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Contents

Preface

v vii

Problems Solved in Student Solutions Manual

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Matrices, Vectors, and Vector Calculus Newtonian Mechanics—Single Particle Oscillations 79 127

1 29

Nonlinear Oscillations and Chaos Gravitation 149

Some Methods in The Calculus of Variations

165 181

Hamilton’s Principle—Lagrangian and Hamiltonian Dynamics Central-Force Motion 233 277 333

Dynamics of a System of Particles

Motion in a Noninertial Reference Frame Dynamics of Rigid Bodies Coupled Oscillations 397 435 461 353

Continuous Systems; Waves Special Theory of Relativity

iii

iv

CONTENTS

CHAPTER

0

Preface

This Instructor’s Manual contains the solutions to all the end-of-chapter problems (but not the appendices) from Classical Dynamics of Particles and Systems, Fifth Edition, by Stephen T. Thornton and Jerry B. Marion. It is intended for use only by instructors using Classical Dynamics as a textbook, and it is not available to students in any form. A Student Solutions Manual containing solutions to about 25% of the end-of-chapter problems is available for sale to students. The problem numbers of those solutions in the Student Solutions Manual are listed on the next page. As a result of surveys received from users, I continue to add more worked out examples in the text and add additional problems. There are now 509 problems, a significant number over the 4th edition. The instructor will find a large array of problems ranging in difficulty from the simple “plug and chug” to the type worthy of the Ph.D. qualifying examinations in classical mechanics. A few of the problems are quite challenging. Many of them require numerical methods. Having this solutions manual should provide a greater appreciation of what the authors intended to accomplish by the statement of the problem in those cases where the problem statement is not completely clear. Please inform me when either the problem statement or solutions can be improved. Specific help is encouraged. The instructor will also be able to pick and choose different levels of difficulty when assigning homework problems. And since students may occasionally need hints to work some problems, this manual will allow the instructor to take a quick peek to see how the students can be helped. It is absolutely forbidden for the students to have access to this manual. Please do not give students solutions from this manual. Posting these solutions on the Internet will result in widespread distribution of the solutions and will ultimately result in the decrease of the usefulness of the text. The author would like to acknowledge the assistance of Tran ngoc Khanh (5th edition), Warren Griffith (4th edition), and Brian Giambattista (3rd edition), who checked the solutions of previous versions, went over user comments, and worked out solutions for new problems. Without their help, this manual would not be possible. The author would appreciate receiving reports of suggested improvements and suspected errors. Comments can be sent by email to stt@virginia.edu, the more detailed the better. Stephen T. Thornton Charlottesville, Virginia

v

vi

PREFACE

CHAPTER

1

Matrices, Vectors, and Vector Calculus

1-1. x2 = x2′ x1′ 45˚ x1

45˚

x3

x3′

Axes x′ and x′ lie in the x1 x3 plane. 1 3 The transformation equations are:

x1 = x1 cos 45° − x3 cos 45° ′ x2 = x2 ′ x3 = x3 cos 45° + x1 cos 45° ′ x1 = ′

1 1 x1 − x3 2 2

x2 = x2 ′ x3 = ′

So the transformation matrix is: 1 1 x1 − x3 2 2

1 2 0 1 2

0 − 1 0

1 2 0 1 2

1

2 1-2. a) x3 CHAPTER 1

D

E

γ O α A x1

β B C

x2

From this diagram, we have

OE cos α = OA OE cos β = OB OE cos γ = OD Taking the square of each equation in (1) and adding, we find OE cos 2 α + cos 2 β + cos 2 γ = OA + OB + OD But

OA + OB = OC

2 2 2

(1)

2

2

2

2

(2)

(3)

and

OC + OD = OE

2 2 2

(4)

2

Therefore,

OA + OB + OD = OE

2 2 2

(5)

Thus, cos 2 α + cos 2 β + cos 2 γ = 1

b)

x3 D E

D′ E′

(6)

O A A′ x1

θ

B C

C′

B′

x2

First, we have the following trigonometric relation:

OE + OE′ − 2OE OE′ cos θ = EE′

2 2 2

(7)

MATRICES, VECTORS, AND VECTOR CALCULUS

3

But,

2 EE′ = OB′ − OB + OA′ − OA + OD′ − OD 2 2 2 2

= OE′ cos β ′ − OE cos β + OE′ cos α ′ − OE cos α + OE′ cos γ ′ − OE cos γ or,

2

2

(8)

2 2 2 EE′ = OE′ cos 2 α ′ + cos 2 β ′ + cos 2 γ ′ + OE cos 2 α + cos 2 β + cos 2 γ

− 2OE′ OE cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′ = OE′ 2 + OE2 − 2OE OE′ cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′ Comparing (9) with (7), we find cos θ = cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′ (10) (9)

1-3. x3 e3 O e1 x1 A e3 e2′

e2

x2 e1 e2

e1′

e3′

Denote the original axes by x1 , x2 , x3 , and the corresponding unit vectors by e1 , e2 , e3 . Denote the new axes by x′ , x′ , x′ and the corresponding unit vectors by e1 , e2 , e3 . The effect of the ′ ′ ′ 1 2 3 rotation is e1 → e3 , e2 → e1 , e3 → e2 . Therefore, the transformation matrix is written as: ′ ′ ′

cos ( e1 , e1 ) cos ( e1 , e2 ) cos ( e1 , e3 ) 0 1 0 ′ ′ ′ λ = cos ( e′ , e1 ) cos ( e′ , e2 ) cos ( e′ , e3 ) = 0 0 1 2 2 2 1 0 0 3 3 3 cos ( e′ , e1 ) cos ( e′ , e2 ) cos ( e′ , e3 )

1-4. a) Let C = AB where A, B, and C are matrices. Then,

Cij = ∑ Aik Bkj k (1)

(C ) t ij

= C ji = ∑ Ajk Bki = ∑ Bki Ajk k k

4 Identifying Bki = Bt

CHAPTER 1

( )

ik

and Ajk = At

( )

kj

,

(C ) = ∑ (B ) ( A ) t t t ij k ik

kj

(2)

or,

C t = ( AB) = Bt At t (3)

b) To show that ( AB) = B−1 A−1 ,

−1

( AB) B−1 A−1 = I = ( B−1 A−1 ) AB

That is,

(4)

( AB) B−1 A−1 = AIA−1 = AA−1 = I

(5) (6)

(B

1-5.

−1

A−1 ( AB) = B−1 IB = B−1B = I

)

Take λ to be a two-dimensional matrix:

λ =

Then,

λ11 λ12 = λ11λ 22 − λ12 λ 21 λ 21 λ 22

(1)

2 2 2 2 2 2 2 2 2 2 2 2 λ = λ11λ 22 − 2λ11λ 22 λ12 λ 21 + λ12 λ 21 + ( λ11λ 21 + λ12 λ 22 ) − ( λ11λ 21 + λ12λ 22 ) 2 2 2 2 2 2 2 2 2 2 2 = λ 22 λ11 + λ12 + λ 21 λ11 + λ12 − λ11λ 21 + 2λ11λ 22λ12 λ 21 + λ12λ 22 2 2 2 2 = λ11 + λ12 λ 22 + λ 21 − ( λ11λ 21 + λ12 λ 22 )

(

)

(

) (

)

(2)

(

)(

)

2

But since λ is an orthogonal transformation matrix, Thus,

∑λ λ ij j

kj

= δ ik .

2 2 2 2 λ11 + λ12 = λ 21 + λ 22 = 1

λ11λ 21 + λ12 λ 22 = 0

Therefore, (2) becomes

(3)

λ =1

2

(4)

1-6.

The lengths of line segments in the x j and x ′ systems are j

L=

∑x j 2 j

; L′ =

∑ x′ i i

2

(1)

MATRICES, VECTORS, AND VECTOR CALCULUS

5

If L = L ′ , then

∑ x = ∑ x′

2 j i j i

2

(2)

The transformation is

xi′ = ∑ λ ij x j j (3)

Then,

∑ x = ∑∑λ

2 j j i k

ik

xk ∑ λ i x

(4)

= ∑ xk x ∑ λik λ i i k,

But this can be true only if

∑λ i ik

λi = δ k

(5)

which is the desired result.

1-7.

x3

(0,0,1) (0,1,1)

(1,0,1)

(1,1,1)

(0,0,0)

(0,1,0)

x2

x1

(1,0,0)

(1,1,0)

There are 4 diagonals: D1 , from (0,0,0) to (1,1,1), so (1,1,1) – (0,0,0) = (1,1,1) = D1 ; D 2 , from (1,0,0) to (0,1,1), so (0,1,1) – (1,0,0) = (–1,1,1) = D 2 ; D 3 , from (0,0,1) to (1,1,0), so (1,1,0) – (0,0,1) = (1,1,–1) = D 3 ; and D 4 , from (0,1,0) to (1,0,1), so (1,0,1) – (0,1,0) = (1,–1,1) = D 4 . The magnitudes of the diagonal vectors are

D1 = D 2 = D 3 = D 4 = 3

The angle between any two of these diagonal vectors is, for example,

D1 ⋅ D 2 (1,1,1) ⋅ ( −1,1,1) 1 = cos θ = = D1 D 2 3 3

6 so that

1 θ = cos−1 = 70.5° 3

CHAPTER 1

Similarly, D1 ⋅ D 3 D ⋅D D ⋅D D ⋅D D ⋅D 1 = 1 4 = 2 3 = 2 4 = 3 4 =± D1 D 3 D1 D 4 D2 D3 D2 D 4 D3 D 4 3 Let θ be the angle between A and r. Then, A ⋅ r = A2 can be written as

Ar cos θ = A2

1-8.

or, r cos θ = A

(1)

This implies

QPO =

π

2

(2)

Therefore, the end point of r must be on a plane perpendicular to A and passing through P.

1-9. a)

A = i + 2j − k

B = −2i + 3j + k

A − B = 3i − j − 2k

2 2 A − B = ( 3) + ( −1) + (−2)2 12

A − B = 14

b)

B θ A component of B along A

The length of the component of B along A is B cos θ.

A ⋅ B = AB cos θ

B cos θ = A ⋅ B −2 + 6 − 1 3 6 = = or A 2 6 6

The direction is, of course, along A. A unit vector in the A direction is 1 ( i + 2j − k ) 6

MATRICES, VECTORS, AND VECTOR CALCULUS

7

So the component of B along A is 1 ( i + 2j − k ) 2

c)

cos θ =

A⋅B 3 3 3 = = ; θ = cos −1 AB 6 14 2 7 2 7

θ i j k

71°

d)

2 −1 1 −1 1 2 −j +k A × B = 1 2 −1 = i 3 1 −2 1 −2 3 −2 3 1 A × B = 5i + j + 7 k

e)

A − B = 3i − j − 2k i −1

A + B = −i + 5j j 5 k 0

( A − B ) × ( A + B ) = 3 −1 − 2

( A − B) × ( A + B) = 10i + 2j + 14k

1-10.

r = 2b sin ω t i + b cos ω t j

v = r = 2bω cos ω t i − bω sin ω t j

a)

a = v = −2bω 2 sin ω t i − bω 2 cos ω t j = −ω 2 r speed = v = 4b 2ω 2 cos 2 ω t + b 2ω 2 sin 2 ω t = bω 4 cos 2 ω t + sin 2 ω t

12 12

speed = bω 3 cos 2 ω t + 1

b)

12

At t = π 2ω , sin ω t = 1 , cos ω t = 0

So, at this time, v = − bω j , a = −2bω 2 i So, θ 90°

8

1-11. a) Since ( A × B) i = ∑ ε ijk Aj Bk , we have jk CHAPTER 1

(A × B) ⋅ C = ∑∑ ε ijk Aj Bk Ci i j,k

= C1 ( A2 B3 − A3 B2 ) − C2 ( A1B3 − A3 B1 ) + C3 ( A1B2 − A2 B1 )

C1 = A1 B1 C2 A2 B2 C3 A1 A3 = − C1 B3 B1 A2 C2 B2 A3 A1 C3 = B1 B3 C1 A2 B2 C2 A3 B3 = A ⋅ ( B × C ) C3

(1)

We can also write

C1 C2 B2 A2 C3 B1 B2 C2 A2

(A × B) ⋅ C = − B1 A1

B3 = C1 A3 A1

C3 = B ⋅ ( C × A ) A3

B3

(2)

We notice from this result that an even number of permutations leaves the determinant unchanged.

b) Consider vectors A and B in the plane defined by e1 , e2 . Since the figure defined by A, B, C is a parallelepiped, A × B = e3 × area of the base, but e3 ⋅ C = altitude of the parallelepiped. Then,

C ⋅ ( A × B) = ( C ⋅ e3 ) × area of the base = altitude × area of the base = volume of the parallelepiped

1-12.

