Discuss these (very practical) results in your groups. Things you might want to talk over include:
Although cars typically do use more gasoline to drive the same distance at 100 kph (∼ 62 mph) than they do at 50 kph, it isn’t four times as much, or even twice as much. Why not?
Things to think about include gears, engine efficiency, fuel wasted idly, friction, streamlining
(dropping to Fd = −bv Stokes’ drag).

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Week 3: Work and Energy
Advanced Problem 12.

+y v r

r sin θ(t)

θ(t) r cos θ (t)

+x

This is a guided exercise in calculus exploring the kinematics of circular motion and the relation between Cartesian and Plane Polar coordinates. It isn’t as intuitive as the derivation given in the first two weeks, but it is much simpler and is formally correct.
In the figure above, note that: r = r cos (θ(t)) x + r sin (θ(t)) y
ˆ
ˆ where r is the radius of the circle and θ(t) is an arbitrary continuous function of time describing where a particle is on the circle at any given time. This is equivalent to: x(t) = r cos(θ(t))

y(t) = r sin(θ(t))
(going from (r, θ) plane polar coordinates to (x, y) cartesian coordinates and the corresponding: r =

θ(t) =

x(t)2 + y(t)2 y tan−1 x You will find the following two definitions useful: ω =

α

=

dθ dt dω d2 θ
= 2 dt dt

The first you should already be familiar with as the angular velocity, the second is the angular acceleration. Recall that the tangential speed vt = rω; similarly the tangential acceleration is at = rα as we shall see below.
Work through the following exercises:
a) Find the velocity of the particle v in cartesian vector coordinates.
b) Form the dot product v · r and show that it is zero. This proves that the velocity vector is perpendicular to the radius vector for any particle moving on a circle!

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Week 4: Systems of Particles, Momentum and Collisions

c) Show that the total acceleration of the particle a in cartesian vector coordinates can be written as: α a = −ω 2 r + v ω Since the direction of v is tangent to the circle of motion, we can identify these two terms as the results: v2 ar = −ω 2 r = − t r (now derived in terms of its cartesian components) and at = αr.

Optional Problems
The following problems are not required or to be handed in, but are provided to give you some extra things to work on or test yourself with after mastering the required problems and concepts above and to prepare for quizzes and exams.
No optional problems (yet) this week.

Week 4: Systems of Particles,
Momentum and Collisions
Summary
• The center of mass of a system of particles is given by: xcm =

i

mi x i
1
=
Mtot
i mi

mi x i i One can differentiate this expression once or twice with respect to time to get the two corollary expressions: mi v i
1
= mi v i v cm = i
Mtot i i mi and acm =

i

mi ai
1
= mi Mtot i mi a i i All three expressions may be summed up in the useful forms:
Mtot xcm

=

mi x i i Mtot v cm

=

mi v i i Mtot acm

=

mi a i i The center of mass coordinates are truly weighted averages of the coordinates – weighted with the actual weights of the particles91 .
• The mass density of a solid object in one, two, or three dimensions is traditionally written in physics as: λ σ ρ dm
∆m
=
∆x→0 ∆x dx ∆m dm = lim
=
∆x→0 ∆A dA dm
∆m
=
= lim
∆x→0 ∆V dV =

lim

In each of these expressions, ∆m is the mass in a small “chunk” of the material, one of length
∆x, area ∆A, or volume ∆V . The mass distribution of an object is in general a complicated
91 Near

the Earth’s surface where the weight only depends on the mass, of course. Really they are weighted with

the mass.

181

182

Week 4: Systems of Particles, Momentum and Collisions function of the coordinates 92 . However we will usually work only with very simple mass distributions that we can easily integrate/sum over in this class. When doing so we are likely to use these definitions backwards: dm = λ dx

1 dimension

dm

= σ dA

2 dimensions

dm

= ρ dV

3 dimensions

Use the following ritual incantation (which will be useful to you repeatedly for both semesters of this course!) when working with mass (or later, charge) density distributions:
The mass of the chunk is the mass per unit (length, area, volume) times the (length, area, volume) of the chunk!
• The Center of Mass of a solid object (continuous mass distribution) is given by: xcm =

x dm
=
dm

x ρ(x) dV
1
=
Mtot
ρ(x) dV

x ρ(x) dV

This can be evaluated one component at a time, e.g.: xcm =

x dm
=
dm

x ρ(x) dV
1
=
Mtot
ρ(x) dV

x ρ(x) dV

(and similarly for ycm and zcm ).
It also can be written (componentwise) for mass distributions in one and two dimensions: xcm =

x dm
=
dm

x λ dx
1
=
Mtot
λ dx

x λ dx

xcm =

x dm
=
dm

x σ dA
1
=
Mtot
σ dA

x σ dA

ycm =

y dm
=
dm

y σ dA
1
=
Mtot
σ dA

y σ dA

(in one dimension) or

and

(in two dimensions).
• The Momentum of a particle is defined to be: p = mv
The momentum of a system of particles is the sum of the momenta of the individual particles: ptot =

mi v i =

mi v cm = Mtot v cm

i

where the last expression follows from the expression for the velocity of the center of mass above. 92 Think

about how mass is distributed in the human body! Or, for that matter, think about the Universe itself, which can be thought of at least partially as a great big mass density distribution ρ(x)...

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Week 4: Systems of Particles, Momentum and Collisions
• The Kinetic Energy in Terms of the Momentum of a particle is easily written as:
2

K=

(mv)
1
m p2 1
=
mv 2 = mv 2
=
2
2
m
2m
2m

or (for a system of particles):
1
2 mi v i =
2

Ktot = i i

p2 i 2mi

These forms are very useful in collision problems where momentum is known and conserved; they will often save you a step or two in the algebra if you express kinetic energies in terms of momenta from the beginning.
• Newton’s Second Law for a single particle can be expressed (and was so expressed, originally, by Newton) as: dp F tot = dt where F tot is the total force acting on the particle.
For a system of particles one can sum this:
F tot =

Fi = i i

d i pi dpi dptot
=
= dt dt dt In this expression the internal forces directed along the lines between particles of the system cancel (due to Newton’s Third Law) and: ext F tot =

Fi i =

dptot dt where the total force in this expression is the sum of only the total external forces acting on the various particles of the system.
• The Law of Conservation of Momentum states (following the previous result) that:
If and only if the total external force acting on a system of particles vanishes, then the total momentum of that system is a constant vector. or (in equationspeak):
If and only if F tot = 0 then ptot = pi = pf , a constant vector where pi and pf are the initial and final momenta across some intervening process or time interval where no external forces acted. Momentum conservation is especially useful in collision problems because the collision force is internal and hence does not change the total momentum. • The Center of Mass Reference Frame is a convenient frame for solving collision problems.
It is the frame whose origin lies at the center of mass and that moves at the constant velocity
(relative to “the lab frame”) of the center of mass. That is, it is the frame wherein: x′ = xi − xcm = xi − v cm t i and (differentiating once): v ′ = v i − v cm i In this frame, p′ = tot mi v ′ = i i

i

mi v i −

i

mi v cm = ptot − ptot = 0

which is why it is so very useful. The total momentum is the constant value 0 in the center of mass frame of a system of particles with no external forces acting on it!

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Week 4: Systems of Particles, Momentum and Collisions
• The Impulse of a collision is defined to be the total momentum transferred during the collision, where a collision is an event where a very large force is exerted over a very short time interval ∆t. Recalling that F = dp/dt, it’s magnitude is:
∆t

I = |∆p| =

F dt = |F avg |∆t

0

and it usually acts along the line of the collision. Note that this the impulse is directly related to the average force exerted by a collision that lasts a very short time ∆t:
F a vg =

1
∆t

∆t

F (t) dt
0

• An Elastic Collision is by definition a collision in which both the momentum and the total kinetic energy of the particles is conserved across the collision. That is: pi = pf

Ki

= Kf

This is actually four independent conservation equations (three components of momentum and kinetic energy).
In general we will be given six “initial values” for a three-dimensional collision – the three components of the initial velocity for each particle. Our goal is to find the six final values – the three components of the final velocity of each particle. However, we don’t have enough simultaneous equations to accomplish this and therefore have to be given two more pieces of information in order to solve a general elastic collision problem in three dimensions.
In two dimensional collisions we are a bit better off – we have three conservation equations
(two momenta, one energy) and four unknowns (four components of the final velocity) and can solve the collision if we know one more number, say the angle at which one of the particles emerges or the impact parameter of the collision93 , but it is still pretty difficult.
In one dimension we have two conservation equations – one momentum, one energy, and two unknowns (the two final velocities) and we can (almost) uniquely solve for the final velocities given the initial ones. In this latter case only, when the initial state of the two particles is given by m1 , v1i , m2 , v2i then the final state is given by: v1f v2f

= −v1i + 2vcm

= −v2i + 2vcm

• An Inelastic Collision is by definition not an elastic collision, that is, a collision where kinetic energy is not conserved. Note well that the term “elastic” therefore refers to conservation of energy which may or may not be present in a collision, but that MOMENTUM
IS ALWAYS CONSERVED IN A COLLISION in the impact approximation, which we will universally make in this course.
A fully inelastic collision is one where the two particles collide and stick together to move as one after the collision. In three dimensions we therefore have three conserved quantities (the components of the momentum) and three unknown quantities (the three components of the final velocity and therefore fully inelastic collisions are trivial to solve! The solution is simply to find:
P tot = P i = m1 v 1,i + m2 v 2,i and set it equal to P f : m1 v 1,i + m2 v 2,i = (m1 + m2 )v f = (m1 + m2 )v cm
93 Wikipedia:

http://www.wikipedia.org/wiki/impact parameter.

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Week 4: Systems of Particles, Momentum and Collisions or P tot m1 v 1,i + m2 v 2,i
=
m1 + m2
Mtot
The final velocity of the stuck together masses is the (constant) velocity of the center of mass of the system, which makes complete sense. v f = v cm =

Kinetic energy is always lost in an inelastic collision, and one can always evaluate it from:
∆K = Kf − Ki =

2
Ptot

2Mtot

p2 p2 2,i
1,i
+
2m1
2m2

In a partially inelastic collision, the particles collide but don’t quite stick together. One has three (momentum) conservation equations and needs six final velocities, so one in general must be given three pieces of information in order to solve one in three dimensions. Even in one dimension one has only one equation and two unknowns and need at least one additional piece of independent information to solve a problem.
• The Kinetic Energy of a System of Particles can in general be written as:

Ki

Ktot =


+ Kcm = Ktot + Kcm

i

which one should read as “The total kinetic energy of the system is the total kinetic energy in the (primed) center of mass frame plus the kinetic energy of the center of mass frame.”
The latter is just:
Kcm =

1
P2
2
Mtot vcm = tot
2
2Mtot

This theorem will prove very useful to us when we consider rotation, but it also means that the total kinetic energy of a macroscopic object many of many microscopic parts is the sum of its macroscopic kinetic energy – its kinetic energy where we treat it as a “particle” located at its center of mass – and its internal microscopic kinetic energy. The latter is essentially related to heat and temperature. Inelastic collisions that “lose kinetic energy” of their macroscopic constituents (e.g. cars) gain it in the increase in temperature of the objects after the collision that results from the greater microscopic kinetic energy of the particles that make them up in the center of mass (object) frame.

4.1: Systems of Particles
The world of one particle is fairly simple. Something pushes on the particle, and it accelerates. Stop pushing, it coasts or remains still. Do work on it and it speeds up. Do negative work on it and it slows down. Increase or decrease its potential energy; decrease or increase its kinetic energy.
However, the real world is not so simple. For one thing, every push works two ways – all forces act symmetrically between objects – no object experiences a force all by itself. For another, real objects are not particles – they are made up of lots of “particles” themselves. Finally, even if we ignore the internal constituents of an object, we seem to inhabit a universe with lots of objects.
Somehow we know intuitively that the details of the motion of every electron in a baseball are irrelevant to the behavior of the baseball as a whole. Clearly, we need to deduce ways of taking a collection of particles and determining its collective behavior. Ideally, this process should be one we can iterate, so that we can treat collections of collections – a box of baseballs, under the right circumstances (falling out of an airplane, for example) might also be expected to behave within reason like a single object independent of the motion of the baseballs inside, or the motion of the atoms in the baseballs, or the motion of the electrons in the atoms.

186

Week 4: Systems of Particles, Momentum and Collisions

Baseball ("particle" of mass M)

Packing (particle of mass m < M)
Atoms in packing

electrons in atom

Figure 45: An object such as a baseball is not really a particle. It is made of many, many particles
– even the atoms it is made of are made of many particles each. Yet it behaves like a particle as far as Newton’s Laws are concerned. Now we find out why.

We will obtain this collective behavior by averaging, or summing over at successively larger scales, the physics that we know applies at the smallest scale to things that really are particles and discover to our surprise that it applies equally well to collections of those particles, subject to a few new definitions and rules.

4.1.1: Newton’s Laws for a System of Particles – Center of Mass

F

F
1
m x 1

x
1

F tot 2

X cm x 3

M m 2

m

2

tot

3
F
3

Figure 46: A system of N = 3 particles is shown above, with various forces F i acting on the masses
(which therefore each their own accelerations ai ). From this, we construct a weighted average acceleration of the system, in such a way that Newton’s Second Law is satisfied for the total mass.
Suppose we have a system of N particles, each of which is experiencing a force. Some (part) of those forces are “external” – they come from outside of the system. Some (part) of them may be “internal” – equal and opposite force pairs between particles that help hold the system together
(solid) or allow its component parts to interact (liquid or gas).
We would like the total force to act on the total mass of this system as if it were a “particle”.
That is, we would like for:
F tot = Mtot A where vA is the “acceleration of the system”. This is easily accomplished.

(341)

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Week 4: Systems of Particles, Momentum and Collisions
Newton’s Second Law for a system of particles is written as:
F tot =

Fi =

mi

i

i

d2 xi dt2 (342)

We now perform the following Algebra Magic:
F tot

=

Fi

(343)

i

=

mi i =

d2 xi dt2 mi i = Mtot

(344)

d2 X dt2 (345)

d2 X
= Mtot A dt2 (346)

Note well the introduction of a new coordinate, X. This introduction isn’t “algebra”, it is a definition. Let’s isolate it so that we can see it better: mi i

d2 xi d2 X
= Mtot 2 dt2 dt

(347)

Basically, if we define an X such that this relation is true then Newton’s second law is recovered for the entire system of particles “located at X” as if that location were indeed a particle of mass
Mtot itself.
We can rearrange this a bit as:
1
dV d2 X
=
=
2
dt dt Mtot

mi i d2 xi
1
=
2
dt
Mtot

mi i dv i dt (348)

and can integrate twice on both sides (as usual, but we only do the integrals formally). The first integral is:
1
dxi
1
dX mi v i + V 0 = mi =V =
+V0
(349) dt Mtot i
Mtot i dt and the second is:
X=

1
Mtot

mi x i + V 0 t + X 0

(350)

i

Note that this equation is exact, but we have had to introduce two constants of integration that are completely arbitrary: V 0 and X 0 .
These constants represent the exact same freedom that we have with our inertial frame of reference – we can put the origin of coordinates anywhere we like, and we will get the same equations of motion even if we put it somewhere and describe everything in a uniformly moving frame. We should have expected this sort of freedom in our definition of a coordinate that describes “the system” because we have precisely the same freedom in our choice of coordinate system in terms of which to describe it.
In many problems, however, we don’t want to use this freedom. Rather, we want the simplest description of the system itself, and push all of the freedom concerning constants of motion over to the coordinate choice itself (where it arguably “belongs”). We therefore select just one (the simplest one) of the infinity of possibly consistent rules represented in our definition above that would preserve Newton’s Second Law and call it by a special name: The Center of Mass!

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We define the position of the center of mass to be:
M X cm =

mi xi

(351)

mi xi

(352)

i

or:
1
M

X cm =

i

(with M = i mi ). If we consider the “location” of the system of particles to be the center of mass, then Newton’s Second Law will be satisfied for the system as if it were a particle, and the location in question will be exactly what we intuitively expect: the “middle” of the (collective) object or system, weighted by its distribution of mass.
Not all systems we treat will appear to be made up of point particles. Most solid objects or fluids appear to be made up of a continuum of mass, a mass distribution. In this case we need to do the sum by means of integration, and our definition becomes:
M X cm =

xdm

(353)

1
M

xdm

(354)

or
X cm =

(with M = dm). The latter form comes from treating every little differential chunk of a solid object like a “particle”, and adding them all up. Integration, recall, is just a way of adding them up. Of course this leaves us with the recursive problem of the fact that “solid” objects are really made out of lots of point-like elementary particles and their fields. It is worth very briefly presenting the standard “coarse-graining” argument that permits us to treat solids and fluids like a continuum of smoothly distributed mass – and the limitations of that argument.

Example 4.1.1: Center of Mass of a Few Discrete Particles y 1 kg

2m

2 kg

1m

2 kg

1m

2m

3 kg

x

Figure 47: A system of four massive particles.
In figure 47 above, a few discrete particles with masses given are located at the positions indicated.
We would like to find the center of mass of this system of particles. We do this by arithmetically evaluating the algebraic expressions for the x and y components of the center of mass separately:

Week 4: Systems of Particles, Momentum and Collisions

189

4.1.2: Coarse Graining: Continuous Mass Distributions
Suppose we wish to find the center of mass of a small cube of some uniform material – such as gold, why not? We know that really gold is made up of gold atoms, and that gold atoms are make up of
(elementary) electrons, quarks, and various massless field particles that bind the massive particles together. In a cube of gold with a mass of 197 grams, there are roughly 6 × 1023 atoms, each with
79 electrons and 591 quarks for a total of 670 elementary particles per atom. This is then about
4 × 1026 elementary particles in a cube just over 2 cm per side.
If we tried to actually use the sum form of the definition of center of mass to evaluate it’s location, and ran the computation on a computer capable of performing one trillion floating point operations per second, it would take several hundred trillion seconds (say ten million years) and – unless we knew the exacly location of every quark – would still be approximate, no better than a guess.
We do far better by averaging. Suppose we take a small chunk of the cube of gold – one with cube edges 1 millimeter long, for example. This still has an enormous number of elementary particles in it
– so many that if we shift the boundaries of the chunk a tiny bit many particles – many whole atoms are moved in or out of the chunk. Clearly we are justified in talking about the ”average number of atoms” or ”average amount of mass of gold” in a tiny cube like this.
A millimeter is still absurdly large on an atomic scale. We could make the cube 1 micron (1×10−6 meter, a thousandth of a millimeter) and because atoms have a “generic” size around one Angstrom
– 1 × 10−10 meters – we would expect it to contain around (10−6 /10−10 )3 = 1012 atoms. Roughly a trillion atoms in a cube too small to see with the naked eye (and each atom still has almost 700 elementary particles, recall). We could go down at least 1-2 more orders of magnitude in size and still have millions of particles in our chunk!
A chunk 10 nanometers to the side is fairly accurately located in space on a scale of meters. It has enough elementary particles in it that we can meaningfully speak of its ”average mass” and use this to define the mass density at the point of location of the chunk – the mass per unit volume at that point in space – with at least 5 or 6 significant figures (one part in a million accuracy). In most real-number computations we might undertake in the kind of physics learned in this class, we wouldn’t pay attention to more than 3 or 4 significant figures, so this is plenty.
The point is that this chunk is now small enough to be considered differentially small for the purposes of doing calculus. This is called coarse graining – treating chunks big on an atomic or molecular scale but small on a macroscopic scale. To complete the argument, in physics we would generally consider a small chunk of matter in a solid or fluid that we wish to treat as a smooth distribution of mass, and write at first:
∆m = ρ∆V
(355)
while reciting the following magical formula to ourselves:
The mass of the chunk is the mass per unit volume ρ times the volume of the chunk. We would then think to ourselves: “Gee, ρ is almost a uniform function of location for chunks that are small enough to be considered a differential as far as doing sums using integrals are concerned.
I’ll just coarse grain this and use integration to evaluation all sums.” Thus: dm = ρ dV

(356)

We do this all of the time, in this course. This semester we do it repeatedly for mass distributions, and sometimes (e.g. when treating planets) will coarse grain on a much larger scale to form the
“average” density on a planetary scale. On a planetary scale, barring chunks of neutronium or the occasional black hole, a cubic kilometer “chunk” is still “small” enough to be considered differentially

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small – we usually won’t need to integrate over every single distinct pebble or clod of dirt on a much smaller scale. Next semester we will do it repeatedly for electrical charge, as after all all of those gold atoms are made up of charged particles so there are just as many charges to consider as there are elementary particles. Our models for electrostatic fields of continuous charge and electrical currents in wires will all rely on this sort of coarse graining.
Before we move on, we should say a word or two about two other common distributions of mass.
If we want to find e.g. the center of mass of a flat piece of paper cut out into (say) the shape of a triangle, we could treat it as a “volume” of paper and integrate over its thickness. However, it is probably a pretty good bet from symmetry that unless the paper is very inhomogeneous across its thickness, the center of mass in the flat plane is in the middle of the “slab” of paper, and the paper is already so thin that we don’t pay much attention to its thickness as a general rule. In this case we basically integrate out the thickness in our minds (by multiplying ρ by the paper thickness t) and get:
∆m = ρ∆V = ρt∆A = σ∆A

(357)

where σ = ρt is the (average) mass per unit area of a chunk of paper with area ∆A. We say our
(slightly modified) magic ritual and poof! We have: dm = σ dA

(358)

for two dimensional areal distributions of mass.
Similarly, we often will want to find the center of mass of things like wires bent into curves, things that are long and thin. By now I shouldn’t have to explain the following reasoning:
∆m = ρ∆V = ρA∆x = λ∆x

(359)

where A is the (small!) cross section of the solid wire and λ = ρA is the mass per unit length of the chunk of wire, magic spell, cloud of smoke, and when the smoke clears we are left with: dm = λ dx

(360)

In all of these cases, note well, ρ, σ, λ can be functions of the coordinates! They are not necessarily constant, they simply describe the (average) mass per unit volume at the point in our object or system in question, subject to the coarse-graining limits. Those limits are pretty sensible ones – if we are trying to solve problems on a length scale of angstroms, we cannot use these averages because the laws of large numbers won’t apply. Or rather, we can and do still use these kinds of averages in quantum theory (because even on the scale of a single atom doing all of the discrete computations proves to be a problem) but then we do so knowing up front that they are approximations and that our answer will be “wrong”.
In order to use the idea of center of mass (CM) in a problem, we need to be able to evaluate it.
For a system of discrete particles, the sum definition is all that there is – you brute-force your way through the sum (decomposing vectors into suitable coordinates and adding them up).
For a solid object that is symmetric, the CM is “in the middle”. But where’s that? To precisely find out, we have to be able to use the integral definition of the CM:
M X cm =
(with M =

xdm

dm, and dm = ρdV or dm = σdA or dm = λdl as discussed above).

Let’s try a few examples:

(361)

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Week 4: Systems of Particles, Momentum and Collisions
M
dm = λ dx =___ dx
L

0

dx

L

Figure 48:

Example 4.1.2: Center of Mass of a Continuous Rod
Let us evaluate the center of mass of a continuous rod of length L and total mass M , to make sure it is in the middle:
L

M X cm =

λxdx

xdm =

(362)

0

where

L

M=

λdx = λL

dm =

(363)

0

(which defines λ, if you like) so that
M X cm = λ

L2
L
=M
2
2

and
X cm =
Gee, that was easy. Let’s try a hard one.

L
.
2

(364)

(365)

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Week 4: Systems of Particles, Momentum and Collisions

dA =r dr dθ

θ

0

dm = σ dA r dθ

r
0

R

dr

Figure 49:

Example 4.1.3: Center of mass of a circular wedge
Let’s find the center of mass of a circular wedge (a shape like a piece of pie, but very flat). It is two dimensional, so we have to do it one coordinate at a time. We start from the same place: σxdA =

xdm =

σrdrdθ = σ

σdA =

dm =
0

0

0

σr2 cos θdrdθ

0

(366)

0

θ0

R

θ0

R

M=

0

0

0

where

θ0

R

θ0

R

M Xcm =

R 2 θ0
2

(367)

(which defines σ, if you like) so that
M Xcm = σ

R3 sin θ0
3

(368)

from which we find (with a bit more work than last time but not much) that:
Xcm =

2R3 sin θ0
.
3R2 θ

(369)

Amazingly enough, this has units of R (length), so it might just be right. To check it, do Ycm on your own!

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Week 4: Systems of Particles, Momentum and Collisions

Example 4.1.4: Breakup of Projectile in Midflight m = m 1+ m

2

v0 m1 θ

m2

x1
R

x2

Figure 50: A projectile breaks up in midflight. The center of mass follows the original trajectory of the particle, allowing us to predict where one part lands if we know where the other one lands, as long as the explosion exerts no vertical component of force on the two particles.
Suppose that a projectile breaks up horizontally into two pieces of mass m1 and m2 in midflight.
Given θ, v0 , and x1 , predict x2 .
The idea is: The trajectory of the center of mass obeys Newton’s Laws for the entire projectile and lands in the same place that it would have, because no external forces other than gravity act.
The projectile breaks up horizontally, which means that both pieces will land at the same time, with the center of mass in between them. We thus need to find the point where the center of mass would have landed, and solve the equation for the center of mass in terms of the two places the projectile fragments land for one, given the other. Thus:
Find R. As usual:

1 y = (v0 sin θ)t − gt2
2
1 tR (v0 sin θ − gtR ) = 0
2
2v0 sin θ) tR = g 2v 2 sin θ cos θ
R = (v0 cos θ)tR = 0
.
g

(370)
(371)
(372)
(373)

R is the position of the center of mass. We write the equation making it so: m1 x1 + m2 x2 = (m1 + m2 )R and solve for the unknown x2 . x2 =

(m1 + m2 )R − m1 x1 m2 (374)

(375)

From this example, we see that it is sometimes easiest to solve a problem by separating the motion of the center of mass of a system from the motion in a reference frame that “rides along” with the center of mass. The price we may have to pay for this convenience is the appearance of pseudoforces in this frame if it happens to be accelerating, but in many cases it will not be accelerating, or the acceleration will be so small that the pseudoforces can be neglected compared to the much larger forces of interest acting within the frame. We call this (at least approximately) inertial reference frame the Center of Mass Frame and will discuss and define it in a few more pages.
First, however, we need to define an extremely useful concept in physics, that of momentum, and discuss the closely related concept of impulse and the impulse approximation that permits us to treat the center of mass frame as being approximately inertial in many problems even when it is accelerating.

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Week 4: Systems of Particles, Momentum and Collisions

4.2: Momentum
Momentum is a useful idea that follows naturally from our decision to treat collections as objects.
It is a way of combining the mass (which is a characteristic of the object) with the velocity of the object. We define the momentum to be: p = mv

(376)

Thus (since the mass of an object is generally constant):
F = ma = m

d dp dv
=
(mv) = dt dt dt (377)

is another way of writing Newton’s second law. In fact, this is the way Newton actually wrote
Newton’s second law – he did not say “F = ma” the way we have been reciting. We emphasize this connection because it makes the path to solving for the trajectories of constant mass particles a bit easier, not because things really make more sense that way.
Note that there exist systems (like rocket ships, cars, etc.) where the mass is not constant. As the rocket rises, its thrust (the force exerted by its exhaust) can be constant, but it continually gets lighter as it burns fuel. Newton’s second law (expressed as F = ma) does tell us what to do in this case – but only if we treat each little bit of burned and exhausted gas as a “particle”, which is dp a pain. On the other hand, Newton’s second law expressed as F = dt still works fine and makes perfect sense – it simultaneously describes the loss of mass and the increase of velocity as a function of the mass correctly.
Clearly we can repeat our previous argument for the sum of the momenta of a collection of particles: P tot = pi = mv i
(378)
i

i

so that dP tot
=
dt

i

dpi
=
dt

F i = F tot

(379)

i

Differentiating our expression for the position of the center of mass above, we also get: d mi xi
=
dt

i

mi i dxi
=
dt

pi = P tot = Mtot v cm

(380)

i

4.2.1: The Law of Conservation of Momentum
We are now in a position to state and trivially prove the Law of Conservation of Momentum.
It reads94 :
If and only if the total external force acting on a system is zero, then the total momentum of a system (of particles) is a constant vector.
You are welcome to learn this in its more succinct algebraic form:
If and only if F tot = 0 then P tot = P initial = P final = a constant vector.

(381)

Please learn this law exactly as it is written here. The condition F tot = 0 is essential – otherwise, as you can see, F tot = dP tot ! dt 94 The

“if and only if” bit, recall, means that if the total momentum of a system is a constant vector, it also implies that the total force acting on it is zero, there is no other way that this condition can come about.

Week 4: Systems of Particles, Momentum and Collisions

195

The proof is almost a one-liner at this point:
F tot =

Fi = 0

(382)

i

implies dP tot
=0
dt

(383)

so that P tot is a constant if the forces all sum to zero. This is not quite enough. We need to note that for the internal forces (between the ith and jth particles in the system, for example) from
Newton’s third law we get:
F ij = −F ji
(384)
so that
F ij + F ji = 0

(385)

pairwise, between every pair of particles in the system. That is, although internal forces may not be zero (and generally are not, in fact) the changes the cause in the momentum of the system cancel.
We can thus subtract:
F internal =
F ij = 0
(386)
i,j

from F tot = F external + F internal to get:
F external =

dP tot
=0
dt

(387)

and the total momentum must be a constant (vector).
This can be thought of as the “bootstrap law” – You cannot lift yourself up by your own bootstraps! No matter what force one part of you exerts on another, those internal forces can never alter the velocity of your center of mass or (equivalently) your total momentum, nor can they overcome or even alter any net external force (such as gravity) to lift you up.
As we shall see, the idea of momentum and its conservation greatly simplify doing a wide range of problems, just like energy and its conservation did in the last chapter. It is especially useful in understanding what happens when one object collides with another object.
Evaluating the dynamics and kinetics of microscopic collisions (between, e.g. electrons, protons, neutrons and targets such as atoms or nuclei) is a big part of contemporary physics – so big that we call it by a special name: Scattering Theory95 . The idea is to take some initial (presumed known) state of an about-to-collide “system”, to let it collide, and to either infer from the observed scattering something about the nature of the force that acted during the collision, or to predict, from the measured final state of some of the particles, the final state of the rest.
Sound confusing? It’s not, really, but it can be complicated because there are lots of things that might make up an initial and final state. In this class we have humbler goals – we will be content simply understanding what happens when macroscopic objects like cars or billiard96 balls collide, where (as we will see) momentum conservation plays an enormous role. This is still the first
95 Wikipedia: http://www.wikipedia.org/wiki/Scattering Theory. This link is mostly for more advanced students,
e.g. physics majors, but future radiologists might want to look it over as well as it is the basis for a whole lot of radiology... 96 Wikipedia: http://www.wikipedia.org/wiki/Billiards. It is always dangerous to assume the every student has had any given experience or knows the same games or was raised in the same culture as the author/teacher, especially nowadays when a significant fraction of my students, at least, come from other countries and cultures, and when this book is in use by students all over the world outside of my own classroom, so I provide this (and various other) links.
In this case, as you will see, billiards or “pool” is a game played on a table where the players try to knock balls in holes by poking one ball (the “cue ball”) with a stick to drive another identically sized ball into a hole. Since the balls are very hard and perfectly spherical, the game is an excellent model for two-dimensional elastic collisions.

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Week 4: Systems of Particles, Momentum and Collisions

step (for physics majors or future radiologists) in understanding more advanced scattering theory but it provides a lot of direct insight into everyday experience and things like car safety and why a straight on shot in pool often stops one ball cold while the other continues on with the original ball’s velocity.
In order to be able to use momentum conservation in a collision, however, no external force can act on the colliding objects during the collision. This is almost never going to precisely be the case, so we will have to idealize by assuming that a “collision” (as opposed to a more general and leisurely force interaction) involves forces that are zero right up to where the collision starts, spike up to very large values (generally much larger than the sum of the other forces acting on the system at the time) and then drop quickly back to zero, being non-zero only in a very short time interval ∆t.
In this idealization, collisions will (by assumption) take place so fast that any other external forces cannot significantly alter the momentum of the participants during the time ∆t. This is called the impulse approximation. With the impulse approximation, we can neglect all other external forces (if any are present) and use momentum conservation as a key principle while analyzing or solving collisions. All collision problems solved in this course should be solved using the impulse approximation. Let’s see just what “impulse” is, and how it can be used to help solve collision problems and understand things like the forces exerted on an object by a fluid that is in contact with it.

4.3: Impulse
Let us imagine a typical collision: one pool ball approaches and strikes another, causing both balls to recoil from the collision in some (probably different) directions and at different speeds. Before they collide, they are widely separated and exert no force on one another. As the surfaces of the two
(hard) balls come into contact, they “suddenly” exert relatively large, relatively violent, equal and opposite forces on each other over a relatively short time, and then the force between the objects once again drops to zero as they either bounce apart or stick together and move with a common velocity. “Relatively” here in all cases means compared to all other forces acting on the system during the collision in the event that those forces are not actually zero.
For example, when skidding cars collide, the collision occurs so fast that even though kinetic friction is acting, it makes an ignorable change in the momentum of the cars during the collision compared to the total change of momentum of each car due to the collision force. When pool balls collide we can similarly ignore the drag force of the air or frictional force exerted by the table’s felt lining for the tiny time they are in actual contact. When a bullet embeds itself in a block, it does so so rapidly that we can ignore the friction of the table on which the block sits. Idealizing and ignoring e.g. friction, gravity, drag forces in situations such as this is known as the impulse approximation, and it greatly simplifies the treatment of collisions.
Note that we will frequently not know the detailed functional form of the collision force, F coll (t) nor the precise amount of time ∆t in any of these cases. The “crumpling” of cars as they collide is a very complicated process and exerts a completely unique force any time such a collision occurs – no two car collisions are exactly alike. Pool balls probably do exert a much more reproducible and understandable force on one another, one that we we could model if we were advanced physicists or engineers working for a company that made billiard tables and balls and our livelihoods depended on it but we’re not and it doesn’t. Bullets embedding themselves in blocks again do so with a force that is different every time that we can never precisely measure, predict, or replicate.
In all cases, although the details of the interaction force are unknown (or even unknowable in any meaningful way), we can obtain or estimate or measure some approximate things about the forces in any given collision situation. In particular we can put reasonable limits on ∆t and make ‘before’ and ‘after’ measurements that permit us to compute the average force exerted over this time.

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Week 4: Systems of Particles, Momentum and Collisions

Let us begin, then, by defining the average force over the (short) time ∆t of any given collision, assuming that we did know F = F 21 (t), the force one object (say m1 ) exerts on the other object
(m2 ). The magnitude of such a force (one perhaps appropriate to the collision of pool balls) is sketched below in figure 51 where for simplicity we assume that the force acts only along the line of contact and is hence effectively one dimensional in this direction97 .

F (t)

Area is ∆p = I (impulse)

F avg

t

∆t

Figure 51: A “typical” collision force that might be exerted by the cue ball on the eight ball in a game of pool, approximately along the line connecting the two ball centers. In this case we would expect a fairly symmetric force as the two balls briefly deform at the point of contact. The time of contact ∆t has been measured to be on the order of a tenth of a millisecond for colliding pool balls.
The time average of this force is computed the same way the time average of any other timedependent quantity might be:
∆t
1
F (t) dt
(388)
F avg =
∆t 0
We can evaluate the integral using Newton’s Second Law expressed in terms of momentum:
F (t) =

dp dt (389)

so that (multiplying out by dt and integrating):
∆t

p2f − p2i = ∆p2 =

F (t) dt

(390)

0

This is the total vector momentum change of the second object during the collision and is also the area underneath the F (t) curve (for each component of a general force – in the figure above we assume that the force only points along one direction over the entire collision and the change in the momentum component in this direction is then the area under the drawn curve). Note that the momentum change of the first ball is equal and opposite. From Newton’s Third Law,
F 12 (t) = −F 21 (t) = F and:
∆t

p1f − p1i = ∆p1 = −

0

F (t) dt = −∆p2

(391)

The integral of a force F over an interval of time is called the impulse 98 imparted by the force t2 t2

F (t) dt =

I= t1 t1

dp dt = dt p2 p1 dp = p2 − p1 = ∆p

(392)

97 This is, as anyone who plays pool knows from experience, an excellent assumption and is in fact how one most generally “aims” the targeted ball (neglecting all of the various fancy tricks that can alter this assumption and the outcome). 98 Wikipedia: http://www.wikipedia.org/wiki/Impulse (physics).

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Week 4: Systems of Particles, Momentum and Collisions

This proves that the (vector) impulse is equal to the (vector) change in momentum over the same time interval, a result known as the impulse-momentum theorem. From our point of view, the impulse is just the momentum transferred between two objects in a collision in such a way that the total momentum of the two is unchanged.
Returning to the average force, we see that the average force in terms of the impulse is just:
F avg =

pf − pi
I
∆p
=
=
∆t
∆t
∆t

(393)

If you refer again to figure 51 you can see that the area under Favg is equal the area under the actual force curve. This makes the average force relatively simple to compute or estimate any time you know the change in momentum produced by a collision and have a way of measuring or assigning an effective or average time ∆t per collision.

Example 4.3.1: Average Force Driving a Golf Ball
A golf ball leaves a 1 wood at a speed of (say) 70 meters/second (this is a reasonable number – the world record as of this writing is 90 meters/second). It has a mass of 45 grams. The time of contact has been measured to be ∆t = 0.0005 seconds (very similar to a collision between pool balls). What is the magnitude of the average force that acts on the golf ball during this “collision”?
This one is easy:
Favg =

I mvf − m(0)
3.15
=
=
= 6300
∆t
∆t
0.0005

Newtons

(394)

Since I personally have a mass conveniently (if embarrassingly) near 100 kg and therefore weigh 1000
Newtons, the golf club exerts an average force of 6.3 times my weight, call it 3/4 of a ton. The peak force, assuming an impact shape for F (t) not unlike that pictured above is as much as two English tons (call it 17400 Newtons).

Note Well! Impulse is related to a whole spectrum of conceptual mistakes students often make!
Here’s an example that many students would get wrong before they take mechanics and that no student should ever get wrong after they take mechanics! But many do. Try not to be one of them...

Example 4.3.2: Force, Impulse and Momentum for Windshield and Bug
There’s a song by Mary Chapin Carpenter called “The Bug” with the refrain:
Sometimes you’re the windshield,
Sometimes you’re the bug...
In a collision between (say) the windshield of a large, heavily laden pickup truck and a teensy little yellowjacket wasp, answer the following qualitative/conceptual questions:
a) Which exerts a larger (magnitude) force on the other during the collision?
b) Which changes the magnitude of its momentum more during the collision?
c) Which changes the magnitude of its velocity more during the collision?

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Week 4: Systems of Particles, Momentum and Collisions

Think about it for a moment, answer all three in your mind. Now, compare it to the correct answers below99 . If you did not get all three perfectly correct then go over this whole chapter until you do – you may want to discuss this with your favorite instructor as well.

4.3.1: The Impulse Approximation
When we analyze actual collisions in the real world, it will almost never be the case that there are no external forces acting on the two colliding objects during the collision process. If we hit a baseball with a bat, if two cars collide, if we slide two air-cushioned disks along a tabletop so that they bounce off of each other, gravity, friction, drag forces are often present. Yet we will, in this textbook, uniformly assume that these forces are irrelevant during the collision.

F (t)

∆pc (collision)

∆p b(background forces)

∆t

t

Figure 52: Impulse forces for a collision where typical external forces such as gravity or friction or drag forces are also present.
Let’s see why (and when!) we can get away with this. Figure 52 shows a typical collision force (as before) for a collision, but this time shows some external force acting on the mass at the same time.
This force might be varying friction and drag forces as a car brakes to try to avoid a collision on a bumpy road, for example. Those forces may be large, but in general they are very small compared to the peak, or average, collision force between two cars. To put it in perpective, in the example above we estimated that the average force between a golf ball and a golf club is over 6000 newtons during the collision – around six times my (substantial) weight. In contrast, the golf ball itself weighs much less than a newton, and the drag force and friction force between the golf ball and the tee are a tiny fraction of that.
If anything, the background forces in this figure are highly exaggerated for a typical collision, compared to the scale of the actual collision force!
The change in momentum resulting from the background force is the area underneath its curve, just as the change in momentum resulting from the collision force alone is the area under the collision force curve.
Over macroscopic time – over seconds, for example – gravity and drag forces and friction can make a significant contribution to the change in momentum of an object. A braking car slows down.
A golf ball soars through the air in a gravitational trajectory modified by drag forces. But during
99 Put here so you can’t see them while you are thinking so easily. The force exerted by the truck on the wasp is exactly the same as the force exerted by the wasp on the truck (Newton’s Third Law!). The magnitude of the momentum (or impulse) transferred from the wasp to the truck is exactly the same as the magnitude of the momentum transferred from the truck to the wasp. However, the velocity of the truck does not measurably change
(for the probable impulse transferred from any normal non-Mothra-scale wasp) while the wasp (as we will see below) bounces off going roughly twice the speed of the truck...

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the collision time ∆t they are negligible, in the specific sense that:
∆p = ∆pc + ∆pb ≈ ∆pc

(395)

(for just one mass) over that time only. Since the collision force is an internal force between the two colliding objects, it cancels for the system making the momentum change of the system during the collision approximately zero.
We call this approximation ∆p ≈ ∆pc (neglecting the change of momentum resulting from background external forces during the collision) the impulse approximation and we will always assume that it is valid in the problems we solve in this course. It justifies treating the center of mass reference frame (discussed in the next section) as an inertial reference frame even when technically it is not for the purpose of analyzing a collision or explosion.
It is, however, useful to have an understanding of when this approximation might fail. In a nutshell, it will fail for collisions that take place over a long enough time ∆t that the external forces produce a change of momentum that is not negligibly small compared to the momentum exchange between the colliding particles, so that the total momentum before the collision is not approximately equal to the total momentum after the collision.
This can happen because the external forces are unusually large (comparable to the collision force), or because the collision force is unusually small (comparable to the external force), or because the collision force acts over a long time ∆t so that the external forces have time to build up a significant ∆p for the system. None of these circumstances are typical, however, although we can imagine setting up an problem where it is true – a collision between two masses sliding on a rough table during the collision where the collision force is caused by a weak spring (a variant of a homework problem, in other words). We will consider this sort of problem (which is considerably more difficult to solve) to be beyond the scope of this course, although it is not beyond the scope of what the concepts of this course would permit you to set up and solve if your life or job depended on it.

4.3.2: Impulse, Fluids, and Pressure
Another valuable use of impulse is when we have many objects colliding with something – so many that even though each collision takes only a short time ∆t, there are so many collisions that they exert a nearly continuous force on the object. This is critical to understanding the notion of pressure exerted by a fluid, because microscopically the fluid is just a lot of very small particles that are constantly colliding with a surface and thereby transferring momentum to it, so many that they exert a nearly continuous and smooth force on it that is the average force exerted per particle times the number of particles that collide. In this case ∆t is conveniently considered to be the inverse of the rate (number per second) with which the fluid particles collide with a section of the surface.
To give you a very crude idea of how this works, let’s review a small piece of the kinetic theory of gases. Suppose you have a cube with sides of length L containing N molecules of a gas. We’ll imagine that all of the molecules have a mass m and an average speed in the x direction of vx , with
(on average) one half going left and one half going right at any given time.
In order to be in equilibrium (so vx doesn’t change) the change in momentum of any molecule that hits, say, the right hand wall perpendicular to x is ∆px = 2mvx . This is the impulse transmitted to the wall per molecular collision. To find the total impulse in the time ∆t, one must multiply this by one half the number of molecules in in a volume L2 vx ∆t. That is,
∆ptot =

1
2

N
L3

L2 vx ∆t (2mvx )

(396)

Let’s call the volume of the box L3 = V and the area of the wall receiving the impulse L2 = A. We

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P =

Favg
∆ptot
=
=
A
A ∆t

N
V

1 mv 2
2 x

=

N
V

Kx,avg

(397)

where the average force per unit area applied to the wall is the pressure, which has SI units of
Newtons/meter2 or Pascals.
If we add a result called the equipartition theorem 100 :
Kx,avg =

1
1
2 mvx = kb T 2
2
2

(398)

where kb is Boltzmann’s constant and T is the temperature in degrees absolute, one gets:
P V = N kT

(399)

which is the Ideal Gas Law 101 .
This all rather amazing and useful, and is generally covered and/or derived in a thermodynamics course, but is a bit beyond our scope for this semester. It’s an excellent use of impulse, though, and the homework problem involving bouncing of a stream of beads off of the pan of a scale is intended to be “practice” for doing it then, or at least reinforcing the understanding of how pressure arises for later on in this course when we treat fluids.
In the meantime, the impulse approximation reduces a potentially complicated force of interaction during a collision to its most basic parameters – the change in momentum it causes and the (short) time over which it occurs. Life is simple, life is good. Momentum conservation (as an equation or set of equations) will yield one or more relations between the various momentum components of the initial and final state in a collision, and with luck and enough additional data in the problem description will enable us to solve them simultaneously for one or more unknowns. Let’s see how this works.