O a h c C a–c b A b–a c–b

B

The distance h from the origin O to the plane defined by A, B, C is

MATRICES, VECTORS, AND VECTOR CALCULUS

9

h=

a ⋅ ( b − a ) × ( c − b)

( b − a ) × ( c − b)

= =

a ⋅ ( b × c − a × c + a × b) b×c−a×c+a×b a⋅b× c a×b+b×c+c×a

(1)

The area of the triangle ABC is:

A=

1 1 1 ( b − a ) × ( c − b) = ( a − c ) × ( b − a ) = ( c − b ) × ( a − c ) 2 2 2

(2)

1-13.

Using the Eq. (1.82) in the text, we have

A × B = A × ( A × X ) = ( X ⋅ A ) A − ( A ⋅ A ) X = φ A − A2 X

from which

X=

( B × A ) + φA

A2

1-14.

a)

1 2 −1 2 1 0 1 −2 1 AB = 0 3 1 0 −1 2 = 1 −2 9 2 0 1 1 1 3 5 3 3 −2 9 1 9 1 −2 +2 +1 3 3 5 3 5 3

AB = −104

Expand by the first row.

AB = 1

b)

1 2 −1 2 1 9 7 AC = 0 3 1 4 3 = 13 9 2 0 1 1 0 5 2

9 7 AC = 13 9 5 2

10 5 1 2 −1 8 ABC = A ( BC ) = 0 3 1 −2 −3 2 0 1 9 4 −5 −5 ABC = 3 −5 25 14

CHAPTER 1

c)

d)

AB − Bt At = ? 1 −2 1 AB = 1 −2 9 5 3 3

(from part a )

2 0 1 1 0 2 1 1 5 B A = 1 −1 1 2 3 0 = −2 −2 3 0 2 3 −1 1 1 1 9 3 t t

0 −3 −4 AB − B A = 3 0 6 4 −6 0 t t

1-15.

If A is an orthogonal matrix, then

At A = 1 1 0 0 1 0 0 1 0 0 0 a a 0 a − a = 0 1 0 0 − a a 0 a a 0 0 1 1 0 2 0 2a 0 0 a= 1 2

0 1 0 0 0 = 0 1 0 2a 2 0 0 1

MATRICES, VECTORS, AND VECTOR CALCULUS

11

1-16. x3 r θ a x2 x1 P r θ a r cos θ

r ⋅ a = constant ra cos θ = constant It is given that a is constant, so we know that r cos θ = constant But r cos θ is the magnitude of the component of r along a. The set of vectors that satisfy r ⋅ a = constant all have the same component along a; however, the component perpendicular to a is arbitrary. Thus the surface represented by r ⋅ a = constant is a plane perpendicular to a.

1-17. a B c θ A C b

Consider the triangle a, b, c which is formed by the vectors A, B, C. Since

C= A−B C = ( A − B) ⋅ ( A − B)

2

(1)

= A 2 − 2A ⋅ B + B 2 or, C = A2 + B2 − 2 AB cos θ

2

(2)

which is the cosine law of plane trigonometry.

1-18.

Consider the triangle a, b, c which is formed by the vectors A, B, C. a B c α γ A C β b

12

CHAPTER 1

C=A−B so that

(1)

C × B = ( A − B) × B but the left-hand side and the right-hand side of (2) are written as:

(2)

C × B = BC sin α e3 and (3)

( A − B) × B = A × B − B × B = A × B = AB sin γ e3 where e3 is the unit vector perpendicular to the triangle abc. Therefore, BC sin α = AB sin γ or, C A = sin γ sin α Similarly, C A B = = sin γ sin α sin β which is the sine law of plane trigonometry.

1-19.

x2 a2 b2 α a1 β b1 a b x1

(4)

(5)

(6)

a) We begin by noting that

a − b = a 2 + b 2 − 2ab cos (α − β )

2

(1)

We can also write that

a − b = ( a1 − b1 ) + ( a2 − b2 )

2 2 2

2

= ( a cos α − b cos β ) + ( a sin α − b sin β )

2

= a 2 sin 2 α + cos 2 α + b 2 sin 2 β + cos 2 β − 2 ab ( cos α cos β + sin α sin β ) = a 2 + b 2 − 2ab ( cos α cos β + sin α sin β )

(2)

(

)

(

)

MATRICES, VECTORS, AND VECTOR CALCULUS

13

Thus, comparing (1) and (2), we conclude that cos (α − β ) = cos α cos β + sin α sin β

b)

(3)

Using (3), we can find sin (α − β ) :

sin (α − β ) = 1 − cos 2 (α − β ) = 1 − cos 2 α cos 2 β − sin 2 α sin 2 β − 2cos α sin α cos β sin β = 1 − cos 2 α 1 − sin 2 β − sin 2 α 1 − cos 2 β − 2cos α sin α cos β sin β = sin 2 α cos 2 β − 2sin α sin β cos α cos β + cos 2 α sin 2 β = so that sin (α − β ) = sin α cos β − cos α sin β (5)

(

)

(

)

( sin α cos β − cos α sin β )2

(4)

1-20. a) Consider the following two cases:

When i ≠ j When i = j Therefore,

δ ij = 0 δ ij ≠ 0

but ε ijk ≠ 0 . but ε ijk = 0 .

∑ε ij ijk

δ ij = 0

(1)

b) We proceed in the following way:

When j = k, ε ijk = ε ijj = 0 . Terms such as ε j11 ε

11

= 0 . Then,

12

∑ε jk ijk

ε jk = ε i12 ε

+ ε i13 ε

13

+ ε i 21 ε

21

+ ε i 31 ε

31

+ ε i 32 ε

32

+ ε i 23 ε

23

Now, suppose i = = 1 , then,

∑= ε jk 123

ε 123 + ε 132 ε 132 = 1 + 1 = 2

14

CHAPTER 1

for i = = 2 ,

= 2 gives

∑= ε jk jk

213

ε 213 + ε 231 ε 231 = 1 + 1 = 2 . For i = = 3 ,

= 1 ; i = 1,

∑= ε jk 312

ε 312 + ε 321 ε 321 = 2 . But i = 1,

= 3 ; i = 3, = 2.

∑ = 0 . Likewise for i = 2,

= 3 ; i = 3,

= 1 ; i = 2,

Therefore,

∑ε j,k ijk

ε jk = 2δ i

(2)

c)

∑ε ijk ijk

ε ijk = ε 123 ε 123 + ε 312 ε 312 + ε 321 ε 321 + ε 132 ε 132 + ε 213 ε 213 + ε 231 ε 231

= 1 ⋅ 1 + 1 ⋅ 1 + ( −1) ⋅ ( −1) + ( −1) ⋅ ( −1) + ( −1) ⋅ ( −1) + (1) ⋅ (1)

or,

∑ε ijk ijk

ε ijk = 6

(3)

1-21.

( A × B)i = ∑ ε ijk Aj Bk jk (1)

( A × B) ⋅ C = ∑ i ∑ε jk ijk

Aj Bk Ci

(2)

By an even permutation, we find

ABC = ∑ ε ijk Ai Bj Ck ijk (3)

1-22. a) b)

To evaluate

∑ε k k

ijk

ε mk we consider the following cases: mk i= j: i= :

∑ε k k

ijk

ε mk = ∑ ε iik ε

= 0 for all i , , m

∑ε

ijk

ε mk = ∑ ε ijk ε imk = 1 for j = m and k ≠ i , j k = 0 for j ≠ m

c)

i = m:

∑ε k ijk

ε mk = ∑ ε ijk ε k ik

= 0 for j ≠ = −1 for j = and k ≠ i , j

d)

j= :

∑ε k ijk

ε mk = ∑ ε ijk ε jmk = 0 for m ≠ i k = −1 for m = i and k ≠ i , j

MATRICES, VECTORS, AND VECTOR CALCULUS

15

e)

j = m:

∑ε k ijk

ε mk = ∑ ε ijk ε k jk

= 0 for i ≠ = 1 for i = and k ≠ i , j

f) g)

= m:

∑ε k ijk

ε mk = ∑ ε ijk ε k k

= 0 for all i , j , k

i ≠ or m : This implies that i = k or i = j or m = k.

Then,

h)

∑ε k ijk

ε mk = 0 for all i , j , , m

j ≠ or m :

∑ε k ijk

ε mk = 0 for all i , j , , m

Now, consider δ i δ jm − δ im δ j and examine it under the same conditions. If this quantity behaves in the same way as the sum above, we have verified the equation

∑ε k ijk

ε mk = δ i δ jm − δ im δ j

a) b)

i = j : δ i δ im − δ im δ i = 0 for all i , , m i = : δ ii δ jm − δ im δ ji = 1 if j = m, i ≠ j , m = 0 if j ≠ m

c)

i = m : δ i δ ji − δ ii δ j = −1 if j = , i ≠ j , = 0 if j ≠

d)

j = : δi δ

m

− δ im δ = −1 if i = m, i ≠ = 0 if i ≠ m

e)

j = m : δ i δ mm − δ im δ m = 1 if i = , m ≠ = 0 if i ≠

f) g) h)

= m : δ i δ j − δ il δ j = 0 for all i , j , i ≠ , m : δ i δ jm − δ im δ j = 0 for all i , j , , m j ≠ , m : δ i δ jm − δ im δ i = 0 for all i , j , , m

Therefore,

∑ε k ijk

ε

mk

= δ i δ jm − δ im δ j

(1)

Using this result we can prove that

A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B) C

16

First ( B × C ) i = ∑ ε ijk Bj Ck . Then, jk CHAPTER 1

[ A × (B × C) ] = ∑ ε mn Am ( B × C ) n = ∑ ε mn Am ∑ ε njk BjCk mn mn jk

=

jkmn

∑ε

mn

ε njk Am Bj Ck =

jkmn

∑ε

mn

ε jkn Am Bj Ck

= ∑ ∑ ε lmn ε jkn Am Bj Ck jkm n = ∑ δ jlδ km − δ k δ jm Am Bj Ck jkm (

)

= ∑ Am B Cm − ∑ Am BmC = B ∑ AmCm − C ∑ Am Bm m m m m = ( A ⋅ C ) B − ( A ⋅ B) C

Therefore,

A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B) C

(2)

1-23.

Write

( A × B) j = ∑ ε j m A Bm m ( C × D) k = ∑ ε krs Cr Ds rs Then,

MATRICES, VECTORS, AND VECTOR CALCULUS

17

[ ( A × B) × (C × D)]i = ∑ ε ijk ∑ ε j m A Bm ∑ ε krs Cr Ds jk m rs

=

jk mrs

∑ε

ijk

ε j m ε krs A BmCr Ds

∑ ε ijk ε rsk A BmCr Ds k ir = =

j mrs

∑ε

j m

j mrs

∑ ε (δ j m

δ js − δ is δ jr A BmCr Ds

)

= ∑ ε j m A BmCi Dj − A Bm Di C j j m

(

)

= ∑ ε j m Dj A Bm Ci − ∑ ε j m C j A Bm Di jm jm = (ABD)Ci − (ABC)Di

Therefore,

[( A × B) × (C × D) ] = (ABD)C − (ABC)D

1-24.

Expanding the triple vector product, we have e × ( A × e) = A ( e ⋅ e) − e ( A ⋅ e)

(1)

But,

A ( e ⋅ e) = A

(2)

Thus,

A = e ( A ⋅ e) + e × ( A × e) e(A · e) is the component of A in the e direction, while e × (A × e) is the component of A perpendicular to e.