100 Wikipedia:

http://www.wikipedia.org/wiki/Equipartition Theorem. http://www.wikipedia.org/wiki/Ideal Gas Law. The physicist version of it anyway. Chemists have the pesky habit of converting the number of molecules into the number of moles using Avogadro’s number N = 6 × 1023 and expressing it as P V = nRT instead, where R = kb NA . then using truly horrendous units such as liter-atmospheres in 101 Wikipedia:

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4.4: Center of Mass Reference Frame lab frame
CM frame v 1 m x

1

x’
1
v cm

1 x cm

m

2

x’
2
x

v
2

2

Figure 53: The coordinates of the “center of mass reference frame”, a very useful inertial reference frame for solving collisions and understanding rigid rotation.
In the “lab frame” – the frame in which we actually live – we are often in some sense out of the picture as we try to solve physics problems, trying to make sense of the motion of flies buzzing around in a moving car as it zips by us. In the Center of Mass Reference Frame we are literally in the middle of the action, watching the flies in the frame of the moving car, or standing a ground zero for an impending collision. This makes it a very convenient frame for analyzing collisions, rigid rotations around an axis through the center of mass (which we’ll study next week), static equilibrium (in a couple more weeks). At the end of this week, we will also derive a crucial result connecting the kinetic energy of a system of particles in the lab to the kinetic energy of the same system evaluated in the center of mass frame that will help us understand how work or mechanical energy can be transformed without loss into enthalpy (the heating of an object) during a collision or to rotational kinetic energy as an object rolls!
Recall from Week 2 the Galilean transformation between two inertial references frames where the primed one is moving at constant velocity v frame compared to the unprimed (lab) reference frame, equation 197. x′ = xi − v frame t
(400)
i
We choose our lab frame so that at time t = 0 the origins of the two frames are the same for simplicity. Then we take the time derivative of this equation, which connects the velocity in the lab frame to the velocity in the moving frame: v ′ = v i − v frame i (401)

I always find it handy to have a simple conceptual metaphor for this last equation: The velocity of flies observed within a moving car equals the velocity of the flies as seen by an observer on the ground minus the velocity of the car, or equivalently the velocity seen on the ground is the velocity of the car plus the velocity of the flies measured relative to the car. That helps me get the sign in the transformation correct without having to draw pictures or do actual algebra.
Let’s define the Center of Mass Frame to be the particular frame whose origin is at the center of mass of a collection of particles that have no external force acting on them, so that the total momentum of the system is constant and the velocity of the center of mass of the system is also constant: P tot = Mtot v cm = a constant vector
(402)
or (dividing by Mtot and using the definition of the velocity of the center of mass): v cm =

1
Mtot

mi v i = a constant vector. i (403)

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Then the following two equations define the Galilean transformation of position and velocity coordinates from the (unprimed) lab frame into the (primed) center of mass frame: x′ = xi − xcm = xi − v cm t i (404)

v ′ = v i − v cm i (405)

An enormously useful property of the center of mass reference frame follows from adding up the total momentum in the center of mass frame:


P tot

mi v ′ = i = i =

i

( i mi v i ) − (

mi (v i − v cm ) mi )v cm i = Mtot v cm − Mtot v cm = 0 (!)

(406)

The total momentum in the center of mass frame is identically zero! In retrospect, this is obvious.
The center of mass is at the origin, at rest, in the center of mass frame by definition, so its velocity

v ′ is zero, and therefore it should come as no surprise that P tot = Mtot v ′ = 0. cm cm
As noted above, the center of mass frame will be very useful to us both conceptually and computationally. Our first application of the concept will be in analyzing collisions. Let’s get started!

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4.5: Collisions
A “collision” in physics occurs when two bodies that are more or less not interacting (because they are too far apart to interact) come “in range” of their mutual interaction force, strongly interact for a short time, and then separate so that they are once again too far apart to interact. We usually think of this in terms of “before” and “after” states of the system – a collision takes a pair of particles from having some known initial “free” state right before the interaction occurs to an unknown final
“free” state right after the interaction occurs. A good mental model for the interaction force (as a function of time) during the collision is the impulse force sketched above that is zero at all times but the short time ∆t that the two particles are in range and strongly interacting.
There are three general “types” of collision:
• Elastic
• Fully Inelastic
• Partially Inelastic
In this section, we will first indicate a single universal assumption we will make when solving scattering problems using kinematics (conservation laws) as opposed to dynamics (solving the actual equations of motion for the interaction through the collision). Next, we will briefly define each type of collision listed above. Finally, in the following sections we’ll spend some time studying each type in some detail and deriving solutions where it is not too difficult.

4.5.1: Momentum Conservation in the Impulse Approximation
All collisions that occur rapidly enough to be treated in the impulse approximation conserve momentum even if the particles are not exactly free before and after (because they are moving in a gravitational field, experiencing drag, etc). The validity of the impulse approximation will be our default assumption in the collisions we treat in this course, and hence we will assume that all collisions conserve total momentum through the collision. That is, the total vector momentum of the colliding particles right before the collision will equal the total vector momentum of the colliding particles right after the collision.
Because momentum is a three-dimensional vector, this yields one to three (relevant) independent equations that constrain the solution, depending on the number of dimensions in which the collision occurs. 4.5.2: Elastic Collisions
By definition, an elastic collision is one that also conserves total kinetic energy so that the total scalar kinetic energy of the colliding particles before the collision must equal the total kinetic energy after the collision. This is an additional independent equation that the solution must satisfy.
It is assumed that all other contributions to the total mechanical energy (for example, gravitational potential energy) are identical before and after if not just zero, again this is the impulse approximation that states that all of these forces are negligible compared to the collision force over the time ∆t. However, two of your homework problems will treat exceptions by explicitly giving you a conservative, “slow” interaction force (gravity and an inclined plane slope, and a spring) that mediates the “collision”. You can use these as mental models for what really happens in elastic collisions on a much faster and more violent time frame.

Week 4: Systems of Particles, Momentum and Collisions

205

4.5.3: Fully Inelastic Collisions
For inelastic collisions, we will assume that the two particles form a single “particle” as a final state with the same total momentum as the system had before the collision. In these collisions, kinetic energy is always lost. Since energy itself is technically conserved, we can ask ourselves: Where did it go? The answer is: Into heat102 !
One important characteristic of fully inelastic collisions, and the property that distinguishes them from partially inelastic collisions, is that the energy lost to heat in a fully inelastic collision is the maximum energy that can be lost in a momentum-conserving collision, as will be proven and discussed below.
Inelastic collisions are much easier to solve than elastic (or partially inelastic) ones, because there are fewer degrees of freedom in the final state (only one velocity, not two).

4.5.4: Partially Inelastic Collisions
As suggested by their name, a partially inelastic collision is one where some kinetic energy is lost in the collision (so it isn’t elastic) but not the maximum amount. The particles do not stick together, so there are in general two velocities that must be solved for in the “after” picture, just as there are for elastic collisions. In general, since any energy from zero (elastic) to some maximum amount
(fully inelastic) can be lost during the collision, you will have to be given more information about the problem (such as the velocity of one of the particles after the collision) in order to be able to solve for the remaining information and answer questions.

4.5.5: Dimension of Scattering and Sufficient Information
Given an actual force law describing a collision, one can in principle always solve the dynamical differential equations that result from applying Newton’s Second Law to all of the masses and find their final velocities from their initial conditions and a knowledge of the interaction force(s).
However, the solution of collisions involving all but the simplest interaction forces is beyond the scope of this course (and is usually quite difficult).
The reason for defining the collision types above is because they all represent kinematic (math with units) constraints that are true independent of the details of the interaction force beyond it being either conservative (elastic) or non-conservative (fully or partially inelastic). In some cases the kinematic conditions alone are sufficient to solve the entire scattering problem! In others, however, one cannot obtain a final answer without knowing the details of the scattering force as well as the initial conditions, or without knowing some of the details of the final state.
To understand this, consider only elastic collisions. If the collision occurs in three dimensions, one has four equations from the kinematic relations – three independent momentum conservation equations (one for each component) plus one equation representing kinetic energy conservation.
However, the outgoing particle velocities have six numbers in them – three components each. There simply aren’t enough kinematic constraints to be able to predict the final state from the initial state without knowing the interaction.
Many collisions occur in two dimensions – think about the game of pool, for example, where the cue ball “elastically” strikes the eight ball. In this case one has two momentum conservation equations and one energy conservation equation, but one needs to solve for the four components of two final velocities in two dimensions. Again we either need to know something about the velocity
102 Or

more properly, into Enthalpy, which is microscopic mechanical energy distributed among the atoms and molecules that make up an object.

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of one of the two outgoing particles – say, its x-component – or we cannot solve for the remaining components without a knowledge of the interaction.
Of course in the game of pool103 we do know something very important about the interaction.
It is a force that is exerted directly along the line connecting the centers of the balls at the instant they strike one another! This is just enough information for us to be able to mentally predict that the eight ball will go into the corner pocket if it begins at rest and is struck by the cue ball on the line from that pocket back through the center of the eight ball. This in turn is sufficient to predict the trajectory of the cue ball as well.
Two dimensional elastic collisions are thus almost solvable from the kinematics. This makes them too difficult for students who are unlikely to spend much time analyzing actual collisions (although it is worth it to look them over in the specific context of a good example, one that many students have direct experience with, such as the game of pool/billiards). Physics majors should spend some time here to prepare for more difficult problems later, but life science students can probably skip this without any great harm.
One dimensional elastic collisions, on the other hand, have one momentum conservation equation and one energy conservation equation to use to solve for two unknown final velocities. The number of independent equations and unknowns match! We can thus solve one dimensional elastic collision problems without knowing the details of the collision force from the kinematics alone.
Things are somewhat simpler for fully elastic collisions. Although one only has one, two, or three momentum conservation equations, this precisely matches the number of components in the final velocity of the masses after they have stuck together. Fully inelastic collisions are thus the easiest collision problems to solve.
Partially inelastic collisions in any number of dimensions are the most difficult to solve. There one loses the energy conservation equation – one cannot even solve the one dimensional partially inelastic collision problem without either being given some additional information about the final state – typically the final velocity of one of the two particles so that the other can be found from momentum conservation – or solving the dynamical equations of motion, which is generally even more difficult.
This explains why this textbook focuses on only four relatively simple collision problems. We first study elastic collisions in one dimension, solving them in two slightly different ways that provide different insights into how the physics works out. I then talk briefly about elastic collisions in two dimensions in an “elective” section that can safely be omitted by non-physics majors (but is quite readable, I hope). We then cover inelastic collisions, concentrating on the easiest case (fully inelastic) but providing a simple example of a partially inelastic collision as well.

4.6: 1-D Elastic Collisions
In figure 54 above, we see a typical one-dimensional collision between two masses, m1 and m2 . m1 has a speed in the x-direction v1i > 0 and m2 has a speed v2i < 0, but our solution should not only handle the specific picture above, it should also handle the (common) case where m2 is initially at rest (v2i = 0) or even the case where m2 is moving to the right, but more slowly than m1 so that m1 overtakes it and collides with it, v1i > v2i > 0. Finally, there is nothing special about the labels
“1” and “2” – our answer should be symmetric (still work if we label the mass on the left 2 and the mass on the right 1).
We seek final velocities that satisfy the two conditions that define an elastic collision.
103 Or

“billiards”.

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Before Collision

m m 1

v

v
1i

2

2i

Xcm

After Collision

v

m
1f

m
1

2

v

2f

Xcm

Figure 54: Before and after snapshots of an elastic collision in one dimension, illustrating the important quantities.
Momentum Conservation: p1i + p2i

= p1f + p2f

m1 v1i x + m2 v2i x = m1 v1f x + m2 v2f x
ˆ
ˆ
ˆ
ˆ m1 v1i + m2 v2i

= m1 v1f + m2 v2f

(x-direction only)

(407)
(408)

Kinetic Energy Conservation:
Ek1i + Ek2i
1
1
2
2 m1 v1i + m2 v2i
2
2

= Ek1f + Ek2f
1
1
2
2
=
m1 v1f + m2 v2f
2
2

(409)

Note well that although our figure shows m2 moving to the left, we expressed momentum conservation without an assumed minus sign! Our solution has to be able to handle both positive and negative velocities for either mass, so we will assume them to be positive in our equations and simply use a negative value for e.g. v2i if it happens to be moving to the left in an actual problem we are trying to solve.
The big question now is: Assuming we know m1 , m2 , v1i and v2i , can we find v1f and v2f , even though we have not specified any of the details of the interaction between the two masses during the collision? This is not a trivial question! In three dimensions, the answer might well be no, not without more information. In one dimension, however, we have two independent equations and two unknowns, and it turns out that these two conditions alone suffice to determine the final velocities.
To get this solution, we must solve the two conservation equations above simultaneously. There are three ways to proceed.
One is to use simple substitution – manipulate the momentum equation to solve for (say) v2f in terms of v1f and the givens, substitute it into the energy equation, and then brute force solve the energy equation for v1f and back substitute to get v2f . This involves solving an annoying quadratic
(and a horrendous amount of intermediate algebra) and in the end, gives us no insight at all into the conceptual “physics” of the solution. We will therefore avoid it, although if one has the patience and care to work through it it will give one the right answer.
The second approach is basically a much better/smarter (but perhaps less obvious) algebraic

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solution, and gives us at least one important insight. We will treat it – the “relative velocity” approach – first in the subsections below.
The third is the most informative, and (in my opinion) the simplest, of the three solutions – once one has mastered the concept of the center of mass reference frame outlined above. This “center of mass frame” approach (where the collision occurs right in front of your eyes, as it were) is the one
I suggest that all students learn, because it can be reduced to four very simple steps and because it yields by far the most conceptual understanding of the scattering process.

4.6.1: The Relative Velocity Approach
As I noted above, using direct substitution openly invites madness and frustration for all but the most skilled young algebraists. Instead of using substitution, then, let’s rearrange the energy conservation equation and momentum conservation equations to get all of the terms with a common mass on the same side of the equals signs and do a bit of simple manipulation of the energy equation as well:
2
2 m1 v1i − m1 v1f

2
2
m1 (v1i − v1f )

m1 (v1i − v1f )(v1i + v1f )

2
2
= m2 v2f − m2 v2i
2
2
= m2 (v2f − v2i )

= m2 (v2f − v2i )(v2i + v2f )

(410)

(from energy conservation) and m1 (v1i − v1f ) = m2 (v2f − v2i )

(411)

(from momentum conservation).
When we divide the first of these by the second (subject to the condition that v1i = v1f and v2i = v2f to avoid dividing by zero, a condition that incidentally guarantees that a collision occurs as one possible solution to the kinematic equations alone is always for the final velocities to equal the initial velocities, meaning that no collision occured), we get:
(v1i + v1f ) = (v2i + v2f )

(412)

(v2f − v1f ) = −(v2i − v1i )

(413)

or (rearranging):
This final equation can be interpreted as follows in English: The relative velocity of recession after a collision equals (minus) the relative velocity of approach before a collision. This is an important conceptual property of elastic collisions.
Although it isn’t obvious, this equation is independent from the momentum conservation equation and can be used with it to solve for v1f and v2f , e.g. – v2f m1 v1i + m2 v2i
(m1 + m2 )v1f

= v1f − (v2i − v1i )

= m1 v1f + m2 (v1f − (v2i − v1i ))

=

(m1 − m2 )v1i + 2m2 v2i

(414)

Instead of just solving for v1f and either backsubstituting or invoking symmetry to find v2f we now work a bit of algebra magic that you won’t see the point of until the end. Specifically, let’s add zero to this equation by adding and subtracting m1 v1i :
(m1 + m2 )v1f

=

(m1 − m2 )v1i + 2m2 v2i + (m1 v1i − m1 v1i )

(m1 + m2 )v1f

= −(m1 + m2 )v1i + 2 (m2 v2i + m1 v1i )

(415)

(check this on your own). Finally, we divide through by m1 + m2 and get: v1f = −v1i + 2

m1 v1i + m2 v2i m1 + m2

(416)

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The last term is just two times the total initial momentum divided by the total mass, which we should recognize to be able to write: v1f = −v1i + 2vcm

(417)

There is nothing special about the labels “1” and “2”, so the solution for mass 2 must be identical : v2f = −v2i + 2vcm

(418)

although you can also obtain this directly by backsubstituting v1f into equation 413.
This solution looks simple enough and isn’t horribly difficult to memorize, but the derivation is difficult to understand and hence learn. Why do we perform the steps above, or rather, why should we have known to try those steps? The best answer is because they end up working out pretty well, a lot better than brute force substitutions (the obvious thing to try), which isn’t very helpful.
We’d prefer a good reason, one linked to our eventual conceptual understanding of the scattering process, and while equation 413 had a whiff of concept and depth and ability to be really learned in it (justifying the work required to obtain the result) the “magical” appearance of vcm in the final answer in a very simple and symmetric way is quite mysterious (and only occurs after performing some adding-zero-in-just-the-right-form dark magic from the book of algebraic arts).
To understand the collision and why this in particular is the answer, it is easiest to put everything into the center of mass (CM) reference frame, evaluate the collision, and then put the results back into the lab frame! This (as we will see) naturally leads to the same result, but in a way we can easily understand and that gives us valuable practice in frame transformations besides!

4.6.2: 1D Elastic Collision in the Center of Mass Frame
Here is a bone-simple recipe for solving the 1D elastic collision problem in the center of mass frame.
a) Transform the problem (initial velocities) into the center of mass frame.
b) Solve the problem. The “solution” in the center of mass frame is (as we will see) trivial :
Reverse the center of mass velocities.
c) Transform the answer back into the lab/original frame.
Suppose as before we have two masses, m1 and m2 , approaching each other with velocities v1i and v2i , respectively. We start by evaluating the velocity of the CM frame: vcm =

m1 v1i + m2 v2i m1 + m2

(419)

and then transform the initial velocities into the CM frame:

v1i

v2i

= v1i − vcm

(420)

= v2i − vcm

(421)

We know that momentum must be conserved in any inertial coordinate frame (in the impact approximation). In the CM frame, of course, the total momentum is zero so that the momentum conservation equation becomes:

′ m1 v1i + m2 v2i

p′
1i

+

p′
2i



= m1 v1f + m2 v2f

=

p′
1f

+

p′
2f

=0

(422)
(423)

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Thus p′ = p′ = −p′ and p′ = p′ = −p′ . The energy conservation equation (in terms of the i 1i
2i
f
1f
2f p’s) becomes:

p′2 i p′2 p′2 i
+ i
2m1
2m2
1
1
+
2m1
2m2
p′2 i p′2 p′2 f f + or 2m1
2m2
1
1
= p′2
+
f
2m1
2m2

=

= p′2 f so that
(424)

Taking the square root of both sides (and recalling that p′ refers equally well to mass 1 or 2): i p′
1f
p′
2f

= ±p′
1i

(425)

±p′
2i

=

(426)

The + sign rather obviously satisfies the two conservation equations. The two particles keep on going at their original speed and with their original energy! This is, actually, a perfectly good solution to the scattering problem and could be true even if the particles “hit” each other. The more interesting case (and the one that is appropriate for “hard” particles that cannot interpenetrate) is for the particles to bounce apart in the center of mass frame after the collision. We therefore choose the minus sign in this result:

p′ = m1 v1f
1f

p′
2f

=

′ m2 v2f


= −m1 v1i = −p′
1i

=


−m2 v2i

=

−p′
2i

(427)
(428)

Since the masses are the same before and after we can divide them out of each equation and obtain the solution to the elastic scattering problem in the CM frame as:

v1f

v2f


= −v1i

=

(429)


−v2i

(430)

or the velocities of m1 and m2 reverse in the CM frame.
This actually makes sense. It guarantees that if the momentum was zero before it is still zero, and since the speed of the particles is unchanged (only the direction of their velocity in this frame changes) the total kinetic energy is similarly unchanged.
Finally, it is trivial to put the these solutions back into the lab frame by adding vcm to them: v1f ′
= v1f + vcm

= −v1i + vcm

v1f v2f = −(v1i − vcm ) + vcm

= −v1i + 2vcm

= −v2i + 2vcm

or

and similarly

(431)
(432)

These are the exact same solutions we got in the first example/derivation above, but now they have considerably more meaning. The “solution” to the elastic collision problem in the CM frame is that the velocities reverse (which of course makes the relative velocity of approach be the negative of the relative velocity of recession, by the way). We can see that this is the solution in the center of mass frame in one dimension without doing the formal algebra above, it makes sense!
That’s it then: to solve the one dimensional elastic collision problem all one has to do is transform the initial velocities into the CM frame, reverse them, and transform them back. Nothing to it.
Note that (however it is derived) these solutions are completely symmetric – we obviously don’t care which of the two particles is labelled “1” or “2”, so the answer should have exactly the same

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form for both. Our derived answers clearly have that property. In the end, we only need one equation (plus our ability to evaluate the velocity of the center of mass): vf = −vi + 2vcm

(433)

valid for either particle.
If you are a physics major, you should be prepared to derive this result one of the various ways it can be derived (I’d strongly suggest the last way, using the CM frame). If you are e.g. a life science major or engineer, you should derive this result for yourself at least once, at least one of the ways
(again, I’d suggest that last one) but then you are also welcome to memorize/learn the resulting solution well enough to use it.
Note well! If you remember the three steps needed for the center of mass frame derivation, even if you forget the actual solution on a quiz or a test – which is probably quite likely as I have little confidence in memorization as a learning tool for mountains of complicated material – you have a prayer of being able to rederive it on a test.

4.6.3: The “BB/bb” or “Pool Ball” Limits
In collision problems in general, it is worthwhile thinking about the “ball bearing and bowling ball
(BB) limits”104 . In the context of elastic 1D collision problems, these are basically the asymptotic results one obtains when one hits a stationary bowling ball (large mass, BB) with rapidly travelling ball bearing (small mass, bb).
This should be something you already know the answer to from experience and intuition. We all know that if you shoot a bb gun at a bowling ball so that it collides elastically, it will bounce back off of it (almost) as fast as it comes in and the bowloing ball will hardly recoil105 . Given that vcm in this case is more or less equal to vBB , that is, vcm ≈ 0 (just a bit greater), note that this is exactly what the solution predicts.
What happens if you throw a bowling ball at a stationary bb? Well, we know perfectly well that the BB in this case will just continue barrelling along at more or less vc m (still roughly equal to the velocity of the more massive bowling ball) – ditto, when your car hits a bug with the windshield, it doesn’t significantly slow down. The bb (or the bug) on the other hand, bounces forward off of the
BB (or the windshield)!
In fact, according to our results above, it will bounce off the BB and recoil forward at approximately twice the speed of the BB. Note well that both of these results preserve the idea derived above that the relative velocity of approach equals the relative velocity of recession, and you can transform from one to the other by just changing your frame of reference to ride along with BB or bb – two different ways of looking at the same collision.
Finally, there is the “pool ball limit” – the elastic collision of roughly equal masses. When the cue ball strikes another ball head on (with no English), then as pool players well know the cue ball stops (nearly) dead and the other ball continues on at the original speed of the cue ball. This, too, is exactly what the equations/solutions above predict, since in this case vcm = v1i /2.
Our solutions thus agree with our experience and intuition in both the limits where one mass is much larger than the other and when they are both roughly the same size. One has to expect that they are probably valid everywhere. Any answer you derive (such as this one) ultimately has to pass the test of common-sense agreement with your everyday experience. This one seems to, however difficult the derivation was, it appears to be correct!
104 Also

known as the “windshield and bug limits”... you’ll put your eye out – kids, do not try this at home!

105 ...and

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As you can probably guess from the extended discussion above, pool is a good example of a game of “approximately elastic collisions” because the hard balls used in the game have a very elastic coefficient of restitution, another way of saying that the surfaces of the balls behave like very small, very hard springs and store and re-release the kinetic energy of the collision from a conservative impulse type force.
However, it also opens up the question: What happens if the collision between two balls is not along a line? Well, then we have to take into account momentum conservation in two dimensions.
So alas, my fellow human students, we are all going to have to bite the bullet and at least think a bit about collisions in more than one dimension.

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4.7: Elastic Collisions in 2-3 Dimensions
As we can see, elastic collisions in one dimension are “good” because we can completely solve them using only kinematics – we don’t care about the details of the interaction between the colliding entities; we can find the final state from the initial state for all possible elastic forces and the only differences that will depend on the forces will be things like how long it takes for the collision to occur. In 2+ dimensions we at the very least have to work much harder to solve the problem. We will no longer be able to use nothing but vector momentum conservation and energy conservation to solve the problem independent of most of the details of the interaction. In two dimensions we have to solve for four outgoing components of velocity (or momentum), but we only have conservation equations for two components of momentum and kinetic energy. Three equations, four unknowns means that the problem is indeterminate unless we are told at least one more thing about the final state, such as one of the components of the velocity or momentum of one of the outgoing masses. In three dimensions it is even worse – we must solve for six outgoing components of velocity/momentum but have only four conservation equations (three momentum, one energy) and need at least two additional pieces of information. Kinematics alone is simply insufficient to solve the scattering problem – need to know the details of the potential/force of interaction and solve the equations of motion for the scattering in order to predict the final/outgoing state from a knowledge of the initial/incoming state.
The dependence of the outoing scattering on the interaction is good and bad. The good thing is that we can learn things about the interaction from the results of a collision experiment (in one dimension, note well, our answers didn’t depend on the interaction force so we learn nothing at all about that force aside from the fact that it is elastic from scattering data). The bad is that for the most part the algebra and calculus involved in solving multidimensional collisions is well beyond the scope of this course. Physics majors, and perhaps a few other select individuals in other majors or professions, will have to sweat blood later to work all this out for a tiny handful of interaction potentials where the problem is analytically solvable, but not yet!
Still, there are a few things that are within the scope of the course, at least for majors. These involve learning a bit about how to set up a good coordinate frame for the scattering, and how to treat “hard sphere” elastic collisions which turn out to be two dimensional, and hence solvable from kinematics plus a single assumption about recoil direction in at least some simple cases. Let’s look at scattering in two dimensions in the case where the target particle is at rest and the outgoing particles lie (necessarily) in a plane.
We expect both energy and momentum to be conserved in any elastic collision. This gives us the following set of equations: p0x = p1x + p2x

(434)

p0y

= p1y + p2y

(435)

p0z

= p1z + p2z

(436)

(for momentum conservation) and p2 p2 p2 0
= E 0 = E1 + E 2 = 1 + 2
2m1
2m1 m2 (437)

for kinetic energy conservation.
We have four equations, and four unknowns, so we might hope to be able to solve it quite generally. However, we don’t really have that many equations – if we assume that the scattering plane is the x − y plane, then necessarily p0z = p1z = p2z = 0 and this equation tells us nothing useful. We need more information in order to be able to solve the problem.
Let’s see what we can tell in this case. Examine figure 55. Note that we have introduced two angles: θ and φ for the incident and target particle’s outgoing angle with respect to the incident

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Week 4: Systems of Particles, Momentum and Collisions p 2

m2 m1 p0

p

2y

p

2x

φ θ p m1 1x p p
1

1y

Figure 55: The geometry for an elastic collision in a two-dimensional plane. direction. Using them and setting p0y = p0z = 0 (and assuming that the target is at rest initially and has no momentum at all initially) we get:

p0x

= p1x + p2x = p1 cos(θ) + p2 cos(φ)

(438)

p0y

= p1y + p2y = 0 = −p1 sin(θ) + p2 sin(φ)

(439)

In other words, the momentum in the x-direction is conserved, and the momentum in the ydirection (after the collision) cancels. The latter is a powerful relation – if we know the y-momentum of one of the outgoing particles, we know the other. If we know the magnitudes/energies of both, we know an important relation between their angles.
This, however, puts us no closer to being able to solve the general problem (although it does help with a special case that is on your homework). To make real progress, it is necessarily to once again change to the center of mass reference frame by subtracting v cm from the velocity of both particles.
We can easily do this: p′ i1

= m1 (v 0 − v cm ) = m1 u1

p′ i2 (440)

= −m2 v cm = m2 u2

(441)

so that p′ + p′ = p′ = 0 in the center of mass frame as usual. The initial energy in the center i1 i2 tot of mass frame is just:

′ p2 p2
(442)
Ei = i1 + i2
2m1
2m2
Since p′ = p′ = p′ (the magnitudes are equal) we can simplify this a bit further: i1 i2 i ′

Ei =



p2 p2 p2 i + i = i
2m1
2m2
2



1
1
+ m1 m2

=

p2 i 2



m1 + m2 m1 m2

(443)

After the collision, we can see by inspection of




p2 p2 p2 f f f Ef =
+
=
2m1
2m2
2



1
1
+ m1 m2

p2 f =
2



m1 + m2 m1 m2

= Ei

(444)

that p′ 1 = p′ 2 = p′ = p′ will cause energy to be conserved, just as it was for a 1 dimensional i f f f collision. All that can change, then, is the direction of the incident momentum in the center of mass frame. In addition, since the total momentum in the center of mass frame is by definition zero before and after the collision, if we know the direction of either particle after the collision in the center of mass frame, the other is the opposite: p′ 1 = −p′ 2
(445)
f f We have then “solved” the collision as much as it can be solved. We cannot uniquely predict the direction of the final momentum of either particle in the center of mass (or any other) frame without knowing more about the interaction and e.g. the incident impact parameter. We can predict the

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magnitude of the outgoing momenta, and if we know the outgoing direction alone of either particle we can find everything – the magnitude and direction of the other particle’s momentum and the magnitude of the momentum of the particle whose angle we measured.
As you can see, this is all pretty difficult, so we’ll leave it at this point as a partially solved problem, ready to be tackled again for specific interactions or collision models in a future course.

4.8: Inelastic Collisions
A fully inelastic collision is where two particles collide and stick together. As always, momentum is conserved in the impact approximation, but now kinetic energy is not! In fact, we will see that macroscopic kinetic energy is always lost in an inelastic collision, either to heat or to some sort of mechanism that traps and reversibly stores the energy.
These collisions are much easier to understand and analyze than elastic collisions. That is because there are fewer degrees of freedom in an inelastic collision – we can easily solve them even in 2 or 3 dimensions. The whole solution is developed from pi,ntot = m1 v 1i + m2 v 2i = (m1 + m2 )v f = (m1 + m2 )v cm = pf,tot

(446)

In other words, in a fully inelastic collision, the velocity of the outgoing combined particle is the velocity of the center of mass of the system, which we can easily compute from a knowledge of the initial momenta or velocities and masses. Of course! How obvious! How easy!
From this relation you can easily find v f in any number of dimensions, and answer many related questions. The collision is “solved”. However, there are a number of different kinds of problems one can solve given this basic solution – things that more or less tag additional physics problems on to the end of this initial one and use its result as their starting point, so you have to solve two or more subproblems in one long problem, one of which is the “inelastic collision”. This is best illustrated in some archetypical examples.

Example 4.8.1: One-dimensional Fully Inelastic Collision (only)

m1

v0

m2

m1 m2

vf

Figure 56: Two blocks of mass m1 and m2 collide and stick together on a frictionless table.
In figure 56 above, a block m1 is sliding across a frictionless table at speed v0 to strike a second block m2 initially at rest, whereupon they stick together and move together as one thereafter at some final speed vf .
Before, after, and during the collision, gravity acts but is opposed by a normal force. There is no friction or drag force doing any work. The only forces in play are the internal forces mediating the collision and making the blocks stick together. We therefore know that momentum is conserved in this problem independent of the features of that internal interaction. Even if friction or drag forces did act, as long as the collision took place “instantly” in the impact approximation, momentum would

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still be conserved from immediately before to immediately after the collision, when the impulse ∆p of the collision force would be much, much larger than any change in momentum due to the drag over the same small time ∆t.
Thus:
pi = m1 v0 = (m1 + m2 )vf = pf or vf =

(447)

mv0
( = vcm )
(m1 + m2 )

(448)

A traditional question that accompanies this is: How much kinetic energy was lost in the collision?
We can answer this by simply figuring it out.
∆K

p2 p2 f
− i
2(m1 + m2 ) 2m1
2
1
1
pi

2 (m1 + m2 ) m1 p2 m1 − (m1 + m2 ) i 2 m1 (m1 + m2 ) m2 p2
− i
2m1 (m1 + m2 ) m2 −
Ki
(m1 + m2 )

= Kf − Ki =
=
=
=
=

(449)

where we have expressed the result as a fraction of the initial kinetic energy!

Before Collision

m m 1

v

v
1i

2

2i

Xcm

After Collision

m +m
1
2

Xcm

Figure 57: Two blocks collide and stick together on a frictionless table – in the center of mass frame.
After the collision they are both at rest at the center of mass and all of the kinetic energy they had before the collision in this frame is lost.
There is a different way to think about the collision and energy loss. In figure 57 you see the same collision portrayed in the CM frame. In this frame, the two particles always come together and stick to remain, at rest, at the center of mass after the collision. All of the kinetic energy in

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217

the CM frame is lost in the collision! That’s exactly the amount we just computed, but I’m leaving the proof of that as an exercise for you.
Note well the BB limits: For a light bb (m1 ) striking a massive BB (m2 ), nearly all the energy is lost. This sort of collision between an asteroid (bb) and the earth (BB) caused at least one of the mass extinction events, the one that ended the Cretaceous and gave mammals the leg up that they needed in a world dominated (to that point) by dinosaurs. For a massive BB (m1 ) stricking a light bb (m2 ) very little of the energy of the massive object is lost. Your truck hardly slows when it smushes a bug “inelastically” against the windshield. In the equal billiard ball bb collision
(m1 = m2 ), exactly one half of the initial kinetic energy is lost.
A similar collision in 2D is given for your homework, where a truck and a car inelastically collide and then slide down the road together. In this problem friction works, but not during the collision!
Only after the “instant” (impact approximation) collision do we start to worry about the effect of friction. Example 4.8.2: Ballistic Pendulum

θf
R
v m M

Figure 58: The “ballistic pendulum”, where a bullet strikes and sticks to/in a block, which then swings up to a maximum angle θf before stopping and swinging back down.
The classic ballistic pendulum question gives you the mass of the block M , the mass of the bullet m, the length of a string or rod suspending the “target” block from a free pivot, and the initial velocity of the bullet v0 . It then asks for the maximum angle θf through which the pendulum swings after the bullet hits and sticks to the block (or alternatively, the maximum height H through which it swings). Variants abound – on your homework you might be asked to find the minimum speed v0 the bullet must have in order the the block whirl around in a circle on a never-slack string, or on the end of a rod. Still other variants permit the bullet to pass through the block and emerge with a different (smaller) velocity. You should be able to do them all, if you completely understand this example (and the other physics we have learned up to now, of course).
There is an actual lab that is commonly done to illustrate the physics; in this lab one typically measures the maximum horizontal displacement of the block, but it amounts to the same thing once one does the trigonometry.
The solution is simple:
• During the collision momentum is conserved in the impact approximation, which in this case basically implies that the block has no time to swing up appreciably “during” the actual collision. 218

Week 4: Systems of Particles, Momentum and Collisions
• After the collision mechanical energy is conserved. Mechanical energy is not conserved during the collision (see solution above of straight up inelastic collision).

One can replace the second sub-problem with any other problem that requires a knowledge of either vf or Kf immediately after the collision as its initial condition. Ballistic loop-the-loop problems are entirely possible, in other words!
At this point the algebra is almost anticlimactic: The collision is one-dimensional (in the xdirection). Thus (for block M and bullet m) we have momentum conservation: pm,0 = mv0 = pM +m,f

(450)

Now if we were foolish we’d evaluate vM +m,f to use in the next step: mechanical energy conservation. Being smart, we instead do the kinetic part of mechanical energy conservation in terms of momentum: E0 =

p2
B+b,f
2(M + m)

p2 b,0 2(M + m)
= Ef = (M + m)gH
=

=
Thus:
θf = cos−1 (1 −

(M + m)gR(1 − cos θf )

(451)

(mv0 )2
)
2(M + m)2 gR

(452)

which only has a solution if mv0 is less than some maximum value. What does it mean if it is greater than this value (there is no inverse cosine of an argument with magnitude bigger than 1)? Will this answer “work” if θ > π/2, for a string? For a rod? For a track?
Don’t leave your common sense at the door when solving problems using algebra!

Example 4.8.3: Partially Inelastic Collision
Let’s briefly consider the previous example in the case where the bullet passes through the block and emerges on the far side with speed v1 < v0 (both given). How is the problem going to be different?
Not at all, not really. Momentum is still conserved during the collision, mechanical energy after.
The only two differences are that we have to evaluate the speed vf of the block M after the collision from this equation: p0 = m1 v0 = M vf + mv1 = pm + p1 = pf
(453)
so that:

∆p m(v0 − v1 )
=
(454)
M
M
We can read this as “the momentum transferred to the block is the momentum lost by the bullet” because momentum is conserved. Given vf of the block only, you should be able to find e.g. the kinetic energy lost in this collision or θf or whatever in any of the many variants involving slightly different “after”-collision subproblems. vf =

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4.9: Kinetic Energy in the CM Frame
Finally, let’s consider the relationship between kinetic energy in the lab frame and the CM frame, using all of the velocity relations we developed above as needed. We start with:
Ktot = i 1
2
mi vi =
2

i

p2 i .
2mi

(455)

in the lab/rest frame.
We recall (from above) that v i = v ′ + v cm i (456)

pi = mi v i = mi v ′ + v cm i (457)

so that:

Then
Ki =

p2 m2 v ′2
2m2 v ′ · v 2 m2 v 2 cm i i i
= i i +
+ i cm .
2mi
2mi
2mi
2mi

(458)

If we sum this as before to construct the total kinetic energy:
Ktot =

Ki = i i

P2 p′2 i
+ tot + v cm ·
2mi
2Mtot

mi v ′ i i
=

P

i

p′ i ·v cm

(459)

= 0

or
Ktot = K(in cm) + K(of cm)

(460)

We thus see that the total kinetic energy in the lab frame is the sum of the kinetic energy of all the particles in the CM frame plus the kinetic energy of the CM frame (system) itself (viewed as single “object”).
To conclude, at last we can understand the mystery of the baseball – how it behaves like a particle itself and yet also accounts for all of the myriad of particles it is made up of. The Newtonian motion of the baseball as a system of particles is identical to that of a particle of the same mass experiencing the same total force. The “best” location to assign the baseball (of all of the points inside) is the center of mass of the baseball. In the frame of the CM of the baseball, the total momentum of the parts of the baseball is zero (but the baseball itself has momentum Mtot v relative to the ground).
Finally, the kinetic energy of a baseball flying through the air is the kinetic energy of the “baseball itself” (the entire system viewed as a particle) plus the kinetic energy of all the particles that make up the baseball measured in the CM frame of the baseball itself. This is comprised of rotational kinetic energy (which we will shortly treat) plus all the general vibrational (atomic) kinetic energy that is what we would call heat.
We see that we can indeed break up big systems into smaller/simpler systems, solve the smaler problems, and reassemble the solutions into a big solution, even as we can combine many, many small problems into one bigger and simpler problem and ignore or average over the details of what goes on “inside” the little problems. Treating many bodies at the same time can be quite complex, and we’ve only scratched the surface here, but it should be enough to help you understand both many things in your daily life and (just as important) the rest of this book.
Next up (after the homework) we’ll pursue this idea of motion in plus motion of a bit further in the context of torque and rotating systems.

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Homework for Week 4

Problem 1.

Physics Concepts: Make this week’s physics concepts summary as you work all of the problems in this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s) they were key to, and include concepts from previous weeks as necessary. Do the work carefully enough that you can (after it has been handed in and graded) punch it and add it to a three ring binder for review and study come finals!

Problem 2.

y
M

x
R

This problem will help you learn required concepts such as:
• Center of Mass
• Integrating a Distribution of Mass so please review them before you begin.
In the figure above, a uniformly thick piece of wire is bent into 3/4 of a circular arc as shown.
Find the center of mass of the wire in the coordinate system given, using integration to find the xcm and ycm components separately.

Week 4: Systems of Particles, Momentum and Collisions

221

Problem 3.
Suppose we have a block of mass m sitting initially at rest on a table. A massless string is attached to the block and to a motor that delivers a constant power P to the block as it pulls it in the x-direction. a) Find the tension T in the string as a function of v, the speed of the block in the x-direction, initially assuming that the table is frictionless.
b) Find the acceleration of the block as a function of v.
c) Solve the equation of motion to find the velocity of the block as a function of time. Show that the result is the same that you would get by evaluating: t P dt =
0

1
1
mv 2 − mv 2
2 f 2 0

with v0 = 0.
d) Suppose that the table exerts a constant force of kinetic friction on the block in the opposite direction to v, with a coefficient of kinetic friction µk . Find the “terminal velocity” of the system after a very long time has passed. Hint: What is the total power delivered to the block by the motor and friction combined at that time?

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Problem 4.

U(x)

E1 x E2
E3

This problem will help you learn required concepts such as:
• Potential energy/total energy diagrams
• Finding the force from the potential energy
• Identifying turning points and stable/unstable equilibribum points on a graph of the potential energy • Identifying classically forbidden versus allowed domains of motion (in one dimension) on an energy diagram so please review them before you begin.
a) On (a large copy of) the diagram above, place a small letter ‘u’ to mark points of unstable equilibrium. b) Place the letter ‘s’ to mark points of stable equilibrium.
c) On the curve itself, place a few arrows in each distinct region indicating the direction of the force. Try to make the lengths of the arrows proportional in a relative way to the arrow you draw for the largest magnitude force.
d) For the three energies shown, mark the turning points of motion with the letter ‘t’.
e) For energy E2 , place might be found. Place can never be found.

allowed

forbidden

to mark out the classically allowed region where the particle to mark out the classically forbidden region where the particle

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Problem 5.

30

20

U(x)

10

0

-10

-20

-30
0

0.5

1

1.5

2

x
This problem will help you learn required concepts such as:
• Energy conservation and the use of E = K + U in graphs.
• Finding the force from the potential energy.
• Finding turning points and stable/unstable equilibrium points from algebraic expressions for the potential as well as visualizing the result on a graph. so please review them before you begin.
An object moves in the force produced by a potential energy function:
U (x) =

1
10
− 6
12
x x This is a one-dimensional representation of an actual important physical potential, the LennardJones potential. This one-dimensional “12-6” Lennard-Jones potential models the dipole-induced dipole Van der Waals interaction between two atoms or molecules in a gas, but the “12-10” form can also model hydrogen bonds in physical chemistry. Note well that the force is strongly repulsive inside the E = 0 turning point xt (which one can think of as where the atoms “collide”) but weakly attractive for all x > x0 , the position of stable equilibrium.
a) Write an algebraic expression for Fx (x).
b) Find x0 , the location of the stable equilibrium distance predicted by this potential.
c) Find U (x0 ) the binding energy for an object located at this distance.
d) Find xt , the turning point distance for E = 0. This is essentially the sum of the radii of the two atoms (in suitable coordinates – the parameters used in this problem are not intended to be physical).

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Problem 6.

(a)

m

v = 0 (for both)

H
M

(b)

vM

M

m

vm

This problem will help you learn required concepts such as:
• Newton’s Third Law
• Momentum Conservation
• Fully Elastic Collisions so please review them before you begin.
A small block with mass m is sitting on a large block of mass M that is sloped so that the small block can slide down the larger block. There is no friction between the two blocks, no friction between the large block and the table, and no drag force. The center of mass of the small block is located a height H above where it would be if it were sitting on the table, and both blocks are started at rest (so that the total momentum of this system is zero, note well!)
a) Are there any net external forces acting in this problem? What quantities do you expect to be conserved?
b) Using suitable conservation laws, find the velocities of the two blocks after the small block has slid down the larger one and they have separated.
c) To check your answer, consider the limiting case of M → ∞ (where one rather expects the larger block to pretty much not move). Does your answer to part b) give you the usual result for a block of mass m sliding down from a height H on a fixed incline?
d) This problem doesn’t look like a collision problem, but it easily could be half of one. Look carefully at your answer, and see if you can determine what initial velocity one should give the two blocks so that they would move together and precisely come to rest with the smaller block a height H above the ground. If you put the two halves together, you have solved a fully elastic collision in one dimension in the case where the center of mass velocity is zero!

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Week 4: Systems of Particles, Momentum and Collisions
Problem 7.

(2 at rest)

(a)
1

vo

2

vcm
(b)
1

2

(c) v1 1

2

v2

This problem will help you learn required concepts such as:
• Newton’s Third Law
• Momentum Conservation
• Energy conservation
• Impulse and average force in a collision so please review them before you begin.
This problem is intended to walk you through the concepts associated with collisions in one dimension. In (a) above, mass m1 approaches mass m2 at velocity v0 to the right. Mass m2 is initially at rest. An ideal massless spring with spring constant k is attached to mass m2 and we will assume that it will not be fully compressed from its uncompressed length in this problem.
Begin by considering the forces that act, neglecting friction and drag forces. What will be conserved throughout this problem?
a) In (b) above, mass m1 has collided with mass m2 , compressing the spring. At the particular instant shown, both masses are moving with the same velocity to the right. Find this velocity.
What physical principle do you use?
b) Also find the compression ∆x of the spring at this instant. What physical principle do you use? c) The spring has sprung back, pushing the two masses apart. Find the final velocities of the two masses. Note that the diagram assumes that m2 > m1 to guess the final directions, but in general your answer should make sense regardless of their relative mass.
d) So check this. What are the two velocities in the “BB limits” – the m1 ≫ m2 (bowling ball strikes ball bearing) and m1 ≪ m2 (ball bearing strikes bowling ball) limits? In other words, does your answer make dimensional and intuitive sense?

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e) In this particular problem one could in principle solve Newton’s second law because the elastic collision force is known. In general, of course, it is not known, although for a very stiff spring this model is an excellent one to model collisions between hard objects. Assuming that the spring is sufficiently stiff that the two masses are in contact for a very short time ∆t, write a simple expression for the impulse imparted to m2 and qualitatively sketch Fav over this time interval compared to Fx (t).

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Problem 8.

vf vo θ

m

vo

M

This problem will help you learn required concepts such as:
• Newton’s Third Law
• Momentum Conservation
• Fully Inelastic Collisions so please review them before you begin.
In the figure above, a large, heavy truck with mass M speeds through a red light to collide with a small, light compact car of mass m. Both cars fail to brake and are travelling at the speed limit
(v0 ) at the time of the collision, and their metal frames tangle together in the collision so that after the collision they move as one big mass.
a) Which exerts a larger force on the other, the car or the truck?
b) Which transfers a larger momentum to the other, the car or the truck?
c) What is the final velocity of the wreck immediately after the collision (please give (vf , θ))?
d) How much kinetic energy was lost in the collision?
e) If the tires blow and the wreckage has a coefficient of kinetic friction µk with the ground after the collision, set up an expression in terms of vf that will let you solve for how far the wreck slides before coming to a halt. You do not need to substitute your expression from part c) into this and get a final answer, but you should definitely be able to do this on a quiz or exam.

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Problem 9.

xj

mj

xb

mb

xr

mr

This problem will help you learn required concepts such as:
• Center of Mass
• Momentum Conservation
• Newton’s Third Law so please review them before you begin.
Romeo and Juliet are sitting in a boat at rest next to a dock, looking deeply into each other’s eyes. Juliet, overcome with emotion, walks at a constant speed v relative to the water from her end of the boat to sit beside him and give Romeo a kiss. Assume that the masses and initial positions of Romeo, Juliet and the boat are (mr , xr ), (mj , xj ), (mb , xb ), where x is measured from the dock as shown, and D = xj − xr is their original distance of separation.
a) While she is moving, the boat and Romeo are moving at speed v ′ in the opposite direction.
What is the ratio v ′ /v?
b) What is vrel = dD/dt, their relative speed of approach in terms of v.
c) How far has the boat moved away from the dock when she reaches him?
d) Make sure that your answer makes sense by thinking about the following “BB” limits: mb ≫ mr = mj ; mb ≪ mr = mj .

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Week 4: Systems of Particles, Momentum and Collisions
Problem 10.

m

H
M

This problem will help you learn required concepts such as:
• Newton’s Second Law
• Gravitation
• Newton’s Third Law
• Impulse and Average Force
• Fully Elastic Collisions so please review them before you begin.
In the figure above, a feeder device provides a steady stream of beads of mass m that fall a distance H and bounce elastically off of one of the hard metal pans of a beam balance scale (and then fall somewhere else into a hopper and disappear from our problem). N beads per second come out of the feeder to bounce off of the pan. Our goal is to derive an expression for M , the mass we should put on the other pan to balance the average force exerted by this stream of beads106
a) First, the easy part. The beads come off of the feeder with an initial velocity of v = v0x x in
ˆ
the x-direction only. Find the y-component of the velocity vy when a single bead hits the pan after falling a height H.
b) Since the beads bounce elastically, the x-component of their velocity is unchanged and the y-component reverses. Find the change of the momentum of this bead ∆p during its collision.
c) Compute the average force being exerted on the stream of beads by the pan over a second
(assuming that N ≫ 1, so that many beads strike the pan per second).
d) Use Newton’s Third Law to deduce the average force exerted by the beads on the pan, and from this determine the mass M that would produce the same force on the other pan to keep the scale in balance.

106 This is very similar (conceptually) to the way a gas microscopically exerts a force on a surface that confines it; we will later use this idea to understand the pressure exerted by a fluid and to derive the kinetic theory of gases and the ideal gas law P V = N kT , which is why I assign it in particular now.

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Problem 11.

R

v m M

This problem will help you learn required concepts such as:
• Conservation of Momentum
• Conservation of Energy
• Disposition of energy in fully inelastic collisions
• Circular motion
• The different kinds of constraint forces exerted by rigid rods versus strings. so please review them before you begin.
A block of mass M is attached to a rigid massless rod of length R (pivot to center-of-mass of the block/bullet distance at collision) and is suspended from a frictionless pivot. A bullet of mass m travelling at velocity v strikes it as shown and is quickly stopped by friction in the hole so that the two masses move together as one thereafter. Find:
a) The minimum speed vr that the bullet must have in order to swing through a complete circle after the collision. Note well that the pendulum is attached to a rod !
b) The energy lost in the collision when the bullet is incident at this speed.

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Advanced Problem 12.

λ(x) = 2M x
L2

0

L

+x

This problem will help you learn required concepts such as:
• Center of Mass of Continuous Mass Distributions
• Integrating Over Distributions so please review them before you begin.
In the figure above a rod of total mass M and length L is portrayed that has been machined so that it has a mass per unit length that increases linearly along the length of the rod: λ(x) =

2M x L2

This might be viewed as a very crude model for the way mass is distributed in something like a baseball bat or tennis racket, with most of the mass near one end of a long object and very little near the other (and a continuum in between).
Treat the rod as if it is really one dimensional (we know that the center of mass will be in the center of the rod as far as y or z are concerned, but the rod is so thin that we can imagine that y ≈ z ≈ 0) and:
a) verify that the total mass of the rod is indeed M for this mass distribution;
b) find xcm , the x-coordinate of the center of mass of the rod.

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Optional Problems
The following problems are not required or to be handed in, but are provided to give you some extra things to work on or test yourself with after mastering the required problems and concepts above and to prepare for quizzes and exams.

Optional Problem 13.

µk

vi m M vb vf

(block at rest)
D

This problem will help you learn required concepts such as:
• Momentum Conservation
• The Impact Approximation
• Elastic versus Inelastic Collisions
• The Non-conservative Work-Mechanical Energy Theorem so please review them before you begin.
In the figure above a bullet of mass m is travelling at initial speed vi to the right when it strikes a larger block of mass M that is resting on a rough horizontal table (with coefficient of friction between block and table of µk ). Instead of “sticking” in the block, the bullet blasts its way through the block (without changing the mass of the block significantly in the process). It emerges with the smaller speed vf , still to the right.
a) Find the speed of the block vb immediately after the collision (but before the block has had time to slide any significant distance on the rough surface).
b) Find the (kinetic) energy lost during this collision. Where did this energy go?
c) How far down the rough surface D does the block slide before coming to rest?

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Optional Problem 14.

vt m θu m vo

θb

vb
This problem will help you learn required concepts such as:
• Vector Momentum Conservation
• Fully Elastic Collisions in Two Dimensions so please review them before you begin.
In the figure above, two identical billiard balls of mass m are sitting in a zero gravity vacuum (so that we can neglect drag forces and gravity). The one on the left is given a push in the x-direction so that it elastically strikes the one on the right (which is at rest) off center at speed v0 . The top ball recoils along the direction shown at a speed vt and angle θt relative to the direction of incidence of the bottom ball, which is deflected so that it comes out of the collision at speed vb at angle θb relative to this direction.
a) Use conservation of momentum to show that in this special case that the two masses are equal: v0 = vu + vb and draw this out as a triangle.
b) Use the fact that the collision was elastic to show that
2
2
2
v0 = vu + v b

(where these speeds are the lengths of the vectors in the triangle you just drew).
c) Identify this equation and triangle with the pythagorean theorem proving that in this case v t ⊥ v b (so that θu = θb + π/2
Using these results, one can actually solve for v u and v b given only v0 and either of θu or θb .
Reasoning very similar to this is used to analyze the results of e.g. nuclear scattering experiments at various laboratories around the world (including Duke)!

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Week 5: Torque and Rotation in
One Dimension
Summary
• Rotations in One Dimension are rotations of a solid object about a single axis. Since we are free to choose any arbitrary coordinate system we wish in a problem, we can without loss of generality select a coordinate system where the z-axis represents the (positive or negative) direction or rotation, so that the rotating object rotates “in” the xy plane. Rotations of a rigid body in the xy plane can then be described by a single angle θ, measured by convention in the counterclockwise direction from the positive x-axis.
• Time-dependent Rotations can thus be described by:
a) The angular position as a function of time, θ(t).
b) The angular velocity as a function of time, ω(t) =

dθ dt c) The angular acceleration as a function of time, α(t) =

dω d2 θ
= 2 dt dt

Hopefully the analogy between these “one dimensional” angular coordinates and their one dimensional linear motion counterparts is obvious.
• Forces applied to a rigid object perpendicular to a line drawn from an axis of rotation exert a torque on the object. The torque is given by: τ = rF sin(φ) = rF⊥ = r⊥ F
• The torque (as we shall see) is a vector quantity and by convention its direction is perpendicular to the plane containing r and F in the direction given by the right hand rule. Although we won’t really work with this until next week, the “proper” definition of the torque is: τ =r×F
• Newton’s Second Law for Rotation in one dimension is: τ = Iα where I is the moment of inertia of the rigid body being rotated by the torqe about a given/specified axis of rotation. The direction of this (one dimensional) rotation is the righthanded direction of the axis – the direction your right handed thumb points if you grasp the axis with your fingers curling around the axis in the direction of the rotation or torque.
235

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Week 5: Torque and Rotation in One Dimension
• The moment of inertia of a point particle of mass m located a (fixed) distance r from some axis of rotation is:
I = mr2
• The moment of inertia of a rigid collection of point particles is:
2
mi ri

I= i • the moment of inertia of a continuous solid rigid object is:
I=

r2 dm

• The rotational kinetic energy of a rigid body (total kinetic energy of all of the chunks of mass that make it up) is:
1
Krot = Iω 2
2
• The work done by a torque as it rotates a rigid body through some angle dθ is: dW = τ dθ
Hence the work-kinetic energy theorem becomes:
W =

τ dθ = ∆Krot

• Consequently rotational work, rotational potential energy, and rotational kinetic energy call all be simply added in the appropriate places to our theory of work and energy. The total mechanical energy includes both the total translational kinetic energy of the rigid body treated as if it is a total mass located at its center of mass plus the kinetic energy of rotation around its center of mass:
Ktot = Kcm + Krot
This is a special case of the last theorem we proved last week.
• If we know the moment of inertia Icm of a rigid body about a given axis through its center of mass, the Parallel Axis Theorem permits us to find the moment of inertia of a rigid body of mass m around a new axis parallel to this axis and displaced from it by a distance rcm :
2
Inew = Icm + mrcm

• For a distribution of mass with planar symmetry (mirror symmetry about the plane of rotation or distribution only in the plane of rotation), if we let z point in the direction of an axis of rotation perpendicular to this plane and x and y be perpendicular axes in the plane of rotation, then the Perpendicular Axis Theorem states that:
Iz = Ix + Iy

5.1: Rotational Coordinates in One Dimension
In the last week/chapter, you learned how a collection of particles can behave like a “particle” of the same total mass located at the center of mass as far as Newton’s Second Law is concerned. We also saw at least four examples of how problems involving systems of particles can be decomposed into two separate problems – one the motion of the center of mass, which generally obeys Newtonian

Week 5: Torque and Rotation in One Dimension

237

dynamics as if the whole system is “a particle”, and the other the motion in the center of mass system107 .
This decomposition is useful (as we saw) even if the system has many particles in it and is fluid or non-interacting, but it is very useful in helping us to describe the motion of rigid bodies. This is because the most general motion of a rigid object is the translation of (the center of mass of) the object according to the total force acting on it and Newton’s Second Law (as demonstrated last week), plus the rotation of that body about its center of mass as unbalanced forces exert a torque on the object.
The first part we are very very familiar with at this point and we’ll take it for granted that you can solve for the motion of the center of mass of a rigid object given any reasonable net force.
The second we are not familiar with at all, and we will now take the next two weeks to study it in detail and understand it, as rotation is just as important and common as translation when it comes to understanding the motion of nearly everything we see on a daily basis. Doors rotate about hinges, tires rotate about axles, electrons and protons “just rotate” because of their intrinsic spin, our fingers and toes and head and arms and legs rotate about their joints, our whole bodies rotate about their center of mass when we get up in the morning, when we do a twirl on ice skates, when we summersault on a trampoline, the entire Earth rotates around its axis while revolving around the sun which rotates on its axis while revolving around the Galactic center which... just goes to show that rotation really is ubiquitous, and pretending that it isn’t important or worthy of understanding is not an option, even for future physicians or non-rocket-scientist bio majors.
It will take two weeks (and maybe even longer, for some of you) because rotation is a wee bit complicated. For many of you, it will be the most difficult single topic we cover this semester, if only because rotation is best described by means of the Evil Cross Product108 . Just as we started our study of coordinate motion with motion in only one dimension, so we will start our study or rotation with “one dimensional rotation” of a rigid body, that is, the rotation of a rigid object through an angle θ about a single fixed axis109 .
Eventually we want to be able to treat arbitrary rigid objects, ones that have their mass symmetrically but non-uniformly distributed (e.g. basketballs or ninja stars) or non-uniformly and not particularly symmetrically distributed (e.g. the human body, automobiles, blobs of putty of arbitrary shape). But at the moment even the rotation of a basketball on the tip of a player’s finger seems like too much for us to handle
We therefore start with the simplest possible example – a “rigid” system with all of its mass concentrated in a single point that rotates around some fixed axis. Consider a small “pointlike” ball of mass m on a rigid massless unstretchable rod, portrayed in figure 59. The rod itself is pivoted on a frictionless axle in the center so that the mass is constrained to move only on the dashed circle in the plane of the picture. The mass therefore maintains a constant distance from the pivot – r is a constant – but the angle θ can vary in time as external forces act on the system.
The very first things we need to do are to bring to mind the set of rotational coordinates that
107 In

particular, we solved elastic collisions in the center of mass frame (where they were easy) while the center of mass of the colliding system obeyed (trivial) Newtonian dynamics, we looked at the exploding rocket where the center of mass followed the parabolic/Newtonian trajectory, we saw that inelastic collisions turn all of the kinetic energy in the center of mass frame into heat, and we proved that in general the kinetic energy of a system in the lab is the sum of the kinetic energy of the system (treated as a particle moving at speed vcm ) plus the kinetic energy of all of the particles in the center of mass frame – this latter being the energy lost in a completely inelastic collision or conserved in an elastic one!
108 Wikipedia: http://www.wikipedia.org/wiki/Cross Product. Something that is covered both in this Wikipedia article and in the online Math Review supplement, so now is a really, really great time to pause in reading this chapter and skip off to refresh your memory of it. It is a memory, we hope, isn’t it? If not, then by all means skip off to learn it...
109 The “direction” of a rotation is considered to be along the axis of its rotation in a right handed sense described later below. So a “one dimensional rotation” is the rotation of any object about a single axis – it does not imply that the object being rotated is in any sense one dimensional.

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y

s v

r θ r

x

Figure 59: A small ball of mass m rotates about a frictionless pivot, moving in a circle of radius r. we have already introduced for doing kinematics of a rotating object. Since r is fixed, the position of the particle is uniquely determined by the positive angle θ(t), measured by convention as a counterclockwise rotation about the z-axis from the +x-axis as drawn in figure 59. We call θ the angular position of the particle.
We can easily relate r and θ to the real position of the particle. The distance the particle must move in the counterclockwise direction from the standard reference position at (x = r, y = 0) around the circular arc to an arbitrary position on the circle is s = rθ. s (the arc length) is a one dimensional coordinate that describes its motion on the arc of the circle itself, and if we know r and s (the latter measured from the +x-axis) we know exactly where the particle is in the x-y plane.
We recall that the tangential velocity of the particle on this circle is then vt =

ds d(rθ) dθ
=
=r
= rω dt dt dt (461)

where we remind you of the angular velocity ω = dθ . Note that for a rigid body vr = dr = 0, that dt dt is, the particle is constrained by the rigid rod or solidity of the body to move in circles of constant radius r about the center of rotation or pivot so its speed moving towards or away from the circle is zero. Similarly, we can differentiate one more time to find the tangential acceleration: at = where α =

dω dt =

d2 θ dt2 dvt dω d2 θ
=r
= r 2 = rα dt dt dt (462)

is the angular accleration of the particle.

Although the magnitude of vr = 0, we note well that the direction of v t is constantly changing and we know that ar = −v 2 /r = −rω 2 which we derived in the first couple of weeks and by now have used repeatedly to solve many problems.
All of this can reasonably be put in a small table that lets us compare and contrast the one dimensional arc coordinates with the associated angular coordinates:

5.2: Newton’s Second Law for 1D Rotations
With these coordinates in hand, we can now consider the angular version of Newton’s Second Law for a force F applied to this particle as portrayed in figure 60. This is an example of a “rigid” body

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Angular

Arc Length

θ

s = rθ

ω=

dθ dt vt =

ds
= rω dt α=

dω dt at =

dvt
= rα dt Table 2: Coordinates used for angular/rotational kinemetics in one dimension. Note that θ is the rotation angle around a given fixed axis, in our picture above the z-axis, and that θ must be given in (dimensionless) radians for these relations to be true. Remember C = 2πr is the formula for the circumference of a circle and is a special case of the general relation s = rθ, but only when θ = 2π radians. y

r

F

Ft

φ
Τ

Fr

x

Figure 60: A force F is applied at some angle φ (relative to r) to the ball on the pivoted massless rod. rotation, but because we aren’t yet ready to tackle extended objects all of the mass is concentrated in the ball at radius r. We’ll handle true, extended rigid objects shortly, once we understand a few basic things well.
Since the rod is rigid, and pivoted by an unmovable frictionless axle of some sort in the center, the tension in the rod opposes any motion along r. If the particle is moving around the circle at some speed vt (not shown), we expect that:
Fr − T = F cos(φ) − T = −mar = −m

2 vt = −mrω 2 r (463)

(where r is an outward directed radius, note that the acceleration is in towards the center) as usual.
The rotational motion is what we are really interested in. Newton’s Law tangent to the circle is just: Ft = F sin(φ) = mat = mrα
(464)
For reasons that will become clear in a bit, we will find it very useful to multiply this whole equation by r and redefine rFt to be a new quantity called the torque, given the symbol τ . We will also collect the factors of r and multiply them by the m to make a new quantity called the moment of intertia and give it the symbol I: τ = rFt = rF sin(φ) = mr2 α = Iα

(465)

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Week 5: Torque and Rotation in One Dimension

In particular, this equation contains the moment of inertia of a point mass m moving in a circle of radius r:
Ipoint mass = mr2
(466)
This looks like, and of course is, Newton’s Second Law for a rigid rotating system in one dimension, where force is replaced by torque, mass is replaced by moment of inertia, and linear acceleration is replaced by angular acceleration.
Although to us so far this looks just like a trivial algebraic rewrite of something we could have worked with just as easily as the real thing in the s coordinates, it is actually far more general and powerful. To completely understand this, we need to understand two things. One of them is how applying the force F to (for example) the rod at different radii rF changes the angular acceleration
α. The other is how a force F applied at some radius rF to the massless rod internally redistributes to muliple masses attached to the rod at different radii so that all the masses experience the same angular acceleration. These are the subjects of the next two sections.

5.2.1: The r-dependence of Torque
Let’s see how the angular acceleration of this mass will scale with the point of application of the force along the rod, and in the process justify our “inspired decision” to multiply Ft by r in our definition of the torque in the previous section. To accomplish this we need a new figure, one where the massless rigid rod extends out past/through the mass m so it can act as a lever arm on the mass no matter where we choose to apply the force F .

y dl = r F dθ

rod dθ F

Fp pivot m rm φ
Fr

Ft = F sin( φ) x rF

Figure 61: The force F is applied to the pivoted rod at an angle φ at the point r F with the mass m attached to the rod at radius rm .
This is displayed in figure 61. A massless rod as long or longer than both rm and rF is pivoted at one end so it can swing freely (no friction). The mass m is attached to the rod at the position rm . A force F is applied to the rod at the position r F (on the rod) and at an angle φ with respect to the direction of r F .
Turning this into a suitable angular equation of motion is a bit of a puzzle. The force F is not applied directly to the mass – it is applied to the massless rigid rod which in turn transmits some of the force to the mass. However, the external force F is not the only force acting on the rod!
In the previous example the pivot only exerted a radial force F p = −T , and exerted no tangential force on m at all. We could even compute T (and hence F p ) if we knew θ, vt and F from rotational kinematics and some vector geometry. In this case, however, if F exerts a force on the rod that can

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be transmitted to and act tangentially upon the mass m, it rather seems that the unknown pivot force F p can as well, but we don’t know F p !
Alas, without knowing all of the forces that act tangentially on m, we cannot use Newton’s
Second Law directly. This motivates us to consider using work and energy to obtain a dynamical principle (basically working the derivation of the WKE theorem backwards) because the pivot does not move and therefore the force F p does no work! Consequently, the fact that we do not yet know F p will not matter!
So to work. Let us suppose that the force F is applied to the rod for a time dt, and that during that time the rod rotates through an angle dθ as shown. In this case we can easily find the work done by the force F . The point on the rod where the force F is applied moves a distance dℓ = rF dθ.
The work is done only by the tangential component of the force moved through this distance Ft so that: dW = F · dℓ = Ft rF dθ
(467)
The WKE theorem tells us that this work must equal the change in the kinetic energy over that time: 2 d 1 mvt
2
Ft rF dθ = dK = dt = mvt dvt
(468)
dt
We make a few useful substitutions from table 3 above:
Ft rF dθ = m rm

dθ dt 2
(rm α dt) = mrm αdθ

(469)

and cancel dθ (and reorder a bit) to get:
2
τ = rF Ft = rF F sin(φ) = mrm α = Iα

(470)

This formally proves that my “guess” of τ = Iα as being the correct form of Newton’s Second
Law for this simple rotating rigid body holds up pretty well no matter where we apply the force(s) that make up the torque, as long as we define the torque: τ = rF Ft = rF F sin(φ)

(471)

It is left as an exercise for the student to draw a picture like the one above but involving many independent and arbitrary forces, F 1 acting at r 1 , F 2 acting at r 2 , ..., you get:
2
ri Fi sin(φi ) = mrm α = Iα

τtot =

(472)

i

for a single point-like mass on the rod at position r m . Note well that each φi is the angle between r i and F i , and you should make the (massless, after all) rod long enough for all of the forces to be able to act on it and also pass through m.
In a bit we will pay attention to the fact that rF sin(φ) is the magnitude of the cross product 110 of r F and F , and that if we assign the direction of the rotation to be parallel to the z-axis of a right handed coordinate system when φ is drawn in the sense shown, we can even make this a vector relation. For the moment, though, we will stick with our simple 1D “scalar” formulation and ask a different question: what if we have a more complicated object than a single mass on a pivoted rigid rod that is being driven by a torque (or sum of torques).
The answer is: We have to sum up the object’s total moment of inertia around the pivot axis.
Let’s prove this.
110 Wikipedia: http://www.wikipedia.org/wiki/Cross Product. Making this a gangbusters good time to go review – or learn – cross products, at least well enough to be able evaluate their magnitude and direction (using the right hand rule). 242

Week 5: Torque and Rotation in One Dimension

5.2.2: Summing the Moment of Inertia
Suppose we have a mass m1 attached to our massless rod pivoted at the origin at the position r1 , and a second mass m2 attached at position r2 . We will then apply the force F at an angle φ to the (extended) rod at position r as shown in figure 62, and duplicate our reasoning from the last chapter (because we still do not known the unknown force exerted by the pivot, but as long as we consider work we don’t have to.

y
F
m1

m2

φ x r1 r2 rF

Figure 62: A single torque τ = rF F sin(φ) is applied to a rod with two masses, m1 at r1 and m2 at r2 .
The WKE theorem for this picture is now (note that v1 and v2 are both necessarily tangential): dW = rF F sin(φ) dθ = τ dθ τ dθ

= dK = m1 v1 dv1 + m2 v2 dv2 so as usual dθ dθ
= m1 r1
(r1 α dt) + m2 r2
(r2 α dt) dt dt

τ

2
2
= m1 r1 α + m2 r2 α

τ

=

2
2
m1 r1 + m2 r2 α = Iα

(473)

where we have now defined
2
2
I = m1 r1 + m2 r2

(474)

That is, the total moment of inertia of the two point masses is just the sum of their individual moments of inertia. From the derivation it should be clear that if we added 3,4,...,N point masses along the massless rod the total moment of inertia would just be the sum of their individual moments of inertia.
Indeed, as we add more forces acting at different points and directions (in the plane) on the rod and add more masses at different points on the rod, everything we did above clearly scales up linearly – we simply have to sum the total torque on the right hand side and sum the total moment of inertia on the left hand side. We therefore conclude that Newton’s Second Law for a system constrained to rotate in (one dimension in a) a plane about a fixed pivot is just: τtot =

2 mj rj α = Itot α

ri Fi sin(φi ) = i (475)

j

So much for discrete forces and discrete masses. However, most rigid bodies that we experience every day are, on a coarse-grained macroscopic scale, made up of a continuous distribution of mass, and instead of a mythical idealized “massless rigid rod” all of this mass is glued together by means of internal forces.
It is pretty clear that our expression τ = Iα will generalize to this case where we will (probably) replace: I= j 2 mj rj →

r2 dm

(476)

but we will need to do just a teensy bit of work to show that this is true and extract any essential conceptual insight to be found along the way.

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5.3: The Moment of Inertia
We begin with a specific example to help smooth the way.

Example 5.3.1: The Moment of Inertia of a Rod Pivoted at One End y pivot

dm = λ dx

L

x

x

dx

Figure 63: A solid rod of length L with a mass M uniformly pivoted about one end. One can think of such a rod as being the “massless rod” of the previous section with an infinite number of masses mi uniformly distributed along its length, that sum to the total mass M .
In figure 63 above a massive rod pivoted about one end is drawn. We would like to determine how this particular rod will rotationally accelerate when we (for example) attach a force to it and apply a torque. We therefore must characterize this rod as having a specific mass M , a specific length L, and we need to say something about the way the mass is distributed, because the rod could be made of aluminum (not very dense) at one end and tungsten (very dense indeed) at the other and still “look” the same. We will assume that this rod is uniformly distributed, and that it is very thin and symmetrical in cross-section – shaped like a piece of wire or perhaps a wooden dowel rod. In a process that should be familiar to you from last week and from the previous section, we know that the moment of inertia of a sum of discrete point masses hung on a “massless” rod (that only serves to assemble them into a rigid structure) is just:
2
mi ri

Itot =

(477)

i

the sum of the moments of inertia of the point masses.
We can clearly approximate the moment of inertia of the continuous rod by dividing it up into N pieces, each of length ∆x = L/N and mass ∆M = M/N , and treating each small piece as a “point mass” located at xi = i ∗ ∆x:
N

Irod ≈

i=1

M
N

N

i∗L
N

∆M x2 i =

(478)

i=1

As before, the limit of this sum as N → ∞ is by definition the integral, and in this limit the sum exactly represents the moment of inertia of the rod.
We can easily evaluate this. To do so, we chant our ritual expression: “The mass of the chunk is the mass per unit length of the chunk times the length of the chunk”, or dm = λdx = M dx, so:
L
L

x2 dm =

Irod =
0

M
L

L

x2 dx =
0

M L2
3

(479)

5.3.1: Moment of Inertia of a General Rigid Body
This specific result can easily be generalized. If we consider a blob-shaped distribution of mass, the differential moment of inertia of a tiny chunk of the mass in the distribution about some fixed axis

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pivot axis

dm r Figure 64: A “blob-shaped” chunk of mass, perhaps a piece of modelling clay, constrained to rotate about an axis through the blob, perhaps a straight piece of nearly massless coat-hanger wire. of rotation is clearly: dI = r2 dm

(480)

By now you should be getting the idea that summing up all of the little chunks that make up the object is just integrating: r2 dm

Iblob =

(481)

blob

where it is quite one thing to write down this formal expression, quite another to be able to actually do the integral over all of the chunks of mass that make up an object.
It isn’t too difficult to do this integral for certain simple distributions of mass, and we will need a certain “stock repertoire” of moments of inertia in order to solve problems. Which ones you should learn to do depends on the level of the course – math/physics majors should learn to integrate over spheres (and maybe engineers as well), but everybody else can probably get away learning to evaluate the moment of inertia of a disk. In practice, for any really complicated mass distribution
(like the blob of clay pictured above) one would either measure the moment of inertia or use a computer to actually break the mass up into a very large number of discrete (but small/point-like) chunks and do the sum.
First let’s do an example that is even simpler than the rod.

Example 5.3.2: Moment of Inertia of a Ring z M dθ R ds = Rd θ
Figure 65: A ring of mass M and radius R in the x-y plane rotates freely about the z-axis.
We would like to find the moment of inertia of the ring of uniformly distributed mass M and radius R portrayed in figure 65 above. A differential chunk of the ring has length ds = R dθ. It’s mass is thus (say the ritual words!): dm = λds =

M
M
R dθ = dθ 2πR


(482)

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Week 5: Torque and Rotation in One Dimension and its moment of inertia is very simple:


r2 dm =

Iring =

0

M 2
R dθ = M R2


(483)

In fact, we could have guessed this. All of the mass M in the ring is at the same distance R from the axis of rotation, so its moment of inertia (which only depends on the mass times the distance and has no “vector” character) is just M R2 just like a point mass at that distance.
Because it is so important, we will do the moment of inertia of a disk next. The disk will be many things to us – a massive pulley, a wheel or tire, a yo-yo, a weight on a grandfather clock (physical) pendulum. Here it is.

Example 5.3.3: Moment of Inertia of a Disk dA = rd θ dr rd θ dr M


R
r

Figure 66: A disk of mass M and radius R is pivoted to spin freely around an axis through its center.
In figure 66 a disk of uniformly distributed mass M and radius R is drawn. We would like to find its moment of inertia. Consider the small chunk of disk that is shaded of area dA. In plane polar coordinates (the only ones we could sanely hope to integrate over) the differential area of this chunk is just its differential height dr times the width of the arc at radius r subtended by the angle dθ, r dθ. The area is thus dA = r drdθ.
This little chunk was selected because the mass dm in it moves in a circle of radius r around the pivot axis. We need to find dm in units we can integrate to cover the disk. We use our litany to set: dm = σdA =

M r drdθ πR2 (484)

dI = r2 dm =

M 3 r drdθ πR2 (485)

and then write down:

We integrate both sides to get (note that the integrals are independent one dimensional integrals that precisely cover the disk):
Idisk

=
=
=

M πR2 2π

R

r3 dr

M
R4
πR2
4
1
M R2
2


0

0

(2π)

This is a very important and useful result, so keep it in mind.

(486)

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5.3.2: Table of Useful Moments of Inertia
Finally, here is a table of a few useful moments of inertia of simple uniform objects. In each case
I indicate the value about an axis through the symmetric center of mass of the object, because we can use the parallel axis theorem and the perpendicular axis theorem to find the moments of inertia around at least some alternative axes.
Shape

Icm

Rod from −L/2 to L/2

1
M L2
12

Ring

M R2

Disk

1
M R2
2

Sphere

2
M R2
5

Spherical Shell

2
M R2
3

Generic “Round” Mass of Mass M and radius R

βM R2

Table 3: A few useful moments of inertia of symmetric objects around an axis of symmetry through their center of mass. You should probably know all of the moments in this table and should be able to evaluate the first three by direct integration.

5.4: Torque as a Cross Product
This section will be rather abbreviated this week; next week we will cover it in gory detail as a vector relation. For the moment, however, we need to make a number of observations that will help us solve problems. First, we know that the one-dimensional torque produced by any single force acting on a rigid object a distance r from a pivot axis is just: τ = rF⊥ = rF sin(φ)

(487)

where F⊥ is just the component of the force perpendicular to the (shortest) vector r from the pivot axis to the point of application. This is really just one component of the total torque, mind you, but it is the one we have learned so far and are covering this week.
First, let’s make an important observation. Provided that r and F lie in a plane (so that the one dimension is the right dimension) the magnitude of the torque is the magnitude of the cross product of r and F : τ = |r × F | = rF sin(φ) = rF⊥ = r⊥ F
(488)
I’ve used the fact that I can move the sin(φ) around to write this in terms of: r⊥ = r sin(φ)

(489)

which is the component of r perpendicular to F , also known as the moment arm of the torque. This is a very useful form of the torque in many problems. It it equally well expressible in terms of the familiar:
F⊥ = F sin(φ),
(490)

Week 5: Torque and Rotation in One Dimension

247

the component of F perpendicular to r. This form, too, is often useful. In fact, both forms may be useful (to evaluate different parts of the total torque) in a single problem!
If we let the vector torque be defined by: τ =r×F

(491)

marvelous things will happen. Next week we will learn about them, and will learn about how to evaluate this a variety of ways. For now let’s just learn one.
The vector torque τ has a magnitude |r × F | = rF sin(φ) and points in the direction given by the right hand rule.
The right hand rule, in turn, is the following:
The direction of the vector cross product A × B is in the direction the thumb of your right hand points when you begin with the fingers of your right hand lined up with the vector A and then curl them naturally through the angle φ < π into the direction of B.
That is, if you imagine “grasping” the axis in the direction of the torque with your right hand, your fingers will curl around in the direction from r to F through the smaller of the two angles in between them (the one less than π).
You will get lots of practice with this rule, but be sure to practice with your right right hand, not your wrong right hand. Countless students (and physics professors and TAs!) have been embarrassed be being caught out evaluating the direction of cross products with their left hand111 . Don’t be one of them!
The direction of the torque matters, even in one dimension. There is no better problem to demonstrate this than the following one, determine what direction a spool of rope resting on a table will roll when one pulls on the rope.

Example 5.4.1: Rolling the Spool
I’m not going to quite finish this one for you, as there are a lot of things one can ask and it is a homework problem. But I do want you to get a good start.
The spool in figure 67 is wrapped many times around with string. It is sitting on a level, rough table so that for weak forces F it will freely roll without slipping (although for a large enough F of course it will slip or even rise up off of the table altogether).
The question is, which direction will it roll (or will it not roll until it slides) for each of the three directions in which the string is pulled.
The answer to this question depends on the direction of the total torque, and the relevant pivot is the point that does not move when it rolls, where the (unknown!) force of static friction acts.
If we choose the pivot to be the point where the spool touches the table, then gravity, the normal force and static friction all exert no torque! The only source of torque is F .
So, what is the direction of the torque for each of the three forces drawn, and will a torque in that direction make the mass roll to the left, the right, or slide (or not move)?
Think about it.
111 Let

he or she who is without sin cast the first stone, I always say. As long as it is cast with the right hand...

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Week 5: Torque and Rotation in One Dimension

F pivot F

F

Figure 67: What direction does one expect the spool in each of the figures to roll (or will it roll at all)? 5.5: Torque and the Center of Gravity
We will often wish to solve problems involving (for example) rods pivoted at one end swinging down under the influence of near-Earth gravity, or a need to understand the trajectory and motion of a spinning basketball. To do this, we need the idea of the center of gravity of a solid object.
Fortunately, this idea is very simple:
The center of gravity of a solid object in an (approximately) uniform near-Earth gravitational field is located at the center of mass of the object. For the purpose of evaluating the torque and angular motion or force and coordinate motion of center of mass, we can consider that the entire force of gravity acting on the object is equivalent to the the force that would be exerted by the entire mass located as a point mass at the center of gravity.
The proof for this is very simple. We’ve already done the Newtonian part of it – we know that the total force of gravity acting on an object makes the center of mass move like a particle with the same mass located there. For torque, we recall that: τ = rF⊥ = r⊥ F

(492)

If we consider the torque acting on a small chunk of mass in near-Earth gravity, the force (down) acting on that chunk is: dFy = −gdm
(493)
The torque (relative to the pivot) is just: dτ = −gr⊥ dm or (494)

L

τ = −g

0

r⊥ dm = −M grcm

(495)

where rcm is (the component of) the position of the center of mass of the object perpendicular to gravity. The torque due to gravity acting on the object around the selected pivot axis is the same

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Week 5: Torque and Rotation in One Dimension

as the torque that would be produced by the entire weight of the object pulling down at the center of mass/gravity. Q.E.D.
Note that this proof is valid for any shape or distribution of mass in a uniform gravitational field, but in a non-uniform field it would fail and the center of gravity would be different from the center of mass as far as computing torque is concerned. This is the realm of tides and will be discussed more in the week where we cover gravity.

Example 5.5.1: The Angular Acceleration of a Hanging Rod

θ

L/2

M
Mg

θ

L

Figure 68: A rod of mass M and length L is suspended/pivoted from one end. It is pulled out to some initial angle θ0 and released.
This is your first example of what we will later learn to call a physical pendulum. A rod is suspended from a pivot at one end in near-Earth gravity. We wish to find the angular acceleration α as a function of θ. This is basically the equation of motion for this rotational system – later we will learn how to (approximately) solve it.
As shown above, for the purpose of evaluating the torque, the force due to gravity can be considered to be M g straight down, acting at the center of mass/gravity of the rod (at L/2 in the middle). The torque exerted at this arbitrary angle θ (positive as drawn, note that it is swung out in the counterclockwise/right-handed direction from the dashed line) is therefore: τ = rFt = −

M gL sin(θ) 2

(496)

It is negative because it acts to make θ smaller ; it exerts a “twist” that is clockwise when θ is counterclockwise and vice versa.
From above, we know that I = M L2 /3 for a rod pivoted about one end, therefore: τ = −

M L2
M gL sin(θ) = α = Iα
2
3
3g
d2 θ
= − α = sin(θ) dt2
2L

or:
(497)

independent of the mass!

5.6: Solving Newton’s Second Law Problems Involving Rolling
One of the most common applications of one dimensional torque and angular momentum is solving rolling problems. Rolling problems include things like:

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Week 5: Torque and Rotation in One Dimension
• A disk rolling down an inclined plane.
• An Atwood’s Machine, but with a massive pulley.
• An unwinding spool of line, either falling or being pulled.

These problems are all solved by using a combination of Newton’s Second Law for the motion of the center of mass of the rolling object (if appropriate) or other masses involved (in e.g. Atwood’s
Machine) and Newton’s Second Law for 1 dimensional rotation, τ = Iα. In general, they will also involve using the rolling constraint:
If a round object of radius r is rolling without slipping, the distance x it travels relative to the surface it is rolling on equals rθ, where θ is the angle it rolls through.
That is, all three are equivalently the “rolling constraint” for a ball of radius r rolling on a level floor, started from a position at x = 0 where also θ = 0: x = rθ

(498)

v

= rω

(499)

a = rα

(500)
(501)

These are all quite familiar results – they look a lot like our angular coordinate relations – but they are not the same thing! These are constraints, not coordinate relations – for a ball skidding along the same floor they will be false, and for certain rolling pulley problems on your homework you’ll have to figure out one appropriate for the particular radius of contact of spool-shaped or yo-yo shaped rolling objects (that may not be the radius of the object!)
It is easier to demonstrate how to proceed for specific examples than it is to expound on the theory any further. So let’s do the simplest one.

Example 5.6.1: A Disk Rolling Down an Incline r N fs m φ mg

φ
Figure 69: A disk of mass M and radius r sits on a plane inclined at an angle φ with respect to the horizontal. It rolls without slipping down the incline.
In figure 69 above, a disk of mass M and radius r sits on an inclined plane (at an angle φ) as shown. It rolls without slipping down the incline. We would like to find its acceleration ax down the incline, because if we know that we know pretty much everything about the disk at all future times that it remains on the incline. We’d also like to know what fs (the force exerted by static friction)

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Week 5: Torque and Rotation in One Dimension

when it is so accelerating, so we can check to see if our assumption of rolling without slipping is justified. If φ is too large, we are pretty sure intuitively that the disk will slip instead of roll, since if φ ≥ π/2 we know the disk will just fall and not roll at all.
As always, since we expect the disk to physically translate down the incline (so a and F tot will point that way) we choose a coordinate system with (say) the x-axis directed down the inline and y directed perpendicular to the incline.
Since the disk rolls without slipping we know two very important things:
1) The force fs exerted by static friction must be less than (or marginally equal to) µs N . If, in the end, it isn’t, then our solution is invalid.
2) If it does roll, then the distance x it travels down the incline is related to the angle θ it rolls through by x = rθ. This also means that vx = rω and ax = rα.
We now proceed to write Newton’s Laws three times: Once for the y-direction, once for the x-direction and once for one dimensional rotation (the rolling). We start with the:
Fy = N − mg cos(φ) = may = 0

(502)

which leads us to the familar N = mg cos(φ).
Next:
Fx = mg sin(φ) − fs

τ = rfs

= max

(503)

= Iα

(504)

Pay attention here, because we’ll do the following sort of things fairly often in problems. We use
1
I = Idisk = 2 mr2 and α = ax /r and divide the last equation by r on both sides. This gives us: mg sin(φ) − fs fs = max
1
= max 2

(505)
(506)

If we add these two equations, the unknown fs cancels out and we get: mg sin(φ) = or: ax =

3 max 2

2 g sin(φ)
3

(507)

(508)

We can then substitute this back into the equation for fs above to get: fs =

1
1
max = mg sin(φ)
2
3

(509)

In order to roll without slipping, we know that fs ≤ µs N or
1
mg sin(φ) ≤ µs mg cos(φ)
3
or µs ≥

1 tan(φ) 3

(510)

(511)

If µs is smaller than this (for any given incline angle φ) then the disk will slip as it rolls down the incline, which is a more difficult problem.
We’ll solve this problem again shortly to find out how fast it is going at the bottom of an incline of length L using energy, but in order to this we have to address rotational energy. First, however, we need to do a couple more examples.

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Week 5: Torque and Rotation in One Dimension

M r T
2

T1

T1

T2

m1

m2 m 1g m 2g
Figure 70: Atwood’s Machine, but this time with a massive pulley of mass M and radius R. The massless unstretchable string connecting the two masses rolls without slipping on the pulley, exerting a torque on the pulley as the masses accelerate to match. Assume that the pulley has a moment of inertia βM r2 for some β. Writing it this way let’s us use β ≈ 1 to approximate the pulley
2
with a disk, or use observations of a to measure β and hence tell something about the distribution of mass in the pulley!

Example 5.6.2: Atwood’s Machine with a Massive Pulley
Our solution strategy is almost identical to that of our first solution back in week 1 – choose an
”around the corner” coordinate system where if we make moving m2 down “positive”, then moving m1 up is also “positive”. To this we add that a positive rotation of the pulley is clockwise, and that the rolling constraint is therefore a = rα.
Now we again write Newton’s Second Law once for each mass and once for the rotating pulley
(as τ = Iα): m2 g − T 2

T 1 − m1 g

τ = rT2 − rT1

= m2 a

(512)

= m1 a

(513)

a
= βM r2 = Iα r (514)

Divide the last equation by r on both sides, then add all three equations to eliminate both unknown tensions T1 and T2 . You should get:
(m2 − m1 )g = (m1 + m2 + βM )a or: a=

(m2 − m1 )g
(m1 + m2 + βM )

(515)

(516)

This is almost like the previous solution – and indeed, in the limit M → 0 is the previous solution – but the net force between the two masses now must also partially accelerate the mass of the pulley. Partially because only the mass near the rim of the pulley is accelerated at the full rate a – most of the mass near the middle of the pulley has a much lower acceleration.
Note also that if M = 0, T1 = T2 ! This justifies – very much after the fact – our assertion early on that for a massless pulley, the tension in the string is everywhere constant. Here we see why that is true – because in order for the tension in the string between two points to be different, there has to be some mass in between those points for the force difference to act upon! In this problem, that mass is the pulley, and to keep the pulley accelerating up with the string, the string has to exert a torque on the pulley due to the unequal forces.

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Week 5: Torque and Rotation in One Dimension

One last thing to note. We are being rather cavalier about the normal force exerted by the pulley on the string – all we can easily tell is that the total force acting up on the string must equal
T1 + T2 , the force that the string pulls down on the pulley with. Similarly, since the center of mass of the pulley does not move, we have something like Tp − T1 − T2 − M g = 0. In other words, there are other questions I could always ask about pictures like this one, and by now you should have a good idea how to answer them.

5.7: Rotational Work and Energy ω Center of Mass v r

dm

Figure 71: A blob of mass rotates about an axis through the center of mass, with an angular velocity as shown.
We have already laid the groundwork for studying work and energy in rotating systems. Let us consider the kinetic energy of an object rotating around its center of mass as portrayed in figure 71.
The center of mass is at rest in this figure, so this is a center of mass inertial coordinate system.
It is easy for us to write down the kinetic energy of the little chunk of mass dm drawn into the figure at a distance r from the axis of rotation. It is just: dKin CM =

1
1
dmv 2 = dmr2 ω 2
2
2

(517)

To find the total, we integrate over all of the mass of the blob:
Kin CM =

1
2

dmr2 ω 2 = blob 1 2

2

(518)

which works because ω is the same for all chunks dm in the blob and is hence a constant that can be taken out of the integral, leaving us with the integral for I.
If we combine this with the theorem proved at the end of the last chapter we at last can precisely describe the kinetic energy of a rotating baseball in rest frame of the ground:
K=

1
1
2
M vcm + Iω 2
2
2

(519)

That is, the kinetic energy in the lab is the kinetic energy of the (mass moving as if it is all at the)
1
center of mass plus the kinetic energy in the center of mass frame, 2 Iω 2 . We’ll have a bit more to say about this when we prove the parallel axis theorem later.