(3)

18

1-25.

er eφ θ φ eθ

CHAPTER 1

The unit vectors in spherical coordinates are expressed in terms of rectangular coordinates by eθ = ( cos θ cos φ , cos θ sin φ , − sin θ ) eφ = ( − sin φ , cos φ , 0 ) er = ( sin θ cos φ , sin θ sin φ , cos θ )

(1)

Thus, eθ = −φ cos θ sin φ − θ sin θ cos φ , φ cos θ cos φ − θ sin θ sin φ , − θ cos θ

= −θ er + φ cos θ eφ

(

)

(2)

Similarly, eφ = −φ cos φ , − φ sin φ , 0

(

)

(3) (4)

= −φ cos θ eθ − φ sin θ er

er = φ sin θ eφ + θ eθ Now, let any position vector be x. Then, x = rer

(5)

x = rer + rer = r φ sin θ eφ + θ eθ + rer = rφ sin θ eφ + rθ eθ + rer x = rφ sin θ + rθφ cos θ + rφ sin θ eφ + rφ sin θ eφ + rθ + rθ eθ + rθ eθ + rer + rer = 2rφ sin θ + 2rθφ cos θ + rφ sin θ eφ + r − rφ 2 sin 2 θ − rθ 2 er + 2rθ + rθ − rφ 2 sin θ cos θ eθ or,

(

)

(6)

( (

)

(

)

)

(

)

(

)

(7)

MATRICES, VECTORS, AND VECTOR CALCULUS

19

1 d 2 r θ − rφ 2 sin θ cos θ eθ x = a = r − rθ 2 − rφ 2 sin 2 θ er + r dt 1 d 2 + r φ sin 2 θ eφ r sin θ dt

( )

(

)

(8)

1-26.

When a particle moves along the curve r = k (1 + cos θ ) (1)

we have 2 r = − k θ cos θ + θ sin θ Now, the velocity vector in polar coordinates is [see Eq. (1.97)] r = − kθ sin θ v = rer + rθ eθ so that v 2 = v = r 2 + r 2θ 2

2

(2)

(3)

= k 2θ 2 sin 2 θ + k 2 1 + 2 cos θ + cos 2 θ θ 2 = k 2θ 2 2 + 2 cos θ and v 2 is, by hypothesis, constant. Therefore, (4)

(

)

θ=

Using (1), we find

v2 2k (1 + cos θ )

2

(5)

θ=

v 2kr

(6)

Differentiating (5) and using the expression for r , we obtain

θ=

v 2 sin θ v 2 sin θ = 2 2 4r 2 4 k (1 + cos θ )

(7)

The acceleration vector is [see Eq. (1.98)] a = r − rθ 2 er + rθ + 2rθ eθ

(

)

(

)

(8)

so that

20

CHAPTER 1

a ⋅ er = r − rθ 2 = − k θ 2 cos θ + θ sin θ − k (1 + cos θ ) θ 2 θ 2 sin 2 θ = − k θ 2 cos θ + + (1 + cos θ ) θ 2 2 (1 + cos θ ) 1 − cos 2 θ = − kθ 2 2 cos θ + + 1 2 (1 + cos θ ) 3 = − kθ 2 (1 + cos θ ) 2 or, a ⋅ er = −

(

)

(9)

3 v2 4 k

(10)

In a similar way, we find a ⋅ eθ = −

3 v 2 sin θ 4 k 1 + cos θ

(11)

From (10) and (11), we have a =

( a ⋅ er ) 2 + ( a ⋅ eθ ) 2

3 v2 4 k 2 1 + cos θ

(12)

or, a =

(13)

1-27.

Since r × (v × r) = (r ⋅ r) v − (r ⋅ v) r

we have d d [ r × ( v × r ) ] = dt [ ( r ⋅ r ) v − ( r ⋅ v ) r ] dt = (r ⋅ r) a + 2 (r ⋅ v) v − (r ⋅ v) v − ( v ⋅ v) r − (r ⋅ a) r = r 2a + ( r ⋅ v ) v − r v2 + r ⋅ a Thus, d [ r × ( v × r )] = r 2a + ( r ⋅ v ) v − r r ⋅ a + v 2 dt

(

)

( )

(1)

(2)

MATRICES, VECTORS, AND VECTOR CALCULUS

21

∂ ( ln r ) ei ∂x i

1-28.

grad ( ln r ) = ∑ i (1)

where

r =

Therefore,

∑x i 2 i

(2)

∂ 1 ( ln r ) = r ∂x i = so that xi

∑x i 2

xi

2 i

r

(3)

grad ( ln r ) = or, 1 xe 2 ∑ i i r i

(4)

grad ( ln r ) =

r r2

(5)

1-29. Let r 2 = 9 describe the surface S1 and x + y + z 2 = 1 describe the surface S2 . The angle θ between S1 and S2 at the point (2,–2,1) is the angle between the normals to these surfaces at the point. The normal to S1 is

grad ( S1 ) = grad r 2 − 9 = grad x 2 + y 2 + z 2 − 9

= ( 2xe1 + 2 ye2 + 2ze3 ) = 4e1 − 4e2 + 2e3 In S2 , the normal is: x = 2, y = 2, z = 1

(

)

(

)

(1)

grad ( S2 ) = grad x + y + z 2 − 1

= ( e1 + e2 + 2ze3 ) = e1 + e2 + 2e3 Therefore,

(

)

(2)

x = 2, y =−2, z = 1

22

CHAPTER 1

cos θ =

grad ( S1 ) ⋅ grad ( S2 ) grad ( S1 ) grad ( S2 )

= or,

( 4e1 − 4e2 + 2e3 ) ⋅ (e1 + e2 + 2e3 )

6 6

(3)

cos θ = from which

4 6 6

(4)

θ = cos −1

6 = 74.2° 9

(5)

1-30.

grad ( φψ ) = ∑ ei i =1

3

∂ψ ∂ φ ∂ ( φψ ) ψ = ∑ ei φ + ∂x i i ∂x i ∂x i ∂ψ ∂φ ψ + ∑ ei ∂x i ∂x i i

= ∑ ei φ i Thus, grad ( φψ ) = φ grad ψ + ψ grad φ

1-31. a)

12 ∂rn ∂ n 2 ∑ xj = ∑ ei grad r = ∑ ei ∂x i ∂x i j i =1 3

n

2 n = ∑ ei 2 xi ∑ x 2 j 2 j i 2 = ∑ ei x i n ∑ x 2 j j i = ∑ ei x i n r ( n − 2) i n

n

−1

−1

(1)

Therefore, grad r n = nr ( n − 2) r (2)

MATRICES, VECTORS, AND VECTOR CALCULUS

23

b)

grad f ( r ) = ∑ ei i =1

3

∂f ( r ) 3 ∂ f ( r ) ∂ r = ∑ ei ∂x i ∂ r ∂x i i =1

12

∂ 2 = ∑ ei ∑ xj ∂x i j i = ∑ ei xi ∑ x 2 j j i = ∑ ei i ∂f ( r ) ∂r ∂f ( r ) ∂r (3)

−1 2

x i ∂f r dr

Therefore, grad f ( r ) =

c)

12 ∂ 2 ln r ∂2 2 ∇ ( ln r ) = ∑ = ∑ 2 ln ∑ x j ∂xi2 ∂x i j i 2 −1 2 1 ⋅ 2 xi ∑ x 2 j 2 j ∂ =∑ 12 i ∂x i 2 ∑ xj j

r ∂f ( r ) r ∂r

(4)

∂ =∑ i ∂x i

−1 xi ∑ x 2 j j −2

= ∑ ( − xi )( 2xi ) ∑ x 2 j j i = ∑ −2x 2 r 2 j i ∂x + ∑ i ∑ x2 j i ∂x i j

−1

(

)( )

−2

1 + 3 2 r (5)

=− or,

2r 2 3 1 + 2 = 2 4 r r r

∇ 2 ( ln r ) =

1 r2

(6)

24

1-32.

CHAPTER 1

Note that the integrand is a perfect differential: 2ar ⋅ r + 2br ⋅ r = a

d d (r ⋅ r) + b (r ⋅ r) dt dt

(1)

Clearly,

∫ ( 2ar ⋅ r + 2br ⋅ r ) dt = ar

1-33.

2

+ br 2 + const.

(2)

Since

d r rr − rr r rr = = − 2 dt r r2 r r we have

(1)

∫ from which

d r r rr r − r 2 dt = ∫ dt r dt

(2)

∫

r r rr r − r 2 dt = r + C

(3)

where C is the integration constant (a vector).

1-34.

First, we note that

d A×A =A×A+A×A dt

But the first term on the right-hand side vanishes. Thus,

(

)

(1)

∫ ( A × A) dt = ∫ dt ( A × A) dt d so that

(2)

∫ ( A × A) dt = A × A + C where C is a constant vector.

(3)

MATRICES, VECTORS, AND VECTOR CALCULUS

25

1-35. y x

z

We compute the volume of the intersection of the two cylinders by dividing the intersection volume into two parts. Part of the common volume is that of one of the cylinders, for example, the one along the y axis, between y = –a and y = a:

V1 = 2 π a 2 a = 2π a 3

The rest of the common volume is formed by 8 equal parts from the other cylinder (the one along the x-axis). One of these parts extends from x = 0 to x = a, y = 0 to y = a 2 − x 2 , z = a to

( )

(1)

z = a 2 − x 2 . The complementary volume is then

V2 = 8 ∫ dx ∫

0

a

a2 − x 2

0

dy ∫

a2 − x 2 a

dz

= 8 ∫ dx a 2 − x 2 a 2 − x 2 − a 0 a x 3 a3 x = 8 a2 x − − sin −1 3 2 a 0 = Then, from (1) and (2):

V = V1 + V2 =

a

16 3 a − 2π a 3 3

(2)

16 a 3 3

(3)

26

1-36.

z

CHAPTER 1

d

y x c2 = x2 + y2

The form of the integral suggests the use of the divergence theorem.

∫

S

A ⋅ da = ∫ ∇ ⋅ A dv

V

(1)

Since ∇ ⋅ A = 1 , we only need to evaluate the total volume. Our cylinder has radius c and height d, and so the answer is

∫

1-37.

V

dv = π c 2 d

(2)

z

R y

x

To do the integral directly, note that A = R3er , on the surface, and that da = daer .

∫

S

A ⋅ da = R 3

∫

S

da = R3 × 4π R 2 = 4π R 5

(1)

To use the divergence theorem, we need to calculate ∇ ⋅ A . This is best done in spherical coordinates, where A = r 3er . Using Appendix F, we see that ∇⋅A = Therefore, 1 ∂ 2 r A r = 5r 2 r 2 ∂r

(

)

(2)

∫

V

∇ ⋅ A dv = ∫ sin θ dθ ∫

0

π

2π 0

dφ ∫ r 2 5r 2 dr = 4π R5

0

R

( )

(3)

Alternatively, one may simply set dv = 4π r 2 dr in this case.

MATRICES, VECTORS, AND VECTOR CALCULUS

27

1-38. z z = 1 – x2 – y2

y C x x2 + y2 = 1

By Stoke’s theorem, we have

∫ (∇ × A) ⋅ da = ∫

S

C

A ⋅ ds

(1)

The curve C that encloses our surface S is the unit circle that lies in the xy plane. Since the element of area on the surface da is chosen to be outward from the origin, the curve is directed counterclockwise, as required by the right-hand rule. Now change to polar coordinates, so that we have ds = dθ eθ and A = sin θ i + cos θ k on the curve. Since eθ ⋅ i = − sin θ and eθ ⋅ k = 0 , we have

∫

1-39.

C

A ⋅ ds = ∫

2π 0

( − sin θ ) dθ = −π

2

(2)

a) Let’s denote A = (1,0,0); B = (0,2,0); C = (0,0,3). Then AB = (−1, 2, 0) ; AC = (−1, 0, 3) ; and

AB × AC = (6, 3, 2) . Any vector perpendicular to plane (ABC) must be parallel to AB × AC , so the unit vector perpendicular to plane (ABC) is n = (6 7 , 3 7 , 2 7 )

b) Let’s denote D = (1,1,1) and H = (x,y,z) be the point on plane (ABC) closest to H. Then DH = ( x − 1, y − 1, z − 1) is parallel to n given in a); this means

x−1 6 = =2 y−1 3

and

x −1 6 = =3 z−1 2

Further, AH = ( x − 1, y , z) is perpendicular to n so one has 6( x − 1) + 3 y + 2 z = 0 . Solving these 3 equations one finds H = ( x , y , z) = (19 49 , 34 49 , 39 49) and DH = 5 7

1-40. a)

At the top of the hill, z is maximum; 0= ∂z = 2 y − 6 x − 18 ∂x and 0= ∂z = 2 x − 8 y + 28 ∂y

28

CHAPTER 1

so x = –2 ; y = 3, and the hill’s height is max[z]= 72 m. Actually, this is the max value of z, because the given equation of z implies that, for each given value of x (or y), z describes an upside down parabola in term of y ( or x) variable.

b) At point A: x = y = 1, z = 13. At this point, two of the tangent vectors to the surface of the hill are

t1 = (1, 0,

∂z ) = (1, 0, −8) ∂x (1,1)

and

t2 = (0,1,

∂z ) = (0,1, 22) ∂y (1,1)

Evidently t1 × t2 = (8, −22,1) is perpendicular to the hill surface, and the angle θ between this and Oz axis is cos θ = (0, 0,1) ⋅ (8, −22,1) 8 + 22 + 1

2 2 2

=

1 23.43

so θ = 87.55 degrees.

c) Suppose that in the α direction ( with respect to W-E axis), at point A = (1,1,13) the hill is steepest. Evidently, dy = (tan α )dx and

dz = 2xdy + 2 ydx − 6 xdx − 8 ydy − 18 dx + 28 dy = 22(tan α − 1)dx

then tan β = dx 2 + dy 2 dz

=

dx cos α −1 = 22(tan α − 1)dx 22 2 cos (α + 45)

The hill is steepest when tan β is minimum, and this happens when α = –45 degrees with respect to W-E axis. (note that α = 135 does not give a physical answer).