5.7.1: Work Done on a Rigid Object
We have already done rotational work. Indeed we began with rotational work in order to obtain
Newton’s Second Law for one dimensional rotations above! However, there is much to be gained by considering the total work done by an arbitrary force acting on an arbitrary extended rigid mass.
Consider the force F in figure 72, where I drew a regular shape (a disk) only to make it easy to see

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Week 5: Torque and Rotation in One Dimension

Fr

F

φ
F
r

R

M

Figure 72: A force F is applied in an arbitrary direction at an arbitrary point on an arbitrary rigid object, decomposed in a center of mass coordinate frame. A disk is portrayed only because it makes it easy to see where the center of mass is. and draw an useful center of mass frame – it could just as easy be a force applied to the blob-shaped mass above in figure 71.
I have decomposed the force F into two components:
Fr

= F cos(φ)

(520)

F⊥

= F sin(φ)

(521)
(522)

Suppose that this force acts for a short time dt, beginning (for convenience) with the mass at rest.
We expect that the work done will consist of two parts: dW = Fr dr + F⊥ ds

(523)

dW = Wr + Wθ = Fr dr + rF⊥ dθ = Fr dr + τ dθ

(524)

where ds = r dθ. This is then:

We know that: dWr = Fr dr

= dKr

dWθ = τ dθ = dKθ

(525)
(526)

and if we integrate these independently, we get:
Wtot = Wcm + Wθ = ∆Kcm + ∆Kθ

(527)

or the work decomposes into two parts! The work done by the component of the force through the center of mass accelerates the center of mass and changes the kinetic energy of the center of mass of the system as if it is a particle! The work done by the component of the force perpendicular to the line connecting the center of mass to the point where the force is applied to the rigid object increases the rotational kinetic energy, the kinetic energy in the center of mass frame.
Hopefully this is all making a certain amount of rather amazing, terrifying, sense to you. One reason that torque and rotational physics is so important is that we can cleanly decompose the physics of rotating rigid objects consistently, everywhere into the physics of the motion of the center of mass and rotation about the center of mass. Note well that we have also written the WKE

Week 5: Torque and Rotation in One Dimension

255

theorem in rotational terms, and are now justified in using all of the results of the work and energy chapter/week in (fully or partially) rotational problems!
Before we start, though, let’s think a teeny bit about the rolling constraint and work, as we will be solving many rolling problems.

5.7.2: The Rolling Constraint and Work
A car is speeding down the highway at 50 meters per second (quite fast!) being chased by the police.
Its tires hum as they roll down the highway without sliding. Fast as it is going, there are four points on this car that are not moving at all relative to the ground! Where are they?
The four places where the tires are in contact with the pavement, of course. Those points aren’t sliding on the pavement, they are rolling, and “rolling” means that they are coming down at rest onto the pavement and then lifting up again as the tire rolls on.
If the car is travelling at a constant speed (and we neglect or arrange for their to be no drag/friction) we expect that the road will exert no force along the direction of motion of the car – the force exerted by static friction will be zero. Indeed, that’s why wheels were invented – an object that is rolling at constant speed on frictionless bearings requires no force to keep it going – wheels are a way to avoid kinetic/sliding friction altogether!
More reasonably, the force exerted by static friction will not be zero, though, when the car speeds up, slows down, climbs or descends a hill, goes around a banked turn, overcomes drag forces to maintain a constant speed.
What happens to the energy in all of these cases, when the only force exerted by the ground is static friction at the points where the tire touches the ground? What is the work done by the force of static friction acting on the tires?
Zero! The force of static friction does no work on the system.
If you think about this for a moment, this result is almost certain to make your head ache. On the one hand, it is obvious: dWs = Fs dx = 0
(528)
because dx = 0 in the frame of the ground – the place where the tires touch the ground does not move, so the force of static friction acts through zero distance and does no work.
Um, but if static friction does no work, how does the car speed up (you might ask)? What else could be doing work on the car? Oooo, head-starting-to-huuuurrrrrt...
Maybe, I dunno, the motor?
In fact, the car’s engine exerts a torque on the wheels that is opposed by the pivot force at the road – the point of contact of a rolling object is a natural pivot to use in a problem, because forces exerted there, in addition to doing no work, exert no torque about that particular pivot. By fixing that pivot point, the car’s engine creates a net torque that accelerates the wheels and, since they are fixed at the pivot, propels the car forward. Note well that the actual source of energy, however, is the engine, not the ground. This is key.
In general, in work/energy problems below, we will treat the force of static friction in rolling problems (disks, wheels, tires, pulleys) where there is rolling without slipping as doing no work and hence acting like a normal force or other force of constraint – not exactly a ”conservative force” but one that we can ignore when considering the Generalized Non-Conservative Work – Mechanical
Energy theorem or just the plain old WKE theorem solving problems.
Later, when we consider pivots in collisions, we will see that pivot forces often cause momentum not to be conserved – another way of saying that they can cause energy to enter or leave a system

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Week 5: Torque and Rotation in One Dimension

– but that they are generally not the source of the energy. Like a skilled martial artist, they do not provide energy themselves, but they are very effective at diverting energy from one form to another.
In fact, very much like a skilled martial artist, come to think about it.
It isn’t really a metaphor...

Example 5.7.1: Work and Energy in Atwood’s Machine
M
r T
2

T1

T1 m1 T2 m2 m 1g m 2g

H

Figure 73: Atwood’s Machine, but this time with a massive pulley of mass M and radius R (and moment of inertia I = βM R2 ), this time solving a “standard” conservation of mechanical energy problem. We would like to find the speed v of m1 and m2 (and the angular speed ω of mass M ) when mass m2 > m1 falls a height H, beginning from rest, when the massless unstretchable string connecting the masses rolls without slipping on the massive pulley. We could do this problem using a from the solution to the example above, finding the time t it takes to reach H, and backsubstituting to find v, but by now we know quite well that it is a lot easier to use energy conservation (since no non-conservative forces act if the string does not slip) which is already time-independent.
Figure ?? shows the geometry of the problem. Note well that mass m1 will go up a distance H at the same time m2 goes down a distance H.
Again our solution strategy is almost identical to that of the conservation of mechanical energy problems of two weeks ago. We simply evaluate the initial and final total mechanical energy including the kinetic energy of the pulley and using the rolling constraint and solve for v.
We can choose the zero of potential energy for the two mass separately, and choose to start m2 a height H above its final position, and we start mass m1 at zero potential. The final potential energy of m2 will thus be zero and the final potential energy of m1 will be m1 gH. Also, we will need to substitute the rolling constraint into the expression for the rotational kinetic energy of the pulley in the little patch of algebra below: v ω=
(529)
r
Thus:
Ei
Ef
m2 gH

= m2 gH
1
= m1 gH + m1 v 2 +
2
1
= m1 gH + m1 v 2 +
2

1 m2 v 2 +
2
1 m2 v 2 +
2

1 βM R2 ω 2
2
1 βM R2 ω 2
2

or

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Week 5: Torque and Rotation in One Dimension and now we substitute the rolling constraint:
(m2 − m1 )gH

=

(m2 − m1 )gH

=

1 v2 1
(m1 + m2 )v 2 + βM R2 2
2
2
R
1
2
(m1 + m2 + βM )v
2

(530)

to arrive at
2(m2 − m1 )gH m1 + m2 + βM

v=

(531)

You can do this! It isn’t really that difficult (or that different from what you’ve done before).
Note well that the pulley behaves like an extra mass βM in the system – all of this mass has to be accelerated by the actual force difference between the two masses. If β = 1 – a ring of mass
– then all of the mass of the pulley ends up moving at v and all of its mass counts. However, for a disk or ball or actual pulley, β < 1 because some of the rotating pulley’s mass is moving more slowly than v and has less kinetic energy when the pulley is rolling.
Also note well that the strings do no net work in the system. They are internal forces, with T2 doing negative work on m2 but equal and opposite positive work on M , with T1 doing negative work on M , but doing equal and opposite positive work on m1 . Ultimately, the tensions in the string serve only to transfer energy between the masses and the pulley so that the change in potential energy is correctly shared by all of the masses when the string rolls without slipping.

Example 5.7.2: Unrolling Spool

T
M

R

Mg

H

Figure 74: A spool of fishing line is tied to a pole and released from rest to fall a height H, unrolling as it falls.
In figure 74 a spool of fishing line that has a total mass M and a radius R and is effectively a disk is tied to a pole and released from rest to fall a height H. Let’s find everything: the acceleration of the spool, the tension T in the fishing line, the speed with which it reaches H.
We start by writing Newton’s Second Law for both the translational and rotational motion. We’ll make down y-positive. Why not! First the force:
Fy = M g − T = M a

(532)

and then the torque: τ = RT = Iα =

1
M R2
2

a
R

(533)

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Week 5: Torque and Rotation in One Dimension

We use the rolling constraint (as shown) to rewrite the second equation, and divide both sides by R. Writing the first and second equation together:
Mg − T

= Ma
1
=
Ma
2

T we add them:

(536)

2 g 3

(537)

1
Mg
3

(538)

and solve for a: a= T =

(535)

3
Ma
2

Mg =

We back substitute to find T :

(534)

Next, we tackle the energy conservation problem. I’ll do it really fast and easy:
Ei = M gH =

1
1
M v2 +
2
2

or
M gH =

1
M R2
2

v
R

2

= Ef

(539)

3
M v2
4

(540)

4gH
3

(541)

and v= Example 5.7.3: A Rolling Ball Loops-the-Loop m,r H

R

Figure 75: A ball of mass m and radius r rolls without slipping to loop the loop on the circular track of radius R.
Let’s redo the “Loop-the-Loop” problem, but this time let’s consider a solid ball of mass m and radius r going around the track of radius R. This is a tricky problem to do precisely because as the normal force decreases (as the ball goes around the track) at some point the static frictional force required to “keep the ball rolling” on the track may well become greater than µs N , at which point the ball will slip. Slipping dissipates energy, so one would have to raise the ball slightly at the

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Week 5: Torque and Rotation in One Dimension

beginning to accommodate this. Of course, raising the ball slightly at the beginning also increases
N so maybe it doesn’t ever slip. So the best we can solve for is the minimum height Hmin it must have to roll without slipping assuming that it doesn’t ever actually slip, and “reality” is probably a bit higher to accommodate or prevent slipping, overcome drag forces, and so on.
With that said, the problem’s solution is exactly the same as before except that in the energy conservation step one has to use:
1 2
1
2 mgHmin = 2mgR + mvmin + Iωmin
2
2

(542)

plus the rolling constraint ωmin = vmin /r to get: mgHmin =
=

1
1
2
2
2mgR + mvmin + mvmin
2
5
7
2
2mgR + mvmin
10

(543)

Combine this with the usual:

2 mvmin R so that the ball “barely” loops the loop and you get:

mg =

(544)

Hmin = 2.7R

(545)

only a tiny bit higher than needed for a block sliding on a frictionless track.
Really, not all that difficult, right? All it takes is some practice, both redoing these examples on your own and doing the homework and it will all make sense.

5.8: The Parallel Axis Theorem
As we have seen, the moment of inertia of an object or collection of point-like objects is just
2
mi ri

I=

(546)

i

where ri is the distance between the axis of rotation and the point mass mi in a rigid system, or
I=

r2 dm

(547)

where r is the distance from the axis of rotation to “point mass” dm in the rigid object composed of continuously distributed mass.
However, in the previous section, we saw that the kinetic energy of a rigid object relative to an arbitrary origin can be written as the sum of the kinetic energy of the object itself treated as a total mass located at the (moving) center of mass plus the kinetic energy of the object in the moving center of mass reference frame.
For the particular case where a rigid object rotates uniformly around an axis that is parallel to an axis through the center of mass of the object, that is, in such a way that the angular velocity of the center of mass equals the angular velocity around the center of mass we can derive a theorem, called the Parallel Axis Theorem, that can greatly simplify problem solving while embodying the previous result for the kinetic energies. Let’s see how.
Suppose we want to find the moment of inertia of the arbitrary “blob shaped” rigid mass distribution pictured above in figure 76 about the axis labelled “New (Parallel) Axis”. This is, by

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Week 5: Torque and Rotation in One Dimension

dm r’ CM Axis

r rcm New (Parallel) Axis
Figure 76: An arbitrary blob of total mass M rotates around the axis at the origin as shown. Note well the geometry of r cm , r ′ , and r = r cm + r ′ . definition (and using the fact that r = r cm + r ′ from the triangle of vectors shown in the figure):
I

=

r2 dm

=

r cm + r ′ · r cm + r ′

dm

=

2 rcm + r′2 + 2r cm · r ′

dm

=

2 rcm dm +

2
= rcm

dm +

r′2 dm + 2r cm · r′2 dm + 2M r cm ·

r ′ dm
1
M

r ′ dm

2
= M rcm + Icm + 2M r cm · (0)

(548)

or
2
I = Icm + M rcm

(549)

In case that was a little fast for you, here’s what I did. I substituted r cm + r ′ for r. I distributed out that product. I used the linearity of integration to write the integral of the sum as the sum of the integrals (all integrals over all of the mass of the rigid object, of course). I noted that r cm is a constant and pulled it out of the integral, leaving me with the integral M = dm. I noted that r′2 dm is just Icm , the moment of intertia of the object about an axis through its center of mass. I noted that (1/M ) r ′ dm is the position of the center of mass in center of mass coordinates, which is zero – by definition the center of mass is at the origin of the center of mass frame.
The result, in words, is that the moment of inertia of an object that uniformly rotates around any axis is the moment of inertia of the object about an axis parallel to that axis through the center of mass of the object plus the moment of inertia of the total mass of the object treated as a point mass located at the center of mass as it revolves!
This sounds a lot like the kinetic energy theorem; let’s see how the two are related.
As long as the object rotates uniformly – that is, the object goes around its own center one time for every time it goes around the axis of rotation, keeping the same side pointing in towards the center as it goes – then its kinetic energy is just:
K=

1 2
1
1
2
Iω = (M rcm )ω 2 + Icm ω 2
2
2
2

(550)

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Week 5: Torque and Rotation in One Dimension
A bit of algebraic legerdemain:
K=

1
2
(M rcm )
2

vcm rcm 2

1
+ Icm ω 2 = K(ofcm) + K(incm)
2

(551)

as before!
Warning! This will not work if an object is revolving many times around its own center of mass for each time it revolves around the parallel axis.

Example 5.8.1: Moon Around Earth, Earth Around Sun
This is a conceptual example, not really algebraic. You may have observed that the Moon always keeps the same face towards the Earth – it is said to be “gravitationally locked” by tidal forces so that this is true. This means that the Moon revolves once on its axis in exactly the same amount of time that the Moon itself revolves around the Earth. We could therefore compute the total angular kinetic energy of the Moon by assuming that it is a solid ball of mass M , radius r, in an orbit around the Earth of radius R, and a period of 28.5 days:
2
Imoon = M R2 + M r2
5

(552)

(from the parallel axis theorem),


T
(you’ll need to find T in seconds, 86400 × 28.5) and then: ω= K=

1
Imoon ω 2
2

(553)

(554)

All that’s left is the arithmetic.
Contrast this with the Earth rotating around the Sun. It revolves on its own axis 365.25 times during the period in which it revolves around the Sun. To find it’s kinetic energy we could not use the parallel axis theorem, but we can still use the theorem at the end of the previous chapter. Here we would find two different angular velocities: ωday =


Tday

(555)

ωyear =


Tyear

(556)

and

(again, 1 day = 86400 seconds is a good number to remember). Then if we let M be the mass of the Earth, r be its radius, and R be the radius of its orbit around the Sun (all numbers that are readily available on Wikipedia112 we could find the total kinetic energy (relative to the Sun) as:
K=

1
1
2
M R2 ωyear +
2
2

2
2
M r2 ωday
5

(557)

which is somewhat more complicated, no?
Let’s do a more readily evaluable example:

Example 5.8.2: Moment of Inertia of a Hoop Pivoted on One Side
In figure ?? a hoop of mass M and radius R is pivoted at a point on the side, on the hoop itself, not in the middle. We already know the moment of inertia of the hoop about its center of mass. What
112 Wikipedia:

http://www.wikipedia.org/wiki/Earth.

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Week 5: Torque and Rotation in One Dimension

pivot
M
R

Figure 77: A hoop of mass M and radius R is pivoted on the side – think of it as being hung on a nail from a barn door. is the moment of inertia of the hoop about this new axis parallel to the one through the center of mass that we used before?
It’s so simple:
Iside pivot = M R2 + Icm = M R2 + M R2 = 2M R2

(558)

and we’re done!
For your homework, you get to evaluate the moment of inertia of a rod about an axis through its center of mass and about one end of the rod and compare the two, both using direct integration and using the parallel axis theorem. Good luck!

5.9: Perpendicular Axis Theorem
In the last section we saw how a bit of geometry and math allowed us to prove a very useful theorem
– useful because we can now learn a short table of moments of inertia about a given axis through the center of mass and then easily extend them to find the moments of inertia of these same shapes when they uniformly rotate around a parallel axis.
In this section we will similarly derive a theorem that is very useful for relating moments of inertia of planar distributions of mass (only) around axes that are perpendicular to one another
– the Perpendicular Axis Theorem. Here’s how it goes.

y dm M y r x x

z

Figure 78: A planar blob of mass and the geometry needed to prove the Perpendicular Axis Theorem

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Week 5: Torque and Rotation in One Dimension

Suppose that we wish to evaluate Ix , the moment of inertia of the plane mass distribution M shown in figure 78. That’s quite easy:
Ix =

y 2 dm

(559)

Iy =

x2 dm

(560)

Similarly,

We add them, and presto chango!
Ix + Iy =

y 2 dm +

x2 dm =

r2 dm = Iz

(561)

This is it, the Perpendicular Axis Theorem:
Iz = Ix + Iy

(562)

I’ll give a single example of its use. Let’s find the moment of inertia of a hoop about an axis through the center in the plane of the loop!

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Week 5: Torque and Rotation in One Dimension

Example 5.9.1: Moment of Inertia of Hoop for Planar Axis y M

R

x

z

Figure 79: A hoop of mass M and radius R is drawn. What is the moment of inertia about the x-axis? This one is really very, very easy. We use the Perpendicular Axis theorem backwards to get the answer. In this case we know Iz = M R2 , and want to find Ix . We observe that from symmetry,
Ix = Iy so that:
Iz = M R2 = Ix + Iy = 2Ix
(563)
or
Ix =

1
1
Iz = M R2
2
2

(564)

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Week 5: Torque and Rotation in One Dimension

Homework for Week 5

Problem 1.

Physics Concepts: Make this week’s physics concepts summary as you work all of the problems in this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s) they were key to, and include concepts from previous weeks as necessary. Do the work carefully enough that you can (after it has been handed in and graded) punch it and add it to a three ring binder for review and study come finals!

Problem 2.
This problem will help you learn required concepts such as:
• Definition/Evaluation of Moment of Inertia
• Parallel Axis Theorem so please review them before you begin.
a) Evaluate the moment of inertia of a uniform rod of mass M and length L about its center of mass by direct integration.

pivot y −L/2

L/2

x

b) Evaluate the moment of inertia of a uniform rod of mass M and length L about one end by direct integration. Also evaluate it using the parallel axis theorem and the result you just obtained. Which is easier?

pivot y L

x

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Week 5: Torque and Rotation in One Dimension

Problem 3.
This problem will help you learn required concepts such as:
• Definition/Evaluation of Moment of Inertia
• Parallel Axis Theorem so please review them before you begin.
a) Evaluate the moment of inertia of a uniform disk of mass M and radius R about its axis of symmetry by direct integration (this can be set up as a “one dimensional integral” and hence is not too difficult).

pivot

c)

b) Evaluate the moment of inertia of this disk around a pivot at the edge of the disk (for example, a thin nail stuck through a hole at the outer edge) using the parallel axis theorem. Would you care to do the actual integral to find the moment of inertia of the disk in this case?

pivot

d)

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Week 5: Torque and Rotation in One Dimension
Problem 4.

M,R
H
θ

This problem will help you learn required concepts such as:
• Newton’s Second Law for Translation and Rotation
• Moment of Inertia
• Conservation of Mechanical Energy
• Static Friction so please review them before you begin.
A round object with mass m and radius R is released from rest to roll without slipping down an inclined plane of height H at angle θ relative to horizontal. The object has a moment of inertia
2
I = βmR2 (where β is a dimensionless number such as 1 or 5 , that might describe a disk or a solid
2
ball, respectively).
a) Begin by relating v (the speed of the center of mass) to the angular velocity (for the rolling object). You will use this (and the related two equations for s and θ and a and α) repeatedly in rolling problems.
b) Using Newton’s second law in both its linear and rotational form plus the rolling constraint, show that the acceleration of the object is: a= g sin(θ)
1+β

c) Using conservation of mechanical energy, show that it arrives at the bottom of the incline with a velocity:
2gH
v=
1+β
d) Show that the condition for the greatest angle for which the object will roll without slipping is that: 1 tan(θ) ≤ (1 + )µs β where µs is the coefficient of static friction between the object and the incline.

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Week 5: Torque and Rotation in One Dimension

Problem 5.

m,R

rough
H
icy

H’

This problem will help you learn required concepts such as:
• Conservation of Mechanical Energy
• Rotational Kinetic Energy
• Rolling Constraint. so please review them before you begin.
A disk of mass m and radius R rolls without slipping down a rough slope of height H onto an icy (frictionless) track at the bottom that leads up a second icy (frictionless) hill as shown.
a) How fast is the disk moving at the bottom of the first incline? How fast is it rotating (what is its angular velocity)?
b) Does the disk’s angular velocity change as it leaves the rough track and moves onto the ice (in the middle of the flat stretch in between the hills)?
c) How far up the second hill (vertically, find H ′ ) does the disk go before it stops rising?

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Week 5: Torque and Rotation in One Dimension
Problem 6.

I (about cm)

mass m

R r F

F

F

This problem will help you learn required concepts such as:
• Direction of torque
• Rolling Constraint so please review them before you begin.
In the figure above, a spool of mass m is wrapped with string around the inner spool. The spool is placed on a rough surface (with coefficient of friction µs = 0.5) and the string is pulled with force
F ≪ 0.5 mg in the three directions shown. The spool, if it rolls at all, rolls without slipping.
(Note that if pulled too hard, the spool can both slip and/or roll.)
a) For each picture, indicate the direction that the spool will accelerate.
b) For each picture, indicate the direction that static friction will point.
c) For each picture, find the magnitude of the force exerted by static friction and the magnitude of the acceleration of the spool. To do this, you will need to use symbols to indicate the radius of the inner spool r, the radius of the outer spool R, and the moment of inertia of the spool about its center of mass, Icm = βmR2 . Use β ≈ 1/3 (reasonable for an arrangement like this) to see if your answer for a “makes sense”.
Note that you do not need to know r, R or Icm to answer the first two questions beyond what you can directly see in the geometry of of the problem. Note also that you can use either the center of mass or the point of contact with the ground (with the parallel axis theorem) as a pivot, the latter being slightly easier.

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Week 5: Torque and Rotation in One Dimension

Problem 7.

R
M

m1

m2

H

This problem will help you learn required concepts such as:
• Newton’s Second Law
• Newton’s Second Law for Rotating Systems (torque and angular acceleration)
• Moments of Inertia
• The Rolling Constraint
• Conservation of Mechanical Energy so please review them before you begin.
In the figure above Atwood’s machine is drawn – two masses m1 and m2 hanging over a massive pulley which you can model as a disk of mass M and radius R, connected by a massless unstretchable string. The string rolls on the pulley without slipping.
a) Draw three free body diagrams (isolated diagrams for each object showing just the forces acting on that object) for the three masses in the figure above.
b) Convert each free body diagram into a statement of Newton’s Second Law (linear or rotational) for that object.
c) Using the rolling constraint (that the pulley rolls without slipping as the masses move up or down) find the acceleration of the system and the tensions in the string on both sides of the pulley in terms of m1 , m2 , M , g, and R.
d) Suppose mass m2 > m1 and the system is released from rest with the masses at equal heights.
When mass m2 has descended a distance H, use conservation of mechanical energy to find velocity of each mass and the angular velocity of the pulley.

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Week 5: Torque and Rotation in One Dimension
Problem 8.

pivot θ R

r
M

This problem will help you learn required concepts such as:
• Newton’s Second Law for rotating objects
• Moment of Inertia
• Parallel Axis Theorem
• Work in rotating systems
• Rotational Kinetic Energy
• Kinetic Energy in Lab versus CM so please review them before you begin.
In the picture above, a physical pendulum is constructed by hanging a disk of mass M and radius r on the end of a massless rigid rod in such a way that the center of mass of the disk is a distance
R away from the pivot and so that the whole disk pivots with the rod. The pendulum is pulled to an initial angle θ0 (relative to vertically down) and then released.
a) Find the torque about the pivot exerted on the pendulum by gravity at an arbitrary angle θ.
b) Integrate the torque from θ = θ0 to θ = 0 to find the total work done by the gravitational torque as the pendulum disk falls to its lowest point. Note that your answer should be M gR(1− cos(θ0 )) = M gH where H is the initial height above this lowest point.
c) Find the moment of inertia of the pendulum about the pivot (using the parallel axis theorem).
1
d) Set the work you evaluated in b) equal to the rotational kinetic energy of the disk 2 Iω 2 using dθ the moment of inertia you found in c). Solve for ω = dt when the disk is at its lowest point.

e) Show that this kinetic energy is equal to the kinetic energy of the moving center of mass of the
1
disk 2 M v 2 plus the kinetic energy of the disk’s rotation about its own center of mass, 1 Icm ω 2 ,
2
at the lowest point.
As an optional additional step, write Newton’s Second Law in rotational coordinates τ = Iα
(using the values for the magnitude of the torque and moment of inertia you determined above) and solve for the angular acceleration as a function of the angle θ. In a few weeks we will learn to solve this equation of motion for small angle oscillations, so it is good to practice obtaining it from the basic physics now.

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Week 5: Torque and Rotation in One Dimension

Problem 9.

m,r

H

R

This problem will help you learn required concepts such as:
• Newton’s Second Law
• Moments of Inertia
• Rotational Kinetic Energy
• The Rolling Constraint
• Conservation of Mechanical Energy
• Centripetal Acceleration
• Static Friction so please review them before you begin.
A solid ball of mass M and radius r sits at rest at the top of a hill of height H leading to a circular loop-the-loop. The center of mass of the ball will move in a circle of radius R if it goes
2
around the loop. Recall that the moment of inertia of a solid ball is Iball = 5 M R2 .
a) Find the minimum height Hmin for which the ball barely goes around the loop staying on the track at the top, assuming that it rolls without slipping the entire time independent of the normal force.
b) How does your answer relate to the minimum height for the earlier homework problem where it was a block that slid around a frictionless track? Does this answer make sense? If it is higher, where did the extra potential energy go? If it is lower, where did the extra kinetic energy come from?
c) Discussion question for recitation: The assumption that the ball will roll around the track without slipping if released from this estimated minimum height is not a good one. Why not?

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Week 5: Torque and Rotation in One Dimension
Advanced Problem 10.

m,R

F

M

θ

M

F

This problem will help you learn required concepts such as:
• Newton’s Second Law (linear and rotational)
• Rolling Constraint
• Static and Kinetic Friction
• Changing Coordinate Frames so please review them before you begin.
A disk of mass m is resting on a slab of mass M , which in turn is resting on a frictionless table. The coefficients of static and kinetic friction between the disk and the slab are µs and µk , respectively. A small force F to the right is applied to the slab as shown, then gradually increased.
a) When F is small, the slab will accelerate to the right and the disk will roll on the slab without slipping. Find the acceleration of the slab, the acceleration of the disk, and the angular acceleration of the disk as this happens, in terms of m, M , R, and the magnitude of the force
F.
b) Find the maximum force Fmax such that it rolls without slipping.
c) If F is greater than this, solve once again for the acceleration and angular acceleration of the disk and the acceleration of the slab.
Hint: The hardest single thing about this problem isn’t the physics (which is really pretty straightfoward). It is visualizing the coordinates as the center of mass of the disk moves with a different acceleration as the slab. I have drawn two figures above to help you with this – the lower figure represents a possible position of the disk after the slab has moved some distance to the right and the disk has rolled back (relative to the slab! It has moved forward relative to the ground!
Why?) without slipping. Note the dashed radius to help you see the angle through which it has rolled and the various dashed lines to help you relate the distance the slab has moved xs , the distance the center of the disk has moved xd , and the angle through which it has rolled θ. Use this relation to connect the acceleration of the slab to the acceleration and angular acceleration of the disk.
If you can do this one, good job!

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Week 5: Torque and Rotation in One Dimension

Optional Problems
The following problems are not required or to be handed in, but are provided to give you some extra things to work on or test yourself with after mastering the required problems and concepts above and to prepare for quizzes and exams.

Optional Problem 11.

F
R
r

A cable spool of mass M , radius R and moment of inertia I = βM R2 is wrapped around its
OUTER disk with fishing line and set on a rough rope as shown. The inner spool has a radius r.
The fishing line is then pulled with a force F to the right so that it rolls down the rope without slipping. a) Find the magnitude and direction of the acceleration of the spool.
b) Find the force (magnitude and direction) exerted by the friction of the rope on the spool.
c) For one particular value of r, the frictional force is zero!. Find that value. For larger values, which way does friction point? For smaller values, which way does friction point? At this value, if the rope were not there would the motion be any different?
In analyzing the “walking the spool” problem in class and in the text above, students often ask how they can predict which direction that static friction acts on a rolling spool, and I reply that they can’t! I can’t, not always, because in this problem it can point either way and which way it ultimately points depends on the details of R, r and β! The best you can do is make a reasonable guess as to the direction and let the algebra speak – if your answer comes out negative, friction points the other way.

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Week 5: Torque and Rotation in One Dimension
Optional Problem 12.

M

M
R

R

M

M

This problem will help you learn required concepts such as:
• Moment of Inertia
• Newton’s Second Law (translational and rotational)
• Conservation of Mechanical Energy so please review them before you begin.
A block of mass M sits on a smooth (frictionless) table. A mass M is suspended from an Acme
(massless, unstretchable, unbreakable) rope that is looped around the two pulleys, each with the same mass M and radius R as shown and attached to the support of the rightmost pulley. The
1
two pulleys each have the moment of inertia of a disk, I = 2 M R2 and the rope rolls on the pulleys without slipping. At time t = 0 the system is released at rest.
a) Draw free body diagrams for each of the four masses. Don’t forget the forces exerted by the bar that attaches the pulley to the mass on the left or the pulley on the right to the table!
Note that some of these forces will cancel due to the constraint that the center of mass of certain masses moves only in one direction or does not move at all. Note also that the torque exerted by the weight of the pulley around the near corder of the mass on the table is not not enough to tip it over.
b) Write the relevant form(s) of Newton’s Second Law for each mass, translational and rotational as needed, separately, wherever the forces in some direction do not cancel. Also write the constraint equations that relate the accelerations (linear and angular) of the masses and pulleys.
c) Find the acceleration of the hanging mass M (only) in terms of the givens. Note that you can’t quite just add all of the equations you get after turning the torque equations into force equations, but if you solve the equations simultaneously, systematically eliminating all internal forces and tensions, you’ll get a simple enough answer.
d) Find the speed of the hanging mass M after it has fallen a height H, using conservation of total mechanical energy. Show that it is consistent with the “usual” constant-acceleration-from-rest

answer v = 2aH for the acceleration found in c).

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Week 6: Vector Torque and Angular Momentum

Optional Problem 13.

dx x0 pivot x L
This problem will help you learn required concepts such as:
• Finding the Center of Mass using Integration
• Finding the Moment of Inertia using Integration so please review them before you begin.

A simple model for the one-dimensional mass distribution of a human leg of length L and mass
M is: λ(x) = C · (L + x0 − x)
Note that this quantity is maximum at x = 0, varies linearly with x, and vanishes smoothly at x = L + x0 . That means that it doesn’t reach λ = 0 when x = L, just as the mass per unit length of your leg doesn’t reach zero at your ankles.
a) Find the constant C in terms of M , L, and x0 by evaluating:
L

λ(x) dx

M=
0

and solving for C.
b) Find the center of mass of the leg (as a distance down the leg from the hip/pivot at the origin).
You may leave your answer in terms of C (now that you know it) or you can express it in terms of L and x0 only as you prefer.
c) Find the moment of inertia of the leg about the hip/pivot at the origin. Again, you may leave it in terms of C if you wish or express it in terms of M , L and x0 . Do your answers all have the right units?
d) How might one improve the estimate of the moment of inertia to take into account the foot
(as a lump of “extra mass” mf out there at x = L that doesn’t quite fit our linear model)?
This is, as you can see, something that an orthopedic specialist might well need to actually do with a much better model in order to e.g. outfit a patient with an artificial hip. True, they might use a computer to do the actual computations required, but is it plausible that they could possibly do what they need to do without knowing the physics involved in some detail?

Week 6: Vector Torque and
Angular Momentum
Summary
• The vector torque acting on a point particle or rigid body is: τ =r×F where r is the vector from the pivot point (not axis!) to the point where the force is applied.
• The vector angular momentum of a point particle is:
L = r × p = m(r × v) where as before, r is a vector from the pivot point to the location of the particle and v is the particle’s velocity.
• The vector form for Newton’s Second Law for Rotation for a point particle is: τ =

dL dt • All of these relations generalize when computing the total vector torque acting on a collection of particles (that may or may not form a rigid body) with a total angular momentum.
Provided that all the internal forces F ij = −F ji act along the lines r ij connecting the particles, there is no net torque due to the internal forces between particles and we get the series of results: τ ext i τ tot = i Ltot = i r i × pi

and τ tot =

dLtot dt • The Law of Conservation of Angular Momentum is:
If (and only if ) the total torque acting on a system is zero, then the total angular momentum of the system is a constant vector (conserved). or in equationspeak:
If (and only if ) τ tot = 0, then Ltot is a constant vector.
277

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Week 6: Vector Torque and Angular Momentum
• For rigid objects (or collections of point particles) that have mirror symmetry across the axis of rotation and/or mirror symmetry across the plane of rotation, the vector angular momentum can be written in terms of the scalar moment of inertia about the axis of rotation (defined and used in week 5) and the vector angular velocity as:
L = Iω
• For rigid objects or collections of point particles that lack this symmetry with respect to an axis of rotation (direction of ω)
L = Iω for any scalar I. In general, L precesses around the axis of rotation in these cases and requires a constantly varying nonzero torque to drive the precession.
• When two (or more) isolated objects collide, both momentum and angular momentum is conserved. Angular momentum conservation becomes an additional equation (set) that can be used in analyzing the collision.
• If one of the objects is pivoted, then angular momentum about this pivot is conserved but in general momentum is not conserved as the pivot itself will convey a significant impulse to the system during the collision.
• Radial forces – any force that can be written as F = Fr r – exert no torque on the masses that they act on. Those object generally move in not-necessarily-circular orbits with constant angular momentum.
• When a rapidly spinning symmetric rotator is acted on by a torque of constant magnitude that is (always) perpendicular to the plane formed by the angular momentum and a vector in second direction, the angular momentum vector precesses around the second vector. In particular, for a spinning top with angular momentum L tipped at an angle θ to the vertical, the magnitude of the torque exerted by gravity and the normal force on the top is: τ = |D × mg z | = mgD sin(θ) =
ˆ
or ωp =

dL
= L sin(θ)ωp dt mgD
L

In this expression, ωp is the angular precession frequency of the top and D is the vector from the point where the tip of the top rests on the ground to the center of mass of the top.
The direction of precession is determined by the right hand rule.

6.1: Vector Torque
In the previous chapter/week we saw that we could describe rigid bodies rotating about a single axis quite accurately by means of a modified version of Newton’s Second Law: τ = rF F sin(φ) = |r F × F | = Iα

(565)

where I is the moment of inertia of the rigid body, evaluated by summing/integrating:
2
mi ri =

I=

r2 dm

(566)

i

In the torque expression r F is a was a vector in the plane perpendicular to the axis of rotation leading from the axis of rotation to the point where the force was applied. r in the moment of inertia

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Week 6: Vector Torque and Angular Momentum

I was similarly the distance from the axis of rotation of the particular mass m or mass chunk dm.
We considered this to be one dimensional rotation because the axis of rotation did not change, all rotation was about that one fixed axis.
This is, alas, not terribly general. We started to see that at the end when we talked about the parallel and perpendicular axis theorem and the possibility of several “moments of inertia” for a single rigid object around different rotation axes. However, it is really even worse than that. Torque
(as we shall see) is a vector quantity, and it acts to change another vector quantity, the angular momentum of not just a rigid object, but an arbitrary collection of particles, much as force did when we considered the center of mass. This will have a profound effect on our understanding of certain kinds of phenomena. Let’s get started.
We have already identified the axis of rotation as being a suitable “direction” for a one-dimensional torque, and have adopted the right-hand-rule as a means of selecting which of the two directions along the axis will be considered “positive” by convention. We therefore begin by simply generalizing this rule to three dimensions and writing: τ =r×F

(567)

where (recall) r × F is the cross product of the two vectors and where r is the vector from the origin of coordinates or pivot point, not the axis of rotation to the point where the force
F is being applied.
Time for some vector magic! Let’s write F = dp/dt or113 : τ =r×F =r×

dp d dr d =
(r × mv) −
× mv = r × p dt dt dt dt

(568)

The last term vanishes because v = dr/dt and v × mv = 0 for any value of the mass m.
Recalling that F = dp/dt is Newton’s Second Law for vector translations, let us define:
L=r×p

(569)

as the angular momentum vector of a particle of mass m and momentum p located at a vector position r with respect to the origin of coordinates.
In that case Newton’s Second Law for a point mass being rotated by a vector torque is: τ =

dL dt (570)

which precisely resembles Newtson’s Second Law for a point mass being translated by a vector torque. This is good for a single particle, but what if there are many particles? In that case we have to recapitulate our work at the beginning of the center of mass chapter/week.

6.2: Total Torque
In figure 80 a small collection of (three) particles is shown, each with both “external” forces F i and
“internal” forces F ij portrayed. The forces and particles do not necessarily live in a plane – we simply cannot see their z-components. Also, this picture is just enough to help us visualize, but be thinking 3, 4...N as we proceed.
113 I’m

using

d r dt

dp
× p = dr × p + r × dt to get this, and subtracting the first term over to the other side. dt 280

Week 6: Vector Torque and Angular Momentum

F
2

F
1
F
12

m1 r1 F31

r2

F
13

F
31
r3

m2
F
23

F
32
m3
F3

Figure 80: The coordinates of a small collection of particles, just enough to illustrate how internal torques work out.
Let us write τ = dL/dt for each particle and sum the whole thing up, much as we did for
F = dp/dt in chapter/week 4:

 d dLtot τ tot = r i × F i +
F ij  = r i × pi = dt i dt i j=i 

d r i × F ij  = ri × F i +  r i × pi
(571)
dt i i i j=i Consider the term that sums the internal torques, the torques produced by the internal forces between the particles, for a particular pair (say, particles 1 and 2) and use good old N3, F 21 = −F 12 : r 1 × F 12 + r 2 × F 21 = r 1 × F 12 − r 2 × F 12 = (r 1 − r 2 ) × F 12 = 0

(!)

(572)

because r 12 = r 1 − r 2 is parallel or antiparallel to F 12 and the cross product of two vectors that are parallel or antiparallel is zero.
Obviously, the same algebra holds for any internal force pair so that:




and τ tot = i i

j=i

r i × F ij  = 0

ri × F i =

d dt i

r i × pi =

dLtot dt (573)

(574)

where τ tot is the sum of only the external torques – the internal torques cancel.
The physical meaning of this cancellation of internal torques is simple – just as you cannot lift yourself up by your own bootstraps, because internal opposing forces acting along the lines connecting particles can never alter the velocity of the center of mass or the total momentum of the system, you cannot exert a torque on yourself and alter your own total angular momentum – only the total external torque acting on a system can alter its total angular momentum.
Wait, what’s that? An isolated system (one with no net force or torque acting) must have a constant angular momentum? Sounds like a conservation law to me...

6.2.1: The Law of Conservation of Angular Momentum
We’ve basically done everything but write this down above, so let’s state it clearly in both words and algebraic notation. First in words:

Week 6: Vector Torque and Angular Momentum

281

If and only if the total vector torque acting on any system of particles is zero, then the total angular momentum of the system is a constant vector.
In equations it is even more succinct:
If and only if τ tot = 0 then Ltot = Linitial = Lfinal = a constant vector

(575)

Note that (like the Law of Conservation of Momentum) this is a conditional law – angular momentum is conserved if and only if the net torque acting on a system is zero (so if angular amomentum is conserved, you may conclude that the total torque is zero as that is the only way it could come about).
Just as was the case for Conservation of Momentum, our primary use at this point for Conservation of Angular Momentum will be to help analyze collisions. Clearly the internal forces in two-body collisions in the impulse approximation (which allows us to ignore the torques exerted by external forces during the tiny time ∆t of the impact) can exert no net torque, therefore we expect both linear momentum and angular momentum to be conserved during a collision.
Before we proceed to analyze collisions, however, we need to understand angular momentum
(the conserved quantity) in more detail, because it, like momentum, is a very important quantity in nature. In part this is because many elementary particles (such as quarks, electrons, heavy vector bosons) and many microscopic composite particles (such as protons and neutrons, atomic nuclei, atoms, and even molecules) can have a net intrinsic angular momentum, called spin 114 .
This spin angular momentum is not classical and does not arise from the physical motion of mass in some kind of path around an axis – and hence is largely beyond the scope of this class, but we certainly need to know how to evaluate and alter (via a torque) the angular momentum of macroscopic objects and collections of particles as they rotate about fixed axes.

6.3: The Angular Momentum of a Symmetric Rotating Rigid
Object
One very important aspect of both vector torque and vector angular momentum is that r in the definition of both is measured from a pivot that is a single point, not measured from a pivot axis as we imagined it to be last week when considering only one dimensional rotations.
We would very much like to see how the two general descriptions of rotation are related, though, especially as at this point we should intuitively feel (given the strong correspondance between onedimensional linear motion equations and one-dimensional angular motion equations) that something like Lz = Iωz ought to hold to relate angular momentum to the moment of inertia. Our intuition is mostly correct, as it turns out, but things are a little more complicated than that.
From the derivation and definitions above, we expect angular momentum L to have three components just like a spatial vector. We also expect ω to be a vector (that points in the direction of the right-handed axis of rotation that passes through the pivot point). We expect there to be a linear relationship between angular velocity and angular momentum. Finally, based on our observation of an extremely consistent analogy between quantities in one dimensional linear motion and one dimensional rotation, we expect the moment of inertia to be a quantity that transforms the angular velocity into the angular momentum by some sort of multiplication.
To work out all of these relationships, we need to start by indexing the particular axes in the coordinate system we are considering with e.g. a = x, y, z and label things like the components of
114 Wikipedia: http://www.wikipedia.org/wiki/Spin (physics). Physics majors should probably take a peek at this link, as well as chem majors who plan to or are taking physical chemistry. I foresee the learning of Quantum Theory in Your Futures, and believe me, you want to preload your neocortex with lots of quantum cartoons and glances at the algebra of angular momentum in quantum theory ahead of time...

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Week 6: Vector Torque and Angular Momentum

L, ω and I with a. Then La=z is the z-component of L, ωa=x is the x-component of ω and so on.
This is simple enough.
It is not so simple, however, to generalize the moment of inertia to three dimensions. Our simple one-dimensional scalar moment of inertia from the last chapter clearly depends on the particular
2
axis of rotation chosen! For rotations around the (say) z-axis we needed to sum up I = i mi ri
2 + y 2 , and these components were clearly all different for a rotation
(for example) where ri = xi i around the x-axis or a z axis through a different pivot (perpendicular or parallel axis theorems).
These were still the easy cases – as we’ll see below, things get really complicated when we rotate even a symmetric object around an axis that is not an axis of symmetry of the object!
Indeed, what we have been evaluating thus far is more correctly called the scalar moment of inertia, the moment of inertia evaluated around a particular “obvious” one-dimensional axis of rotation where one or both of two symmetry conditions given below are satisfied. The moment of inertia of a general object in some coordinate system is more generally described by the moment of inertia tensor Iab . Treating the moment of inertia tensor correctly is beyond the scope of this course, but math, physics or engineering students are well advised to take a peek at the Wikipedia article on the moment of inertia115 to at least get a glimpse of the mathematically more elegant and correct version of what we are covering here.
Here are the two conditions and the result. Consider a particular pivot point at the origin of coordinates and right handed rotation around an axis in the ath direction of a coordinate frame with this origin. Let the plane of rotation be the plane perpendicular to this axis that contains the pivot/origin. There isn’t anything particularly mysterious about this – think of the a = z-axis being the axis of rotation, with positive in the right-handed direction of ω, and with the x-y plane being the plane of rotation.
In this coordinate frame, if the mass distribution has:
• Mirror symmetry across the axis of rotation and/or
• Mirror symmetry across the plane of rotation, we can write:
L = La = Iaa ωa = Iω

(576)

where ω = ωa a points in the (right handed) direction of the axis of rotation and where:
ˆ
2 mi ri

I = Iaa =

or

r2 dm

(577)

i

with ri or r the distance from the a-axis of rotation as usual.
Note that “mirror symmetry” just means that if there is a chunk of mass or point mass in the rigid object on one side of the axis or plane of rotation, there is an equal chunk of mass or point mass in the “mirror position” on the exact opposite of the line or plane, for every bit of mass that makes up the object. This will be illustrated in the next section below, along with why these rules are needed.
In other words, the scalar moments of inertia I we evaluated last chapter are just the diagonal parts of the moment of inertia tensor I = Iaa for the coordinate direction a corresponding to the axis of rotation. Since we aren’t going to do much – well, we aren’t going to do anything – with the non-diagonal parts of I in this course, from now on I will just write the scalar moment I where I
115 Wikipedia: http://www.wikipedia.org/wiki/Moment of inertia#Moment of inertia tensor. This is a link to the middle of the article and the tensor part, but even introductory students may find it useful to review the beginning of this article.