1-41.

A ⋅ B = 2a( a − 1) then A ⋅ B = 0 if only a = 1 or a = 0.

CHAPTER

2

(1)

Newtonian Mechanics— Single Particle

2-1.

The basic equation is

F = mi xi

a) b)

F ( xi , t ) = f ( xi ) g ( t ) = mi xi : Not integrable

(2)

F ( xi , t ) = f ( xi ) g ( t ) = mi xi mi dxi = f ( xi ) g ( t ) dt

g (t) dxi = dt : Integrable f ( xi ) mi

c)

(3) (4)

F ( xi , xi ) = f ( xi ) g ( xi ) = mi xi : Not integrable

2-2.

Using spherical coordinates, we can write the force applied to the particle as

F = Fr er + Fθ eθ + Fφ eφ

(1)

But since the particle is constrained to move on the surface of a sphere, there must exist a reaction force − Fr er that acts on the particle. Therefore, the total force acting on the particle is

Ftotal = Fθ eθ + Fφ eφ = mr

(2)

The position vector of the particle is r = Re r

(3)

where R is the radius of the sphere and is constant. The acceleration of the particle is a = r = Re r

(4)

29

30

CHAPTER 2

We must now express er in terms of er , eθ , and eφ . Because the unit vectors in rectangular coordinates, e1 , e2 , e3 , do not change with time, it is convenient to make the calculation in terms of these quantities. Using Fig. F-3, Appendix F, we see that er = e1 sin θ cos φ + e2 sin θ sin φ + e3 cos θ eθ = e1 cos θ cos φ + e2 cos θ sin φ − e3 sin θ eφ = −e1 sin φ + e2 cos φ

(5)

Then er = e1 −φ sin θ sin φ + θ cos θ cos φ + e2 θ cos θ sin φ + φ sin θ cos φ − e3 θ sin θ

= eφ φ sin θ + eθ θ

(

)

(

)

(6)

Similarly, eθ = −er θ + eφ φ cos θ eφ = −er φ sin θ − eθ φ cos θ And, further, er = −er φ 2 sin 2 θ + θ 2 + eθ θ − φ 2 sin θ cos θ + eφ 2θφ cos θ + φ sin θ which is the only second time derivative needed. The total force acting on the particle is Ftotal = mr = mRer and the components are

Fθ = mR θ − φ 2 sin θ cos θ

(7) (8)

(

)

(

)

(

)

(9)

(10)

( (

) )

(11)

Fφ = mR 2θφ cos θ + φ sin θ

NEWTONIAN MECHANICS—SINGLE PARTICLE

31

2-3. y v0 α

P

β x

The equation of motion is

F=ma

The gravitational force is the only applied force; therefore,

Fx = mx = 0 Fy = my = − mg Integrating these equations and using the initial conditions, x ( t = 0 ) = v0 cos α y ( t = 0 ) = v0 sin α We find x ( t ) = v0 cos α y ( t ) = v0 sin α − gt

(1)

(2)

(3)

(4)

So the equations for x and y are

x ( t ) = v0 t cos α

1 2 y ( t ) = v0 t sin α − gt 2

(5)

Suppose it takes a time t0 to reach the point P. Then, cos β = v0 t0 cos α 1 2 sin β = v0 t0 sin α − gt0 2 Eliminating between these equations, 2v sin α 2v0 1 + cos α tan β = 0 gt0 t0 − 0 2 g g from which (7)

(6)

32

2v0 ( sin α − cos α tan β ) g

CHAPTER 2

t0 =

(8)

2-4. One of the balls’ height can be described by y = y0 + v0 t − gt 2 2 . The amount of time it takes to rise and fall to its initial height is therefore given by 2v0 g . If the time it takes to cycle the ball through the juggler’s hands is τ = 0.9 s , then there must be 3 balls in the air during that

time τ. A single ball must stay in the air for at least 3τ, so the condition is 2v0 g ≥ 3τ , or v0 ≥ 13.2 m ⋅ s −1 .

2-5.

flightpath N er plane point of maximum acceleration mg

a) From the force diagram we have N − mg = mv 2 R er . The acceleration that the pilot feels is

N m = g + mv R er , which has a maximum magnitude at the bottom of the maneuver.

2

(

)

(

)

b) If the acceleration felt by the pilot must be less than 9g, then we have

3 ⋅ 330 m ⋅ s −1 v2 R≥ = 8g 8 ⋅ 9.8 m ⋅ s −2

(

)

12.5 km

(1)

A circle smaller than this will result in pilot blackout.

2-6.

Let the origin of our coordinate system be at the tail end of the cattle (or the closest cow/bull).

a) The bales are moving initially at the speed of the plane when dropped. Describe one of these bales by the parametric equations

x = x 0 + v0 t

(1)

NEWTONIAN MECHANICS—SINGLE PARTICLE

33

y = y0 −

1 2 gt 2

(2)

where y0 = 80 m , and we need to solve for x0 . From (2), the time the bale hits the ground is

τ = 2 y0 g . If we want the bale to land at x (τ ) = −30 m , then x0 = x (τ ) − v0τ . Substituting v0 = 44.4 m ⋅ s -1 and the other values, this gives x0 210 m behind the cattle. −210 m . The rancher should drop the bales

b) She could drop the bale earlier by any amount of time and not hit the cattle. If she were late by the amount of time it takes the bale (or the plane) to travel by 30 m in the x-direction, then she will strike cattle. This time is given by ( 30 m ) v0 0.68 s .

2-7.

Air resistance is always anti-parallel to the velocity. The vector expression is:

W=

1 1 v cw ρ Av 2 − = − cw ρ Avv 2 2 v

(1)

Including gravity and setting Fnet = ma , we obtain the parametric equations x = − bx x 2 + y 2 y = −by x 2 + y 2 − g where b = cw ρ A 2m . Solving with a computer using the given values and ρ = 1.3 kg ⋅ m -3 , we find that if the rancher drops the bale 210 m behind the cattle (the answer from the previous problem), then it takes 4.44 s to land 62.5 m behind the cattle. This means that the bale should be dropped at 178 m behind the cattle to land 30 m behind. This solution is what is plotted in the figure. The time error she is allowed to make is the same as in the previous problem since it only depends on how fast the plane is moving.

80 60

(2) (3)

y (m)

40 20 0

–200

–180

–160

–140

–120 x (m)

–100

–80

–60

–40

With air resistance No air resistance

34

2-8.

y v0 P α h Q x

CHAPTER 2

From problem 2-3 the equations for the coordinates are x = v0 t cos α y = v0 t sin α − 1 2 gt 2 (1) (2)

In order to calculate the time when a projective reaches the ground, we let y = 0 in (2): v0 t sin α − t= 1 2 gt = 0 2

(3)

(4)

2v0 sin α g

Substituting (4) into (1) we find the relation between the range and the angle as x= The range is maximum when 2α =

2 v0 sin 2α g

(5)

π

2

, i.e., α =

π

4

. For this value of α the coordinates become

v 1 x = 0 t − gt 2 2 2 x= v0 t 2 Eliminating t between these equations yields x2 −

2 v0 v2 x+ 0 y=0 g g

(6)

(7)

We can find the x-coordinate of the projectile when it is at the height h by putting y = h in (7): x2 − This equation has two solutions: x1 =

2 2 v0 v0 2 − v0 − 4 gh 2g 2g 2 2 v0 v0 2 x2 = v0 − 4 gh + 2g 2g

2 v0 v2 h x+ 0 =0 g g

(8)

(9)

NEWTONIAN MECHANICS—SINGLE PARTICLE

35

where x1 corresponds to the point P and x2 to Q in the diagram. Therefore, d = x2 − x1 = v0 g

2 v0 − 4 gh

(10)

2-9. a) Zero resisting force ( Fr = 0 ):

The equation of motion for the vertical motion is: F = ma = m Integration of (1) yields v = − gt + v0 where v0 is the initial velocity of the projectile and t = 0 is the initial time. The time tm required for the projectile to reach its maximum height is obtained from (2). Since tm corresponds to the point of zero velocity, v ( tm ) = 0 = v0 − gtm , we obtain tm = v0 g (4) (3) (2) dv = − mg dt (1)

b) Resisting force proportional to the velocity ( Fr = − kmv ) :

The equation of motion for this case is: F=m dv = − mg − kmv dt (5)

where –kmv is a downward force for t < tm and is an upward force for t > tm . Integrating, we ′ ′ obtain v (t) = − For t = tm , v(t) = 0, then from (6), v0 = which can be rewritten as kv ktm = ln 1 + 0 g Since, for small z (z 1) the expansion (8) g ktm e −1 k g kv0 + g − kt e + k k (6)

(

)

(7)

36

1 1 ln (1 + z ) = z − z 2 + z 3 2 3 is valid, (8) can be expressed approximately as

CHAPTER 2

(9)

v tm = 0 g

2 kv 1 kv 1 − 0 + 0 − … 2g 3 2g

(10)

which gives the correct result, as in (4) for the limit k → 0.

2-10. The differential equation we are asked to solve is Equation (2.22), which is x = − kx . Using the given values, the plots are shown in the figure. Of course, the reader will not be able to distinguish between the results shown here and the analytical results. The reader will have to take the word of the author that the graphs were obtained using numerical methods on a computer. The results obtained were at most within 10 −8 of the analytical solution.

10

v (m/s)

v vs t

5

0

5

10

15 t (s) x vs t

20

25

30

100 x (m)

50

0

0

5

10

15 t (s)

20

25

30

10 v (m/s)

v vs x

5

0

0

20

40 x (m)

60

80

100

2-11.

The equation of motion is m d2 x = − kmv 2 + mg dt 2

(1)

This equation can be solved exactly in the same way as in problem 2-12 and we find

NEWTONIAN MECHANICS—SINGLE PARTICLE

37

x=

2 g − kv0 1 log 2 2k g − kv

(2)

where the origin is taken to be the point at which v = v0 so that the initial condition is x ( v = v0 ) = 0 . Thus, the distance from the point v = v0 to the point v = v1 is

2 g − kv0 1 s ( v0 → v1 ) = log 2 2k g − kv1

(3)

2-12.

The equation of motion for the upward motion is m d2 x = − mkv 2 − mg 2 dt (1)

Using the relation d 2 x dv dv dx dv = = =v 2 dt dt dx dt dx we can rewrite (1) as v dv = − dx kv 2 + g Integrating (3), we find 1 log kv 2 + g = − x + C 2k where the constant C can be computed by using the initial condition that v = v0 when x = 0: C= Therefore, x= kv 2 + g 1 log 0 kv 2 + g 2k 1 2 log kv0 + g 2k (3) (2)

(

)

(4)

(

)

(5)

(6)

Now, the equation of downward motion is m d2 x = − mkv 2 + mg dt 2 v dv = dx − kv 2 + g

(7)

This can be rewritten as (8)

Integrating (8) and using the initial condition that x = 0 at v = 0 (w take the highest point as the origin for the downward motion), we find

38 g 1 log g − kv 2 2k

CHAPTER 2

x=

(9)

At the highest point the velocity of the particle must be zero. So we find the highest point by substituting v = 0 in (6): xh = kv 2 + g 1 log 0 g 2k

(10)

Then, substituting (10) into (9), kv 2 + g 1 g 1 = log 0 log 2k 2k g g − kv 2

(11)

Solving for v, g 2 v0 k v= g 2 v0 + k

(12)

We can find the terminal velocity by putting x → ∞ in(9). This gives vt = g k

(13)

Therefore, v= v0 vt

2 v0 + vt2

(14)

2-13.

The equation of motion of the particle is m dv = − mk v 3 + a 2 v dt

(

)

(1)

Integrating,

∫ v(v and using Eq. (E.3), Appendix E, we find

2

dv = − k ∫ dt + a2

)

(2)

v2 1 ln 2 = − kt + C 2 2 2a a + v

(3)

Therefore, we have v2 = C ′ e − At a2 + v 2

(4)

NEWTONIAN MECHANICS—SINGLE PARTICLE

39

where A ≡ 2a 2 k and where C′ is a new constant. We can evaluate C′ by using the initial condition, v = v0 at t = 0:

C′ =

2 v0 2 a 2 + v0

(5)

Substituting (5) into (4) and rearranging, we have

a 2C ′e − At v= − At 1 − C ′e

12

=

dx dt

(6)

Now, in order to integrate (6), we introduce u ≡ e − At so that du = –Au dt. Then, x= ∫

a 2C ′e − At 1 − C e − At ′

12

a dt = A

∫

C ′u 1 − C ′u

12

du u

=−

a C′ A

∫

du −C ′u2 + u

(7)

Using Eq. (E.8c), Appendix E, we find x= a sin −1 (1 − 2C ′u) + C ′′ A

(8)

Again, the constant C″ can be evaluated by setting x = 0 at t = 0; i.e., x = 0 at u = 1:

C ′′ = − a sin −1 (1 − 2C ′ ) A

(9)

Therefore, we have x= a sin −1 −2C ′ e − At + 1 − sin −1 ( −2C ′ + 1) A

(

)

Using (4) and (5), we can write x= −v 2 + a2 1 −1 −v 2 + a 2 − sin −1 20 2 sin 2 2 2ak v +a v0 + a (10)

From (6) we see that v → 0 as t → ∞. Therefore, −v 2 + a2 π lim sin −1 2 = sin −1 (1) = 2 t →∞ 2 v +a Also, for very large initial velocities,

−v 2 + a2 π lim sin −1 20 2 = sin −1 ( −1) = − v0 →∞ 2 v0 + a Therefore, using (11) and (12) in (10), we have (12)

(11)

40

CHAPTER 2

x (t → ∞) =

π

2ka

(13)

and the particle can never move a distance greater than π 2ka for any initial velocity.

2-14.

α

β

d y x

a) The equations for the projectile are

x = v0 cos α t y = v0 sin α t − 1 2 gt 2

Solving the first for t and substituting into the second gives y = x tan α − Using x = d cos β and y = d sin β gives d sin β = d cos β tan α − gd 2 cos 2 β 2 2v0 cos 2 α gx 2 1 2 2 v0 cos 2 α

gd cos 2 β − cos β tan α + sin β 0 = d 2 2 2v0 cos α Since the root d = 0 is not of interest, we have d=

2 2 ( cos β tan α − sin β ) v0 cos 2 α

g cos 2 β

2 2v0 cos α ( sin α cos β − cos α sin β )

=

g cos 2 β d=

2 2v0 cos α sin (α − β ) g cos 2 β

(1)

NEWTONIAN MECHANICS—SINGLE PARTICLE

41

b) Maximize d with respect to α

2 2v0 d − sin α sin (α − β ) + cos α cos (α − β ) cos ( 2α − β ) ( d) = 0 = dα g cos 2 β

cos ( 2α − β ) = 0 2α − β =

π

2

α=

c) Substitute (2) into (1)

π

4

+

β

2

dmax = Using the identity

2 2v0 π β π β 2 cos 4 + 2 sin 4 − 2 g cos β

sin A − sin B = 2 cos we have

1 1 ( A + B) sin ( A − B) 2 2

dmax

sin − sin β v 2 1 − sin β 2 2v0 2 = ⋅ = 0 2 g cos β 2 g 1 − sin 2 β dmax =

2 v0 g (1 + sin β )

π

2-15. mg sin θ θ

mg

The equation of motion along the plane is m Rewriting this equation in the form

1 dv = dt k g sin θ − v 2 k (2)

dv = mg sin θ − kmv 2 dt

(1)

42

CHAPTER 2

We know that the velocity of the particle continues to increase with time (i.e., dv dt > 0 ), so that

( g k ) sin θ > v2 . Therefore, we must use Eq. (E.5a), Appendix E, to perform the integration. We

1 k v tanh −1 =t+C g g sin θ sin θ k k 1

find

(3)

The initial condition v(t = 0) = 0 implies C = 0. Therefore, v= g sin θ tanh k

(

gk sin θ t =

)

dx dt

(4)

We can integrate this equation to obtain the displacement x as a function of time: x= g sin θ ∫ tanh k

(

gk sin θ t dt

)

Using Eq. (E.17a), Appendix E, we obtain x= ln cosh gk sin θ t g sin θ + C′ k gk sin θ

(

)

(5)

The initial condition x(t = 0) = 0 implies C′ = 0. Therefore, the relation between d and t is d= From this equation, we can easily find t= cosh −1 e dk 1 ln cosh k

(

gk sin θ t

)

(6)

( )

gk sin θ

(7)

2-16. The only force which is applied to the article is the component of the gravitational force along the slope: mg sin α. So the acceleration is g sin α. Therefore the velocity and displacement along the slope for upward motion are described by:

v = v0 − ( g sin α ) t x = v0 t − 1 ( g sin α ) t 2 2

(1) (2)

where the initial conditions v ( t = 0 ) = v0 and x ( t = 0 ) = 0 have been used. At the highest position the velocity becomes zero, so the time required to reach the highest position is, from (1), t0 = At that time, the displacement is v0 g sin α (3)

NEWTONIAN MECHANICS—SINGLE PARTICLE

43

2 1 v0 2 g sin α

x0 =

(4)

For downward motion, the velocity and the displacement are described by v = ( g sin α ) t x= 1 ( g sin α ) t 2 2 (5) (6)

where we take a new origin for x and t at the highest position so that the initial conditions are v(t = 0) = 0 and x(t = 0) = 0. We find the time required to move from the highest position to the starting position by substituting (4) into (6): t′ = Adding (3) and (7), we find t= 2v0 g sin α (8) v0 g sin α (7)

for the total time required to return to the initial position.

2-17.

v0 35˚

60 m

0.7 m Fence

The setup for this problem is as follows: x = v0 t cos θ y = y0 + v0 t sin θ − 1 2 gt 2 (1) (2)

where θ = 35 and y0 = 0.7 m . The ball crosses the fence at a time τ = R ( v0 cos θ ) , where R = 60 m. It must be at least h = 2 m high, so we also need h − y0 = v0τ sin θ − gτ 2 2 . Solving for v0 , we obtain

2 v0 =

2 cos θ R sin θ − ( h − y0 ) cos θ

gR 2

(3)

which gives v0

25.4 m ⋅ s −1 .

44

2-18.

CHAPTER 2

a) The differential equation here is the same as that used in Problem 2-7. It must be solved for many different values of v0 in order to find the minimum required to have the ball go over the fence. This can be a computer-intensive and time-consuming task, although if done correctly is easily tractable by a personal computer. This minimum v0 is 35.2 m ⋅ s −1 , and the trajectory is

shown in Figure (a). (We take the density of air as ρ = 1.3 kg ⋅ m −3 .)

15

10 y (m)

5

0

0

10

20

30 x (m)

40

50

60

With air resistance No air resistance fence height fence range

b) The process here is the same as for part (a), but now we have v0 fixed at the result just obtained, and the elevation angle θ must be varied to give the ball a maximum height at the fence. The angle that does this is 0.71 rad = 40.7°, and the ball now clears the fence by 1.1 m. This trajectory is shown in Figure (b).

20

15 y (m)

10

5

0

0

10

Flight Path fence height fence range

20

30 x (m)

40

50

60

NEWTONIAN MECHANICS—SINGLE PARTICLE

45

2-19.

The projectile’s motion is described by x = ( v0 cos α ) t 1 y = ( v0 sin α ) t − gt 2 2

(1)

where v0 is the initial velocity. The distance from the point of projection is r = x2 + y2 Since r must always increase with time, we must have r > 0 : (2)

r= Using (1), we have xx + yy =

xx + yy >0 r

(3)

1 2 3 3 2 g t − g ( v0 sin α ) t 2 + v0 t 2 2

(4)

Let us now find the value of t which yields xx + yy = 0 (i.e., r = 0 ): t= 3 v0 sin α v0 ± 9 sin 2 α − 8 2 g 2g (5)

For small values of α, the second term in (5) is imaginary. That is, r = 0 is never attained and the value of t resulting from the condition r = 0 is unphysical. Only for values of α greater than the value for which the radicand is zero does t become a physical time at which r does in fact vanish. Therefore, the maximum value of α that insures r > 0 for all values of t is obtained from 9 sin 2 α max − 8 = 0 or, sin α max = so that 2 2 3 (7) (6)

α max ≅ 70.5°

(8)

2-20.

If there were no retardation, the range of the projectile would be given by Eq. (2.54): R= (1)

2 v0 sin 2θ 0 g The angle of elevation is therefore obtained from

46

Rg 2 v0

CHAPTER 2

sin 2θ 0 =

(1000 m ) × ( 9.8 m/sec 2 ) = (140 m/sec ) 2

= 0.50 so that (2) (3)

θ 0 = 15°

Now, the real range R′, in the linear approximation, is given by Eq. (2.55):

4 kV R′ = R 1 − 3g =

2 v0 sin 2θ 4 kv0 sin θ 1 − 3g g

(4)

Since we expect the real angle θ to be not too different from the angle θ 0 calculated above, we can solve (4) for θ by substituting θ 0 for θ in the correction term in the parentheses. Thus,

sin 2θ =

g R′ 4kv0 sin θ 0 v 1 − 3g

2 0

(5)

Next, we need the value of k. From Fig. 2-3(c) we find the value of km by measuring the slope of the curve in the vicinity of v = 140 m/sec. We find km ≅ (110 N ) ( 500 m/s ) ≅ 0.22 kg/s . The curve is that appropriate for a projectile of mass 1 kg, so the value of k is k 0.022 sec −1 (6) Substituting the values of the various quantities into (5) we find θ = 17.1° . Since this angle is somewhat greater than θ 0 , we should iterate our solution by using this new value for θ 0 in (5). We then find θ = 17.4° . Further iteration does not substantially change the value, and so we conclude that

θ = 17.4°

If there were no retardation, a projectile fired at an angle of 17.4° with an initial velocity of 140 m/sec would have a range of R=

(140 m/sec ) 2 sin 34.8°

9.8 m/sec 2 1140 m

NEWTONIAN MECHANICS—SINGLE PARTICLE

47

2-21. x3 v0 α x2

x1

Assume a coordinate system in which the projectile moves in the x2 − x3 plane. Then, x2 = v0 t cos α 1 x3 = v0 t sin α − gt 2 2 or, r = x 2 e2 + x 3 e 3

(1)

1 = ( v0 t cos α ) e2 + v0 t sin α − gt 2 e3 2 The linear momentum of the projectile is p = mr = m ( v0 cos α ) e2 + ( v0 sin α − gt ) e3 and the angular momentum is

(2)

(3)

L = r × p = ( v0 t cos α ) e2 + v0 t sin α − gt 2 e3 × m ( v0 cos α ) e2 + ( v0 sin α − gt ) e3 Using the property of the unit vectors that ei × e j = e3 ε ijk , we find

L=

(

)

(4)

1 mg v0 t 2 cos α e1 2

(

)

(5)

This gives

L = − ( mg v0 t cos α ) e1

(6)

Now, the force acting on the projectile is

F = − mg e3

(7)

so that the torque is 1 N = r × F = ( v0 t cos α ) e2 + v0 t sin α − gt 2 e3 ( − mg ) e3 2 = − ( mg v0 t cos α ) e1 which is the same result as in (6).

48

2-22.

z B E

CHAPTER 2

y e x

Our force equation is

F = q ( E + v × B)

(1)

a) Note that when E = 0, the force is always perpendicular to the velocity. This is a centripetal acceleration and may be analyzed by elementary means. In this case we have also v ⊥ B so that v × B = vB .

macentripetal = Solving this for r r= with ω c ≡ qB/m .

mv 2 = qvB r

(2)

mv v = qB ω c

(3)

b) Here we don’t make any assumptions about the relative orientations of v and B, i.e. the velocity may have a component in the z direction upon entering the field region. Let r = xi + yj + zk , with v = r and a = r . Let us calculate first the v × B term.

v × B = x y z = B ( yi − xj) 0 0 B

i

j

k

(4)

The Lorentz equation (1) becomes

F = mr = qByi + q Ey − Bx j + qEz k

(

)

(5)

Rewriting this as component equations: x= qB y = ωc y m qEy Ey qB x+ = −ω c x − m m B (6) (7) (8)

y=− z=

qEz m

NEWTONIAN MECHANICS—SINGLE PARTICLE

49

The z-component equation of motion (8) is easily integrable, with the constants of integration given by the initial conditions in the problem statement. z ( t ) = z0 + z0 t + qEz 2 t 2m (9)

c) We are asked to find expressions for x and y , which we will call vx and vy , respectively.