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Week 6: Vector Torque and Angular Momentum

really mean Iaa for some axis a such that at least one of the two conditions above are satisfied, but math/physics/engineering students, at least, should try to remember that it really ain’t so116 .
All of the (scalar) I’s we computed in the last chapter satisfied these symmetry conditions:
• A ring rotating about an axis through the center perpendicular to the plane of the ring has both symmetries. So does a disk.
• A rod rotating about one end in the plane perpendicular to ω has mirror symmetry in the plane but not mirror symmetry across the axis of rotation.
• A hollow or filled sphere have both symmetries.
• A disk around an axis off to the side (evaluated using the parallel axis theorem) has the plane symmetry. • A disk around an axis that lies in the plane of the disk with a pivot in the perpendicular plane through the center of the disk has at least the planar symmetry relative to the perpendicular plane of rotation. and so on. Nearly all of the problems we consider in this course will be sufficiently symmetric that we can use:
L = Iω
(578)
with the pivot and direction of the rotation and the symmetry of the object with respect to the axis and/or plane of rotation “understood”.
Let us take a quick tour, then, of the angular momentum we expect in these cases. A handful of examples should suffice, where I will try to indicate the correct direction as well as show the
“understood” scalar result.

Example 6.3.1: Angular Momentum of a Point Mass Moving in a Circle
For a point mass moving in a circle of radius r in the x-y plane, we have the planar symmetry. ω = ω z is in the z-direction, and I = Izz = mr2 . The angular momentum in this direction is:
ˆ
L = Lz = (r × p)z = mvr = mr2

v2
= Iω r2 (579)

The direction of this angular momentum is most easily found by using a variant of the right hand rule. Let the fingers of your right hand curl around the axis of rotation in the direction of the motion of the mass. Then your thumb points out the direction. You should verify that this gives the same result as using L = r × p, always, but this “grasp the axis” rule is much easier and faster to use, just grab the axis with your fingers curled in the direction of rotation and your thumb has got it.

Example 6.3.2: Angular Momentum of a Rod Swinging in a Circle
To compute the angular momentum of a rod rotating in a plane around a pivot through one end, we choose coordinates such that the rod is in the x-y plane, rotating around z, and has mass M
116 Just

FYI, in case you care: The correct rule for computing L from ω is
X
Iab ωb
La = b for a, b = x, y, z.

284

Week 6: Vector Torque and Angular Momentum

+y

M

v

l dm dr

dL z

+x

r

Figure 81: The geometry of a rod of mass M and length L, rotating around a pivot through the end in the x-y plane. and length l (note that it is now tricky to call its length L as that’s also the symbol for angular momentum, sigh). From the previous example, each little “point-like” bit of mass in the rod dm has an angular momentum of: dLz = |r × dp| = r(dm v) = r2 dm

v r = dI ω

(580)

so that if we integrate this as usual from 0 to l, we get:
|L| = Lz =

1
M l2 ω = Iω
3

(581)

Example 6.3.3: Angular Momentum of a Rotating Disk
Suppose a disk is rotating around its center of mass in the x-y plane of the disk. Then using exactly the same argument as before:
L = Lz =

r2 dmω = Iω =

1
M R2 ω
2

(582)

The disk is symmetric, so if we should be rotating it like a spinning coin or poker chip around
(say) the x axis, we can also find (using the perpendicular axis theorem to find Ix ):
L = Lx = Ix ω =

1
M R2 ω
4

(583)

and you begin to see why the direction labels are necessary. A disk has a different scalar moment of inertia about different axes through the same pivot point. Even when the symmetry is obvious, we may still need to label the result or risk confusing the previous two results!
We’re not done! If we attach the disk to a massless string and swing it around the z axis at a distance ℓ from the center of mass, we can use the parallel axis theorem and find that:
1
new
L = Lnew = Iz (M ℓ2 + M R2 )ω z 2

(584)

That’s three results for a single object, and of course we can apply the parallel axis theorem to the x-rotation or y-rotation as well! The L = Iω result works for all of these cases, but the direction of L and ω as well as the value of the scalar moment of inertia I used will vary from case to case, so you may want to carefully label things just to avoid making mistakes!

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Week 6: Vector Torque and Angular Momentum

Example 6.3.4: Angular Momentum of Rod Sweeping out Cone
Ha! Caught you! This is a rotation that does not satisfy either of our two conditions. As we shall see below, in this case we cannot write L = Iω or L = Iω – they simply are not correct!

6.4: Angular Momentum Conservation
We have derived (trivially) the Law of Conservation of Angular Momentum: When the total external torque acting on a systems is zero, the total angular momentum of the system is constant, that is, conserved. As you can imagine, this is a powerful concept we can use to understand many everyday phenomena and to solve many problems, both very simple conceptual ones and very complex and difficult ones.
The simplest application of this concept comes, now that we understand well the relationship between the scalar moment of inertia and the angular momentum, in systems where the moment of inertia of the system can change over time due to strictly internal forces. We will look at two particular example problems in this genre, deriving a few very useful results along the way.

Example 6.4.1: The Spinning Professor ω0 D

ωf

D

D/2

D/2

Figure 82: A professor stands on a freely pivoted platform at rest (total moment of inertia of professor and platform I0 ) with two large masses m held horizontally out at the side a distance D from the axis of rotation, initially rotating with some angular velocity ω0 .
A professor stands on a freely pivoted platform at rest with large masses held horizontally out at the side. A student gives the professor a push to start the platform and professor and masses rotating around a vertical axis. The professor then pulls the masses in towards the axis of rotation, reducing their contribution to the total moment of inertia as illustrated in figure 82
If the moment of inertia of the professor and platform is I0 and the masses m (including the arms’ contribution) are held at a distance D from the axis of rotation and the initial angular velocity is ω0 , what is the final angular velocity of the system ωf when the professor has pulled the masses in to a distance D/2?
The platform is freely pivoted so it exerts no external torque on the system. Pulling in the masses exerts no external torque on the system (although it may well exert a torque on the masses themselves as they transfer angular momentum to the professor). The angular momentum of the system is thus conserved.
Initially it is (in this highly idealized description)
Li = Ii ω0 = (2mD2 + I0 )ω0

(585)

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Week 6: Vector Torque and Angular Momentum

Finally it is:
Lf = If ωf = (2m

D
2

Solving for ωf : ωf =

2

+ I0 )ωf = (2mD2 + I0 )ω0 = Li
(2mD2 + I0 )
(2m

D 2
2

+ I0 )

ω0

(586)

(587)

From this all sorts of other things can be asked and answered. For example, what is the initial kinetic energy of the system in terms of the givens? What is the final? How much work did the professor do with his arms?
Note that this is exactly how ice skaters speed up their spin when performing their various nifty moves – start spinning with arms and legs spread out, then draw them in to spin up, extend them to slow down again. It is how high-divers control their rotation. It is how neutron stars spin up as their parent stars explode. It is part of the way cats manage to always land on their feet – for a value of the world “always” that really means “usually” or “mostly”117 .
Although there are more general ways of a system of particles altering its own moment of inertia, a fairly common way is indeed through the application of what we might call radial forces. Radial forces are a bit special and worth treating in the context of angular momentum conservation in their own right.

6.4.1: Radial Forces and Angular Momentum Conservation
One of the most important aspects of torque and angular momentum arises because of a curious feature of two of the most important force laws of nature: gravitation and the electrostatic force.
Both of these force laws are radial, that is, they act along a line connecting two masses or charges. Just for grins (and to give you a quick look at them, first in a long line of glances and repetitions that will culminate in your knowing them, here is the simple form of the gravitational force on a
“point-like” object (say, the Moon) being acted on by a second “point-like” object (say, the Earth) where for convenience we will locate the Earth at the origin of coordinates:
F m = −G

Mm Me r ˆ r2 (588)

In this expression, r = rˆ is the position of the moon in a spherical polar coordinate system (the r direction is actually specified by two angles, neither of which affects the magnitude of the force). G is called the gravitational constant and this entire formula is a special case of Newton’s Law of
Gravitation, currently believed to be a fundamental force law of nature on the basis of considerable evidence. A similar expression for the force on a charged particle with charge q located at position r = rˆ r exerted a charged particle with charge Q located at the origin is known as a (special case of)
Coulomb’s Law and is also held to be a fundamental force law of nature. It is the force that binds electrons to nuclei (while making the electrons themselves repel one another) and hence is the dominant force in all of chemistry – it, more than any other force of nature, is “us”118 . Coulomb’s
Law is just: qQ F q = −ke 2 r
ˆ
(589) r 117 I’ve seen some stupid cats land flat on their back in my lifetime, and a single counterexample serves to disprove the absolute rule...
118 Modulated by quantum principles, especially the notion of quantization and the Pauli Exclusion Principle, both beyond the scope of this course. Pauli is arguably co-equal with Coulomb in determining atomic and molecular structure. 287

Week 6: Vector Torque and Angular Momentum where ke is once again a constant of nature.

Both of these are radial force laws. If we compute the torque exerted by the Earth on the moon: τm = r × −G

Mm Me r2 r=0
ˆ

(590)

If we compute the torque exerted by Q on q: τq = r × ke

qQ r2 r=0
ˆ

(591)

Indeed, for any force law of the form F (r) = F (r)ˆ the torque exerted by the force is: r τ = r × F (r)ˆ = 0 r (592)

and we can conclude that radial forces exert no torque!
In all problems where those radial forces are the only (significant) forces that act:
A radial force exerts no torque and the angular momentum of the object upon which the force acts is conserved.
Note that this means that the angular momentum of the Moon in its orbit around the Earth is constant – this will have important consequences as we shall see in two or three weeks. It means that the electron orbiting the nucleus in a hydrogen atom has a constant angular momentum, at least as far as classical physics is concerned (so far). It means that if you tie a ball to a rubber band fastened to a pivot and then throw it so that the band remains stretches and shrinks as it moves around the pivot, the angular momentum of the ball is conserved. It means that when an exploding star collapses under the force of gravity to where it becomes a neutron star, a tiny fraction of its original radius, the angular momentum of the original star is (at least approximately, allowing for the mass it cast off in the radial explosion) conserved. It means that a mass revolving around a center on the end of a string of radius r has an angular momentum that is conserved, and that this angular momentum will remain conserved as the string is slowly pulled in or let out while the particle “orbits”.
Let’s understand this further using one or two examples.

Example 6.4.2: Mass Orbits On a String r m v F
Figure 83:
A particle of mass m is tied to a string that passes through a hole in a frictionless table and held. The mass is given a push so that it moves in a circle of radius r at speed v. Here are several questions that might be asked – and their answers:
a) What is the torque exerted on the particle by the string? Will angular momentum be conserved if the string pulls the particle into “orbits” with different radii?

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Week 6: Vector Torque and Angular Momentum
This is clearly a radial force – the string pulls along the vector r from the hole (pivot) to the mass. Consequently the tension in the string exerts no torque on mass m and its angular momentum is conserved. It will still be conserved as the string pulls the particle in to a new “orbit”.
This question is typically just asked to help remind you of the correct physics, and might well be omitted if this question were on, say, the final exam (by which point you are expected to have figured all of this out).

b) What is the magnitude of the angular momentum L of the particle in the direction of the axis of rotation (as a function of m, r and v)?
Trivial:
L = |r × p| = mvr = mr2 (v/r) = mr2 ω = Iω

(593)

By the time you’ve done your homework and properly studied the examples, this should be instantaneous. Note that this is the initial angular momentum, and that – from the previous question – angular momentum is conserved! Bear this in mind!
c) Show that the magnitude of the force (the tension in the string) that must be exerted to keep the particle moving in a circle is:
L2
F =T = mr3 This is a general result for a particle moving in a circle and in no way depends on the fact that the force is being exerted by a string in particular.
As a general result, we should be able to derive it fairly easily from what we know. We know two things – the particle is moving in a circle with a constant v, so that:
F =

mv 2 r (594)

We also know that L = mvr from the previous question! All that remains is to do some algebra magic to convert one to the other. If we had one more factor of m on to, and a factor of r2 on top, the top would magically turn into L2 . However, we are only allowed to multiply by one, so: mr2 m2 v 2 r 2
L2
mv 2
×
=
=
(595)
F = r mr2 mr3 mr3 as desired, Q.E.D., all done, fabulous.
d) Show that the kinetic energy of the particle in terms of its angular momentum is:
K=

L2
2mr2

More straight up algebra magic of exactly the same sort:
K=

mv 2 mr2 L2 mv 2
=
×
=
2
2
mr2
2mr2

(596)

Now, suppose that the radius of the orbit and initial speed are ri and vi , respectively. From under the table, the string is slowly pulled down (so that the puck is always moving in an approximately circular trajectory and the tension in the string remains radial) to where the particle is moving in a circle of radius r2 .
e) Find its velocity v2 using angular momentum conservation.
This should be very easy, and thanks to the results above, it is:
L1 = mv1 r1 = mv2 r2 = L2

(597)

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Week 6: Vector Torque and Angular Momentum or v2 =

r1 v1 r2

(598)

f) Compute the work done by the force from part c) above and identify the answer as the workkinetic energy theorem. Use this to to find the velocity v2 . You should get the same answer!
Well, what can we do but follow instructions. L and m are constants and we can take them right out of the integral as soon as they appear. Note that dr points out and F points in along r so that: r2 W

= −
= −

F dr r1 r2 r1 L2
= − m =
=
=

L2 dr mr3 r2 r−3 dr

r1
−2 r2

L2 r m 2

r1

L2
L2
2 − 2mr 2
2mr2
1
∆K

(599)

Not really so difficult after all.
Note that the last two results are pretty amazing – they show that our torque and angular momentum theory so far is remarkably consistent since two very different approaches give the same answer. Solving this problem now will make it easy later to understand the angular momentum barrier, the angular kinetic energy term that appears in the radial part of conservation of mechanical energy in problems involving a central force (such as gravitation and Coulomb’s Law). This in turn will make it easy for us to understand certain properties of orbits from their potential energy curves.
The final application of the Law of Conservation of Angular Momentum, collisions in this text is too important to be just a subsection – it gets its very own topical section, following immediately.

6.5: Collisions
We don’t need to dwell too much on the general theory of collisions at this point – all of the definitions of terms and the general methodology we learned in week 4 still hold when we allow for rotations.
The primary difference is that we can now apply the Law of Conservation of Angular Momentum as well as the Law of Conservation of Linear Momentum to the actual collision impulse.
In particular, in collisions where no external force acts (in the impulse approximation), no external torque can act as well. In these collisions both linear momentum and angular momentum are conserved by the collision. Furthermore, the angular momentum can be computed relative to any pivot, so one can choose a convenient pivot to simply the algebra involved in solving any given problem. This is illustrated in figure 84 above, where a small disk collides with a bar, both sitting (we imagine) on a frictionless table so that there is no net external force or torque acting. Both momentum and angular momentum are conserved in this collision. The most convenient pivot for problems of this sort is usually the center of mass of the bar, or possibly the center of mass of the system at the instant of collision (which continues moving a the constant speed of the center of mass before the collision, of course).

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Week 6: Vector Torque and Angular Momentum

M

M

v

m

m

Figure 84: In the collision above, no physical pivot exist and hence no external force or torque is exerted during the collision. In collisions of this sort both momentum and angular momentum about any pivot chosen are conserved.
All of the terminology developed to describe the energetics of different collisions still holds when we consider conservation of angular momentum in addition to conservation of linear momentum.
Thus we can speak of elastic collisions where kinetic energy is conserved during the collision, and partially or fully inelastic collisions where it is not, with “fully inelastic” as usual being a collision where the systems collide and stick together (so that they have the same velocity of and angular velocity around the center of mass after the collision).

pivot
M

m

v

M

m

Figure 85: In the collision above, a physical pivot exists – the bar has a hinge at one end that prevents its linear motion while permitting the bar to swing freely. In collisions of this sort linear momentum is not conserved, but since the pivot force exerts no torque about the pivot, angular momentum about the pivot is conserved.
We do, however, have a new class of collision that can occur, illustrated in figure 85, one where the angular momentum is conserved but linear momentum is not. This can and in general will occur when a system experiences a collision where a certain point in the system is physically pivoted by means of a nail, an axle, a hinge so that during the collision an unknown force 119 is exerted there as an extra external “impulse” acting on the system. This impulse acting at the pivot exerts no external torque around the pivot so angular momentum relative to the pivot is conserved but linear momentum is, alas, not conserved in these collisions.
119 Often

we can actually evaluate at least the impulse imparted by such a pivot during the collision – it is “unknown” in that it is usually not given as part of the initial data.

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Week 6: Vector Torque and Angular Momentum

It is extremely important for you to be able to analyze any given problem to identify the conserved quantities. To help you out, I’ve made up a a wee “collision type” table, where you can look for the term “elastic” in the problem – if it isn’t explicitly there, by default it is at least partially inelastic unless/until proven otherwise during the solution – and also look to see if there is a pivot force that again by default prevents momentum from being conserved unless/until proven otherwise during the solution. Elastic
Inelastic

Pivot Force
K, L conserved
L conserved

No Pivot Force
K, L and P conserved.
L and P conserved.

Table 4: Table to help you categorize a collision problem so that you can use the correct conservation laws to try to solve it. Note that you can get over half the credit for any given problem simply by correctly identifying the conserved quantities even if you then completely screw up the algebra.
The best way to come to understand this table (and how to proceed to add angular momentum conservation to your repertoire of tricks for analyzing collisions) is by considering the following examples. I’m only doing part of the work of solving them here, so you can experience the joy of solving them the rest of the way – and learning how it all goes – for homework.
We’ll start with the easiest collisions of this sort to solve – fully inelastic collisions.

Example 6.5.1: Fully Inelastic Collision of Ball of Putty with a Free Rod

M

M

x cm
L

vf

ωf

m m v0

Figure 86: A blob of putty of mass m, travelling at initial velocity v0 to the right, strikes an unpivoted rod of mass M and length L at the end and sticks to it. No friction or external forces act on the system. In figure 86 a blob of putty of mass m strikes a stationary rod of mass M at one end and sticks.
The putty and rod recoil together, rotating around their mutual center of mass. Everything is in a vacuum in a space station or on a frictionless table or something like that – in any event there are no other forces acting during the collision or we ignore them in the impulse approximation.
First we have to figure out the physics. We mentally examine our table of possible collision types.
There is no pivot, so there are no relevant external forces. No external force, no external torque, so both momentum and angular momentum are conserved by this collision. However, it is a fully inelastic collision so that kinetic energy is (maximally) not conserved.
Typical questions are:
• Where is the center of mass at the time of the collision (what is xcm )?

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Week 6: Vector Torque and Angular Momentum
• What is vf , the speed of the center of mass after the collision? Note that if we know the answer to these two questions, we actually know xcm (t) for all future times!
• What is ωf , the final angular velocity of rotation around the center of mass? If we know this, we also know θ(t) and hence can precisely locate every bit of mass in the system for all times after the collision.
• How much kinetic energy is lost in the collision, and where does it go?

We’ll answer these very systematically, in this order. Note well that for each answer, the physics knowledge required is pretty simple and well within your reach – it’s just that there are a lot of parts to patiently wade through.
To find xcm :
(M + m)xcm = M or xcm =

L
+ mL
2

M/2 + m
L
M +m

(600)

(601)

where I’m taking it as “obvious” that the center of mass of the rod itself is at L/2.
To find vf , we note that momentum is conserved (and also recall that the answer is going to be vcm : pi = mv0 = (M + m)vf = pf
(602)
or vf = vcm =

m v0 M +m

(603)

To find ωf , we note that angular momentum is going to be conserved. This is where we have to start to actually think a bit – I’m hoping that the previous two solutions are really easy for you at this point as we’ve seen each one (and worked through them in detail) at least a half dozen to a dozen times on homework and examples in class and in this book.
First of all, the good news. The rotation of ball and rod before and after the collision all happens in the plane of rotation, so we don’t have to mess with anything but scalar moments of inertia and
L = Iω. Then, the bad news: We have to choose a pivot since none was provided for us. The answer will be the same no matter which pivot you choose, but the algebra required to find the answer may be quite different (and more difficult for some choices).
Let’s think for a bit. We know the standard scalar moment of inertia of the rod (which applies in this case) around two points – the end or the middle/center of mass. However, the final rotation is around not the center of mass of the rod but the center of mass of the system, as the center of mass of the system itself moves in a completely straight line throughout.
Of course, the angular velocity is the same regardless of our choice of pivot. We could choose the end of the rod, the center of the rod, or the center of mass of the system and in all cases the final angular momentum will be the same, but unless we choose the center of mass of the system to be our pivot we will have to deal with the fact that our final angular momentum will have both a translational and a rotational piece.
This suggests that our “best choice” is to choose xcm as our pivot, eliminating the translational angular momentum altogether, and that is how we will proceed. However, I’m also going to solve this problem using the upper end of the rod at the instant of the collision as a pivot, because I’m quite certain that no student reading this yet understands what I mean about the translational component of the angular momentum!
Using xcm :

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Week 6: Vector Torque and Angular Momentum

We must compute the initial angular momentum of the system before the collision. This is just the angular momentum of the incoming blob of putty at the instant of collision as the rod is at rest.
Li = |r × p| = mv0 r⊥ = mv0 (L − xcm )

(604)

Note that the “moment arm” of the angular momentum of the mass m in this frame is just the perpendicular minimum distance from the pivot to the line of motion of m, L − xcm .
This must equal the final angular momentum of the system. This is easy enough to write down:
Li = mv0 (L − xcm ) = If ωf = Lf

(605)

where If is the moment of inertia of the entire rotating system about the xcm pivot!
Note well: The advantage of using this frame with the pivot at xcm at the instant of collision
(or any other frame with the pivot on the straight line of motion of xcm ) is that in this frame the angular momentum of the system treated as a mass at the center of mass is zero. We only have a rotational part of L in any of these coordinate frames, not a rotational and translational part. This makes the algebra (in my opinion) very slightly simpler in this frame than in, say, the frame with a pivot at the end of the rod/origin illustrated next, although the algebra in the the frame with pivot at the origin almost instantly “corrects itself” and gives us the center of mass pivot result. This now reveals the only point where we have to do real work in this frame (or any other) – finding If around the center of mass! Lots of opportunities to make mistakes, a need to use Our
Friend, the Parallel Axis Theorem, alarums and excursions galore. However, if you have clearly stated Li = Lf , and correctly represented them as in the equation above, you have little to fear – you might lose a point if you screw up the evaluation of If , you might even lose two or three, but that’s out of 10 to 25 points total for the problem – you’re already way up there as far as your demonstrated knowledge of physics is concerned!
So let’s give it a try. The total moment of intertia is the moment of inertia of the rod around the new (parallel) axis through xcm plus the moment of inertia of the blob of putty as a “point mass” stuck on at the end. Sounds like a job for the Parallel Axis Theorem!
If =

1
M L2 + M (xcm − L/2)2 + m(L − xcm )2
12

(606)

Now be honest; this isn’t really that hard to write down, is it?
Of course the “mess” occurs when we substitute this back into the conservation of momentum equation and solve for ωf : ωf =

Li
=
If

1
2
12 M L

mv0 (L − xcm )
+ M (xcm − L/2)2 + m(L − xcm )2

(607)

One could possible square out everything in the denominator and “simplify” this, but why would one want to? If we know the actual numerical values of m, M , L, and v0 , we can compute (in order) xcm , If and ωf as easily from this expression as from any other, and this expression actually means something and can be checked at a glance by your instructors. Your instructors would have to work just as hard as you would to reduce it to minimal terms, and are just as averse to doing pointless work. That’s not to say that one should never multiply things out and simplify, only that it seems unreasonable to count doing so as being part of the physics of the “answer”, and all we really care about is the physics! As a rule in this course, if you are a math, physics, or engineering major I expect you to go the extra mile and finish off the algebra, but if you are a life science major who came into the course terrified of anything involving algebra, well, I’m proud of you already because by

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Week 6: Vector Torque and Angular Momentum

this point in the course you have no doubt gotten much better at math and have started to overcome your fears – there is no need to charge you points for wading through stuff I could make a mistake doing almost as easily as you could. If anything, we’ll give you extra points if you try it and succeed
– after giving us something clear and correct to grade for primary credit first!
Finally, we do need to compute the kinetic energy lost in the collision:
∆K = Kf − Ki =

L2
1
1 f 2
2
+ (m + M )vf − mv0
2If
2
2

(608)

is as easy a form as any. Here there may be some point to squaring everything out to simplify, as one expects an answer that should be “some fraction of Ki ”, and the value of the fraction might be interesting. Again, if you are a physics major you should probably do the full simplification just for practice doing lots of tedious algebra without fear, useful self-discipline. Everybody else that does it will likely get extra credit unless the problem explicitly calls for it.
Now, let’s do the whole thing over, using a different pivot, and see where things are the same and where they are different.
Using the end of the rod:
Obviously there is no change in the computation of xcm and vf – indeed, we really did these in the (given) coordinate frame starting at the end of the rod anyway – that’s the “lab” frame drawn into our figure and the one wherein our answer is finally expressed. All we need to do, then, is compute our angular momenta relative to an origin/pivot at the end of the rod:
Li = |r × p| = r⊥ mv0 = mv0 L

(609)

Lf = (m + M )vf xcm + If ωf

(610)

and:
In this final expression, (m + M )vf xcm is the angular momentum of the entire system treated as a mass moving at speed vf located at xcm right after the collision, plus the angular momentum of the system around the center of mass, which must be computed exactly as before, same If , same ωf (to be found). Thus:
If ωf = mv0 L − (m + M )vf xcm
(611)
Here is a case where one really must do a bit more simplification – there are just too many things that depend on the initial conditions. If we substitute in vf from above, in particular, we get:
If ωf = mv0 L − mv0 xcm = mv0 (L − xcm ) and: ωf =

mv0 (L − xcm )
=
If

1
2
12 M L

mv0 (L − xcm )
+ M (xcm − L/2)2 + m(L − xcm )2

(612)

(613)

as before. The algebra somehow manages the frame change for us, giving us an answer that doesn’t depend on the particular choice of frame once we account for the angular momentum of the center of mass in any frame with a pivot that isn’t on the line of motion of xcm (where it is zero). Obviously, computing ∆K is the same, and so we are done!. Same answer, two different frames!

Example 6.5.2: Fully Inelastic Collision of Ball of Putty with Pivoted Rod
In figure 87 we see the alternative version of the fully inelastic collision between a pivoted rod and a blob of putty. Let us consider the answers to the questions in the previous problem. Some of them will either be exactly the same or in some sense “irrelevant” in the case of a pivoted rod, but either way we need to understand that.

295

Week 6: Vector Torque and Angular Momentum

pivot

M
M

L

ωf m m

v0

Figure 87: A blob of putty of mass m travelling at speed v0 strikes the rod of mass M and length
L at the end and sticks. The rod, however, is pivoted about the other end on a frictionless nail or hinge. First, however, the physics! Without the physics we wither and die! There is a pivot, and during the collision the pivot exerts a large and unknown pivot force on the rod120 and this force cannot be ignored in the impulse approximation. Consequently, linear momentum is not conserved! It is obviously an inelastic collision, so kinetic energy is not conserved.
The force exerted by the physical hinge or nail may be unknown and large, but if we make the hinge the pivot for the purposes of computing torque and angular momentum, this force exerts no torque on the system! Consequently, for this choice of pivot only, angular momentum is conserved. In a sense, this makes this problem much simpler than the last one! We have only one physical principle to work with (plus our definitions of e.g. center of mass), and all of the other answers must be derivable from this one thing. So we might as well get to work:
Li = I0 ω0 = mL2

v0
1
= ( M L2 + mL2 )ωf = If ωf = Lf
L
3

(614)

where I used Lf = If ωf and inserted the scalar moment of inertial If by inspection, as I know the moment of inertia of a rod about its end as well as the moment of inertia of a point mass a distance
L from the pivot axis. Thus: ωf =

mv0 L m m v0 = 1
= 1 ω0 1
( 3 M + m)L2
( 3 M + m) L
( 3 M + m)

(615)

where I’ve written it in the latter form (in terms of the initial angular velocity of the blob of putty relative to this pivot) both to make the correctness of the units manifest and to illustrate how conceptually simple the answer is:
I0
(616) ωf = ω0
If
It is now straightforward to answer any other questions that might be asked. The center of mass is still a distance: m + 1M
2
xcm =
L
(617) m+M 120 Unless

the mass strikes the rod at a particular point called the center of percussion of the rod, such that the velocity of the center of mass right after the collision is exactly the same as that of the free rod found above...

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Week 6: Vector Torque and Angular Momentum

from the pivot, but now it moves in a circular arc after the collision, not in a straight line.
The velocity of the center of mass after the collision is determined from ωf : vf = ωf xcm =

1 m + 2M m(m + 1 M ) m v0
2
×
L=
v0
1
m+M
(m + M )( 3 M + m)
+ m) L

(1M
3

(618)

The kinetic energy lost in the collision is:
∆K = Kf − Ki =

L2
1
f
2
− mv0
2If
2

(619)

(which can be simplified, but the simplification is left as an exercise for everybody – it isn’t difficult).
One can even compute the impulse provided by the pivot hinge during the collision:
Ihinge = ∆p = pf − pi = (m + M )vf − mv0

(620)

using vf from above. (Note well that I have to reuse symbols such as I, in the same problem sorry
– in this context it clearly means “impulse” and not “moment of inertia”.)

6.5.1: More General Collisions
As you can imagine, problems can get to be somewhat more difficult than the previous two examples in several ways. For one, instead of collisions between point masses that stick and rods (pivoted or not) one can have collisions between point masses and rods where the masses do not stick. This doesn’t change the basic physics. Either the problem will specify that the collision is elastic (so
Ktot is conserved) or it will specify something like vf for the point mass so that you can compute
∆K and possibly ∆p or ∆L (depending on what is conserved), much as you did for bullet-passesthrough-block problems in week 4.
Instead of point mass and rod at all you could be given a point mass and a disk or ball that can rotate, or even a collision between two disks. The algebra of things like this will be identical to the algebra above except that one will have to substitute e.g. the moment of inertia of a disk, or ball, or whatever for the moment of inertia of the rod, pivoted or free. The general method of solution will therefore be the same. You have several homework problems where you can work through this method on your own and working in your groups – make sure that you are comfortable with this before the quiz.
Angular momentum isn’t just conserved in the case of symmetric objects rotating, but these are by far the easiest for us to treat. We now need to tackle the various difficulties associated with the rotation of asymmetric rigid objects, and then move on to finally and irrevocably understand how torque and angular momentum are vector quantities and that this matters.

6.6: Angular Momentum of an Asymmetric Rotating Rigid
Object
Consider the single particle in figure 88 that moves in a simple circle of radius r sin(θ) in a plane above the x-y plane! The axis of rotation of the particle (the “direction of the rotation” we considered in week 5 is clearly ω = ω z .
ˆ
Note well that the mass distribution of this rigid object rotation violates both of the symmetry rules above. It is not symmetric across the axis of rotation, nor is it symmetric across the plane of rotation. Consequently, according to our fundamental definition of the vector angular momentum:
L = r × p = r × mv

(621)

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Week 6: Vector Torque and Angular Momentum z ω v m r τ

θ
L
y

x

Figure 88: A point mass m at the end of a massless rod that makes a vector r from the origin to the location of the mass, moving at constant speed v sweeps out a circular path of radius r sin(θ) in a plane above the (point) pivot. The angular momentum of this mass is (at the instant shown)
L = r×mv = r×p as shown. As the particle sweeps out a circle, so does L! The extended massless rigid rod exerts a constantly changing/precessing torque τ on the mass in order to accomplish this. which points up and to the left at the instant shown in figure 88.
Note well that L is perpendicular to both the plane containing r and v, and that as the mass moves around in a circle, so does L! In fact the vector L sweeps out a cone, just as the vector r does. Finally, note that the magnitude of L has the constant value:
L = |r × v| = mvr

(622)

because r and v are mutually perpendicular.
This means, of course, that although L = |L| is constant, L is constantly changing in time. Also, we know that the time rate of change of L is τ , so the rod must be exerting a nonzero torque on the mass! Finally, the scalar moment of inertia I = Izz = mr2 sin2 (θ) for this rotation is a constant
(and so is ω) – manifestly L = Iω! They don’t even point in the same direction!
Consider the following physics. We know that the actual magnitude and direction the force acting on m at the instant drawn is precisely Fc = mv 2 /(r sin(θ) (in towards the axis of rotation) because the mass m is moving in a circle. This force must be exerted by the massless rod because there is nothing else touching the mass (and we are neglecting gravity, drag, and all that). In turn, this force must be transmitted by the rod back to a bearing of some sort located at the origin, that keeps the rod from twisting out to rotate the mass in the same plane as the pivot (it’s “natural” state of rigid rotation).
The rod exerts a torque on the mass of magnitude: τrod = |r × F c | = r cos(θ)Fc = mv 2

cos(θ)
= mω 2 r2 sin(θ) cos(θ) sin(θ) (623)

Now, let’s see how this compares to the total change in angular momentum per unit time. Note that the magnitude of L, L, does not change in time, nor does Lz , the component parallel to the z-axis. Only the component in the x-y plane changes, and that components sweeps out a circle!
The radius of the circle is L⊥ = L cos(θ) (from examining the various right triangles in the figure) and hence the total change in L in one revolution is:
∆L = 2πL⊥ = 2πL cos(θ)

(624)

This change occurs in a time interval ∆t = T , the period of rotation of the mass m. The period of

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Week 6: Vector Torque and Angular Momentum

rotation of m is the distance it travels (circumference of the circle of motion) divided by its speed:
T =

2πr sin(θ) v (625)

Thus the magnitude of the torque exerted over ∆t = T is (using L = mvr as well):
∆L
v cos(θ) = 2πL cos(θ)
= mv 2
= mω 2 r2 sin(θ) cos(θ)
∆t
2πr sin(θ) sin(θ) so that indeed, τ rod =

(626)

∆L
∆t

(627)

for the cycle of motion.
The rod must indeed exert a constantly changing torque on the rod, a torque that remains perpendicular to the angular momentum vector at all times. The particle itself, the angular momentum, and the torque acting on the angular momentum all precess around the z-axis with a period of revolution of T and clearly at no time is it true that L = Iω for any scalar I and constant
ω.
z ω v

z ω m

v

m v L tot

m r L

L tot

r

L

L

y

L y x

x

r

r v m

Figure 89: When the two masses have mirror symmetry across the axis of rotation, their total angular momentum L does line up with ω. When the two masses have mirror symmetry across the plane of rotation, their total angular momentum L also lines up with ω.
Notice how things change if we balance the mass with a second one on an opposing rod as drawn in figure 89, making the distribution mirror symmetric across the axis of rotation. Now each of the two masses has a torque acting on it due to the rod connecting it with the origin, each mass has a vector angular momentum that at right angles to both r and v, but the components of L in the x-y plane phcancel so that the total angular momentum once again lines up with the z-axis!
The same thing happens if we add a second mass at the mirror-symmetric position below the plane of rotation as shown in the second panel of figure 89. In this case as well the components of
L in the x-y plane cancel while the z components add, producing a total vector angular momentum that points in the z-direction, parallel to ω.
The two ways of balancing a mass point around a pivot are not quite equivalent. In the first case, the pivot axis passes through the center of mass of the system, and the rotation can be maintained without any external force as well as torque. In the second case, however, the center of mass of the system itself is moving in a circle (in the plane of rotation). Consequently, while no net external torque is required to maintain the motion, there is a net external force required to maintain the motion. We will differentiate between these two cases below in an everyday example where they matter. 299

Week 6: Vector Torque and Angular Momentum

This is why I at least tried to be careful to assert throughout week 5) that the mass distributions for the one dimensional rotations we considered were sufficiently symmetric. “Sufficiently”, as you should now be able to see and understand, means mirror symmetric across the axis of rotation (best, zero external force or torque required to maintain rotation)) or plane of rotation (sufficient, but need external force to maintain motion of the center of mass in a circle). Only in these two cases is the total angular momentum L is parallel to ω such that L = Iω for a suitable scalar I.

Example 6.6.1: Rotating Your Tires tire mass excess

ω

bearings axle bearing wear

mass excess

bearing wear

Figure 90: Three tires viewed in cross section. The first one is perfectly symmetric and balanced.
The second one has a static imbalance – one side is literally more massive than the other. It will stress the bearings as it rotates as the bearings have to exert a differential centripetal force on the more massive side. The third is dynamically imbalanced – it has a non-planar mass assymetry and the bearings will have to exert a constantly precessing torque on the tire. Both of the latter situations will make the drive train noisy, the car more difficult and dangerous to drive, and will wear your bearings and tires out much faster.
This is why you should regularly rotate your tires and keep them well-balanced. A “perfect tire” is one that is precisely cylindrically symmetric. If we view it from the side it has a uniform mass distribution that has both mirror symmetry across the axis of rotation and mirror symmetry across the plane of rotation. If we mount such a tire on a frictionless bearing, no particular side will be heavier than any other and therefore be more likely to rotate down towards the ground. If we spin it on a frictionless axis, it will spin perfectly symmetrically as the bearings will not have to exert any precessing torque or time-varying force on it of the sort exerted by the massless rod in figure
88.
For a variety of reasons – uneven wear, manufacturing variations, accidents of the road – tires
(and the hubs they are mounted on) rarely stay in such a perfect state for the lifetime of either tire or car. Two particular kinds of imbalance can occur. In figure 90 three tires are viewed in crosssection. The first is our mythical brand new perfect tire, one that is both statically and dynamically balanced. The second is a tire that is statically imbalanced – it has mirror symmetry across the plane of rotation but not across the axis of rotation. One side has thicker tread than the other (and would tend to rotate down if the tire were elevated and allowed to spin freely). When a statically imbalanced tire rotates while driving, the center of mass of the tire moves in a circle around the axle and the bearings on the opposite side from the increased mass are anomalously compressed in order to provide the required centripetal force. The car bearings will wear faster than they should, and the car will have an annoying vibration and make a wubba-wubba noise as you drive (the latter can occur for many reasons but this is one of them).
The third is a tire that is dynamically imbalanced. The surplus masses shown are balanced well enough from left to right – neither side would roll down as in static imbalance, but the mass

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distribution does not have mirror symmetry across the axis of rotation nor does it have mirror symmetry across the plane of rotation through the pivot. Like the unbalanced mass sweeping out a cone, the bearings have to exert a dynamically changing torque on the tires as they rotate because their angular momentum is not parallel to the axis of rotation. At the instant shown, the bearings are stressed in two places (to exert a net torque on the hub out of the page if the direction of ω is up as drawn).
The solution to the problem of tire imbalance is simple – rotate your tires (to maintain even wear) and have your tires regularly balanced by adding compensatory weights on the “light” side of a static imbalance and to restore relative cylindrical symmetry for dynamical imbalance, the way adding a second mass did to our single mass sweeping out a cone.
I should point out that there are other ways to balance a rotating rigid object, and that every rigid object, no matter how its mass is distributed, has at least certain axes through the center of mass that can “diagonalize” the moment of inertia with respect to those axes. These are the axes that you can spin the object about and it will rotate freely without any application of an outside force or torque. Sadly, though, this is beyond the scope of this course121
At this point you should understand how easy it is to evaluate the angular momentum of symmetric rotating rigid objects (given their moment of inertia) using L = Iω, and how very difficult it can be to manage the angular momentum in the cases where the mass distribution is asymmetric and unbalanced. In the latter case we will usually find that the angular momentum vector will precess around the axis of rotation, necessitating the application of a continuous torque to maintain the (somewhat “unnatural”) motion.
There is one other very important context where precession occurs, and that is when a symmetric rigid body is rapidly rotating and has a large angular momentum, and a torque is applied to it in just the right way. This is one of the most important problems we will learn to solve this week, one essential for everybody to know, future physicians, physicists, mathematicians, engineers: The
Precession of a Top (or other symmetric rotator).

6.7: Precession of a Top ωp ω

M

R θ Mg
D

Figure 91:
Nowhere is the vector character of torque and angular momentum more clearly demonstrated than in the phenomenon of precession of a top, a spinning proton, or the Earth itself. This is also an
121 Yes,

I know, I know – this was a joke! I know that you aren’t, in fact, terribly sad about this...;-)

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Week 6: Vector Torque and Angular Momentum

important problem to clearly and quantitatively understand even in this introductory course because it is the basis of Magnetic Resonance Imaging (MRI) in medicine, the basis of understanding quantum phenomena ranging from spin resonance to resonant emission from two-level atoms for physicists, and the basis for gyroscopes to the engineer. Evvybody need to know it, in other words, no fair hiding behind the sputtered “but I don’t need to know this crap” weasel-squirm all too often uttered by frustrated students.
Look, I’ll bribe you. This is one of those topics/problems that I guarantee will be on at least one quiz, hour exam, or the final this semester. That is, mastering it will be worth at least ten points of your total credit, and in most years it is more likely to be thirty or even fifty points.
It’s that important! If you master it now, then understanding the precession of a proton around a magnetic field will be a breeze next semester. Otherwise, you risk the additional 10-50 points it might be worth next semester.
If you are a sensible person, then, you will recognize that I’ve just made it worth your while to invest the time required to completely understand precession in terms of vector torque, and to be able to derive the angular frequency of precession, ωp – which is our goal. I’ll show you three ways to do it, don’t worry, and any of the three will be acceptable (although I prefer that you use the second or third as the first is a bit too simple, it gets the right answer but doesn’t give you a good feel for what happens if the forces that produce the torque change in time).
First, though, let’s understand the phenonenon. If figure ?? above, a simple top is shown spinning at some angular velocity ω (not to be confused with ωp , the precession frequency). We will idealize this particular top as a massive disk with mass M , radius R, spinning around a rigid massless spindle that is resting on the ground, tipped at an angle θ with respect to the vertical.
We begin by noting that this top is symmetric, so we can easily compute the magnitude and direction of its angular momentum:
L = |L| = Iω =

1
M R2 ω
2

(628)

Using the “grasp the axis” version of the right hand rule, we see that in the example portrayed this angular momentum points in the direction from the pivot at the point of contact between the spindle and the ground and the center of mass of the disk along the axis of rotation.
Second, we note that there is a net torque exerted on the top relative to this spindle-ground contact point pivot due to gravity. There is no torque, of course, from the normal force that holds up the top, and we are assuming the spindle and ground are frictionless. This torque is a vector : τ = D × (−M g z )
ˆ

(629)

τ = M gD sin(θ)

(630)

or

into the page as drawn above, where z is vertical.
Third we note that the torque will change the angular momentum by displacing it into the page in the direction of the torque. But as it does so, the plane containing the D and M g z will be
ˆ
rotated a tiny bit around the z (precession) axis, and the torque will also shift its direction to remain perpendicular to this plane. It will shift L a bit more, which shifts τ a bit more and so on.
In the end L will precess around z, with τ precessing as well, always π/2 ahead .
Since τ ⊥ L, the magnitude of L does not change, only the direction. L sweeps out a cone, exactly like the cone swept out in the unbalanced rotation problems above and we can use similar considerations to relate the magnitude of the torque (known above) to the change of angular momentum per period. This is the simplest (and least accurate) way to find the precession frequency.
Let’s start by giving this a try:

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Week 6: Vector Torque and Angular Momentum

Example 6.7.1: Finding ωp From ∆L/∆t (Average)
We already derived above that τavg = τ = M gD sin(θ)

(631)

is the magnitude of the torque at all points in the precession cycle, and hence is also the average of the magnitude of the torque over a precession cycle (as opposed to the average of the vector torque over a cycle, which is obviously zero).
We now compute the average torque algebraically from ∆L/∆t (average) and set it equal to the computed torque above. That is:
∆Lcycle = 2πL⊥ = 2πL sin(θ) and: ∆tcycle = T =

2π ωp (632)

(633)

where ωp is the (desired) precession frequency, or: τavg = M gD sin(θ) =

∆Lcycle
= ωp L sin(θ)
∆tcycle

(634)

We now solve the the precession frequency ωp : ωp =

M gD
2gD
M gD
=
= 2
L

R ω

(635)

Note well that the precession frequencey is independent of θ – this is extremely important next semester when you study the precession of spinning charged particles around an applied magnetic field, the basis of Magnetic Resonance Imaging (MRI).

Example 6.7.2: Finding ωp from ∆L and ∆t Separately
Simple and accurate as it is, the previous derivation has a few “issues” and hence it is not my favorite one; averaging in this way doesn’t give you the most insight into what’s going on and doesn’t help you get a good feel for the calculus of it all. This matters in MRI, where the magnetic field varies in time, although it is a bit more difficult to make gravity vary in a similar way so it doesn’t matter so much this semester.
Nevertheless, to get you off on the right foot, so to speak, for E&M122 , let’s do a second derivation that is in between the very crude averaging up above and a rigorous application of calculus to the problem that we can’t even understand properly until we reach the week where we cover oscillation.
Let’s repeat the general idea of the previous derivation, only instead of setting τ equal to the average change in L per unit time, we will set it equal to the instantaneous change in L per unit time, writing everything in terms of very small (but finite) intervals and then taking the appropriate limits. To do this, we need to picture how L changes in time. Note well that at any given instant, τ is perpendicular to the plane containing −ˆ (the direction of the gravitational force) and L. This y direction is thus always perpendicular to both L and −M g z and cannot change the magnitude
ˆ
of L or its z component Lz . Over a very short time ∆t, the change in L is thus ∆L in the plane perpendicular to z . As the angular momentum changes direction only into the τ direction,
ˆ
the τ direction also changes to remain perpendicular. This should remind you have our long-ago
122 Electricity

and Magnetism

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Week 6: Vector Torque and Angular Momentum

z

y

∆L
L
Lz

τ in

∆ L = L ∆φ = τ ∆t

∆φ

L

L

x

θ y x
Figure 92: The cone swept out by the precession of the angular momentum vector, L, as well as an
“overhead view” of the trajectory of L⊥ , the component of L perpendicular to the z-axis. discussion of the kinematics of circular motion – L sweeps out a cone around the z-axis where Lz and the magnitude of L remain constant.
L⊥ sweeps out a circle as we saw in the previous example, and we can visualize both the cone swept out by L and the change over a short time ∆t, ∆L, in figure 92. We can see from the figure that: ∆L = L⊥ ∆φ = τ ∆t
(636)
where ∆φ is the angle the angular momentum vector precesses through in time ∆t. We substitute
L⊥ = L sin(θ) and τ = M gD sin(θ) as before, and get:
L sin(θ) ∆φ = M gD sin(θ) ∆t

(637)

Finally we solve for:


∆φ
M gD
2gD
= lim
=
= 2
(638)
∆t→0 ∆φ dt L
R ω as before. Note well that because this time we could take the limit as ∆t → 0, we get an expression that is good at any instant in time, even if the top is in a rocket ship (an accelerating frame) and g → g ′ is varying in time! ωp =

This still isn’t the most elegant approach. The best approach, although it does use some real calculus, is to just write down the equation of motion for the system as differential equations and solve them for both L(t) and for ωp .

Example 6.7.3: Finding ωp from Calculus
Way back at the beginning of this section we wrote down Newton’s Second Law for the rotation of the gyroscope directly: τ = D × (−M g z )
ˆ
(639)
Because −M g z points only in the negative z-direction and
ˆ
D=D

L
L

(640)

because L is parallel to D, for a general L = Lx x + Ly y we get precisely two terms out of the cross
ˆ
ˆ product: dLx
= τx dt dLy
= τy dt M gDLy
L
M gDLx
= −M gD sin(θ) = −
L
= M gD sin(θ) =

(641)
(642)

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Week 6: Vector Torque and Angular Momentum

or dLx dt dLy dt

=
=

M gD
Ly
L
M gD
Lx
L

(643)
(644)

If we differentiate the first expression and substitute the second into the first (and vice versa) we transform this pair of coupled first order differential equations for Lx and Ly into the following pair of second order differential equations: d2 Lx dt2 M gD
L

2

=

d2 Ly dt2 M gD
L

2

=

2
Lx = ωp Lx

(645)

2
Ly = ωp Ly

(646)

These (either one) we will learn to recognize as the differential equation of motion for the simple harmonic oscillator. A particular solution of interest (that satisfies the first order equations above) might be:
Lx (t)

= L⊥ cos(ωp t)

(647)

Ly (t)

= L⊥ sin(ωp t)

(648)

Lz (t)

= Lz0

(649)

where L⊥ is the magnitude of the component of L which is perpendicular to z . This is then the
ˆ
exact solution to the equation of motion for L(t) that describes the actual cone being swept out, with no hand-waving or limit taking required.
The only catch to this approach is, of course, that we don’t know how to solve the equation of motion yet, and the very phrase “second order differential equation” strikes terror into our hearts in spite of the fact that we’ve been solving one after another since week 1 in this class!
All of the equations of motion we have solved from Newton’s Second Law have been second order ones, after all – it is just that the ones for constant acceleration were directly integrable where this set is not, at least not easily.
It is pretty easy to solve for all of that, but we will postpone the actual solution until later. In the meantime, remember, you must know how to reproduce one of the three derivations above for ωp , the angular precession frequency, for at least one quiz, test, or hour exam. Not to mention a homework problem, below. Be sure that you master this because precession is important.
One last suggestion before we move on to treat angular collisions. Most students have a lot of experience with pushes and pulls, and so far it has been pretty easy to come up with everyday experiences of force, energy, one dimensional torque and rolling, circular motion, and all that. It’s not so easy to come up with everyday experiences involving vector torque and precession. Yes, may of you played with tops when you were kids, yes, nearly everybody rode bicycles and balancing and steering a bike involves torque, but you haven’t really felt it knowing what was going on.
The only device likely to help you to personally experience torque “with your own two hands” is the bicycle wheel with handles and the string that was demonstrated in class. I urge you to take a turn spinning this wheel and trying to turn it by means of its handles while it is spinning, to spin it and suspend it by the rope attached to one handle and watch it precess. Get to where you can predict the direction of precession given the direction of the spin and the handle the rope is attached to, get so you have experience of pulling it (say) in and out and feeling the handles deflect up and down or vice versa. Only thus can you feel the nasty old cross product in both the torque and the angular momentum, and only thus can you come to understand torque with your gut as well as your head.

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Week 6: Vector Torque and Angular Momentum

Homework for Week 6

Problem 1.

Physics Concepts: Make this week’s physics concepts summary as you work all of the problems in this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s) they were key to, and include concepts from previous weeks as necessary. Do the work carefully enough that you can (after it has been handed in and graded) punch it and add it to a three ring binder for review and study come finals!

Problem 2.

ωf
M

R

v

m

This problem will help you learn required concepts such as:
• Angular Momentum Conservation
• Inelastic Collisions so please review them before you begin.
Satchmo, a dog with mass m runs and jumps onto the edge of a merry-go-round (that is initially at rest) and then sits down there to take a ride as it spins. The merry-go-round can be thought of as a disk of radius R and mass M , and has approximately frictionless bearings in its axle. At the time of this angular “collision” Satchmo is travelling at speed v perpendicular to the radius of the merry-go-round and you can neglect Satchmo’s moment of inertia about an axis through his OWN center of mass compared to that of Satchmo travelling around the merry-go-round axis (because R of the merry-go-round is much larger than Satchmo’s size, so we can treat him like “a particle”).
a) What is the angular velocity ωf of the merry-go-round (and Satchmo) right after the collision?
b) How much of Satchmo’s initial kinetic energy was lost, compared to the final kinetic energy of
Satchmo plus the merry-go-round, in the collision? (Believe me, no matter what he’s lost it isn’t enough – he’s a border collie and he has plenty more!)

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Week 6: Vector Torque and Angular Momentum

Problem 3.

ωp

M.R ω θ

D

This problem will help you learn required concepts such as:
• Vector Torque
• Vector Angular Momentum
• Geometry of Precession so please review them before you begin.
A top is made of a disk of radius R and mass M with a very thin, light nail (r ≪ R and m ≪ M ) for a spindle so that the disk is a distance D from the tip. The top is spun with a large angular velocity ω. When the top is spinning at a small angle θ with the vertical (as shown) what is the angular frequency ωp of the top’s precession? Note that this is a required problem that will be on at least one test in one form or another, so be sure that you have mastered it when you are done with the homework!

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Week 6: Vector Torque and Angular Momentum
Problem 4.

d
L

m v M
This problem will help you learn required concepts such as:
• Angular Momentum Conservation
• Momentum Conservation
• Inelastic Collisions
• Impulse

so please review them before you begin. You may also find it useful to read: Wikipedia: http://www.wikipedia
A rod of mass M and length L rests on a frictionless table and is pivoted on a frictionless nail at one end as shown. A blob of putty of mass m approaches with velocity v from the left and strikes the rod a distance d from the end as shown, sticking to the rod.
• Find the angular velocity ω of the system about the nail after the collision.
• Is the linear momentum of the rod/blob system conserved in this collision for a general value of d? If not, why not?
• Is there a value of d for which it is conserved? If there were such a value, it would be called the center of percussion for the rod for this sort of collision.
All answers should be in terms of M , m, L, v and d as needed. Note well that you should clearly indicate what physical principles you are using to solve this problem at the beginning of the work.

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Week 6: Vector Torque and Angular Momentum

Problem 5.

r

m v F

This problem will help you learn required concepts such as:
• Torque Due to Radial Forces
• Angular Momentum Conservation
• Centripetal Acceleration
• Work and Kinetic Energy so please review them before you begin.
A particle of mass m is tied to a string that passes through a hole in a frictionless table and held.
The mass is given a push so that it moves in a circle of radius r at speed v.
a) What is the torque exerted on the particle by the string?
b) What is the magnitude of the angular momentum L of the particle in the direction of the axis of rotation (as a function of m, r and v)?
c) Show that the magnitude of the force (the tension in the string) that must be exerted to keep the particle moving in a circle is:
L2
F = mr3 d) Show that the kinetic energy of the particle in terms of its angular momentum is:
K=

L2
2mr2

Now, suppose that the radius of the orbit and initial speed are ri and vi , respectively. From under the table, the string is slowly pulled down (so that the puck is always moving in an approximately circular trajectory and the tension in the string remains radial) to where the particle is moving in a circle of radius r2 .
e) Find its velocity v2 using angular momentum conservation. This should be very easy.
f) Compute the work done by the force from part c) above and identify the answer as the workkinetic energy theorem.

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Week 6: Vector Torque and Angular Momentum
Problem 6.

z

vin m r θ pivot

y

x
This problem will help you learn required concepts such as:
• Vector Torque
• Non-equivalence of L and Iω. so please review them before you begin.
In the figure above, a mass m is being spun in a circular path at a constant speed v around the z-axis on the end of a massless rigid rod of length r pivoted at the orgin (that itself is being pushed by forces not shown). All answers should be given in terms of m, r, θ and v. Ignore gravity and friction. a) Find the angular momentum vector of this system relative to the pivot at the instant shown and draw it in on a copy of the figure.
b) Find the angular velocity vector of the mass m at the instant shown and draw it in on the figure above. Is L = Iz ω, where Iz = mr2 sin2 (θ) is the moment of inertia around the z axis?
What component of the angular momentum is equal to Iz ω?
c) What is the magnitude and direction of the torque exerted by the rod on the mass (relative to the pivot shown) in order to keep it moving in this circle at constant speed? (Hint: Consider the precession of the angular momentum around the z-axis where the precession frequency is ω, and follow the method presented in the textbook above.)
d) What is the direction of the force exerted by the rod in order to create this torque? (Hint:
Shift gears and think about the actual trajectory of the particle. What must the direction of the force be?)
e) Set τ = rF⊥ or τ = r⊥ F and determine the magnitude of this torque. Note that it is equal in magnitude to your answer to c) above and has just the right direction!

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Week 6: Vector Torque and Angular Momentum

Optional Problems
The following problems are not required or to be handed in, but are provided to give you some extra things to work on or test yourself with.

Optional Problem 7.

d
L

m v M
A rod of mass M and length L is hanging vertically from a frictionless pivot (where gravity is
“down”). A blob of putty of mass m approaches with velocity v from the left and strikes the rod a distance d from its center of mass as shown, sticking to the rod.
a) Find the angular velocity ωf of the system about the pivot (at the top of the rod) after the collision. b) Find the distance xcm from the pivot of the center of mass of the rod-putty system immediately after the collision.
c) Find the velocity of the center of mass vcm of the system (immediately) after the collision.
d) Find the kinetic energy of the rod and putty immediately after the collision. Was kinetic energy conserved in this collision? If not, how much energy was lost to heat?
e) After the collision, the rod swings up to a maximum angle θmax and then comes momentarily to rest. Find θmax .
f) In general (for most values of d is linear momentum conserved in this collision? At the risk of giving away the answer, why not? What exerts an external force on the system during the collision when momentum is not conserved?
All answers should be in terms of M , m, L, v, g and d as needed.

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Week 7: Statics
Optional Problem 8.

M

L d v m This problem will help you learn required concepts such as:
• Angular Momentum Conservation
• Momentum Conservation
• Inelastic Collisions
• Impulse so please review them before you begin.
A rod of mass M and length L rests on a frictionless table. A blob of putty of mass m approaches with velocity v from the left and strikes the rod a distance d from the end as shown, sticking to the rod. • Find the angular velocity ω of the system about the center of mass of the system after the collision. Note that the rod and putty will not be rotating about the center of mass of the rod!
• Is the linear momentum of the rod/blob system conserved in this collision for a general value of d? If not, why not?
• Is kinetic energy conserved in this collision? If not, how much is lost? Where does the energy go? All answers should be in terms of M , m, L, v and d as needed. Note well that you should clearly indicate what physical principles you are using to solve this problem at the beginning of the work.

312

Week 7: Statics

Week 7: Statics
Statics Summary
• A rigid object is in static equilibrium when both the vector torque and the vector force acting on it are zero. That is:
If F tot = 0 and τ tot = 0, then an object initially at rest will remain at rest and neither accelerate nor rotate.
This rule applies to particles with intrinsic spin as well as “rigid objects”, but this week we will primarily concern ourselves with rigid objects as static force equilibrium for particles was previously discussed.
Note well that the torque and force in the previous problem are both vectors. In many problems they will effectively be one dimensional, but in some they will not and you must establish e.g. the torque equilibrium condition for several different directions.
• A common question that arises in statics is the tipping problem. For an object placed on a slope or pivoted in some way such that gravity opposed by normal forces provides one of the sources of torque that tends to keep the object stable, while some variable force provides a torque that tends to tip the object over the pivot, one uses the condition of marginal static equilibrium to determine, e.g. the lowest value of the variable force that will tip the object over. • A force couple is defined to be a pair of forces that are equal and opposite but that do not necessarily or generally act along the same line upon an object. The point of this definition is that it is easily to see that force couples exert no net force on an object but they will exert a net torque on the object as long as they do not act along the same line.
Furthermore:
• The vector torque exerted on a rigid object by a force couple is the same for all choices of pivot! This (and the frequency with which they occur in problems) is the basis for the definition. As you can see, this is a short week, just perfect to share with the midterm hour exam.

7.1: Conditions for Static Equilibrium
We already know well (I hope) from our work in the first few weeks of the course that an object at rest remains at rest unless acted on by a net external force! After all, this is just Newton’s First
Law! If a particle is located at some position in any inertial reference frame, and isn’t moving, it won’t start to move unless we push on it with some force produced by a law of nature.
313

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Week 7: Statics

Newton’s Second Law, of course, applies only to particles – infinitesimal chunklets of mass in extended objects or elementary particles that really appear to have no finite extent. However, in week 4 we showed that it also applies to systems of particles, with the replacement of the position of the particle by the position of the center of mass of the system and the force with the total external force acting on the entire system (internal forces cancelled), and to extended objects made up of many of those infinitesimal chunklets. We could then extend Newton’s First Law to apply as well to amorphous systems such as clouds of gas or structured systems such as “rigid objects” as long as we considered being “at rest” a statement concerning the motion of their center of mass. Thus a “baseball”, made up of a truly staggering number of elementary microscopic particles, becomes a
“particle” in its own right located at its center of mass.
We also learned that the force equilibrium of particles acted on by conservative force occurred at the points where the potential energy was maximum or minimum or neutral (flat), where we named maxima “unstable equilibrium points”, minima “stable equilibrium points” and flat regions “neutral equilibria”123 .
However, in weeks 5 and 6 we learned enough to now be able to see that force equilibrium alone is not sufficient to cause an extended object or collection of particles to be in equilibrium. We can easily arrange situations where two forces act on an object in opposite directions (so there is no net force) but along lines such that together they exert a nonzero torque on the object and hence cause it to angularly accelerate and gain kinetic energy without bound, hardly a condition one would call
“equilibrium”.
The good news is that Newton’s Second Law for Rotation is sufficient to imply Newton’s First
Law for Rotation:
If, in an inertial reference frame, a rigid object is initially at rotational rest
(not rotating), it will remain at rotational rest unless acted upon by a net external torque.
That is, τ = I α = 0 implies ω = 0 and constant124 . We will call the condition where τ = 0 and a rigid object is not rotating torque equilibrium.
We can make a baseball (initially with its center of mass at rest and not rotating) spin without exerting a net force on it that makes its center of mass move – it can be in force equilibrium but not torque equilibrium. Similarly, we can throw a baseball without imparting any rotational spin – it can be in torque equilibrium but not force equilibrium. If we want the baseball (or any rigid object) to be in a true static equilibrium, one where it is neither translating nor rotating in the future if it is at rest and not rotating initially, we need both the conditions for force equilibrium and torque equilibrium to be true.
Therefore we now define the conditions for the static equilibrium of a rigid body to be:
A rigid object is in static equilibrium when both the vector torque and the vector force acting on it are zero.
That is:
If F tot = 0 and τ tot = 0, then an object initially at translational and rotational rest will remain at rest and neither accelerate nor rotate.
123 Recall that neutral equilibria were generally closer to being unstable than stable, as any nonzero velocity, no matter how small, would cause a particle to move continuously across the neutral region, making no particular point stable to say the least. That same particle would oscillate, due to the restoring force that traps it between the turning points of motion that we previously learned about and at least remain in the “vicinity” of a true stable equilibrium point for small enough velocities/kinetic energies.
124 Whether or not I is a scalar or a tensor form...

Week 7: Statics

315

That’s it. Really, pretty simple.
Needless to say, the idea of stable versus unstable and neutral equilibrium still holds for torques as well as forces. We will consider an equilibrium to be stable only if both the force and the torque are “restoring” – and push or twist the system back to the equilibrium if we make small linear or angular displacements away from it.

7.2: Static Equilibrium Problems
After working so long and hard on solving actual dynamical problems involving force and torque, static equilibrium problems sound like they would be pretty trivial. After all, nothing happens! It seems as though solving for what happens when nothing happens would be easy.
Not so. To put it in perspective, let’s consider why we might want to solve a problem in static equilibrium. Suppose we want to build a house. A properly built house is one that won’t just fall down, either all by itself or the first time the wolf huffs and puffs at your door. It seems as though building a house that is stable enough not to fall down when you move around in it, or load it with furniture in different ways, or the first time a category 2 hurricane roars by overhead and whacks it with 160 kilometer per hour winds is a worthy design goal. You might even want it to survive earthquakes! If you think that building a stable house is easy, I commend trying to build a house out of cards125 . You will soon learn that balancing force loads, taking advantage of friction (or other
“fastening” forces”, avoiding unbalanced torques is all actually remarkably tricky, for all that we can learn to do it without solving actual equations. Engineers who want to build serious structures such as bridges, skyscrapers, radio towers, cars, airplanes, and so on spend a lot of time learning statics (and a certain amount of dynamics, because no structure in a dynamical world filled with
Big Bad Wolves is truly “static”) because it is very expensive when buildings, bridges, and so on fall down, when their structural integrity fails.
Stability is just as important for physicians to understand. The human body is not the world’s most stable structure, as it turns out. If you have ever played with Barbie or G.I. Joe dolls, then you understand that getting them to stand up on their own takes a bit of doing – just a bit of bend at the waist, just a bit too much weight at the side, or the feet not adjusted just so, and they fall right over. Actual humans stabilize their erect stance by constantly adjusting the force balance of their feet, shifting weight without thinking to the heel or to the toes, from the left foot to the right foot as they move their arms or bend at the waist or lift something. Even healthy, coordinated adults who are paying attention nevertheless sometimes lose their balance because their motions exceed the fairly narrow tolerance for stability in some stance or another.
This ability to remain stable standing up or walking rapidly disappears as one’s various sensory feedback mechanisms are impaired, and many, many health conditions impair them. Drugs or alcohol, neuropathy, disorders of the vestibular system, pain and weakness due to arthritis or aging.
Many injuries (especially in the elderly) occur because people just plain fall over.
Then there is the fact that nearly all of our physical activity involves the adjustment of static balance between muscles (providing tension) and bones (providing compression), with our joints becoming stress-points that have to provide enormous forces, painlessly, on demand. In the end, physicians have to have a very good conceptual understanding of static equilibrium in order to help their patients acheive it and maintain it in the many aspects of the “mechanical” operation of the human body where it is essential.
For this reason, no matter who you are taking this course, you need to learn to solve real statics
125 Wikipedia:

actually!

http://www.wikipedia.org/wiki/House of cards. Yes, even this has a wikipedia entry. Pretty cool,

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problems, ones that can help you later understand and work with statics as your life and career demand. This may be nothing more than helping your kid build a stable tree house, but y’know, you don’t want to help them build a tree house and have that house fall out of the tree with your kid in it, nor do you want to deny them the joy of hanging out in their own tree house up in the trees! As has been our habit from the beginning of the course, we start by considering the simplest problems in static equilibrium and then move on to more difficult ones. The simplest problems cannot, alas, be truly one dimensional because if the forces involved are truly one dimensional (and act to the right or left along a single line) there is no possible torque and force equilibrium suffices for both.
The simplest problem involving both force and torque is therefore at least three dimensional – two dimensional as far as the forces are concerned and one dimensional as far as torque and rotation is concerned. In other words, it will involve force balance in some plane of (possible) rotation and torque balance perpendicular to this plane alone a (possible) axis of rotation.

Example 7.2.1: Balancing a See-Saw y F m2 m1 x1 x2

m1g m 2g
Figure 93: You are given m1 , x1 , and x2 and are asked to find m2 and F such that the see-saw is in static equilibrium.
One typical problem in statics is balancing weights on a see-saw type arrangement – a uniform plank supported by a fulcrum in the middle. This particular problem is really only one dimensional as far as force is concerned, as there is no force acting in the x-direction or z-direction. It is one dimensional as far as torque is concerned, with rotation around any pivot one might select either into or out of the paper.
Static equilibrium requires force balance (one equation) and torque balance (one equation) and therefore we can solve for pretty much any two variables (unknowns) visible in figure 93 above. Let’s imagine that in this particular problem, the mass m1 and the distances x1 and x2 are given, and we need to find m2 and F .
We have two choices to make – where we select the pivot and which direction (in or out of the page) we are going to define to be “positive”. A perfectly reasonable (but not unique or necessarily
“best”) choice is to select the pivot at the fulcrum of the see-saw where the unknown force F is exerted, and to select the +z-axis as positive rotation (out of the page as drawn).
We then write:
Fy

= F − m1 g − m2 g = 0

(650)

τz

= x1 m1 g − x2 m2 g = 0

(651)

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Week 7: Statics
This is almost embarrassingly simple to solve. From the second equation: m2 =

m1 gx1
=
gx2

x1 x2 m1

(652)

It is worth noting that this is precisely the mass that moves the center of mass of the system so that it is square over the fulcrum/pivot.
From the first equation and the solution for m2 :
F = m1 g + m2 g = m1 g 1 +

x1 x2 = m1 g

x1 + x2 x2 (653)

That’s all there is to it! Obviously, we could have been given m1 and m2 and x1 and been asked to find x2 and F , etc, just as easily.

Example 7.2.2: Two Saw Horses pivot y

F2
F1

M x mg
L

Mg x Figure 94: Two saw horses separated by a distance L support a plank of mass m symmetrically placed across them as shown. A block of mass M is placed on the plank a distance x from the saw horse on the left.
In figure 94, two saw horses separated by a distance L support a symmetrically placed plank.
The rigid plank has mass m and supports a block of mass M placed a distance x from the left-hand saw horse. Find F1 and F2 , the upward (normal) force exerted by each saw horse in order for this system to be in static equilibribum.
First let us pick something to put into equilibrium. The saw horses look pretty stable. The mass
M does need to be in equilibrium, but that is pretty trivial – the plank exerts a normal force on
M equal to its weight. The only tricky thing is the plank itself, which could and would rotate or collapse if F1 and F2 don’t correctly balance the load created by the weight of the plank plus the weight of the mass M .
Again there are no forces in x or z, so we simply ignore those directions. In the y direction:
F1 + F2 − mg − M g = 0

(654)

or “the two saw horses must support the total weight of the plank plus the block”, F1 + F2 =
(m + M )g. This is not unreasonable or even unexpected, but it doesn’t tell us how this weight is distributed between the two saw horses.
Once again we much choose a pivot. Four possible points – the point on the left-hand saw horse where F1 is applied, the point at L/2 that is the center of mass of the plank (and half-way in between the two saw horses), the point under mass M where its gravitational force acts, and the point on the right-hand saw horse where F2 is applied. Any of these will eliminate the torque due to one of the forces and presumably will simplify the problem relative to more arbitrary points. We select the left-hand point as shown – why not?
Then:
τz = F2 L − mgL/2 − M gx = 0

(655)

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states that the torque around this pivot must be zero. We can easily solve for F2 :
F2 =

mgL/2 + M gx mg x
=
+ Mg
L
2
L

(656)

Finally, we can solve for F1 :
F1 = (m + M )g − F2 =

L−x mg + Mg
2
L

(657)

Does this make sense? Sure. The two saw horses share the weight of the symmetrically placed plank, obviously. The saw horse closest to the block M supports most of its weight, in a completely symmetric way. In the picture above, with x > L/2, that is saw horse 2, but if x < L/2, it would have been saw horse 1. In the middle, where x = L/2, the two saw horses symmetrically share the weight of the block as well!
This picture and solution are worth studying until all of this makes sense. Carrying things like sofas and tables (with the load shared between a person on either end) is a frequent experience, and from the solution to this problem you can see that if the load is not symmetrical, the person closest to the center of gravity will carry the largest load.
Let’s do a slightly more difficult one, one involving equilibrium in two force directions (and one torque direction). This will allow us to solve for three unknown quantities.

Example 7.2.3: Hanging a Tavern Sign

pivot

F y T

Fx

θ

Mechanic−Ale and Physics Beer for sale mg Figure 95: A sign with mass m is hung from a massless rigid pole of length L attached to a post and suspended by means of a wire at an angle θ relative to the horizontal.
Suppose that one day you get tired being a hardworking professional and decide to give it all up and open your own tavern/brewpub. Naturally, you site it in a lovely brick building close to campus (perhaps in Brightleaf Square). To attract passers-by you need a really good sign – the old fashioned sort made out of solid oak that hangs from a pole, one that (with the pole) masses m = 50 kg. However, you really don’t want the sign to either punch through the brick wall or break the suspension wire you are going to use to support the end farthest from the wall and you don’t trust your architect because he seems way too interested in your future wares, so you decide to work out for yourself just what the forces are that the wall and wire have to support, as a function of the angle θ between the support pole and the support wire.
The physical arrangement you expect to end up with is shown in figure 95. You wish to find Fx ,
Fy and T , given m and θ.

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By now the idea should be sinking in. Static equilibrium requires
F = 0 and τ = 0. There are no forces in the z direction so we ignore it. There is only torque in the +z direction. In this problem there is a clearly-best pivot to choose – one at the point of contact with the wall, where the two forces Fx and Fy are exerted. If we choose this as our pivot, these forces will not contribute to the net torque!
Thus:
Fx − T cos(θ)

Fy + T sin(θ) − mg

T sin(θ)L − mgL/2

=

0

(658)

=

0

(659)

=

0

(660)

The last equation involves only T , so we solve it for:
T =

mg
2 sin(θ)

(661)

We can substitute this into the first equation and solve for Fx :
Fx =
Ditto for the second equation:

1 mg cot(θ)
2

(662)

1
1
Fy = mg − mg = mg
2
2

(663)

There are several features of interest in this solution. One is that the wire and the wall support each must support half of the weight of the sign. However, in order to accomplish this, the tension in the wire will be strictly greater than half the weight!
Consider θ = 30◦ . Then T = mg (the entire weight of the sign) and Fx = of the force exerted by the wall on the pole equals the tension.



3
2 mg.

The magnitude

Consider θ = 10◦ . Now T ≈ 2.9mg (which still must equal the magnitude of the force exerted by the wall. Why?). The smaller the angle, the larger the tension (and force exerted by/on the wall).
Make the angle too small, and your pole will punch right through the brick wall!

7.2.1: Equilibrium with a Vector Torque
So far we’ve only treated problems where all of the forces and moment arms live in a single plane (if not in a single direction). What if the moment arms themselves live in a plane? What if the forces exert torques in different directions?
Nothing changes a whole lot, actually. One simply has to set each component of the force and torque to zero separately. If anything, it may give us more equations to work with, and hence the ability to deal with more unknowns, at the cost of – naturally – some algebraic complexity.
Static equilibrium problems involving multiple torque directions are actually rather common.
Every time you sit in a chair, every time food is placed on a table, the legs increase the forces they supply to the seat to maintain force and torque equilibrium. In fact, every time any two-dimensional sheet of mass, such as a floor, a roof, a tray, a table is suspended horizontally, one must solve a problem in vector torque to keep it from rotating around any of the axes in the plane.
We don’t need to solve the most algebraically complex problems in the Universe in order to learn how to balance both multiple force components and multiple torque components, but we do need to do one or two to get the idea, because nearly everybody who is taking this course needs to be able to actually work with static equilibrium in multiple dimensions. Physicists need to be able to understand it both to teach it and to prime themselves for the full-blown theory of angular

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momentum in a more advanced course. Engineers, well, we don’t want those roofs to tip over, those bridges to fall down. Physicians and veterinarians – balancing human or animal bodies so that they don’t tip over one way or another seems like a good idea. Things like canes, four point walkers, special support shoes all are tools you may one day use to help patients retain the precious ability to navigate the world in an otherwise precarious vertical static equilibrium.

Example 7.2.4: Building a Deck

L=6

F2

F3 = 0

Mg (down/in)
W=4

1
F1

mg (down/in)

F4

1
Figure 96:
In figure 96 a very simple deck layout is shown. The deck is 4 meters wide and 6 meters long.
It is supported by four load-bearing posts, one at each corner. You would like to put a hot tub on the deck, one that has a loaded mass of m = 2000 kg, so that its center of mass is 1 meter in and
1 meter up from the lower left corner (as drawn) next to the house. The deck itself is a uniform concrete slab of mass M = 4000 kg with its center of mass is at (3,2) from the lower left corner.
You would like to know if putting the hot tub on the deck will exceed the safe load capacity of the nearest corner support. It seems to you that as it is loaded with the hot tub, it will actually reduce the load on F3 . So find F1 , F2 , and F4 when the deck is loaded in this way, assuming a perfectly rigid plane deck and F3 = 0.
First of all, we need to choose a pivot, and I’ve chosen a fairly obvious one – the lower left corner, where the x axis runs through F1 and F4 and the y axis runs through F1 and F2 .
Second, we need to note that
Fx and
Fy can be ignored – there are no lateral forces at all at work here. Gravity pulls the masses down, the corner beams push the deck itself up. We can solve the normal force and force transfer in our heads – supporting the hot tub, the deck experiences a force equal to the weight of the hot tub right below the center of mass of the hot tub.
This gives us only one force equation:
F1 + F2 + F4 − mg − M g = 0

(664)

That is, yes, the three pillars we’ve selected must support the total weight of the hot tub and deck together, since the F3 pillar refuses to help out.
It gives us two torque equations, as hopefully it is obvious that τz = 0! To make this nice and algebraic, we will set hx = 1, hy = 1 as the position of the hot tub, and use L/2 and W/2 as the

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Week 7: Statics position of the center of mass of the deck. τx τy

W
M g − hy mg = 0
2
L
= hx mg + M g − LF4 = 0
2

= W F2 −

(665)
(666)

Note that if F3 was acting, it would contribute to both of these torques and to the force above, and there would be an infinite number of possible solutions. As it is, though, solving this is pretty easy. Solve the last two equations for F2 and F4 respectively, then substitute the result into F1 .
Only at the end substitute numbers in and see roughly what F1 might be. Bear in mind that 1000 kg is a “metric ton” and weighs roughly 2200 pounds. So the deck and hot tub together, in this not-too-realistic problem, weigh over 6 tons!
Oops, we forgot the people in the hot tub and the barbecue grill at the far end and the furniture and the dog and the dancing. Better make the corner posts twice as strong as they need to be. Or even four times.
Once you see how this one goes, you should be ready to tackle the homework problem involving three legs, a tabletop, and a weight – same problem, really, but more complicated numbers.

7.3: Tipping
Another important application of the ideas of static equilibrium is to tipping problems. A tipping problem is one where one uses the ideas of static equilibrium to identify the particular angle or force combination that will marginally cause some object to tip over. Sometimes this is presented in the context of objects on an inclined plane, held in place by static friction, and a tipping problem can be combined with a slipping problem: determining if a block placed on an incline that is gradually raised tips first or slips first.
The idea of tipping is simple enough. An object placed on a flat surface is typically stable as long as the center of gravity is vertically inside the edges that are in contact with the surface, so that the torque created by the gravitational force around this limiting pivot is opposed by the torque exerted by the (variable) normal force.
That’s all there is to it! Look at the center of gravity, look at the corner or edge intuition tells you the object will “tip over”, done.

Example 7.3.1: Tipping Versus Slipping
In figure 97 a rectangular block of height H and width W is sitting on a rough plank that is gradually being raised at one end (so the angle it makes with the horizontal, θ, is slowly increasing).
At some angle we know that the block will start to slide. This will occur because the normal force is decreasing with the angle (and hence, so is the maximum force static friction can exert) and at the same time, the component of the weight of the object that points down the incline is increasing. Eventually the latter will exceed the former and the block will slide.
However, at some angle the block will also tip over. We know that this will happen because the normal force can only prevent the block from rotating clockwise (as drawn) around the pivot consisting of the lower left corner of the block. Unless the block has a magnetic lower surface, or a lower surface covered with velcro or glue, the plank cannot attract the lower surface of the block and prevent it from rotating counterclockwise.
As long as the net torque due to gravity (about this lower left pivot point) is into the page, the plank itself can exert a countertorque out of the page sufficient to keep the block from rotating down

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Week 7: Statics

W

H

pivot θ Figure 97: A rectangular block either tips first or slips (slides down the incline) first as the incline is gradually increased. Which one happens first? The figure is show with the block just past the tipping angle. through the plank. If the torque due to gravity is out of the page – as it is in the figure 97 above, when the center of gravity moves over and to the vertical left of the pivot corner – the normal force exerted by the plank cannot oppose the counterclockwise torque of gravity and the block will fall over. The tipping point, or tipping angle is thus the angle where the center of gravity is directly over the pivot that the object will “tip” around as it falls over.
A very common sort of problem here is to determine whether some given block or shape will tip first or slip first. This is easy to find. First let’s find the slipping angle θs . Let “down” mean “down the incline”. Then:
Fdown

= mg sin(θ) − Fs = 0

(667)

F⊥

= N − mg cos(θ) = 0

(668)

From the latter, as usual:
N = mg cos(θ)

(669)

max mg sin(θs ) = Fs
= µs cos(θs )

and Fs ≤

max
Fs

(670)

= µs N .

When the force of gravity down the incline precisely balances the force of static friction. We can solve for the angle where this occurs: θs = tan−1 (µs )
(671)
Now let’s determine the angle where it tips over. As noted, this is where the torque to to gravity around the pivot that the object will tip over changes sign from in the page (as drawn, stable) to out of the page (unstable, tipping over). This happens when the center of mass passes directly over the pivot.
From inspection of the figure (which is drawn very close to the tipping point) it should be clear that the tipping angle θt is given by: θt = tan−1

W
H

(672)

So, which one wins? The smaller of the two, θs or θt , of course – that’s the one that happens first as the plank is raised. Indeed, since both are inverse tangents, the smaller of: µs , W/H

(673)

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Week 7: Statics

determines whether the system slips first or tips first, no need to actually evaluate any tangents or inverse tangents!

Example 7.3.2: Tipping While Pushing
W
F
H

µs

h

Figure 98: A uniform rectangular block with dimensions W by H (which has its center of mass at
W/2, H/2) is pushed at a height h by a force F . The block sits on a horizontal smooth table with coefficient of static friction µs .
A uniform block of mass M being pushed by a horizontal outside force (say, a finger) a height h above the flat, smooth surface it is resting on (say, a table) as portrayed in figure 98. If it is pushed down low (small h) the block slides. If pushed up high (large h) it tips. Find a condition for the height h at which it tips and slips at the same time.
The solution here is much the same as the solution to the previous problem. We independently determine the condition for slipping, as that is rather easy, and then using the maximum force that can be applied without it quite slipping, find the height h at which the block barely starts to tip over. “Tipping over” in this case means that all of the normal force will be concentrated right at the pivot corner (the one the block will rotate around as it tips over) because the rest of the bottom surface is barely starting to leave the ground. All of the force of static friction is similarly concentrated at this one point. This is convenient to us, since neither one will therefore contribute to the torque around this point!
Conceptually, then, we seek the point h where, pushing with the maximum non-slipping force, the torque due to gravity alone is exactly equal to the torque exerted by the force F . This seems simple enough.
To find the force we need only to examine the usual force balance equations:
F − Fs

N − Mg

=

0

(674)

=

0

(675)

and hence N = M g (as usual) and: max Fmax = Fs
= µs N = µs M g

(676)

(also as usual). Hopefully by now you had this completely solved in your head before I even wrote it all down neatly and were saying to yourself “Fmax , yeah, sure, that’s µs M g, let’s get on with it...”
So we shall. Consider the torque around the bottom right hand corner of the block (which is clearly and intuitively the “tipping pivot” around which the block will rotate as it falls over when the torque due to F is large enough to overcome the torque of gravity). Let us choose the positive direction for torque to be out of the page. It should then be quite obvious that when the block is barely tipping over, so that we can ignore any torque due to N and Fs :
WMg
WMg
− hcrit Fmax =
− hcrit µs M g = 0
2
2

(677)

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Week 7: Statics

or (solving for hcrit , the critical height where it barely tips over even as it starts to slip): hcrit =

W
2µs

(678)

Now, does this make sense? If µs → 0 (a frictionless surface) we will never tip it before it starts to slide, although we might well push hard enough to tip it over in spite of it sliding. We note that in this limit, hcrit → ∞, which makes sense. On the other hand for finite µs if we let W become very small then hcrit similarly becomes very small, because the block is now very thin and is indeed rather precariously balanced.
The last bit of “sense” we need to worry about is hcrit compared to H. If hcrit is larger than
H, this basically means that we can’t tip the block over before it slips, for any reasonable µs . This limit will always be realized for W ≫ H. Suppose, for example, µs = 1 (the upper limit of “normal” values of the coefficient of static friction that doesn’t describe actual adhesion). Then hcrit = W/2, and if H < W/2 there is no way to push in the block to make it tip before it slips. If µs is more reasonable, say µs = 0.5, then only pushing at the very top of a block that is W × W in dimension marginally causes the block to tip. We can thus easily determine blocks “can” be tipped by a horizontal force and which ones cannot, just by knowing µs and looking at the blocks!

7.4: Force Couples

F2 r12 r2

F
1
r1

pivot
Figure 99: A Force Couple is a pair of equal and opposite forces that may or may not act along the line between the points where they are applied to a rigid object. Force couples exert a torque that is indenpedent of the pivot on an object and (of course) do not accelerate the center of mass of the object.
Two equal forces that act in opposite directions but not necessarily along the same line are called a force couple. Force couples are important both in torque and rotation problems and in static equilibrium problems. One doesn’t have to be able to name them, of course – we know everything we need to be able to handle the physics of such a pair already without a name.
One important property of force couples does stand out as being worth deriving and learning on its own, though – hence this section. Consider the total torque exerted by a force couple in the coordinate frame portrayed in figure ??: τ = r1 × F 1 + r2 × F 2

(679)

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Week 7: Statics
By hypothesis, F 2 = −F 1 , so: τ = r 1 × F 1 − r 2 × F 1 = (r 1 − r 2 ) × F 1 = r 12 × F 1

(680)

This torque no longer depends on the coordinate frame! It depends only on the difference between r 1 and r 2 , which is independent of coordinate system.
Note that we already used this property of couples when proving the law of conservation of angular momentum – it implies that internal Newton’s Third Law forces can exert no torque on a system independent of intertial reference frame. Here it has a slightly different implication – it means that if the net torque produced by a force couple is zero in one frame, it is zero in all frames! The idea of static equilibrium itself is independent of frame!
It also means that equilibrium implies that the vector sum of all forces form force couples in each coordinate direction that are equal and opposite and that ultimately pass through the center of mass of the system. This is a conceptually useful way to think about some tipping or slipping or static equilibrium problems.

Example 7.4.1: Rolling the Cylinder Over a Step
F

pivot

M
R
h

Figure 100:
One classic example of static equilibrium and force couples is that of a ball or cylinder being rolled up over a step. The way the problem is typically phrased is:
a) Find the minimum force F that must be applied (as shown in figure 100) to cause the cylinder to barely lift up off of the bottom step and rotate up around the corner of the next one, assuming that the cylinder does not slip on the corner of the next step.
b) Find the force exerted by that corner at this marginal condition.
The simplest way to solve this is to recognize the point of the term “barely”. When the force F is zero, gravity exerts a torque around the pivot out of the page, but the normal force of the tread of the lower stair exerts a countertorque precisely sufficient to keep the cylinder from rolling down into the stair itself. It also supports the weight of the cylinder. As F is increased, it exerts a torque around the pivot that is into the page, also opposing the gravitational torque, and the normal force decreases as less is needed to prevent rotation down into the step. At the same time, the pivot exerts a force that has to both oppose F (so the cylinder doesn’t translate to the right) and support more and more of the weight of the cylinder as the normal force supports less.
At some particular point, the force exerted by the step up will precisely equal the weight of the step down. The force exerted by the step to the left will exactly equal the force F to the right.
These forces will (vector) sum to zero and will incidentally exert no net torque either, as a pair of opposing couples.

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Week 7: Statics

That is enough that we could almost guess the answer (at least, if we drew some very nice pictures). However, we should work the problem algebraically to make sure that we all understand it. Let us assume that F = Fm , the desired minimum force where N → 0. Then (with out of the page positive): τ = mg R2 − (R − h)2 − Fm (2R − h) = 0
(681)
where I have used the r⊥ F form of the torque in both cases, and used the pythagorean theorem and/or inspection of the figure to determine r⊥ for each of the two forces. No torque due to N is present, so Fm in this case is indeed the minimum force F at the marginal point where rotation just starts to happen: mg R2 − (R − h)2
Fm =
(682)
2R − h
Next summing the forces in the x and y direction and solving for Fx and Fy exerted by the pivot corner itself we get:
Fx

= −Fm = −

Fy

= mg

mg

R2 − (R − h)2
2R − h

(683)
(684)

Obviously, these forces form a perfect couple such that the torques vanish.
That’s all there is to it! There are probably other questions one could ask, or other ways to ask the main question, but the idea is simple – look for the marginal static condition where rotation, or tipping, occur. Set it up algebraically, and then solve!

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Homework for Week 7

Problem 1.

Physics Concepts: Make this week’s physics concepts summary as you work all of the problems in this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s) they were key to, and include concepts from previous weeks as necessary. Do the work carefully enough that you can (after it has been handed in and graded) punch it and add it to a three ring binder for review and study come finals!

Problem 2.

x y F

r

Fy
Fx

In the figure above, a force
F = 2ˆ + 1ˆ x y
Newtons is applied to a disk at the point r = 2ˆ − 2ˆ x y as shown. (That is, Fx = 2 N, Fy = 1 N, x = 2 m, y = −2 m). Find the total torque about a pivot at the origin. Don’t forget that torque is a vector, so specify its direction as well as its magnitude
(or give the answer as a cartesian vector)! Show your work!

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Week 7: Statics

Problem 3.

lift pivot In the figure above, three shapes (with uniform mass distribution and thickness) are drawn sitting on a plane that can be tipped up gradually. Assuming that static friction is great enough that all of these shapes will tip over before they slide, rank them in the order they will tip over as the angle of the board they are sitting on is increased.

Problem 4.

F
M
R h This problem will help you learn required concepts such as:
• Static Equilibrium
• Torque (about selected pivots)
• Geometry of Right Triangles so please review them before you begin.
A cylinder of mass M and radius R sits against a step of height h = R/2 as shown above. A force F is applied at right angles to the line connecting the corner of the step and the center of the cylinder. All answers should be in terms of M , R, g.
a) Find the minimum value of |F | that will roll the cylinder over the step if the cylinder does not slide on the corner.
b) What is the force exerted by the corner (magnitude and direction) when that force F is being exerted on the center?

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Problem 5.

T?

m

M
D
θ

F?

d

This problem will help you learn required concepts such as:
• Static Equilibrium
• Force and Torque so please review them before you begin.
An exercising human person holds their arm of mass M and a barbell of mass m at rest at an angle θ with respect to the horizontal in an isometric curl as shown. Their bicep muscle that supports the suspended weight is connected at right angles to the bone a short distance d up from the elbow joint. The bone that supports the weight has length D.
a) Find the tension T in the muscle, assuming for the moment that the center of mass of the forearm is in the middle at D/2. Note that it is much larger than the weight of the arm and barbell combined, assuming a reasonable ratio of D/d ≈ 25 or thereabouts.
b) Find the force F (magnitude and direction) exerted on the supporting bone by the elbow joint in the geometry shown. Again, note that it is much larger than “just” the weight being supported. 330

Week 7: Statics

Problem 6.

w/2

d

w/3 m d/3

w
Top view

This problem will help you learn required concepts such as:
• Force Balance
• Torque Balance
• Static Equilibrium so please review them before you begin.
The figure below shows a mass m placed on a table consisting of three narrow cylindrical legs at the positions shown with a light (presume massless) sheet of Plexiglas placed on top. What is the vertical force exerted by the Plexiglas on each leg when the mass is in the position shown?

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Problem 7.

m
M
T

L

θ
P

This problem will help you learn required concepts such as:
• Force Balance
• Torque Balance
• Static Equilibrium so please review them before you begin.
A small round mass M sits on the end of a rod of length L and mass m that is attached to a wall with a hinge at point P . The rod is kept from falling by a thin (massless) string attached horizontally between the midpoint of the rod and the wall. The rod makes an angle θ with the ground. Find:
a) the tension T in the string;
b) the force F exerted by the hinge on the rod.

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Problem 8.

Ft d H
M

Fb

d
W

A door of mass M that has height H and width W is hung from two hinges located a distance d from the top and bottom, respectively. Assuming that the vertical weight of the door is equally distributed between the two hinges, find the total force (magnitude and direction) exerted by each hinge.
Neglect the mass of the doorknob and assume that the center of mass of the door is at W/2, H/2.
The force directions drawn for you are NOT likely to be correct or even close.

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Problem 9.