Differentiate (6) once with respect to time, and substitute (7) for v y Ey vx = ω c vy = −ω c2 vx − B or

2 vx + ω c2 vx = ω c

(10)

Ey B

(11)

This is an inhomogeneous differential equation that has both a homogeneous solution (the solution for the above equation with the right side set to zero) and a particular solution. The most general solution is the sum of both, which in this case is vx = C1 cos (ω c t ) + C2 sin (ω c t ) + Ey B (12)

where C1 and C2 are constants of integration. This result may be substituted into (7) to get v y v y = −C1ω c cos (ω c t ) − C2ω c sin (ω c t ) vy = −C1 sin (ω c t ) + C2 cos (ω c t ) + K (13) (14)

where K is yet another constant of integration. It is found upon substitution into (6), however, that we must have K = 0. To compute the time averages, note that both sine and cosine have an average of zero over one of their periods T ≡ 2π ω c . x = Ey B , y =0 (15)

d) We get the parametric equations by simply integrating the velocity equations.

x= y=

ωc

C1 C1

sin (ω c t ) − cos (ω c t ) +

ωc

C2 C2

cos (ω c t ) +

Ey B

t + Dx

(16) (17)

ωc

ωc

sin (ω c t ) + Dy

where, indeed, Dx and Dy are constants of integration. We may now evaluate all the C’s and D’s using our initial conditions x ( 0 ) = − A ω c , x ( 0 ) = Ey B , y ( 0 ) = 0 , y ( 0 ) = A . This gives us C1 = Dx = Dy = 0 , C2 = A and gives the correct answer x (t) = −A

ωc

cos (ω c t ) +

Ey B

t

(18)

50

CHAPTER 2

y(t) =

A

ωc

sin (ω c t )

(19)

These cases are shown in the figure as (i) A > Ey B , (ii) A < Ey B , and (iii) A = Ey B .

(i)

(ii)

(iii)

2-23.

F ( t ) = ma ( t ) = kte − at

(1)

with the initial conditions x(t) = v(t) = 0. We integrate to get the velocity. Showing this explicitly,

∫ v(0)a (t ) dt = m ∫ 0te v( t ) t

k

−α t

dt

(2)

Integrating this by parts and using our initial conditions, we obtain v(t) =

k 1 1 1 − t + e −α t m α 2 α α

(3)

By similarly integrating v(t), and using the integral (2) we can obtain x(t). x (t) = k 2 1 1 2 −α t − α 3 + α 2 + α 2 t + α e m x ( t ) = te − t 2 v ( t ) = 4 − 2 ( t + 2) e − t 2 (4)

To make our graphs, substitute the given values of m = 1 kg, k = 1 N ⋅ s −1 , and α = 0.5 s −1 . (5) (6) (7)

α ( t ) = −16 + 4t + 4 ( t + 4 ) e − t 2

NEWTONIAN MECHANICS—SINGLE PARTICLE

51

100

x(t)

50

0

0

5

10 t

15

20

4

v(t)

2

0

0

5

10 t

15

20

1

a(t)

0.5

0

0

5

10 t

15

20

2-24.

Ff N d = length of incline s = distance skier travels along level ground

sθ

y

N′ mg

mg

co

mg

x θ B

y x

Ff

sin

θ

mg

While on the plane:

∑ Fy = N − mg cosθ = my = 0 ∑ Fx = mg sin θ − Ff ;

So the acceleration down the plane is:

so N = mg cos θ

Ff = µ N = µ mg cos θ

mg sin θ − µ mg cos θ = mx a1 = g ( sin θ − µ cos θ ) = constant

52

CHAPTER 2

While on level ground: N ′ = mg ; Ff = − µ mg So

∑ Fx = mx becomes − µmg = mx a2 = − µ g = constant

The acceleration while on level ground is

For motion with constant acceleration, we can get the velocity and position by simple integration: x=a v = x = at + v0 x − x 0 = v0 t + Solving (1) for t and substituting into (2) gives: v − v0 =t a x − x0 = or

2 2a ( x − x0 ) = v 2 − v0

(1) (2)

1 2 at 2

v0 ( v − v0 ) a

1 ( v − v0 ) + ⋅ 2 a

2

Using this equation with the initial and final points being the top and bottom of the incline respectively, we get: 2a1d = VB2 VB = speed at bottom of incline 2a2 s = −VB2 Thus a1d = − a2 s So gd ( sin θ − µ cos θ ) = µ gs Solving for µ gives d sin θ d cos θ + s a1 = g ( sin θ − µ cos θ ) a2 = − µ g

Using the same equation for motion along the ground: (3)

µ=

Substituting θ = 17°, d = 100 m, s = 70 m gives

µ = 0.18

Substituting this value into (3):

NEWTONIAN MECHANICS—SINGLE PARTICLE

53

−2µ gs = −VB2 VB = 2µ gs VB = 15.6 m/sec

2-25. a)

At A, the forces on the ball are:

N

mg

The track counters the gravitational force and provides centripetal acceleration N − mg = mv 2 R Get v by conservation of energy: Etop = Ttop + Utop = 0 + mgh EA = TA + U A = 1 mv 2 + 0 2

Etop = EA → v = 2 gh So N = mg + m2 gh R 2h N = mg 1 + R

b)

At B the forces are:

N

45˚ mg

N = mv 2 R + mg cos 45° = mv 2 R + mg Get v by conservation of energy. From a), Etotal = mgh . At B, E = 1 mv 2 + mgh ′ 2 2 (1)

54

R 2

CHAPTER 2

R R 45˚

R cos 45˚ = R

2

h′

R= So Etotal = TB + UB becomes:

R + h′ 2

or

1 h′ = R 1 − 2

1 1 2 mgh = mgR 1 − + mv 2 2 Solving for v 2 1 2 2 gh − gR 1 − =v 2 Substituting into (1): 2h 3 N = mg + − 2 R 2

2 c) From b) vB = 2 g h − R + R

2 v = 2g h − R + R

(

2

)

12

d) This is a projectile motion problem

45˚ 45˚ B A

Put the origin at A. The equations: x = x0 + vx 0 t y = y0 + v y 0 t − become x= R v + B t 2 2 vB 1 t − gt 2 2 2 (2) (3) 1 2 gt 2

y = h′ + Solve (3) for t when y = 0 (ball lands).

NEWTONIAN MECHANICS—SINGLE PARTICLE

55

gt 2 − 2 vBt − 2h ′ = 0 t=

2 2 vB ± 2vB + 8 gh′

2g

We discard the negative root since it gives a negative time. Substituting into (2): x=

2 R v 2 vB ± 2vB + 8 gh ′ + B 2g 2 2

Using the previous expressions for vB and h′ yields x=

e)

(

3 2 − 1 R + h + h 2 − R2 + 2 R2 2

)

12

U ( x) = mgy( x ) , with y(0) = h , so U ( x) has the shape of the track.

2-26. All of the kinetic energy of the block goes into compressing the spring, so that mv 2 2 = kx 2 2 , or x = v m k 2.3 m , where x is the maximum compression and the given values have been substituted. When there is a rough floor, it exerts a force µk mg in a direction

that opposes the block’s velocity. It therefore does an amount of work µk mgd in slowing the block down after traveling across the floor a distance d. After 2 m of floor, the block has energy mv 2 2 − µk mgd , which now goes into compressing the spring and still overcoming the friction on the floor, which is kx 2 2 + µk mgx . Use of the quadratic formula gives x=−

µ mg k mv 2 2µ mgd µ mg + + − k k k

2

(1)

Upon substitution of the given values, the result is

1.12 m.

2-27.

0.6 m

To lift a small mass dm of rope onto the table, an amount of work dW = ( dm) g ( z0 − z ) must be done on it, where z0 = 0.6 m is the height of the table. The total amount of work that needs to be done is the integration over all the small segments of rope, giving W = ∫ (µ dz) g( z0 − z) =

0

z0

2 µ gz0

2 0.18 J .

(1)

When we substitute µ = m L = ( 0.4 kg ) ( 4 m ) , we obtain W

56

2-28.

m v v M v3 v4

CHAPTER 2

before collision

after collision

The problem, as stated, is completely one-dimensional. We may therefore use the elementary result obtained from the use of our conservation theorems: energy (since the collision is elastic) and momentum. We can factor the momentum conservation equation m1v1 + m2 v2 = m1v3 + m2 v4 out of the energy conservation equation 1 1 1 1 2 2 2 2 m1v1 + m2 v2 = m1v3 + m2 v4 2 2 2 2 and get v1 + v3 = v2 + v4 (3) (2) (1)

This is the “conservation” of relative velocities that motivates the definition of the coefficient of restitution. In this problem, we initially have the superball of mass M coming up from the ground with velocity v = 2gh , while the marble of mass m is falling at the same velocity. Conservation of momentum gives Mv + m ( −v ) = Mv3 + mv4 and our result for elastic collisions in one dimension gives v + v3 = ( − v ) + v 4 solving for v3 and v4 and setting them equal to hmarble 2 ghitem , we obtain

2

(4)

(5)

3 −α h = 1+ α

2

(6)

1 − 3α hsuperball = h 1+α

(7)

where α ≡ m M . Note that if α < 1 3 , the superball will bounce on the floor a second time after the collision.

NEWTONIAN MECHANICS—SINGLE PARTICLE

57

2-29.

Ff mg cos θ mg mg sin θ θ x N y

θ = tan −1 0.08 = 4.6°

∑F

y

= N − mg cos θ = my = 0

N = mg cos θ

∑F

x

= mg sin θ − Ff = mx

Ff = µ N = µ mg cos θ so mx = mg sin θ − µ mg cos θ x = g ( sin θ − µ cos θ ) Integrate with respect to time x = gt ( sin θ − µ cos θ ) + x0 Integrate again: x = x0 + x0 t + 1 2 gt ( sin θ − µ cos θ ) 2 (2) (1)

Now we calculate the time required for the driver to stop for a given x0 (initial speed) by solving Eq. (1) for t with x = 0 . t′ = − x0 ( sin θ − µ cos θ )−1 g

Substituting this time into Eq. (2) gives us the distance traveled before coming to a stop.

( x ′ − x0 ) = x0t ′ + 2 gt ′ 2 ( sin θ − µ cos θ )

∆x = − ∆x =

2 x0 1 x2 ( sin θ − µ cos θ )−1 + 2 g g02 ( sin θ − µ cos θ )−1 g

1

2 x0 ( µ cos θ − sin θ )−1 2g

58

We have θ = 4.6° , µ = 0.45 , g = 9.8 m/sec 2 . For x0 = 25 mph = 11.2 m/sec , ∆x = 17.4 meters . If the driver had been going at 25 mph, he could only have skidded 17.4 meters. Therefore, he was speeding How fast was he going? ∆x ≥ 30 meters gives x0 ≥ 32.9 mph . T = t1 + t2 T = total time = 4.021 sec. t1 = the time required for the balloon to reach the ground.

CHAPTER 2

2-30.

(1)

where

t2 = the additional time required for the sound of the splash to reach the first student. We can get t1 from the equation y = y0 + y 0 t − 1 2 gt ; 2 y0 = y 0 = 0

When t = t1 , y = –h; so (h = height of building) −h = − t= Substituting into (1): T= 2h h + g v or h 2h + −T =0 v g h . Using the quadratic formula, we get: 2 gT −1 ± 1 + V 1 2 gt1 2 or t1 = 2h g

distance sound travels h = speed of sound v

This is a quadratic equation in the variable − h= 2 ± g 2 v 2 4T + g v

=

v 2g

Substituting

V = 331 m/sec g = 9.8 m/sec 2 T = 4.021 sec

and taking the positive root because it is the physically acceptable one, we get:

NEWTONIAN MECHANICS—SINGLE PARTICLE

59

h = 8.426 m1 2 h = 71 meters

2-31. For x0 ≠ 0 , example 2.10 proceeds as is until the equations following Eq. (2.78). Proceeding from there we have

α B = x0 ≠ 0 α A = z0 so ( x − x0 ) = α0 cos α t + α0 sin α t

z

x

( y − y0 ) = y 0 t

( z − z0 ) = − α0 cos α t + α0 sin α t

Note that

x

z

( x − x 0 ) + ( z − z0 ) =

2 2

α2

2 z0

+

α2

2 x0

Thus the projection of the motion onto the x–z plane is a circle of radius

(x α

1

2 0

2 + z0

)

12

.

So the motion is unchanged except for a change in the m 2 2 12 radius of the helix. The new radius is x 0 + z0 . qB0

(

)

2-32.

The forces on the hanging mass are

T

mg

The equation of motion is (calling downward positive) mg − T = ma The forces on the other mass are or T = m ( g − a) (1)

60

CHAPTER 2

N y x

T 2mg cos θ

Ff 2mg

2mg sin θ

θ

The y equation of motion gives N − 2mg cos θ = my = 0 or N = 2mg cos θ The x equation of motion gives Ff = µ k N = 2µk mg cos θ

(

)

(2)

T − 2mg sin θ − 2µ k mg cos θ = ma Substituting from (1) into (2) mg − 2mg sin θ − 2µ k mg cos θ = 2ma When θ = θ 0 , a = 0. So g − 2 g sin θ 0 − 2µk g cos θ 0 = 0

1 = sin θ 0 + µ k cos θ 0 2 = sin θ 0 + µ k 1 − sin 2 θ 0

(

)

12

Isolating the square root, squaring both sides and rearranging gives

(1 + µ ) sin

2 k

2

2 θ 0 − sin θ 0 − µ k = 0

1 4

Using the quadratic formula gives sin θ 0 =

2 1 ± µk 3 + 4µk 2 2 1 + µk

(

)

2-33.

The differential equation to solve is v 2 cW ρ Av 2 − g = g − 1 y= 2m vt (1)

where vt = 2mg cw ρ A is the terminal velocity. The initial conditions are y0 = 100 m , and v0 = 0 . The computer integrations for parts (a), (b), and (c) are shown in the figure.

NEWTONIAN MECHANICS—SINGLE PARTICLE

61

100

100

100

y (m)

50

50

50

0

0

2 t (s)

4

6

0

0

5 t (s)

10

15

0

0

5 t (s)

10

0

0

0 –5

v (m/s)

–20

–5

–10

0 2 t (s)

0

0

4

6

0

5 t (s)

10

15

0

5 t (s)

10

0

a (m/s2)

–5

–5

–5

0

2 t (s)

4

6

0

5 t (s)

10

15

0

5 t (s)

10

d) Taking ρ = 1.3 kg ⋅ m −3 as the density of air, the terminal velocities are 32.2, 8.0, and 11.0 (all

m ⋅ s -1 ) for the baseball, ping-pong ball, and raindrop, respectively. Both the ping-pong ball and the raindrop essentially reach their terminal velocities by the time they hit the ground. If we rewrite the mass as average density times volume, then we find that vt ∝ ρmaterial R . The differences in terminal velocities of the three objects can be explained in terms of their densities and sizes.

e) Our differential equation shows that the effect of air resistance is an acceleration that is inversely proportional to the square of the terminal velocity. Since the baseball has a higher terminal velocity than the ping-pong ball, the magnitude of its deceleration is smaller for a given speed. If a person throws the two objects with the same initial velocity, the baseball goes farther because it has less drag. f) We have shown in part (d) that the terminal velocity of a raindrop of radius 0.004 m will be larger than for one with radius 0.002 m ( 9.0 m ⋅ s -1 ) by a factor of 2 . 2-34.

FR

y mg

Take the y-axis to be positive downwards. The initial conditions are y = y = 0 at t = 0.

62

a)

CHAPTER 2

FR = α v

The equation of motion is my = m dv = mg − α v dt

m dv = dt mg − α v

Integrating gives: −

m

α

ln ( mg − α v ) = t + C

Evaluate C using the condition v = 0 at t = 0: − So or − − m m

α

ln ( mg ) = C

α

ln ( mg − α v ) +

m

α

ln ( mg ) = t

αt

mg − α v αv = ln = ln 1 − m mg mg

Take the exponential of both sides and solve for v: e −α r m = 1 −

αv mg αv mg v= dy =

= 1 − e −α t m

mg

α mg (1 − e (1 − e

−α t m

)

(1)

−α t m

α

) dt

Integrate again: y+C = y = 0 at t = 0, so: C= y= mg m = m2 g α 2 α α mg m −α t m t + e α α

mg m m −α t m − α + t + α e α

(2)

Solve (3) for t and substitute into (4): 1−

αv mg = e −α t m

(3)

NEWTONIAN MECHANICS—SINGLE PARTICLE

63

t=−

αv ln 1 − α mg m y=

mg m m α v m α v mg v m α v − − ln 1 − − − ln 1 − + 1 − = α α α mg α mg α g α mg y=− mg α v m ln 1 − v + α α mg

(4)

b)

FR = β v 2

The equation of motion becomes: m dv = mg − β v 2 dt

m dv = dt mg − β v 2

Integrate and apply the initial condition v = 0 at t = 0:

∫ mg β From integral tables

dv

=

2

−v

m∫

β

dt

∫a

2

dx 1 x = tanh −1 ; so 2 −x a a

1 v β tanh −1 = t + C where a ≡ a a m 1

mg

β

α so: tanh −1 0 = 0 = 0 + C

1 v β tanh −1 = t a a m Solving for v: v = a tanh aβ t m

(5)

dy aβt = a tanh dt m

From integral tables ∫ tanh u du = ln cosh u So y + C = m β

ln cosh

aβ t m

64

Apply the conditions at y = 0 and t = 0

C= m

CHAPTER 2

β

ln ( cosh 0 ) =

m

β

ln 1 = 0

So y= m

β

ln cosh

aβ t m

(6)

Solving (5) for t: t= m

αβ

tanh −1

v a

Substituting into (6): y= m

ln cosh tanh −1 β , where u < 1 .

v a

Use the identity: tanh −1 u = cosh −1

1 1 − u2

(In our case u < 1 as it should be because

v βv2 = ; and the condition that u < 1 just says that a mg

gravity is stronger than the retarding force, which it must be.) So y= m

ln cosh cosh −1 β y=− m 2 = ln 1 − β v mg 2 β 1 − β v mg 1

(

)

−1 2

m ln 1 − β v 2 mg 2β

(

)

NEWTONIAN MECHANICS—SINGLE PARTICLE

65

2-35.

14 12 10

y (km)

8 6 4 2 0

0

5

10

15 x (km)

20

25

30

35

30

Range (km)

20 10 0

0

0.02

0.04

0.06

0.08

k (1/s)

We are asked to solve Equations (2.41) and (2.42), for the values k = 0, 0.005, 0.01, 0.02, 0.04, and 0.08 (all in s−1 ), with initial speed v0 = 600 m ⋅ s −1 and angle of elevation θ = 60° . The first figure is produced by numerical solution of the differential equations, and agrees closely with Figure 2-8. Figure 2-9 can be most closely reproduced by finding the range for our values of k, and plotting them vs. k. A smooth curve could be drawn, or more ranges could be calculated with more values of k to fill in the plot, but we chose here to just connect the points with straight lines.

66

2-36.

y θ h x R

CHAPTER 2

Put the origin at the initial point. The equations for the x and y motion are then x = v0 ( cos θ ) t y = v0 ( sin θ ) t −

1 2 gt 2

Call τ the time when the projectile lands on the valley floor. The y equation then gives − h = v0 ( sin θ ) τ − Using the quadratic formula, we may find τ

2 v0 sin 2 θ + 2 gh v0 sin θ + τ= g g

1 2 gτ 2

(We take the positive since τ > 0 .) Substituting τ into the x equation gives the range R as a function of θ.

R=

2 v0 cos θ sin θ + sin 2 θ + x 2 g

(1)

2 where we have defined x 2 ≡ 2 gh v0 . To maximize R for a given h and v0 , we set dR dθ = 0 . The equation we obtain is

cos 2 θ − sin 2 θ − sin θ sin 2 θ + x 2 +

sin θ cos 2 θ sin 2 θ + x 2

=0

(2)

Although it can give x = x (θ ) , the above equation cannot be solved to give θ = θ ( x ) in terms of the elementary functions. The optimum θ for a given x is plotted in the figure, along with its 2 respective range in units of v0 g . Note that x = 0, which among other things corresponds to

2 h = 0, gives the familiar result θ = 45° and R = v0 g .

NEWTONIAN MECHANICS—SINGLE PARTICLE

67

50 40 30 45˚

θ

20 10 0 0 1 2 3 4 5 x 10 6 7 8 9 10

R/(v02/g)

5

1 0 0 1 2 3 4 5 x 6 7 8 9 10

2-37.

v=α x

dv dx = −α x 2

Since

dv dv dx dv = = v then dt dx dt dx F=m dv dv α α = mv = m − 2 dt dx x x

F ( x ) = − mα 2 x 3 v ( x ) = ax − n F=m dv dv dx dv =m = mv = m ax − n − nax − n −1 dt dx dt dx F ( x ) = − mna 2 x − ( 2 n + 1)

2-38. a)

(

)(

)

b)

v ( x) =

dx = ax − n dt

x n dx = adt

Integrate: x n+1 = at + C n+1 C = 0 using given initial conditions

68 x n + 1 = ( n + 1) at

CHAPTER 2

x = [ ( n + 1) at ]

c) Substitute x(t) into F(x):

1 ( n + 1)

F ( t ) = − mna 2 {( n + 1) at}

1 ( n + 1)

− ( 2 n + 1)

F ( t ) = − mna 2 [ ( n + 1) at ]

− ( 2 n + 1) ( n + 1)

2-39. a)

F = −α e β v dv α = − e βv dt m

∫e

− v = v0 at t = 0, so

− βv

dv = −

α

m∫

dt

1

β

e − βv = −

α m t+C

− − Solving for v gives v (t) = −

b) Solve for t when v = 0

1

β

e − β v0 = C − e − β v0 = −

(e β

1

− βv

)

α m t

αβ t − β v0 ln +e β m 1

αβ t m t=

c) From a) we have

+ e − β v0 = 1 1 − e − β v0

αβ

m

dx = −

αβ t − β v0 ln +e dt β m 1

NEWTONIAN MECHANICS—SINGLE PARTICLE

69

Using

∫ ln ( ax + b) dx =

ax + b ln ( ax + b ) − x we obtain a

αβ t αβ t − β v0 − β v0 m + e ln m + e 1 − t x + C = − β αβ m

Evaluating C using x = 0 at t = 0 gives C= So x=− mv0 v0 m

αβ

e − β v0

αβ

e − β v0 +

t

β

−

m αβ t − β v0 αβ t − β v0 +e ln m + e αβ 2 m

Substituting the time required to stop from b) gives the distance required to stop x= m 1 − β v0 −e αβ β 1 v0 + β

2-40. y (x(t),y(t)) an at x

Write the velocity as v(t) = v(t)T(t). It follows that

a (t) =

dv dv dT T+v = = at T + an N dt dt dt

(1)

where N is the unit vector in the direction of dT dt . That N is normal to T follows from 0 = d dt ( T ⋅ T ) . Note also an is positive definite.

a) We have v = x 2 + y = Aα 5 − 4 cos α t . Computing from the above equation,

2

at =

dv 2 Aα 2 sin α t = dt 5 − 4 cos α t

2

(2)

We can get an from knowing a in addition to at . Using a = x 2 + y = Aα 2 , we get an = a 2 − at2 = Aα 2 2 cos α t − 1 5 − 4 cos α t (3)

b) Graphing an versus t shows that it has maxima at α t = nπ , where an = Aα 2 .

70

2-41. a)

CHAPTER 2

As measured on the train: Ti = 0 ; Tf = ∆T = 1 mv 2 2

1 mv 2 2

b)

As measured on the ground: Ti = 1 1 2 mu 2 ; T f = m ( v + u ) 2 2 ∆T = 1 mv 2 + mvu 2

c) The woman does an amount of work equal to the kinetic energy gain of the ball as measured in her frame.

W=

1 mv 2 2

d) The train does work in order to keep moving at a constant speed u. (If the train did no work, its speed after the woman threw the ball would be slightly less than u, and the speed of the ball relative to the ground would not be u + v.) The term mvu is the work that must be supplied by the train.

W = mvu

2-42.