M

m

µs θ h

This problem will help you learn required concepts such as:
• Torque Balance
• Force Balance
• Static Equilibrium
• Static Friction so please review them before you begin.
In the figure above, a ladder of mass m and length L is leaning against a wall at an angle θ.
A person of mass M begins to climb the ladder. The ladder sits on the ground with a coefficient of static friction µs between the ground and the ladder. The wall is frictionless – it exerts only a normal force on the ladder.
If the person climbs the ladder, find the height h where the ladder slips.

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Optional Problems
The following problems are not required or to be handed in, but are provided to give you some extra things to work on or test yourself with after mastering the required problems and concepts above and to prepare for quizzes and exams.

Optional Problem 10.

boom
T?
m
30

L

M

45

F?

A crane with a boom (the long support between the body and the load) of mass m and length L holds a mass M suspended as shown. Assume that the center of mass of the boom is at L/2. Note that the wire with the tension T is fixed to the top of the boom, not run over a pulley to the mass
M.
a) Find the tension in the wire.
b) Find the force exerted on the boom by the crane body.
Note:
sin(30◦ ) = cos(60◦ ) =

1
2


3
2

2


sin(45 ) = cos(45 ) =
2

cos(30◦ ) = sin(60◦ ) =

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*

Optional Problem 11.

F
M
R h A cylinder of mass M and radius R sits against a step of height h = R/2 as shown above. A force F is applied parallel to the ground as shown. All answers should be in terms of M , R, g.
a) Find the minimum value of |F | that will roll the cylinder over the step if the cylinder does not slide on the corner.
b) What is the force exerted by the corner (magnitude and direction) when that force F is being exerted on the center?

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III: Applications of Mechanics

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Fluids Summary
• Fluids are states of matter characterized by a lack of long range order. They are characterized by their density ρ and their compressibility. Liquids such as water are (typically) relatively incompressible; gases can be significantly compressed. Fluids have other characteristics, for example viscosity (how “sticky” the fluid is). We will ignore these in this course.
• Pressure is the force per unit area exerted by a fluid on its surroundings:
P = F/A

(685)

Its SI units are pascals where 1 pascal = 1 newton/meter squared. Pressure is also measured in “atmospheres” (the pressure of air at or near sea level) where 1 atmosphere ≈ 105 pascals.
The pressure in an incompressible fluid varies with depth according to:
P = P0 + ρgD

(686)

where P0 is the pressure at the top and D is the depth.
• Pascal’s Principle Pressure applied to a fluid is transmitted undiminished to all points of the fluid.
• Archimedes’ Principle The buoyant force on an object
Fb = ρgVdisp

(687)

where frequency Vdisp is the volume of fluid displaced by an object.
• Conservation of Flow We will study only steady/laminar flow in the absence of turbulence and viscosity.
I = A1 v1 = A2 v2
(688)
where I is the flow, the volume per unit time that passes a given point in e.g. a pipe.
• For a circular smooth round pipe of length L and radius r carrying a fluid in laminar flow with dynamical viscosity 126 µ, the flow is related to the pressure difference across the pipe by the resistance R:
∆P = IR
(689)
126 Wikipedia: http://www.wikipedia.org/wiki/viscosity. We will defer any actual statement of how viscosity is related to forces until we cover shear stress in a couple of weeks. It’s just too much for now. Oh, and sorry about the symbol. Yes, we already have used µ for e.g. static and kinetic friction. Alas, we will use µ for still more things later. Even with both greek and roman characters to draw on, there just aren’t enough characters to cover all of the quantities we want to algebraically work with, so you have to get used to their reuse in different contexts that hopefully make them easy enough to keep straight. I decided that it is better to use the accepted symbol in this textbook rather than make one up myself or steal a character from, say, Urdu or a rune from Ancient Norse.

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It is worth noting that this is the fluid-flow version of Ohm’s Law, which you will learn next semester if you continue. We will generally omit the modifier “dynamical” from the term viscosity in this course, although there is actually another, equivalent measure of viscosity called the kinematic viscosity, ν = µ/ρ. The primary difference is the units – µ has the SI units of pascal-seconds where ν has units of meters square per second.
• The resistance R is given by the follow formula:
R=

8Lµ πr4 (690)

and the flow equation above becomes Poiseuille’s Law 127 :
I=

∆P πr4 ∆P
=
R
8Lµ

(691)

The key facts from this series of definitions are that flow increases linearly with pressure (so to achieve a given e.g. perfusion in a system of capillaries one requires a sufficient pressure difference across them), increases with the fourth power of the radius of the pipe (which is why narrowing blood vessels become so dangerous past a certain point) and decreases with the length (longer blood vessels have a greater resistance).
• If we neglect resistance (an idealization roughly equivalent to neglecting friction) and consider the flow of fluid in a closed pipe that can e.g. go up and down, the work-mechanical energy theorem per unit volume of the fluid can be written as Bernoulli’s Equation:
1
P + ρv 2 + ρgh = constant
2

(692)

• Venturi Effect At constant height, the pressure in a fluid decreases as the velocity of the fluid increases (the work done by the pressure difference is what speeds up the fluid!. This is responsible for e.g. the lift of an airplane wing and the force that makes a spinning baseball or golf ball curve.
• Torricelli’s Rule: If a fluid is flowing through a very small hole (for example at the bottom of a large tank) then the velocity of the fluid at the large end can be neglected in Bernoulli’s
Equation. In that case the exit speed is the same as the speed of a mass dropped the same distance: (693) v = 2gH where H is the depth of the hole relative to the top surface of the fluid in the tank.

8.1: General Fluid Properties
Fluids are the generic name given to two states of matter, liquids and gases128 characterized by a lack of long range order and a high degree of mobility at the molecular scale. Let us begin by visualizing fluids microscopically, since we like to build our understanding of matter from the ground up. 127 Wikipedia:

http://www.wikipedia.org/wiki/Hagen-Poiseuille equation. The derivation of this result isn’t horribly difficult or hard to understand, but it is long and beyond the scope of this course. Physics and math majors are encouraged to give it a peek though, if only to learn where it comes from.
128 We will not concern ourselves with “plasma” as a possible fourth state of matter in this class, viewing it as just an “ionized gas” although a very dense plasma might well be more like a liquid. Only physics majors and perhaps a few engineers are likely to study plasmas, and you have plenty of time to figure them out after you have learned some electromagnetic theory.

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Impulse

Figure 101: A large number of atoms or molecules are confined within in a “box”, where they bounce around off of each other and the walls. They exert a force on the walls equal and opposite the the force the walls exert on them as the collisions more or less elastically reverse the particles’ momenta perpendicular to the walls.
In figure 101 we see a highly idealized picture of what we might see looking into a tiny box full of gas. Many particles all of mass m are constantly moving in random, constantly changing directions
(as the particles collide with each other and the walls) with an average kinetic energy related to the temperature of the fluid. Some of the particles (which might be atoms such as helium or neon or molecules such as H2 or O2 ) happen to be close to the walls of the container and moving in the right direction to bounce (elastically) off of those walls.
When they do, their momentum perpendicular to those walls is reversed. Since many, many of these collisions occur each second, there is a nearly continuous momentum transfer between the walls and the gas and the gas and the walls. This transfer, per unit time, becomes the average force exerted by the walls on the gas and the gas on the walls (see the problem in Week 4 with beads bouncing off of a pan).
Eventually, we will transform this simple picture into the Kinetic Theory of Gases and use it to derive the venerable Ideal Gas Law (physicist style)129 :
P V = N kb T

(694)

but for now we will ignore the role of temperature and focus more on understanding the physical characteristics of the fluid such as its density, the idea of pressure itself and the force exerted by fluids on themselves (internally) and on anything the fluid presses upon along the lines of the particles above and the walls of the box.

8.1.1: Pressure
As noted above, the walls of the container exert an average force on the fluid molecules that confine them by reversing their perpendicular momenta in collisions. The total momentum transfer is proportional to the number of molecules that collide per unit time, and this in turn is (all things being equal) clearly proportional to the area of the walls. Twice the surface area – when confining the same number of molecules over each part – has to exert twice the force as twice the number of collisions occur per unit of time, each transferring (on average) the same impulse. It thus makes sense, when considering fluids, to describe the forces that confine and act on the fluids in terms of pressure, defined to be the force per unit area with which a fluid pushes on a confining wall or
129 Wikipedia:

http://www.wikipedia.org/wiki/Ideal Gas Law.

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the confining wall pushes on the fluid:
P =

F
A

(695)

Pressure gets its own SI units, which clearly must be Newtons per square meter. We give these units their own name, Pascals:
Newton
1 Pascal =
(696)
meter2
A Pascal is a tiny unit of pressure – a Newton isn’t very big, recall (one kilogram weighs roughly ten Newtons or 2.2 pounds) so a Pascal is the weight of a quarter pound spread out over a square meter. Writing out “pascal” is a bit cumbersome and you’ll see it sometimes abbreviated Pa (with the usual power-of-ten modifications, kPa, MPa, GPa, mPa and so on).
A more convenient measure of pressure in our everyday world is a form of the unit called a bar :
1 bar = 105 Pa = 100 kPa

(697)

As it happens, the average air pressure at sea level is very nearly 1 bar, and varies by at most a few percent on either side of this. For that reason, air pressure in the modern world is generally reported on the scale of millibars, for example you might see air pressure given as 959 mbar (characteristic of the low pressure in a major storm such as a hurricane), 1023 mbar (on a fine, sunny day).
The mbar is probably the “best” of these units for describing everyday air pressure (and its temporal and local and height variation without the need for a decimal or power of ten), with
Pascals being an equally good and useful general purpose arbitrary precision unit
The symbol atm stands for one standard atmosphere. The connection between atmospheres, bars, and pascals is:
1 standard atmosphere = 101.325 kPa = 1013.25 mbar

(698)

Note that real air pressure at sea level is most unlikely to be this exact value, and although this pressure is often referred to in textbooks and encylopedias as “the average air pressure at sea level” this is not, in fact, the case. The extra significant digits therefore refer only to a fairly arbitrary value (in pascals) historically related to the original definition of a standard atmosphere in terms of
“millimeters of mercury” or torr :
1 standard atmosphere = 760.00 mmHg = 760.00 torr

(699)

that is of no practical or immediate use. All of this is discussed in some detail in the section on barometers below.
In this class we will use the simple rule 1 bar ≈ 1 atm to avoid having to divide out the extra digits, just as we approximated g ≈ 10 when it is really closer to 9.8. This rule is more than adequate for nearly all purposes and makes pressure arithmetic something you can often do with fingers and toes or the back of an envelope, with around a 1% error if somebody actually gave a pressure in atmospheres with lots of significant digits instead of the superior pascal or bar SI units.
Note well: in the field of medicine blood pressures are given in mm of mercury (or torr) by long standing tradition (largely because for at least a century blood pressure was measured with a mercury-based sphygmomanometer). This is discussed further in the section below on the human heart and circulatory system. These can be converted into atmospheres by dividing by
760, remembering that one is measuring the difference between these pressures and the standard atmosphere (so the actual blood pressure is always greater than one atmosphere).
Pressure isn’t only exerted at the boundaries of fluids. Pressure also describes the internal transmission of forces within a fluid. For example, we will soon ask ourselves “Why don’t fluid molecules all fall to the ground under the influence of gravity and stay there?” The answer is that

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(at a sufficient temperature) the internal pressure of the fluid suffices to support the fluid above upon the back (so to speak) of the fluid below, all the way down to the ground, which of course has to support the weight of the entire column of fluid. Just as “tension” exists in a stretched string at all points along the string from end to end, so the pressure within a fluid is well-defined at all points from one side of a volume of the fluid to the other, although in neither case will the tension or pressure in general be constant.

8.1.2: Density
As we have done from almost the beginning, let us note that even a very tiny volume of fluid has many, many atoms or molecules in it, at least under ordinary circumstances in our everyday lives.
True, we can work to create a vacuum – a volume that has relatively few molecules in it per unit volume, but it is almost impossible to make that number zero – even the hard vacuum of outer space has on average one molecule per cubic meter or thereabouts130 . We live at the bottom of a gravity well that confines our atmosphere – the air that we breathe – so that it forms a relatively thick soup that we move through and breathe with order of Avogadro’s Number (6 × 1023 ) molecules per liter
– hundreds of billions of billions per cubic centimeter.
At this point we cannot possibly track the motion and interactions of all of the individual molecules, so we coarse grain and average. The coarse graining means that we once again consider volumes that are large relative to the sizes of the atoms but small relative to our macroscopic length scale of meters – cubic millimeters or cubic microns, for example – that are large enough to contain many, many molecules (and hence a well defined average number of molecules) but small enough to treat like a differential volume for the purposes of using calculus to add things up.
We could just count molecules in these tiny volumes, but the properties of oxygen molecules and helium molecules might well be very different, so the molecular count alone may not be the most useful quantity. Since we are interested in how forces might act on these small volumes, we need to know their mass, and thus we define the density of a fluid to be: dm ,
(700)
dV the mass per unit volume we are all familiar with from our discussions of the center of mass of continuous objects and moments of inertia of rigid objects. ρ= Although the definition itself is the same, the density of a fluid behaves in a manner that is similar, but not quite identical in its properties, to the density of a solid. The density of a fluid usually varies smoothly from one location to another, because an excess of density in one place will spread out as the molecules travel and collide to smooth out, on average. The particles in some fluids
(or almost any fluid at certain temperatures) are “sticky”, or strongly interacting, and hence the fluid coheres together in clumps where the particles are mostly touching, forming a liquid. In other fluids (or all fluids at higher temperatures) the molecules move so fast that they do not interact much and spend most of their time relatively far apart, forming a gas.
A gas spreads itself out to fill any volume it is placed in, subject only to forces that confine it such as the walls of containers or gravity. It assumes the shape of containers, and forms a (usually nearly spherical) layer of atmosphere around planets or stars when confined by gravity. Liquids also spread themselves out to some extent to fill containers they are place in or volumes they are confined to by a mix of surface forces and gravity, but they also have the property of surface tension that can permit a liquid to exert a force of confinement on itself. Hence water fills a glass, but water also forms nearly spherical droplets when falling freely as surface tension causes the droplet to minimize its surface area relative to its volume, forming a sphere.
130 Wikipedia: http://www.wikipedia.org/wiki/Vacuum. Vacuum is, of course, “nothing”, and if you take the time to read this Wikipedia article on it you will realize that even nothing can be pretty amazing. In man-made vacuums, there are nearly always as many as hundreds of molecules per cubic centimeter.

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Surface chemistry or surface adhesion can also exert forces on fluids and initiate things like capillary flow of e.g. water up into very fine tubes, drawn there by the surface interaction of the hydrophilic walls of the tube with the water. Similarly, hydrophobic materials can actually repel water and cause water to bead up instead of spreading out to wet the surface. We will largely ignore these phenomena in this course, but they are very interesting and are actually useful to physicians as they use pipettes to collect fluid samples that draw themselves up into sample tubes as if by magic. It’s not magic, it’s just physics.

8.1.3: Compressibility
A major difference between fluids and solids, and liquids and gases within the fluids, is the compressibility of these materials. Compressibility describes how a material responds to changes in pressure.
Intuitively, we expect that if we change the volume of the container (making it smaller, for example, by pushing a piston into a confining cylinder) while holding the amount of material inside the volume constant we will change the pressure; a smaller volume makes for a larger pressure. Although we are not quite prepared to derive and fully justify it, it seems at least reasonable that this can be expressed as a simple linear relationship:
∆P = −B

∆V
V

(701)

Pressure up, volume down and vice versa, where the amount it goes up or down is related, not unreasonably, to the total volume that was present in the first place. The constant of proportionality
B is called the bulk modulus of the material, and it is very much like (and closely related to) the spring constant in Hooke’s Law for springs.
Note well that we haven’t really specified yet whether the “material” is solid, liquid or gas.
All three of them have densities, all three of them have bulk moduli. Where they differ is in the qualitative properties of their compressibility.
Solids are typically relatively incompressible (large B), although there are certainly exceptions.
The have long range order – all of the molecules are packed and tightly bonded together in structures and there is usually very little free volume. Atoms themselves violently oppose being “squeezed together” because of the Pauli exclusion principle that forbids electrons from having the same set of quantum numbers as well as straight up Coulomb repulsion that you will learn about next semester. Liquids are also relatively incompressible (large B). They differ from solids in that they lack long range order. All of the molecules are constantly moving around and any small “structures” that appear due to local interaction are short-lived. The molecules of a liquid are close enough together that there is often significant physical and chemical interaction, giving rise to surface tension and wetting properties – especially in water, which is (as one sack of water speaking to another) an amazing fluid!
Gases are in contrast quite compressible (small B). One can usually squeeze gases smoothly into smaller and smaller volumes, until they reach the point where the molecules are basically all touching and the gas converts to a liquid! Gases per se (especially hot gases) usually remain “weakly interacting” right up to where they become a liquid, although the correct (non-ideal) equation of state for a real gas often displays features that are the results of moderate interaction, depending on the pressure and temperature.
Water 131 is, as noted, a remarkable liquid. H2 O is a polar molecules with a permanent dipole moment, so water molecules are very strongly interacting, both with each other and with other materials. It organizes itself quickly into a state of relative order that is very incompressible. The
131 Wikipedia:

http://www.wikipedia.org/wiki/Properties of Water. As I said, water is amazing. This article is well worth reading just for fun.

Week 8: Fluids

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bulk modulus of water is 2.2 × 109 Pa, which means that even deep in the ocean where pressures can be measured in the tens of millions of Pascals (or hundreds of atmospheres) the density of water only varies by a few percent from that on the surface. Its density varies much more rapidly with temperature than with pressure132 . We will idealize water by considering it to be perfectly incompressible in this course, which is close enough to true for nearly any mundane application of hydraulics that you are most unlikely to ever observe an exception that matters.

8.1.4: Viscosity and fluid flow
Fluids, whether liquid or gas, have some internal “stickiness” that resists the relative motion of one part of the fluid compared to another, a kind of internal “friction” that tries to equilibrate an entire body of fluid to move together. They also interact with the walls of any container in which they are confined. The viscosity of a fluid (symbol µ) is a measure of this internal friction or stickiness.
Thin fluids have a low viscosity and flow easily with minimum resistance; thick sticky fluids have a high viscosity and resist flow.
Fluid, when flowing through (say) a cylindrical pipe tends to organize itself in one of two very different ways – a state of laminar flow where the fluid at the very edge of the flowing volume is at rest where it is in contact with the pipe and the speed concentrically and symmetrically increases to a maximum in the center of the pipe, and turbulent flow where the fluid tumbles and rolls and forms eddies as it flows through the pipe. Turbulence and flow and viscosity are properties that will be discussed in more detail below.

8.1.5: Properties Summary
To summarize, fluids have the following properties that you should conceptually and intuitively understand and be able to use in working fluid problems:
• They usually assume the shape of any vessel they are placed in (exceptions are associated with surface effects such as surface tension and how well the fluid adheres to the surface in question). • They are characterized by a mass per unit volume density ρ.
• They exert a pressure P (force per unit area) on themselves and any surfaces they are in contact with.
• The pressure can vary according to the dynamic and static properties of the fluid.
• The fluid has a measure of its “stickiness” and resistance to flow called viscosity. Viscosity is the internal friction of a fluid, more or less. We will treat fluids as being “ideal” and ignore viscosity in this course.
• Fluids are compressible – when the pressure in a fluid is increased, its volume descreases according to the relation:
∆V
∆P = −B
(702)
V where B is called the bulk modulus of the fluid (the equivalent of a spring constant).
• Fluids where B is a large number (so large changes in pressure create only tiny changes in fractional volume) are called incompressible. Water is an example of an incompressible fluid.
132 A fact that impacts my beer-making activities quite significantly, as the specific gravity of hot wort fresh off of the boil is quite different from the specific gravity of the same wort cooled to room temperature. The specific gravity of the wort is related to the sugar content, which is ultimately related to the alcohol content of the fermented beer.
Just in case this interests you...

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Week 8: Fluids
• Below a critical speed, the dynamic flow of a moving fluid tends to be laminar, where every bit of fluid moves parallel to its neighbors in response to pressure differentials and around obstacles. Above that speed it becomes turbulent flow. Turbulent flow is quite difficult to treat mathematically and is hence beyond the scope of this introductory course – we will restrict our attention to ideal fluids either static or in laminar flow.

We will now use these general properties and definitions, plus our existing knowledge of physics, to deduce a number of important properties of and laws pertaining to static fluids, fluids that are in static equilibrium.

Static Fluids
8.1.6: Pressure and Confinement of Static Fluids
∆A

F left

F right

Fluid (density ρ)
∆V
Confining box
Figure 102: A fluid in static equilibrium confined to a sealed rectilinear box in zero gravity.
In figure 102 we see a box of a fluid that is confined within the box by the rigid walls of the box. We will imagine that this particular box is in “free space” far from any gravitational attractor and is therefore at rest with no external forces acting on it. We know from our intuition based on things like cups of coffee that no matter how this fluid is initially stirred up and moving within the container, after a very long time the fluid will damp down any initial motion by interacting with the walls of the container and arrive at static equilibrium 133 .
A fluid in static equilibrium has the property that every single tiny chunk of volume in the fluid has to independently be in force equilibrium – the total force acting on the differential volume chunk must be zero. In addition the net torques acting on all of these differential subvolumes must be zero, and the fluid must be at rest, neither translating nor rotating. Fluid rotation is more complex than the rotation of a static object because a fluid can be internally rotating even if all of the fluid in the outermost layer is in contact with a contain and is stationary. It can also be turbulent – there can be lots of internal eddies and swirls of motion, including some that can exist at very small length scales and persist for fair amounts of time. We will idealize all of this – when we discuss static properties of fluids we will assume that all of this sort of internal motion has disappeared.
We can now make a few very simple observations about the forces exerted by the walls of the container on the fluid within. First of all the mass of the fluid in the box above is clearly:
∆M = ρ∆V
133 This

(703)

state will also entail thermodynamic equilibrium with the box (which must be at a uniform temperature) and hence the fluid in this particular non-accelerating box has a uniform density.

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Week 8: Fluids

where ∆V is the volume of the box. Since it is at rest and remains at rest, the net external force exerted on it (only) by the the box must be zero (see Week 4). We drew a symmetric box to make it easy to see that the magnitudes of the forces exerted by opposing walls are equal Fleft = Fright
(for example). Similarly the forces exerted by the top and bottom surfaces, and the front and back surfaces, must cancel.
The average velocity of the molecules in the box must be zero, but the molecules themselves will generally not be at rest at any nonzero temperature. They will be in a state of constant motion where they bounce elastically off of the walls of the box, both giving and receiving an impulse (change in momentum) from the walls as they do. The walls of any box large enough to contain many molecules thus exerts a nearly continuous nonzero force that confines any fluid not at zero temperature134 .
From this physical picture we can also deduce an important scaling property of the force exerted by the walls. We have deliberately omitted giving any actual dimensions to our box in figure 102.
Suppose (as shown) the cross-sectional area of the left and right walls are ∆A originally. Consider now what we expect if we double the size of the box and at the same time add enough additional fluid for the fluid density to remain the same, making the side walls have the area 2∆A. With twice the area (and twice the volume and twice as much fluid), we have twice as many molecular collisions per unit time on the doubled wall areas (with the same average impulse per collision). The average force exerted by the doubled wall areas therefore also doubles.
From this simple argument we can conclude that the average force exerted by any wall is proportional to the area of the wall. This force is therefore most naturally expressible in terms of pressure, for example:
Fleft = Pleft ∆A = Pright ∆A = Fright

(704)

which implies that the pressure at the left and right confining walls is the same:
Pleft = Pright = P

(705)

, and that this pressure describes the force exerted by the fluid on the walls and vice versa. Again, the exact same thing is true for the other four sides.
There is nothing special about our particular choice of left and right. If we had originally drawn a cubic box (as indeed we did) we can easily see that the pressure P on all the faces of the cube must be the same and indeed (as we shall see more explicitly below) the pressure everywhere in the fluid must be the same!
That’s quite a lot of mileage to get out of symmetry and the definition of static equilibrium, but there is one more important piece to get. This last bit involves forces exerted by the wall parallel to its surface. On average, there cannot be any! To see why, suppose that one surface, say the left one, exerted a force tangent to the surface itself on the fluid in contact with that surface. An important property of fluids is that one part of a fluid can move independent of another so the fluid in at least some layer with a finite thickness near the wall would therefore experience a net force and would accelerate. But this violates our assumption of static equilibrium, so a fluid in static equilibrium exerts no tangential force on the walls of a confining container and vice versa.
We therefore conclude that the direction of the force exerted by a confining surface with an area
∆A on the fluid that is in contact with it is:
F = P ∆Aˆ n (706)

where n is an inward-directed unit vector perpendicular to (normal to) the surface. This final rule
ˆ
permits us to handle the force exerted on fluids confined to irregular amoebic blob shaped containers, or balloons, or bags, or – well, us, by our skins and vascular system.
134 Or

at a temperature low enough for the fluid to freeze and becomes a solid

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Week 8: Fluids

Note well that this says nothing about the tangential force exerted by fluids in relative motion to the walls of the confining container. We already know that a fluid moving across a solid surface will exert a drag force, and later this week we’ll attempt to at least approximately quantify this.
Next, let’s consider what happens when we bring this box of fluid135 down to Earth and consider what happens to the pressure in a box in near-Earth gravity.

8.1.7: Pressure and Confinement of Static Fluids in Gravity
The principle change brought about by setting our box of fluid down on the ground in a gravitational field (or equivalently, accelerating the box of fluid uniformly in some direction to develop a pseudogravitational field in the non-inertial frame of the box) is that an additional external force comes into play: The weight of the fluid. A static fluid, confined in some way in a gravitational field, must support the weight of its many component parts internally, and of course the box itself must support the weight of the entire mass ∆M of the fluid.
As hopefully you can see if you carefully read the previous section. The only force available to provide the necessary internal support or confinement force is the variation of pressure within the fluid. We would like to know how the pressure varies as we move up or down in a static fluid so that it supports its own weight.
There are countless reasons that this knowledge is valuable. It is this pressure variation that hurts your ears if you dive deep into the water or collapses submarines if they dive too far. It is this pressure variation that causes your ears to pop as you ride in a car up the side of a mountain or your blood to boil into the vacuum of space if you ride in a rocket all the way out of the atmosphere without a special suit or vehicle that provides a personally pressurized environment. It is this pressure variation that will one day very likely cause you to have varicose veins and edema in your lower extremities from standing on your feet all day – and can help treat/reverse both if you stand in 1.5 meter deep water instead of air. The pressure variation drives water out of the pipes in your home when you open the tap, helps lift a balloon filled with helium, floats a boat but fails to float a rock.
We need to understand all of this, whether our eventual goal is to become a physicist, a physician, an engineer, or just a scientifically literate human being. Let’s get to it.
Here’s the general idea. If we consider a tiny (eventually differentially small) chunk of fluid in force equilibrium, gravity will pull it down and the only thing that can push it up is a pressure difference between the top and the bottom of the chunk. By requiring that the force exerted by the pressure difference balance the weight, we will learn how the pressure varies with increasing depth.
For incompressible fluids, this is really all there is to it – it takes only a few lines to derive a lovely formula for the increase in pressure as a function of depth in an incompressible liquid.
For gases there is, alas, a small complication. Compressible fluids have densities that increase as the pressure increases. This means that boxes of the same size also have more mass in them as one descends. More mass means that the pressure difference has to increase, faster, which makes the density/mass greater still, and one discovers (in the end) that the pressure varies exponentially with depth. Hence the air pressure drops relatively quickly as one goes up from the Earth’s surface to very close to zero at a height of ten miles, but the atmosphere itself extends for a very long way into space, never quite dropping to “zero” even when one is twenty or a hundred miles high.
As it happens, the calculus for the two kinds of fluids is the same up to a given (very important) common point, and then differs, becoming very simple indeed for incompressible fluids and a bit more difficult for compressible ones. Simple solutions suffice to help us build our conceptual understanding; we will therefore treat incompressible fluids first and everybody is responsible for understanding
135 It’s

just a box of rain. I don’t know who put it there...

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them. Physics majors, math majors, engineers, and people who love a good bit of calculus now and then should probably continue on and learn how to integrate the simple model provided for compressible fluids.

x
P(0) = P0
0

y z ∆y
∆x

Ft
Fl

P(z)
Fr
z + ∆z

∆z
Fb
∆mg

P(z + ∆ z)
)

Fluid (density ρ) z Figure 103: A fluid in static equilibrium confined to a sealed rectilinear box in a near-Earth gravitational field g. Note well the small chunk of fluid with dimensions ∆x, ∆y, ∆z in the middle of the fluid. Also note that the coordinate system selected has z increasing from the top of the box down, so that z can be thought of as the depth of the fluid.
In figure 103 a (portion of) a fluid confined to a box is illustrated. The box could be a completely sealed one with rigid walls on all sides, or it could be something like a cup or bucket that is open on the top but where the fluid is still confined there by e.g. atmospheric pressure.
Let us consider a small (eventually infinitesimal) chunk of fluid somewhere in the middle of the container. As shown, it has physical dimensions ∆x, ∆y, ∆z; its upper surface is a distance z below the origin (where z increases down and hence can represent “depth”) and its lower surface is at depth z + ∆z. The areas of the top and bottom surfaces of this small chunk are e.g. ∆Atb = ∆x∆y, the areas of the sides are ∆x∆z and ∆y∆z respectively, and the volume of this small chunk is
∆V = ∆x∆y∆z.
This small chunk is itself in static equilibrium – therefore the forces between any pair of its horizontal sides (in the x or y direction) must cancel. As before (for the box in space) Fl = Fr in magnitude (and opposite in their y-direction) and similarly for the force on the front and back faces in the x-direction, which will always be true if the pressure does not vary horizontally with variations in x or y. In the z-direction, however, force equilibrium requires that:
Ft + ∆mg − Fb = 0

(707)

(where recall, down is positive).
The only possible source of Ft and Fb are the pressure in the fluid itself which will vary with the depth z: Ft = P (z)∆Atb and Fb = P (z + ∆z)∆Atb . Also, the mass of fluid in the (small) box is ∆m = ρ∆V (using our ritual incantation “the mass of the chunks is...”). We can thus write:
P (z)∆x∆y + ρ(∆x∆y∆z)g − P (z + ∆z)∆x∆y = 0

(708)

or (dividing by ∆x∆y∆z and rearranging):
∆P
P (z + ∆z) − P (z)
=
= ρg
∆z
∆z

(709)

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Week 8: Fluids
Finally, we take the limit ∆z → 0 and identify the definition of the derivative to get: dP = ρg dz (710)

Identical arguments but without any horizontal external force followed by ∆x → 0 and ∆y → 0 lead to: dP dP =
=0
(711) dx dy as well – P does not vary with x or y as already noted136 .
In order to find P (z) from this differential expression (which applies, recall, to any confined fluid in static equilibrium in a gravitational field) we have to integrate it. This integral is very simple if the fluid is incompressible because in that case ρ is a constant. The integral isn’t that difficult if ρ is not a constant as implied by the equation we wrote above for the bulk compressibility. We will therefore first do incompressible fluids, then compressible ones.

8.1.8: Variation of Pressure in Incompressible Fluids
In the case of incompressible fluids, ρ is a constant and does not vary with pressure and/or depth.
Therefore we can easily multiple dP/dz = ρg above by dz on both sides and integrate to find P : dP = ρg dz

dP

=

ρg dz

P (z) = ρgz + P0

(712)

where P0 is the constant of integration for both integrals, and practically speaking is the pressure in the fluid at zero depth (wherever that might be in the coordinate system chosen).

Example 8.1.1: Barometers
Mercury barometers were originally invented by Evangelista Torricelli137 a natural philosopher who acted as Galileo’s secretary for the last three months of Galileo’s life under house arrest. The invention was inspired by Torricelli’s attempt to solve an important engineering problem. The pump makers of the Grand Duke of Tuscany had built powerful pumps intended to raise water twelve or more meters, but discovered that no matter how powerful the pump, water stubbornly refused to rise more than ten meters into a pipe evacuated at the top.
Torricelli demonstrated that a shorter glass tube filled with mercury, when inverted into a dish of mercury, would fall back into a column with a height of roughly 0.76 meters with a vacuum on top, and soon thereafter discovered that the height of the column fluctuated with the pressure of the outside air pressing down on the mercury in the dish, correctly concluding that water would behave exactly the same way138 . Torricelli made a number of other important 17th century discoveries, correctly describing the causes of wind and discovering “Torricelli’s Law” (an aspect of the Bernoulli
Equation we will note below).
In honor of Torricelli, a unit of pressure was named after him. The torr is the pressure required to push the mercury in Torricelli’s barometer up one millimeter. Because mercury barometers
136 Physics and math majors and other students of multivariate calculus will recognize that I should probably be using partial derivatives here and establishing that ∇P = ρg, where in free space we should instead have had ∇P = 0 → P constant. 137 Wikipedia: http://www.wikipedia.org/wiki/Evangelista Torricelli. ,
138 You, too, get to solve Torricelli’s problem as one of your homework problems, but armed with a lot better understanding. 351

Week 8: Fluids

were at one time nearly ubiquitous as the most precise way to measure the pressure of the air, a specific height of the mercury column was the original definition of the standard atmosphere. For better or worse, Torricelli’s original observation defined one standard atmosphere to be exactly “760 millimeters of mercury” (which is a lot to write or say) or as we would now say, “760 torr”139 .
Mercury barometers are now more or less banned, certainly from the workplace, because mercury is a potentially toxic heavy metal. In actual fact, liquid mercury is not biologically active and hence is not particularly toxic. Mercury vapor is toxic, but the amount of mercury vapor emitted by the exposed surface of a mercury barometer at room temperature is well below the levels considered to be a risk to human health by OSHA unless the barometer is kept in a small, hot, poorly ventilated room with someone who works there over years. This isn’t all that common a situation, but with all toxic metals we are probably better safe than sorry140 .
At this point mercury barometers are rapidly disappearing everywhere but from the hands of collectors. Their manufacture is banned in the U.S., Canada, Europe, and many other nations. We had a lovely one (probably more than one, but I recall one) in the Duke Physics Department up until sometime in the 90’s141 , but it was – sanely enough – removed and retired during a renovation that also cleaned up most if not all of the asbestos in the building. Ah, my toxic youth...
Still, at one time they were extremely common – most ships had one, many households had one, businesses and government agencies had them – knowing the pressure of the air is an important factor in weather prediction. Let’s see how they work(ed).

P=0

H

P = P0

Figure 104: A simple fluid barometer consists of a tube with a vacuum at the top filled with fluid supported by the air pressure outside.
A simple mercury barometer is shown in figure 104. It consists of a tube that is completely
139 Not

to be outdone, one standard atmosphere (or atmospheric pressures in weather reporting) in the U.S. is often given as 29.92 barleycorn-derived inches of mercury instead of millimeters. Sigh.
140 The single biggest risk associated with uncontained liquid mercury (in a barometer or otherwise) is that you can easily spill it, and once spilled it is fairly likely to sooner or later make its way into either mercury vapor or methyl mercury, both of which are biologically active and highly toxic. Liquid mercury itself you could drink a glass of and it would pretty much pass straight through you with minimal absorption and little to no damage if you – um –
“collected” it carefully and disposed of it properly on the other side.
141 I used to work in the small, cramped space with poor circulation where it was located from time to time but never very long at a time and besides, the room was cold. But if I seem “mad as a hatter” – mercury nitrate was used in the making of hats and the vapor used to poison the hatters vapor used to poison hat makers – it probably isn’t from this... 352

Week 8: Fluids

filled with mercury. Mercury has a specific gravity of 13.534 at a typical room temperature, hence a density of 13534 kg/m3 ). The filled tube is then inverted into a small reservoir of mercury (although other designs of the bottom are possible, some with smaller exposed surface area of the mercury).
The mercury falls (pulled down by gravity) out of the tube, leaving behind a vacuum at the top.
We can easily compute the expected height of the mercury column if P0 is the pressure on the exposed surface of the mercury in the reservoir. In that case
P = P0 + ρgz

(713)

as usual for an incompressible fluid. Applying this formula to both the top and the bottom,
P (0) = P0

(714)

P (H) = P0 − ρgH = 0

(715)

and

(recall that the upper surface is above the lower one, z = −H). From this last equation:
P0 = ρgH

(716)

and one can easily convert the measured height H of mercury above the top surface of mercury in the reservoir into P0 , the air pressure on the top of the reservoir.
At one standard atmosphere, we can easily determine what a mercury barometer at room temperature will read (the height H of its column of mercury above the level of mercury in the reservoir):
P0 = 13534

kg m × 9.80665 2 × H = 101325Pa m3 sec

(717)

Note well, we have used the precise SI value of g in this expression, and the density of mercury at
“room temperature” around 20◦ C or 293◦ K. Dividing, we find the expected height of mercury in a barometer at room temperature and one standard atmosphere is H = 0.763 meters or 763 torr
Note that this is not exactly the 760 torr we expect to read for a standard atmosphere. This is because for high precision work one cannot just use any old temperature (because mercury has a significant thermal expansion coefficient and was then and continues to be used today in mercury thermometers as a consequence). The unit definition is based on using the density of mercury at 0◦ C or 273.16◦ K, which has a specific gravity (according to NIST, the National Institute of Standards in the US) of 13.595. Then the precise connection between SI units and torr follows from:
P0 = 13595

kg m × 9.80665 2 × H = 101325Pa m3 sec

(718)

Dividing we find the value of H expected at one standard atmosphere:
Hatm = 0.76000 = 760.00 millimeters

(719)

Note well the precision, indicative of the fact that the SI units for a standard atmosphere follow from their definition in torr, not the other way around.
Curiously, this value is invariably given in both textbooks and even the wikipedia article on atmospheric pressure as the average atmospheric pressure at sea level, which it almost certainly is not – a spatiotemporal averaging of sea level pressure would have been utterly impossible during
Torricelli’s time (and would be difficult today!) and if it was done, could not possibly have worked out to be exactly 760.00 millimeters of mercury at 273.16◦ K.

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Week 8: Fluids

Example 8.1.2: Variation of Oceanic Pressure with Depth
The pressure on the surface of the ocean is, approximately, by definition, one atmosphere. Water is a highly incompressible fluid with ρw = 1000 kilograms per cubic meter142 . g ≈ 10 meters/second2 .
Thus:
P (z) = P0 + ρw gz = 105 + 104 z Pa
(720)
or
P (z) = (1.0 + 0.1z) bar = (1000 + 100z) mbar

(721)

Every ten meters of depth (either way) increases water pressure by (approximately) one atmosphere!
Wow, that was easy. This is a very important rule of thumb and is actually fairly easy to remember! How about compressible fluids?

8.1.9: Variation of Pressure in Compressible Fluids
Compressible fluids, as noted, have a density which varies with pressure. Recall our equation for the compressibility:
∆V
∆P = −B
(722)
V
If one increases the pressure, one therefore decreases occupied volume of any given chunk of mass, and hence increases the density. However, to predict precisely how the density will depend on pressure requires more than just this – it requires a model relating pressure, volume and mass.
Just such a model for a compressible gas is provided (for example) by the Ideal Gas Law 143 :
P V = N kb T = nRT

(723)

where N is the number of molecules in the volume V , kb is Boltzmann’s constant144 n is the number of moles of gas in the volume V , R is the ideal gas constant145 and T is the temperature in degrees
Kelvin (or Absolute)146 . If we assume constant temperature, and convert N to the mass of the gas by multiplying by the molar mass and dividing by Avogadro’s Number147 6 × 1023 .
(Aside: If you’ve never taken chemistry a lot of this is going to sound like Martian to you.
Sorry about that. As always, consider visting the e.g. Wikipedia pages linked above to learn enough about these topics to get by for the moment, or just keep reading as the details of all of this won’t turn out to be very important...)
When we do this, we get the following formula for the density of an ideal gas: ρ= M
P
RT

(724)

where M is the molar mass148 , the number of kilograms of the gas per mole. Note well that this result is idealized – that’s why they call it the Ideal Gas Law! – and that no real gases are
“ideal” for all pressures and temperatures because sooner or later they all become liquids or solids due to molecular interactions. However, the gases that make up “air” are all reasonably ideal at temperatures in the ballpark of room temperature, and in any event it is worth seeing how the pressure of an ideal gas varies with z to get an idea of how air pressure will vary with height.
Nature will probably be somewhat different than this prediction, but we ought to be able to make a qualitatively accurate model that is also moderately quantitatively predictive as well.
142 Good

number to remember. In fact, great number to remember. http://www.wikipedia.org/wiki/Ideal Gas Law.
144 Wikipedia: http://www.wikipedia.org/wiki/Boltzmann’s Constant. ,
145 Wikipedia: http://www.wikipedia.org/wiki/Gas Constant. ,
146 Wikipedia: http://www.wikipedia.org/wiki/Temperature.
147 Wikipedia: http://www.wikipedia.org/wiki/Avogadro’s Number.
148 Wikipedia: http://www.wikipedia.org/wiki/Molar Mass.
143 Wikipedia:

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Week 8: Fluids

As mentioned above, the formula for the derivative of pressure with z is unchanged for compressible or incompressible fluids. If we take dP/dz = ρg and multiply both sides by dz as before and integrate, now we get (assuming a fixed temperature T ): dP dP
P
dP
P
ln(P )

= ρg dz =
=
=
=

Mg
P dz
RT

Mg dz RT
Mg
dz
RT
Mg z+C RT

We now do the usual149 – exponentiate both sides, turn the exponential of the sum into the product of exponentials, turn the exponential of a constant of integration into a constant of integration, and match the units:
Mg
P (z) = P e( RT )z
(725)
0

where P0 is the pressure at zero depth, because (recall!) z is measured positive down in our expression for dP/dz.

Example 8.1.3: Variation of Atmospheric Pressure with Height
Using z to describe depth is moderately inconvenient, so let us define the height h above sea level to be −z. In that case P0 is (how about that!) 1 Atmosphere. The molar mass of dry air is M = 0.029 kilograms per mole. R = 8.31 Joules/(mole-K◦ ). Hence a bit of multiplication at
T = 300◦ :
0.029 × 10
Mg
=
= 1.12 × 10−4 meters−1
(726)
RT
8.31 × 300 or: P (h) = 105 exp(−0.00012 h) Pa = 1000 exp(−0.00012 h) mbar

(727)

Note well that the temperature of air is not constant as one ascends – it drops by a fairly significant amount, even on the absolute scale (and higher still, it rises by an even greater amount before dropping again as one moves through the layers of the atmosphere. Since the pressure is found from an integral, this in turn means that the exponential behavior itself is rather inexact, but still it isn’t a terrible predictor of the variation of pressure with height. This equation predicts that air pressure should drop to 1/e of its sea-level value of 1000 mbar at a height of around 8000 meters, the height of the so-called death zone 150 . We can compare the actual (average) pressure at 8000 meters, 356 mbar, to 1000 × e−1 = 368 mbar. We get remarkably good agreement!
This agreement rapidly breaks down, however, and meteorologists actually use a patchwork of formulae (both algebraic and exponential) to give better agreement to the actual variation of air pressure with height as one moves up and down through the various named layers of the atmosphere with the pressure, temperature and even molecular composition of “air” varying all the way. This simple model explains a lot of the variation, but its assumptions are not really correct.
149 Which should be familiar to you both from solving the linear drag problem in Week 2 and from the online Math
Review.
150 Wikipedia: http://www.wikipedia.org/wiki/Effects of high altitude on humans. This is the height where air pressure drops to where humans are at extreme risk of dying if they climb without supplemental oxygen support – beyond this height hypoxia reduces one’s ability to make important and life-critical decisions during the very last, most stressful, part of the climb. Mount Everest (for example) can only be climbed with oxygen masks and some of the greatest disasters that have occurred climbing it and other peaks are associated with a lack of or failure of supplemental oxygen. 355

Week 8: Fluids

8.2: Pascal’s Principle and Hydraulics
We note that (from the above) the general form of P of a fluid confined to a sealed container has the most general form: z ρgdz

P (z) = P0 +

(728)

0

where P0 is the constant of integration or value of the pressure at the reference depth z = 0. This has an important consequence that forms the basis of hydraulics.

F
Fp

Piston

z

P0

P(z)

x z z
Figure 105: A single piston seated tightly in a frictionless cylinder of cross-sectional area A is used to compress water in a sealed container. Water is incompressible and does not significantly change its volume at P = 1 bar (and a constant room temperature) for pressure changes on the order of
0.1-100 bar.
Suppose, then, that we have an incompressible fluid e.g. water confined within a sealed container by e.g. a piston that can be pushed or pulled on to increase or decrease the confinement pressure on the surface of the piston. Such an arrangement is portrayed in figure 105.
We can push down (or pull back) on the piston with any total downward force F that we like that leaves the system in equilibrium. Since the piston itself is in static equilibrium, the force we push with must be opposed by the pressure in the fluid, which exerts an equal and opposite upwards force: F = Fp = P 0 A

(729)

where A is the cross sectional area of the piston and where we’ve put the cylinder face at z = 0, which we are obviously free to do. For all practical purposes this means that we can make P0 “anything we like” within the range of pressures that are unlikely to make water at room temperature change it’s state or volume do other bad things, say P = (0.1, 100) bar.
The pressure at a depth z in the container is then (from our previous work):
P (z) = P0 + ρgz

(730)

where ρ = ρw if the cylinder is indeed filled with water, but the cylinder could equally well be filled with hydraulic fluid (basically oil, which assists in lubricating the piston and ensuring that it remains “frictionless’ while assisting the seal), alcohol, mercury, or any other incompressible liquid.

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Week 8: Fluids

We recall that the pressure changes only when we change our depth. Moving laterally does not change the pressure, because e.g. dP/dx = dP/dy = 0. We can always find a path consisting of vertical and lateral displacements from z = 0 to any other point in the container – two such points at the same depth z are shown in 105, along with a (deliberately ziggy-zaggy151 ) vertical/horizontal path connecting them. Clearly these two points must have the same pressure P (z)!
Now consider the following. Suppose we start with pressure P0 (so that the pressure at these

two points is P (z), but then change F to make the pressure P0 and the pressure at the two points

P (z). Then:
P (z) = P0 + ρgz

P ′ (z) = P0 + ρgz

∆P (z) = P ′ (z) − P (z) = P0 − P0 = ∆P0

(731)

That is, the pressure change at depth z does not depend on z at any point in the fluid! It depends only on the change in the pressure exerted by the piston!
This result is known as Pascal’s Principle and it holds (more or less) for any compressible fluid, not just incompressible ones, but in the case of compressible fluids the piston will move up or down or in or out and the density of the fluid will change and hence the treatment of the integral will be too complicated to cope with. Pascal’s Principle is more commonly given in English words as: Any change in the pressure exerted at a given point on a confined fluid is transmitted, undiminished, throughout the fluid.
Pascal’s principle is the basis of hydraulics. Hydraulics are a kind of fluid-based simple machine that can be used to greatly amplify an applied force. To understand it, consider the following figure:

Example 8.2.1: A Hydraulic Lift

F
2
M m F
1

A1

A2

mg
Mg

Figure 106: A simple schematic for a hydraulic lift of the sort used in auto shops to lift your car.
Figure 106 illustrates the way we can multiply forces using Pascal’s Principle. Two pistons seal off a pair of cylinders connected by a closed tube that contains an incompressible fluid. The two pistons
151 Because we can make the zigs and zags differentially small, at which point this piecewise horizontal-vertical line becomes an arbitrary curve that remain in the fluid. Multivariate calculus can be used to formulate all of these results more prettily, but the reasoning behind them is completely contained in the picture and this text explanation.

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are deliberately given the same height (which might as well be z = 0, then, in the figure, although we could easily deal with the variation of pressure associated with them being at different heights since we know P (z) = P0 + ρgz. The two pistons have cross sectional areas A1 and A2 respectively, and support a small mass m on the left and large mass M on the right in static equilibrium.
For them to be in equilibrium, clearly:
F1 − mg

F2 − M g

=

0

(732)

=

0

(733)

We also/therefore have:
F1

= P0 A1 = mg

(734)

F2

= P0 A2 = M g

(735)

Thus

F1
F2
= P0 =
A1
A2

(736)

or (substituting and cancelling g):
A2
m
(737)
A1
A small mass on a small-area piston can easily balance a much larger mass on an equally larger area piston! M=

Just like a lever, we can balance or lift a large weight with a small one. Also just as was the case with a lever, there ain’t no such thing as a free lunch! If we try to lift (say) a car with a hydraulic lift, we have to move the same volume ∆V = A∆z from under the small piston (as it descends) to under the large one (as it ascends). If the small one goes down a distance z1 and the large one goes up a distance z2 , then: z1 A2
=
(738) z2 A1
The work done by the two cylinders thus precisely balances:
W2 = F2 z2 = F1

A2 A1
A2
z 2 = F1 z1
= F 1 z 1 = W1
A1
A1 A2

(739)

The hydraulic arrangement thus transforms pushing a small force through a large distance into a large force moved through a small distance so that the work done on piston 1 matches the work done by piston 2. No energy is created or destroyed (although in the real world a bit will be lost to heat as things move around) and all is well, quite literally, with the Universe.
This example is pretty simple, but it should suffice to guide you through doing a work-energy conservation problem where (for example) the mass m goes down a distance d (losing gravitational energy) and the mass M goes up a distance D (gaining gravitational energy while the fluid itself also is net moved up above its former level! Don’t forget that last, tricky bit if you ever have a problem like that!

8.3: Fluid Displacement and Buoyancy
First, a story. Archimedes152 was, quite possibly, the smartest person who has ever lived (so far).
His day job was being the “court magician” in the island kingdom of Syracuse in the third century
152 Wikipedia:

http://www.wikipedia.org/wiki/Archimedes. A very, very interesting person. I strongly recommend that my students read this short article on this person who came within a hair of inventing physics and calculus and starting the Enlightenment some 1900 years before Newton. Scary supergenius polymath guy. Would have won multiple Nobel prizes, a Macarthur “Genius” grant, and so on if alive today. Arguably the smartest person who has ever lived – so far.

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BCE, some 2200 years ago; in his free time he did things like invent primitive integration, accurately compute pi, invent amazing machines of war and peace, determine the key principles of both statics and fluid statics (including the one we are about to study and the principles of the lever – “Give me put a place to stand and I can move the world!” is a famous Archimedes quote, implying that a sufficiently long lever would allow the small forces humans can exert to move even something as large as the Earth, although yeah, there are a few problems with that that go beyond just a place to stand153 ).
The king (Hieron II) of Syracuse had a problem. He had given a goldsmith a mass of pure gold to make him a votive crown, but when the crown came back he had the niggling suspicion that the goldsmith had substituted cheap silver for some of the gold and kept the gold. It was keeping him awake at nights, because if somebody can steal from the king and get away with it (and word gets out) it can only encourage a loss of respect and rebellion.
So he called in his court magician (Archimedes) and gave him the task of determining whether or not the crown had been made by adulterated gold – or else. And oh, yeah – you can’t melt down the crown and cast it back into a regular shape whose dimensions can be directly compared to the same shape of gold, permitting a direct comparison of their densities (the density of pure gold is not equal to the density of gold with an admixture of silver). And don’t forget the “or else”.
Archimedes puzzled over this for some days, and decided to take a bath and cool off his overheated brain. In those days, baths were large public affairs – you went to the baths as opposed to having one in your home – where a filled tub was provided, sometimes with attendants happy to help you wash. As the possibly apocryphal story has it, Archimedes lowered himself into the overfull tub and as he did so, water sloshed out as he displaced its volume with his own volume. In an intuitive, instantaneous flash of insight – a “light bulb moment” – he realized that displacement of a liquid by an irregular shaped solid can be used to measure its volume, and that such a measurement of displaced volume would allow the king’s problem to be solved.
Archimedes then leaped out of the tub and ran naked through the streets of Syracuse (which we can only imagine provided its inhabitants with as much amusement then as it would provide now) yelling “Eureka!”, which in Greek means “I have found it!” The test (two possible versions of which are supplied below, one more probable than the other but less instructive for our own purposes) was performed, and revealed that the goldsmith was indeed dishonest and had stolen some of the king’s gold. Bad move, goldsmith! We will draw a tasteful veil over the probable painful and messy fate of the goldsmith.
Archimedes transformed his serendipitous discovery of static fluid displacement into an elaborate physical principle that explained buoyancy, the tendency of fluids to support all or part of the weight of objects immersed in them.
The fate of Archimedes himself is worth a moment more of our time. In roughly 212 BCE, the
Romans invaded Syracuse in the Second Punic War after a two year siege. As legend has it, as the city fell and armed soldiers raced through the streets “subduing” the population as only soldiers can,
Archimedes was in his court chambers working on a problem in the geometry of circles, which he had drawn out in the sand boxes that then served as a “chalkboard”. A Roman soldier demanded that he leave his work and come meet with the conquering general, Marcus Claudius Marcellus.
Archimedes declined, replying with his last words “Do not disturb my circles” and the soldier killed him. Bad move, soldier – Archimedes himself was a major part of the loot of the city and Marcellus had ordered that he was not to be harmed. The fate of the soldier that killed him is unknown, but it wasn’t really a very good idea to anger a conquering general by destroying an object or person of enormous value, and I doubt that it was very good.
Anyway, let’s see the modern version of Archimedes’ discovery and see as well how Archimedes
153 The

“sound bite” is hardly a modern invention, after all. Humans have always loved a good, pithy statement of insight, even if it isn’t actually even approximately true...

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Week 8: Fluids very probably used it to test the crown.

8.3.1: Archimedes’ Principle x P(0) = P0
0

y

Block of mass m z ∆y
∆x

Ft
Fl

P(z)
Fr
z + ∆z

∆z
Fb
mg

P(z + ∆ z)
)

Fluid (density ρ) z Figure 107: A solid chunk of “stuff” of mass m and the dimensions shown is immersed in a fluid of density ρ at a depth z. The vertical pressure difference in the fluid (that arises as the fluid itself becomes static static) exerts a vertical force on the cube.
If you are astute, you will note that figure 107 is exactly like figure 103 above, except that the internal chunk of fluid has been replace by some other material. The point is that this replacemend does not matter – the net force exerted on the cube by the fluid is the same!
Hopefully, that is obvious. The net upward force exerted by the fluid is called the buoyant force Fb and is equal to:
Fb

= P (z + ∆z)∆x∆y − P (z)∆x∆y
=

((P0 + ρg(z + ∆z)) − (P0 + ρgz)) ∆x∆y

= ρg∆z∆x∆y
= ρg∆V

(740)

where ∆V is the volume of the small block.
The buoyant force is thus the weight of the fluid displaced by this single tiny block. This is all we need to show that the same thing is true for an arbitrary immersed shape of object.
In figure 108, an arbitrary blob-shape is immersed in a fluid (not shown) of density ρ. Imagine that we’ve taken a french-fry cutter and cuts the whole blob into nice rectangular segments, one of which (of length h and cross-sectional area ∆A) is shown. We can trim or average the end caps so that they are all perfectly horizontal by making all of the rectangles arbitrarily small (in fact, differentially small in a moment). In that case the vertical force exerted by the fluid on just the two lightly shaded surfaces shown would be:
Fd

= P (z)∆A

(741)

Fu

= P (z + h)∆A

(742)

where we assume the upper surface is at depth z (this won’t matter, as we’ll see in a moment). Since
P (z + h) = P (z) + ρgh, we can find the net upward buoyant force exerted on this little cross-section

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Week 8: Fluids

∆A
Fd = P(z) ∆ A h Fu = P(z + h) ∆ A

∆ Fb = ρ g h ∆A = ρg ∆V (up)
Figure 108: An arbitrary chunk of stuff is immersed in a fluid and we consider a vertical cross section with horizontal ends of area ∆A and height h through the chunk. by subtracting the first from the second:
∆Fb = Fu − Fd = ρg h∆A = ρg ∆V

(743)

where the volume of this piece of the entire blob is ∆V = h ∆A.
We can now let ∆A → dA, so that ∆V − > dV , and write
Fb =

ρg dV = ρgV = mf g

dFb =

(744)

V of blob

where mf = ρV is the mass of the fluid displaced, so that mf g is its weight.
That is:
The total buoyant force on the immersed object is the weight of the fluid displaced by the object.
This is really an adequate proof of this statement, although if we were really going to be picky we’d use the fact that P doesn’t vary in x or y to show that the net force in the x or y direction is zero independent of the shape of the blob, using our differential french-fry cutter mentally in the x direction and then noting that the blob is arbitrary in shape and we could have just as easily labelled or oriented the blob with this direction called y so it must be true in any direction perpendicular to
g.
This statement – in the English or algebraic statement as you prefer – is known as Archimedes’
Principle, although Archimedes could hardly have formulated it quite the way we did algebraically above as he died before he could quite finish inventing the calculus and physics.
This principle is enormously important and ubiquitous. Buoyancy is why boats float, but rocks don’t. It is why childrens’ helium-filled balloons do odd things in accelerating cars. It exerts a subtle force on everything submerged in the air, in water, in beer, in liquid mercury, as long as the fluid itself is either in a gravitational field (and hence has a pressure gradient) or is in an accelerating container with its own “pseudogravity” (and hence has a pressure gradient).
Let’s see how Archimedes could have used this principle to test the crown two ways. The first way is very simple and conceptually instructive; the second way is more practical to us as it illustrates the way we generally do algebra associated with buoyancy problems.

Example 8.3.1: Testing the Crown I
The tools Archimedes probably had available to him were balance-type scales, as these tools for comparatively measuring the weight were well-known in antiquity. He certainly had vessels he could

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fill with water. He had thread or string, he had the crown itself, and he had access to pure gold from the king’s treasury (at least for the duration of the test.
?

Fb (crown)

mg

mg

Fb (gold)

mg

mg

a

b

Figure 109: In a), the crown is balanced against an equal weight/mass of pure gold in air. In b) the crown and the gold are symmetrically submerged in containers of still water.
Archimedes very likely used his balance to first select and trim a piece of gold so it had exactly the same weight as the crown as illustrated in figure 109a. Then all he had to do was submerge the crown and the gold symmetrically in two urns filled with water, taking care that they are both fully underwater. Pure gold is more dense than gold adulterated with silver (the most likely metal the goldsmith would have used; although a few others such as copper might have also been available and/or used they are also less dense than gold). This means that any given mass/weight (in air, with its negligible buoyant force) of adulterated gold would have a greater volume than an equal mass/weight (in air) of pure gold.
If the crown were made of pure gold, then, the buoyant forces acting on the gold and the crown would be equal. The weights of the gold and crown are equal. Therefore the submerged crown and submerged gold would be supported in static equilibrium by the same force on the ropes, and the balance would indicate “equal” (the indicator needle straight up). The goldsmith lives, the king is happy, Archimedes lives, everybody is happy.
If the crown is made of less-dense gold alloy, then its volume will be greater than that of pure gold. The buoyant force acting on it when submerged will therefore also be greater, so the tension in the string supporting it needed to keep it in static equilibrium will be smaller.
But this smaller tension then would fail to balance the torque exerted on the balance arms by the string attached to the gold, and the whole balance would rotate to the right, with the more dense gold sinking relative to the less dense crown. The balance needle would not read “equal”. In the story, it didn’t read equal. So sad – for the goldsmith.

Example 8.3.2: Testing the Crown II
Of course nowadays we don’t do things with balance-type scales so often. More often than not we would use a spring balance to weigh something from a string. The good thing about a spring balance is that you can directly read off the weight instead of having to delicately balance some force or weight with masses in a counterbalance pan. Using such a balance (or any other accurate scale) we can measure and record the density of pure gold once and for all.
Let us imagine that we have done so, and discovered that: ρAu = 19300 kilograms/meter

3

(745)

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Week 8: Fluids
3

For grins, please note that ρAg = 10490 kilograms/meter . This is a bit over half the density of gold, so that adulterating the gold of the crown with 10% silver would have decreased its density by around 5%. If the mass of the crown was (say) a kilogram, the goldsmith would have stolen 100 grams – almost four ounces – of pure gold at the cost of 100 grams of silver. Even if he stole twice that, the 9% increase in volume would have been nearly impossible to directly observe in a worked piece. At that point the color of the gold would have been off, though. This could be remedied
3
by adding copper (ρCu = 8940 kilograms/meter ) along with the silver. Gold-Silver-Copper all three alloy together, with silver whitening and yellowing the natural color of pure gold and copper reddening it, but with the two balanced one can create an alloy that is perhaps 10% each copper and silver that has almost exactly the same color as pure gold. This would harden and strengthen the gold of the crown, but you’d have to damage the crown to discover this.

Fb (crown)

Tw

Ta

mg

mg

a

b

Figure 110: We now measure the effective weight of the one crown both in air (very close to its true weight) and in water, where the measured weight is reduced by the buoyant force.
Instead we hang the crown (of mass m) as before, but this time from a spring balance, both in air and in the water, recording both weights as measured by the balance (which measures, recall, the tension in the supporting string). This is illustrated in figure 110, where we note that the measured weights in a) and b) are Ta , the weight in air, and Tw , the measured weight while immersed in water.
Let’s work this out. a) is simple. In static equilibrium:
Ta − mg

=

0

Ta

= mg

Ta

= ρcrown V g

(746)

so the scale in a) just measures the almost-true weight of the crown (off by the buoyant force exerted by the air which, because the density of air is very small at ρair ≈ 1.2 kilograms/meter3 , which represents around a 0.1% error in the measured weight of objects roughly the density of water, and an even smaller error for denser stuff like gold).
In b):
Tw + Fb − mg
Tw

=

0

= mg − Fb

= ρcrown V g − ρwater V g
=

(ρcrown − ρwater )V g

(747)

We know (we measured ) the values of Ta and Tw , but we don’t know V or ρcrown . We have two equations and two unknowns, and we would like most of all to solve for ρcrown . To do so, we divide

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Week 8: Fluids these two equations by one another to eliminate the V : ρcrown − ρwater
Tw
=
Ta
ρcrown

(748)

Whoa! g went away too! This means that from here on we don’t even care what g is – we could make these weight measurements on the moon or on mars and we’d still get the relative density of the crown (compared to the density of water) right!
A bit of algebra-fu: ρwater = ρcrown (1 −

Tw
Ta − Tw
) = ρcrown
Ta
Ta

or finally: ρcrown = ρwater

Ta
Ta − Tw

(749)

(750)

We are now prepared to be precise. Suppose that the color of the crown is very good. We perform the measurements above (using a scale accurate to better than a hundredth of a Newton or we might end up condemning our goldsmith due to a measurement error! ) and find that Ta = 10.00
Newtons, Tw = 9.45 Newtons. Then ρcrown = 1000

10.00
3
= 18182 kilograms/meter
10.00 − 9.45

(751)

We subtract, 19300 − 18182 = 1118; divide, 1118/19300 × 100 = 6%. Our crown’s material is around six percent less dense than gold which means that our clever goldsmith has adulterated the gold by removing some 12% of the gold (give or take a percent) and replace it with some mixture of silver and copper. Baaaaad goldsmith, bad.
If the goldsmith were smart, of course, he could have beaten Archimedes (and us). What he needed to do is adulterate the gold with a mixture of metals that have exactly the same density as gold! Not so easy to do, but tungsten’s density, ρW = 19300 (to three digits) almost exactly matches that of gold. Alas, it has the highest melting point of all metals at 3684◦ K, is enormously hard, and might or might not alloy with gold or change the color of the gold if alloyed. It is also pretty expensive in its own right. Platinum, Plutonium, Iridium, and Osmium are all even denser then gold, but three of these are very expensive (even more expensive than gold!) and one is very explosive, a transuranic compound used to make nuclear bombs, enormously expensive and illegal to manufacture or own (and rather toxic as well). Not so easy, matching the density via adulteration and making a profit out of it...
Enough of all of this fluid statics. Time to return to some dynamics.

8.4: Fluid Flow
In figure 111 we see fluid flowing from left to right in a circular pipe. The pipe is assumed to be
“frictionless” for the time being – to exert no drag force on the fluid flowing within – and hence all of the fluid is moving uniformly (at the same speed v with no relative internal motion) in a state of dynamic equilibrium.
We are interested in understanding the flow or current of water carried by the pipe, which we will define to be the volume per unit time that passes any given point in the pipe. Note well that we could instead talk about the mass per unit time that passes a point, but this is just the volume per unit time times the density and hence for fluids with a more or less uniform density the two are the same within a constant.
For this reason we will restrict our discussion in the following to incompressible fluids, with constant ρ. This means that the concepts we develop will work gangbusters well for understanding

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Week 8: Fluids

∆V
A

v

v ∆t
Figure 111: Fluid in uniform flow is transported down a pipe with a constant cross-section at a constant speed v. From this we can easily compute the flow, the volume per unit time that passes
(through a surface that cuts the pipe at) a point on the pipe. water flowing in pipes, beer flowing from kegs, blood flowing in veins, and even rivers flowing slowly in not-too-rocky river beds but not so well to describe the dynamical evolution of weather patterns or the movement of oceanic currents. The ideas will still be extensible, but future climatologists or oceanographers will have to work a bit harder to understand the correct theory when dealing the compressibility. We expect a “big pipe” (one with a large cross-sectional area) to carry more fluid per unit time, all other things being equal, than a “small pipe”. To understand the relationship between area, speed and flow we turn our attention to figure 111. In a time ∆t, all of the water within a distance v∆t to the left of the second shaded surface (which is strictly imaginary – there is nothing actually in that pipe at that point but fluid) will pass through this surface and hence past the point indicated by the arrow underneath. The volume of this fluid is just the area of the surface times the height of the cylinder of water:
∆V = Av∆t
(752)
If we divide out the ∆t, we get:
I=

∆V
= Av
∆t

(753)

This, then is the flow, or volumetric current of fluid in the pipe.
This is an extremely important relation, but the picture and derivation itself is arguably even more important, as this is the first time – but not the last time – you have seen it, and it will be a crucial part of understanding things like flux and electric current in the second semester of this course. Physics and math majors will want to consider what happens when they take the quantity v and make it a vector field v that might not be flowing uniformly in the pipe, which might not have a uniform shape or cross section, and thence think still more generally to fluids flowing in arbitrary streamlined patterns. Future physicians, however, can draw a graceful curtain across these meditations for the moment, although they too will benefit next semester if they at least try to think about them now.

8.4.1: Conservation of Flow
Fluid does not, of course, only flow in smooth pipes with a single cross-sectional area. Sometimes it flows from large pipes into smaller ones or vice versa. We will now proceed to derive an important aspect of that flow for incompressible fluids and/or steady state flows of compressible ones.
Figure 112 shows a fluid as it flows from just such a wider pipe down a gently sloping neck into a narrower one. As before, we will ignore drag forces and assume that the flow is as uniform as possible

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Week 8: Fluids

∆V

A1

A2

P
1
P2
V (constant)

v1

v2
2

v1∆t

v2 ∆t
1

Figure 112: Water flows from a wider pipe with a “larger” cros-sectional area A1 into a narrower pipe with a smaller cross-sectional area A2 . The speed of the fluid in the wider pipe is v1 , in the narrower one it is v2 . The pressure in the wider pipe is P1 , in the narrower one it is P2 . as it narrows, while remaining completely uniform in the wider pipe and smaller pipe on either side of the neck. The pressure, speed of the (presumed incompressible) fluid, and cross sectional area for either pipe are P1 , v1 , and A1 in the wider one and P2 , v2 , and A2 in the narrower one.
Pay careful attention to the following reasoning. In a time ∆t then – as before – a volume of fluid ∆V = A1 v1 ∆t passes through the surface/past the point 1 marked with an arrow in the figure.
In the volume between this surface and the next grey surface at the point 2 marked with an arrow no fluid can build up so actual quantity of mass in this volume must be a constant.
This is very important. The argument is simple. If more fluid flowed into this volume through the first surface than escaped through the second one, then fluid would be building up in the volume.
This would increase the density. But the fluid’s density cannot change – it is (by hypothesis) incompressible. Nor can more fluid escape through the second surface than enters through the first one. Note well that this assertion implies that the fluid itself cannot be created or destroyed, it can only flow into the volume through one surface and out through another, and because it is incompressible and uniform and the walls of the vessel are impermeable (don’t leak) the quantity of fluid inside the surface cannot change in any other way.
This is a kind of conservation law which, for a continuous fluid or similar medium, is called a continuity equation. In particular, we are postulating the law of conservation of matter, implying a continuous flow of matter from one place to another! Strictly speaking, continuity alone would permit fluid to build up in between the surfaces (as this can be managed without creating or destroying the mass of the fluid) but we’ve trumped that by insisting that the fluid be incompressible.
This means that however much fluid enters on the left must exit on the right in the time ∆t; the shaded volumes on the left and right in the figure above must be equal. If we write this out algebraically: ∆V
∆V
I =
∆t

= A1 v1 ∆t = A2 v2 ∆t
= A1 v1 = A2 v2

(754)

Thus the current or flow through the two surfaces marked 1 and 2 must be the same:
A1 v1 = A2 v2

(755)

Obviously, this argument would continue to work if it necked down (or up) further into a pipe with cross sectional area A3 , where it had speed v3 and pressure P3 , and so on. The flow of water in the pipe must be uniform, I = Av must be a constant independent of where you are in the pipe!

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There are two more meaty results to extract from this picture before we move on, that combine into one “named” phenomenon. The first is that conservation of flow implies that the fluid speeds up when it flows from a wide tube and into a narrow one or vice versa, it slows down when it enters a wider tube from a narrow one. This means that every little chunk of mass in the fluid on the right is moving faster than it is on the left. The fluid has accelerated!
Well, by now you should very well appreciate that if the fluid accelerates then there must be a net external force that acts on it. The only catch is, where is that force? What exerts it?
The force is exerted by the pressure difference ∆P between P1 and P2 . The force exerted by pressure at the walls of the container points only perpendicular to the pipe at that point; the fluid is moving parallel to the surface of the pipe and hence this “normal” confining force does no work and cannot speed up the fluid.
In a bit we will work out quantitatively how much the fluid speeds up, but even now we can see that since A1 > A2 , it must be true that v2 > v1 , and hence P1 > P2 . This is a general result, which we state in words:
The pressure decreases in the direction that fluid velocity increases.
This might well be stated (in other books or in a paper you are reading) the other way: When a fluid slows down, the pressure in it increases. Either way the result is the same.
This result is responsible for many observable phenomena, notably the mechanism of the lift that supports a frisbee or airplane wing or the Magnus effectWikipedia: http://www.wikipedia.org/wiki/Magnus Effect that causes a spinning thrown baseball ball to curve.
Unfortunately, treating these phenomena quantitatively is beyond, and I do mean way beyond, the scope of this course. To correctly deal with lift for compressible or incompressible fluids one must work with and solve either the Euler equations 154 , which are coupled partial differential equations that express Newton’s Laws for fluids dynamically moving themselves in terms of the local density, the local pressure, and the local fluid velocity, or the Navier-Stokes Equations 155
, ditto but including the effects of viscosity (neglected by Euler). Engineering students (especially those interested in aerospace engineering and real fluid dynamics) and math and physics majors are encouraged to take a peek at these articles, but not too long a peek lest you decide that perhaps majoring in advanced basket weaving really was the right choice after all. They are really, really difficult; on the high end “supergenius” difficult156 .
This isn’t surprising – the equations have to be able to describe every possible dynamical state of a fluid as it flows in every possible environment – laminar flow, rotational flow, turbulence, drag, around/over smooth shapes, horribly not smooth shapes, and everything in between. At that, they don’t account for things like temperature and the mixing of fluids and fluid chemistry – reality is more complex still. That’s why we are stopping with the simple rules above – fluid flow is conserved
(safe enough) and pressure decreases as fluid velocity increases, all things being equal.
All things are, of course, not always equal. In particular, one thing that can easily vary in the case of fluid flowing in pipes is the height of the pipes. The increase in velocity caused by a pressure differential can be interpreted or predicted by the Work-Kinetic Energy theorem, but if the fluid is moving up or down hill then we may discover that gravity is doing work as well!
In this case we should really use the Work-Mechanical Energy theorem to determine how pressure changes can move fluids. This is actually pretty easy to do, so let’s do it.
154 Wikipedia:

http://www.wikipedia.org/wiki/Euler Equations (fluid dynamics). http://www.wikipedia.org/wiki/Navier-Stokes equations.
156 To give you an idea of how difficult they are, note that there is a $1,000,000 prize just for showing that solutions to the 3 dimensional Navier-Stokes equations generally exist and/or are not singular.
155 Wikipedia:

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8.4.2: Work-Mechanical Energy in Fluids: Bernoulli’s Equation
Daniel Bernoulli was a third generation member of the famous Bernoulli family157 who worked on
(among many other things) fluid dynamics, along with his good friend and contemporary, Leonhard
Euler. In 1738 he published a long work wherein he developed the idea of energy conservation to fluid motion. We’ll try to manage it in a page or so.

A2

F2
D

∆V P
2
v2

y2
∆V A 1
F1
v1

P
1

y1

d
Figure 113: A circular cross-sectional necked pipe is arranged so that the pipe changes height between the larger and smaller sections. We will assume that both pipe segments are narrow compared to the height change, so that we don’t have to account for a potential energy difference (per unit volume) between water flowing at the top of a pipe compared to the bottom, but for ease of viewing we do not draw the picture that way.
In figure 113 we see the same pipe we used to discuss conservation of flow, only now it is bent uphill so the 1 and 2 sections of the pipe are at heights y1 and y2 respectively. This really is the only change, otherwise we will recapitulate the same reasoning. The fluid is incompressible and the pipe itself does not leak, so fluid cannot build up between the bottom and the top. As the fluid on the bottom moves to the left a distance d (which might be v1 ∆t but we don’t insist on it as rates will not be important in our result) exactly the same amount fluid must move to the left a distance
D up at the top so that fluid is conserved.
The total mechanical consequence of this movement is thus the disappearance of a chunk of fluid mass: ∆m = ρ∆V = ρA1 d = ρA2 D
(756)
that is moving at speed v1 and at height y1 at the bottom and it’s appearance moving at speed v2 and at height y2 at the top. Clearly both the kinetic energy and the potential energy of this chunk of mass have changed.
What caused this change in mechanical energy? Well, it can only be work. What does the work? The walls of the (frictionless, drag free) pipe can do no work as the only force it exerts is perpendicular to the wall and hence to v in the fluid. The only thing left is the pressure that acts on the entire block of water between the first surface (lightly shaded) drawn at both the top and
157 Wikipedia: http://www.wikipedia.org/wiki/Bernoulli family. The Bernoullis were in on many of major mathematical and physical discoveries of the eighteenth and nineteenth century. Calculus, number theory, polynomial algebra, probability and statistics, fluid dynamics – if a theorem, distribution, principle has the name “Bernoulli” on it, it’s gotta be good...

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the bottom as it moves forward to become the second surface (darkly shaded) drawn at the top and the bottom, effecting this net transfer of mass ∆m.
The force F1 exerted to the right on this block of fluid at the bottom is just F1 = P1 A1 ; the force
F2 exerted to the left on this block of fluid at the top is similarly F2 = P2 A2 . The work done by the pressure acting over a distance d at the bottom is W1 = P1 A1 d, at the top it is W2 = −P2 A2 D.
The total work is equal to the total change in mechanical energy of the chunk ∆m:
Wtot

=

∆Emech

W1 + W2

= Emech (f inal) − Emech (initial)
1
1
2
2
P1 A1 d − P2 A2 D = ( ∆mv2 + ∆mgy2 ) − ( ∆mv1 + ∆mgy1 )
2
2
1
1
2
2
(P1 − P2 )∆V = ( ρ∆V v2 + ρ∆V gy2 ) − ( ρ∆V v1 + ρ∆V gy1 )
2
2
1 2
1 2
(P1 − P2 ) = ( ρv2 + ρgy2 ) − ( ρv1 + ρy1 )
2
2
1 2
1 2
P1 + ρv1 + ρy1 = P2 + ρv2 + ρgy2 = a constant (units of pressure)
2
2

(757)

There, that wasn’t so difficult, was it? This lovely result is known as Bernoulli’s Principle
(or the Bernoulli fluid equation). It contains pretty much everything we’ve done so far except conservation of flow (which is a distinct result, for all that we used it in the derivation) and Archimedes’
Principle.
For example, if v1 = v2 = 0, it describes a static fluid:
P2 − P1 = −ρg(y2 − y1 )

(758)

and if we change variables to make z (depth) −y (negative height) we get the familiar:
∆P = ρg∆z

(759)

for a static incompressible fluid. It also not only tells us that pressure drops where fluid velocity increases, it tells us how much the pressure drops when it increases, allowing for things like the fluid flowing up or downhill at the same time! Very powerful formula, and all it is is the Work-Mechanical
Energy theorem (per unit volume, as we divided out ∆V in the derivation, note well) applied to the fluid! Example 8.4.1: Emptying the Iced Tea
8.4.3: Fluid Viscosity and Resistance
In the discussion above, we have consistently ignored viscosity and drag, which behave like “friction”, exerting a force parallel to the confining walls of the pipe in the opposite direction to the relative motion of fluid and pipe. In laminar flow, a layer of the fluid “sticks” to the wall and does not move, and the viscosity of the fluid describes the way shear velocity builds up as one goes from the wall of the pipe (velocity zero) to the center of the pipe (maximum velocity). If the fluid is “thick and sticky” (has a large viscosity), fluid in the middle is still experiencing significant backwards force slowing it down, transmitted from the walls to the fluid there. If the fluid is thin and slippery (low visocosity) then fluid even a short distance away from the wall is moving rapidly even thought the fluid “on” the wall itself isn’t moving.
In the absence of any driving force, this drag force will bring fluid flowing in a pipe to rest quite rapidly. To keep it moving, then, requires the continuous application of some driving force. This force, as we shall see, has to maintain a pressure gradient that pushes the fluid through the pipe

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from high pressure to low pressure just enough to overcome drag/friction and keep the fluid flowing at a constant speed.
To correctly derive all of this, even for the simplest of geometries, is beyond the scope of this course. It isn’t horribly difficult for a circular pipe, but it requires a treatment of shear stress and viscosity for laminar flow and we don’t yet know what shear stress is. Also, the algebra is a bit involved. Instead I’m going to invoke my instructor’s privilege (again, but this is pretty close tot he last time) to give you an important result, one that I’d like you to learn even though I do not derive it. I only do this two or three times in this course, and this is, sadly, one of them. To make up for it I will wave my hands and squawk a bit out the result instead, so that you at least conceptually understand where the result comes from and how things scale.
In figure 114 a circular pipe is carrying a fluid with viscosity µ from left to right at a constant speed. Once again, this is a sort of dynamic equilibrium; the net force on the fluid in the pipe segment shown must be zero for the speed of the fluid through it to be maintained unabated during the flow.
The fluid is in contact with and interacts with the walls of the pipe, creating a thin layer of fluid at least a few atoms thick that are “at rest”, stuck to the pipe. As fluid is pushed through the pipe, this layer at rest interacts with and exerts an opposing force on the layer moving just above it via the viscosity of the fluid. This layer in turn interacts with and slows the layer above it and so on right up to the center of the pipe, where the fluid flows most rapidly. The fluid flow thus forms cylindrical layers of constant speed, where the speed increases more or less smoothly from zero where it is contact with the pipe to a peak speed in the middle. The layers are the laminae of laminar flow.

F r F
1

F
F
2

v η P
1
A

F

F

P2
F

F

A

L
Figure 114: A circular pipe with friction carrying a fluid with viscosity – in other words, a “ real pipe”. This can be a model for everything from household plumbing to the blood vessels of the human circulatory system, within reason.
The interaction of the surface layer with the fluid, redistributed to the whole fluid via the viscosity, exerts a net opposing force on the fluid as it moves through the pipe. In order for the average speed of the fluid to continue, an outside force must act on it with an equal and opposite force. The only available source of this force in the figure is obviously the fluid pressure; if it is larger on the left than on the right (as shown) it will exert a net force on the fluid in between that can balance the drag force exerted by the walls.
The forces at the ends are F1 = P1 A, F2 = P2 A. The net force acting on the fluid mass is thus:
∆F = F1 − F2 = (P1 − P2 )A

(760)

All things being equal, we expect the flow rate to increase linearly with v, and for laminar flow, the drag force is proportional to v. Therefore we expect that:
∆F = Fd ∝ v ∝ I (the flow)

(761)

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We can then divide out the area and write:
∆P ∝

I
A

(762)

Since we cannot derive the constant of proportionality in this expression without doing Evil
Math Magic158 we’ll save putting the constants of proportionality in until the end, but we can give what’s called a “hand waving argument” and look at how we expect ∆P to scale with the variables that describe the problem. Hand-waving arguments are actually very important in physics, as we can often use a mix of intuition and dimensional analysis and scaling to guess the right form of a relation to a remarkably accurate degree.
Suppose we had two identical segments of pipe, one right after the other, carrying the same fluid flow. The first pipe needs a drop of ∆P in order to maintain flow I; so does the second. When we put one right after the other, then, we expect the total pressure difference required to overcome drag to be twice ∆P . From this, we expect the drag force to scale linearly with the length of the pipe; if the pipe is twice as long we need twice the pressure difference to maintain flow, half as long we need half the pressure difference. This means that we expect an L (to the first power) on the top in the equation for ∆P .
Next, we expect ∆P (for constant I) to increase monotonically with the viscosity µ. Since the viscosity increases the stickiness and distance the static layer’s reaction force is extended into the fluid, we can’t easily imagine any circumstances where the pressure required to push treacle through a pipe would be less than the pressure required to push plain water. Unfortunately there are an infinite number of monotonically increasing functions of µ available, all heuristically possible.
Fortunately all of them have a Taylor Series expansion with a leading linear term so it seems reasonable to at least try a linear dependence (which turns out to work well physically throughout the laminar flow region right up to where the Reynolds number for the fluid flow indicates a – yes, highly nonlinear – transition from laminar/linear drag to nonlinear drag, as we would expect from our Week 2 discussion on drag). So we’ll stick an µ on top.
Finally, suppose we have two identical pipe segments, each carrying identical flow I of identical fluids across identical lengths L through identical cross-sectional areas with identical force difference ∆F (Note well: not ∆P ) across the pipes. If we put them side by side, two pipes carry twice as much fluid – we already know that the flow itself scales with the area of the aperture. We therefore expect the flow itself to double if we double the cross-sectional area at constant force difference, and hence we expect to need half the force to maintain the same flow if we double the area at the same force difference. This puts another factor of A, the cross-sectional area of the pipe, on the bottom.
If we put all of this together, our expression for ∆P will look like:
∆P ∝

ILµ
∝I
A2

Lµ π 2 r4

(763)

All that’s missing is the constant of proportionality. Note that all of our scaling arguments above would work for almost any cross-sectional shape of pipe – there is nothing about them that absolutely requires circular pipes. We rather expect that the constant of proportionality for different shapes would be different, and we have no idea how to compute one – it seems as though it would depend on integrals of some sort over the area or around the perimeter of the pipe but we’d have to have a better model for the relationship between viscosity and force to be able to formulate them.
This, and this only I am going to just give you the answer on. The correct constant of propor158 Defined

as anything requiring more than a page of algebra, serious partial derivatives, and concepts we haven’t covered yet...

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∆P = I

8Lµ πr4 = IR

(764)

where I have introduced the resistance of the pipe to flow :

R=

8Lµ πr4 (765)

Equation 764 is know as Poiseuille’s Law and is a key relation for physicians and plumbers to know because it describes both flow of water in pipes and the flow of blood in blood vessels wherever the flow is slow enough that it is laminar and not turbulent (which is actually “mostly”, so that the expression is useful ).
Before moving on, let me note that ∆P = IR is the fluid-flow version of a named formula in electricity and magnetism (from next semester): Ohm’s Law. The hand-waving “derivation” of Ohm’s Law there will be very similar, with all of the scaling worked out using intuition but a final constant of proportionality that includes the details of the resistive material turned at the last minute into a parameter called the resistivity 159 .

8.4.4: A Brief Note on Turbulence
You will not be responsible for the formulas or numbers in this section, but you should be conceptually aware of the phenomenon of the “onset of turbulence” in fluid flow. The velocity of the flow in a circular pipe (and other parameters such as µ and r) can be transformed into a general dimensionless parameter called the Reynolds Number Re160 . The Reynolds number for a circular pipe is:

Re =

ρvD ρv 2r
=
µ µ (766)

where D = 2r is the hydraulic diameter 161 , which in the case of a circular pipe is the actual diameter. One can (as you can see) express Poiseuille’s Law in terms of the Reynolds number, although there is no particular benefit to doing so.
The one thing the Reynolds number does for us is that it serves as a marker for the transition to turbulent flow. For Re < 2300 flow in a circular pipe is laminar and all of the relations above hold. Turbulent flow occurs for Re > 4000. In between is the region known as the onset of turbulence, where the resistance of the pipe depends on flow in a very nonlinear fashion, and among other things dramatically increases with the Reynolds number. As we will see in a moment (in an example) this means that the partial occlusion of blood vessels can have a profound effect on the human circulatory system.
At this point you should understand fluid statics and dynamics quite well, armed both with equations such as the Bernoulli Equation that describe idealized fluid dynamics and statics as well as with conceptual (but possibly quantitative) ideas such Pascal’s Principle or Archimedes’ principle as the relationship between pressure differences, flow, and the geometric factors that contribute to resistance. Let’s put some of this nascent understanding to the test by looking over and analyzing the human circulatory system.

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Figure 115: A simple cut-away diagram of the human heart.

8.5: The Human Circulatory System
For once, this is a chapter that math majors, physics majors, and engineers may, if they wish, skip, although personally I think that any intellectually curious person would want to learn all sorts of things that sooner or later will impact on their own health and life. To put that rather bluntly, kids, sure now you’re all young and healthy and everything, but in thirty or forty more years (if you survive) you won’t be, and understanding the things taught in this chapter will be extremely useful to you then, if not now as you choose a lifestyle and diet that might get you through to then in reasonably good cardiovascular shape!
Here is a list of True Facts about the human cardiovascular system, in no particular order, that you should now be able to understand at least qualitatively and conceptually if not quantitatively.
• The heart, illustrated in the schematic in figure 115 above162 is the “pump” that drives blood through your blood vessels.
• The blood vessels are differentiated into three distinct types:
– Arteries, which lead strictly away from the heart and which contain a muscular layer that elastically dilates and contracts the arteries in a synchronous way to help carry the surging waves of blood. This acts as a “shock absorber” and hence reduces the peak systolic blood pressure (see below). As people age, this muscular tissue becomes less elastic – “hardening of the arteries” – as collagen repair mechanisms degrade or plaque
(see below) is deposited and systolic blood pressure often increases as a result.
159 The argument won’t quite be identical – the potential difference ∆V there will relate to current more like the way ∆F does to flow in our treatment, rather than ∆P , so electrical resistance will be inversely proportional to A, not A2 .
160 Wikipedia: http://www.wikipedia.org/wiki/Reynolds Number.
161 Wikipedia: http://www.wikipedia.org/wiki/hydraulic diameter.
162 Wikipedia: http://www.wikipedia.org/wiki/Human heart. The diagram itself is borrowed from the wikipedia creative commons, and of course you can learn a lot more of the anatomy and function of the heart and circulation by reading the wikipedia articles on the heart and following links.

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Arteries split up the farther one is from the heart, eventually becoming arterioles, the very small arteries that actually split off into capillaries.
– Capillaries, which are a dense network of very fine vessels (often only a single cell thick) that deliver oxygenated blood throughout all living tissue so that the oxygen can disassociate from the carrying hemoglobin molecules and diffuse into the surrounding cells in systemic circulation, or permit the oxygenation of blood in pulmonary circulation.
– Veins, which lead strictly back to the heart from the capillaries. Veins also have a muscle layer that expand or contract to aid in thermoregulation and regulation of blood pressure as one lies down or stands up. Veins also provide “capacitance” to the circulatory system and store the body’s “spare” blood; 60% of the body’s total blood supply is usually in the veins at any one time. Most of the veins, especially long vertical veins, are equipped with one-way venous valves every 4-9 cm that prevent backflow and pooling in the lower body during e.g. diastoli (see below).
Blood from the capillaries is collected first in venules (the return-side equivalent of arterioles) and then into veins proper.
• There are two distinct circulatory systems in humans (and in the rest of the mammals and birds): – Systemic circulation, where oxygenated blood enters the heart via pulmonary veins from the lungs and is pumped at high pressure into systemic arteries that deliver it through the capillaries and (deoxygenated) back via systemic veins to the heart.
– Pulmonary circulation, where deoxgenated blood that has returned from the system circulation is pumped into pulmonary arteries that deliver it to the lungs, where it is oxygenated and returned to the heart by means of pulmonary veins. These two distinct circulations do not mix and together, form a closed double circulation loop.
• The heart is the pump that serves both systemic and pulmonary circulation. Blood enters into the atria and is expelled into the two circulatory system from the ventricles.
Systemic circulation enters from the pulmonary veins into the left atrium, is pumped into the left ventricle through the one-way mitral valve, which then pumps the blood at high pressure into the systemic arteries via the aorta through the one-way aortic valve. It is eventually returned by the systemic veins (the superior and inferior vena cava) to the right atrium, pumped into the right ventricle through the one-way tricuspid valve, and then pumped at high pressure into the pulmonary artery for delivery to the lungs.
The human heart (as well as the hearts of birds and mammals in general) is thus fourchambered – two atria and two ventricles. The total resistance of the systemic circulation is generally larger than that of the pulmonary circulation and hence systemic arterial blood pressure must be higher than pulmonary arterial blood pressure in order to maintain the same flow. The left ventricle (primary systemic pump) is thus typically composed of thicker and stronger muscle tissue than the right ventricle. Certain reptiles also have four-chambered hearts, but their pulmonary and systemic circulations are not completely distinct and it is thought that their hearts became four-chambered by a different evolutionary pathway.
• Blood pressure is generally measured and reported in terms of two numbers:
– Systolic blood pressure. This is the peak/maximum arterial pressure in the wave pulse generated that drives systemic circulation. It is measured in the (brachial artery of the) arm, where it is supposed to be a reasonably accurate reflection of peak aortic pressure just outside of the heart, where, sadly, it cannot easily be directly measured without resorting to invasive methods that are, in fact, used e.g. during surgery.
– Diastolic blood pressure. This is the trough/minimum arterial pressure in the wave pulse of systemic circulation.

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Blood pressure has historically been measured in millimeters of mercury, in part because until fairly recently a sphygnomometer built using an integrated mercury barometer was by far the most accurate way to measure blood pressure, and it still extremely widely used in situations where high precision is required. Recall that 760 mmHg (torr) is 1 atm or 101325
Pa.

“Normal” Systolic systemic blood pressure can fairly accurately be estimated on the basis of the distance between the heart and the feet; a distance on the order of 1.5 meters leads to a pressure difference of around 0.15 atm or 120 mmHg.
Blood is driven through the relatively high resistance of the capillaries by the difference in arterial pressure and venous pressure. The venous system is e