Rθ b R θ

θ

From the figure, we have h(θ ) = (R + b 2) cos θ + Rθ sin θ , and the potential is U (θ ) = mgh(θ ) . Now compute: dU b = mg − sin θ + Rθ cos θ dθ 2 d 2U b = mg R − cos θ − Rθ sin θ 2 2 dθ (1) (2)

NEWTONIAN MECHANICS—SINGLE PARTICLE

71

The equilibrium point (where dU dθ = 0 ) that we wish to look at is clearly θ = 0. At that point, we have d 2U dθ 2 = mg ( R − b 2) , which is stable for R > b 2 and unstable for R < b/ 2 . We can use the results of Problem 2-46 to obtain stability for the case R = b 2 , where we will find that the first non-trivial result is in fourth order and is negative. We therefore have an equilibrium at θ = 0 which is stable for R > b 2 and unstable for R ≤ b 2 . F = − kx + kx 3 α 2 U ( x ) = − ∫ F dx = 1 2 1 x4 kx − k 2 2 4 α 1 2 kx . For large x, the 2

2-43.

To sketch U(x), we note that for small x, U(x) behaves like the parabola behavior is determined by − 1 x4 k 4 α2

U(x) E0 E1 E2 x4 x5 x1 x2 x3 x E3 = 0

E4

E=

1 mv 2 + U ( x ) 2

For E = E0 , the motion is unbounded; the particle may be anywhere. For E = E1 (at the maxima in U(x)) the particle is at a point of unstable equilibrium. It may remain at rest where it is, but if perturbed slightly, it will move away from the equilibrium. What is the value of E1 ? We find the x values by setting 0 = kx − kx 3 α 2 x = 0, ± α are the equilibrium points U ( ±α ) = E1 = 1 2 1 2 1 2 kα − kα = kα 2 4 4 dU =0. dx

For E = E2 , the particle is either bounded and oscillates between − x2 and x2 ; or the particle comes in from ±∞ to ± x3 and returns to ±∞.

72

CHAPTER 2

For E3 = 0 , the particle is either at the stable equilibrium point x = 0, or beyond x = ± x4 . For E4 , the particle comes in from ±∞ to ± x5 and returns.

2-44.

T

θ

T

T m1 m1 g m2 m2 g

From the figure, the forces acting on the masses give the equations of motion m1 x 1 = m1 g − T m2 x 2 = m2 g − 2T cos θ where x2 is related to x1 by the relation x2 = (1) (2)

( b − x1 )2 − d2

4

(3)

and cos θ = d ( b − x1 ) 2 . At equilibrium, x 1 = x 2 = 0 and T = m1 g . This gives as the equilibrium values for the coordinates x10 = b − x20 = 4 m1 d

2 2 4 m1 − m2

(4)

m2 d

2 2 4 m1 − m2

(5)

We recognize that our expression x10 is identical to Equation (2.105), and has the same requirement that m2 m1 < 2 for the equilibrium to exist. When the system is in motion, the descriptive equations are obtained from the force laws: m1 ( x 1 − g ) = m2 ( b − x1 ) 4 x2 (x 2 − g) (6)

To examine stability, let us expand the coordinates about their equilibrium values and look at their behavior for small displacements. Let ξ1 ≡ x1 − x10 and ξ2 ≡ x2 − x20 . In the calculations, take terms in ξ1 and ξ2 , and their time derivatives, only up to first order. Equation (3) then becomes ξ2 −(m1 m2 )ξ1 . When written in terms of these new coordinates, the equation of motion becomes

ξ1 = −

4 m1m2 ( m1 + m2 ) d

2 2 g 4 m1 − m2

(

)

32

ξ1

(7)

NEWTONIAN MECHANICS—SINGLE PARTICLE

73

which is the equation for simple harmonic motion. The equilibrium is therefore stable, when it exists.

2-45. and 2-46.

Expand the potential about the equilibrium point U ( x) = 1 di u i ∑ i x i = n + 1 i ! dx 0

∞

(1)

The leading term in the force is then F( x ) = − dU 1 d( n + 1)U = − ( n + 1) x n dx n! dx 0 (2)

The force is restoring for a stable point, so we need F ( x > 0 ) < 0 and F ( x < 0 ) > 0 . This is never true when n is even (e.g., U = kx 3 ), and is only true for n odd when d( n + 1)U dx( n + 1) 0 < 0 . We are given U ( x) = U 0 ( a x + x a ) for x > 0 . Equilibrium points are defined by

2-47.

dU dx = 0 , with stability determined by d 2U dx 2 at those points. Here we have dU a 1 = U0 − 2 + dx a x which vanishes at x = a. Now evaluate d 2U 2U0 dx 2 = a 3 > 0 a indicating that the equilibrium point is stable.

25 20 15 10 5 0

(1)

(2)

U(x)/U0

0

0.5

1 x/a 1.5

2

74

CHAPTER 2

2-48. In the equilibrium, the gravitational force and the eccentric force acting on each star must be equal

Gm2 mv 2 = ⇒v= d2 d /2

mG πd ⇒τ = = 2d v

2π d 3 2 mG

2-49.

The distances from stars to the center of mass of the system are respectively r1 = dm2 m1 + m2 and r2 = dm1 m1 + m2

At equilibrium, like in previous problem, we have Gm1m2 m1v12 = ⇒ v1 = d2 r1 Gm2 2 2π r1 2π d 3 2 ⇒τ = = d(m1 + m2 ) v1 G(m1 + m2 )

The result will be the same if we consider the equilibrium of forces acting on 2nd star.

2-50. a)

d m0 v dt v2 1− 2 c

m0 v = F⇒∫d v2 0 1− 2 c

t

m0 v = = Ft ⇒ v(t) = v2 1− 2 c t Ft

2 m0 +

F 2t 2 c2

⇒ x(t) = ∫ v(t)dt =

0

c2 F 2t 2 m0 2 + 2 − m0 F c

b) v t

c) From a) we find

t=

vm0 F 1− v2 c2

Now if

F = 10 , then m0

NEWTONIAN MECHANICS—SINGLE PARTICLE

75

when v = c 2 , we have t =

c = 0.55 year 10 3 99c = 6.67 years 10 199

when v = 99% c, we have t =

2-51. a)

m

mv0 dv dv b = − bv 2 ⇒ ∫ 2 = − ∫ dt ⇒ v(t) = dt v m btv0 + m 999m = 138.7 hours . v0 b

Now let v(t) = v0/1000 , one finds t = v t

b)

x(t ) = ∫ vdt =

0

t

m btv0 + m ln b m

We use the value of t found in question a) to find the corresponding distance x(t ) = m ln(1000) = 6.9 km b

2-52. a) b)

U

F( x ) = −

4U x dU x2 = − 20 1 − 2 dx a a

x

When F = 0, there is equilibrium; further when U has a local minimum (i.e. dF dx < 0 ) it is stable, and when U has a local maximum (i.e. dF dx > 0 ) it is unstable.

76

CHAPTER 2

So one can see that in this problem x = a and x = –a are unstable equilibrium positions, and x = 0 is a stable equilibrium position.

c)

Around the origin, F ≈ −

4U 0 x ≡ − kx ⇒ ω = a2

k = m

4U 0 ma 2

d) To escape to infinity from x = 0, the particle needs to get at least to the peak of the potential,

2 2U 0 mvmin = U max = U 0 ⇒ vmin = m 2

e) From energy conservation, we have

2 mv 2 U 0 x 2 mvmin dx + 2 = ⇒ =v= a dt 2 2

2U 0 x2 1 − 2 m a

We note that, in the ideal case, because the initial velocity is the escape velocity found in d), ideally x is always smaller or equal to a, then from the above expression,

t=

a exp t m dx ma a+x ln = ⇒ x(t) = 2U 0 ∫ 8U 0 a − x x2 0 1 − 2 exp t a x 2

8U 0 − 1 ma 2 8U 0 + 1 ma 2

x

t

2-53.

F is a conservative force when there exists a non-singular potential function U(x) satisfying F(x) = –grad(U(x)). So if F is conservative, its components satisfy the following relations ∂Fx ∂Fy = ∂y ∂x and so on.

a) In this case all relations above are satisfied, so F is indeed a conservative force.

Fx = −

bx 2 ∂U = ayz + bx + c ⇒ U = − ayzx − − cx + f1 ( y , z) 2 ∂x

(1)

where f1 ( y , z) is a function of only y and z Fy = − ∂U = axz + bz ⇒ U = − ayzx − byz + f 2 ( x , z) ∂y (2)

NEWTONIAN MECHANICS—SINGLE PARTICLE

77

where f 2 ( x , z) is a function of only x and z Fz = − ∂U = axy + by + c ⇒ U = − ayzx − byz + f 3 ( y , z) ∂z bx 2 +C 2 (3)

then from (1), (2), (3) we find that U = − axyz − byz − cx 2 − where C is a arbitrary constant.

b) is

Using the same method we find that F in this case is a conservative force, and its potential U = − z exp( − x) − y ln z + C

c) Using the same method we find that F in this case is a conservative force, and its potential is ( using the result of problem 1-31b):

U = − a ln r

2-54. a) Terminal velocity means final steady velocity (here we assume that the potato reaches this velocity before the impact with the Earth) when the total force acting on the potato is zero.

mg = kmv

b)

and consequently

v = g k = 1000 m/s .

dv F dx dv vdv = = −( g + kv) ⇒ = dt = − ⇒ ∫ dx = − ∫ ⇒ dt m v g + kv g + kv 0 v0 xmax = g v0 g + 2 ln = 679.7 m k k g + kv0 where v0 is the initial velocity of the potato.

x

0

2-55.

Let’s denote vx 0 and vy 0 the initial horizontal and vertical velocity of the pumpkin. vx 0 − vxf dvx dv dx = Fx = − mkvx ⇒ − = − dt = x ⇒ x f = dt vx kvx k

Evidently, vx 0 = vy 0 in this problem. m (1)

where the suffix f always denote the final value. From the second equality of (1), we have − dt = Combining (1) and (2) we have xf = vx 0 ( − kt 1− e f ) k (3) dvx − kt ⇒ vxf = vx 0 e f kvx (2)

78

Do the same thing with the y-component, and we have m dvy dt = Fy = − mg − mkvy ⇒ − dvy g + kvyf vy 0 − vyf dy g = − dt = ⇒ 0 = y f = 2 ln + vy g + kvy k g + kvy 0 k − dt = dvy g + kvy ⇒ g + kvyf = ( g + kv0 f ) e

− kt f

CHAPTER 2

(4)

and

(5)

From (4) and (5) with a little manipulation, we obtain 1− e

− kt f

=

gkt f g + kvy 0

(6)

(3) and (6) are 2 equations with 2 unknowns, t f and k. We can eliminate t f , and obtain an equation of single variable k. xf = Putting x f = 142 m and vx 0 = vy 0 = vx 0 ( − kt g + kv ( gv ) 1 − e f ( y0 ) x0 ) k

v0 = 38.2 m/s we can numerically solve for k and obtain 2 K= 0.00246 s −1 .

CHAPTER

3

Oscillations

3-1.

a)

1 ν0 = 2π

k 1 = m 2π

10 4 dyne/cm 10 = 10 2 gram 2π

gram ⋅ cm sec 2 ⋅ cm = 10 sec −1 gram 2π

or,

ν 0 ≅ 1.6 Hz τ0 = or, 1

(1)

ν0

=

2π sec 10

τ 0 ≅ 0.63 sec

b)

(2)

E=

1 2 1 kA = × 10 4 × 32 dyne-cm 2 2

so that

E = 4.5 × 10 4 erg

c) The maximum velocity is attained when the total energy of the oscillator is equal to the kinetic energy. Therefore,

(3)

1 2 mvmax = 4.5 × 10 4 erg 2 v max = 2 × 4.5 × 10 4 100

79

80

CHAPTER 3

or, vmax = 30 cm/sec (4)

3-2. a) The statement that at a certain time t = t1 the maximum amplitude has decreased to onehalf the initial value means that

xen = A0 e − βt1 = or, e − βt1 = so that 1 2

1 A0 2

(1)

(2)

β=

Since t1 = 10 sec ,

ln 2 0.69 = t1 t1

(3)

β = 6.9 × 10 −2 sec −1

b)

(4)

According to Eq. (3.38), the angular frequency is

2 ω1 = ω 0 − β 2

(5)

where, from Problem 3-1, ω 0 = 10 sec −1 . Therefore,

ω 1 = (10 ) − ( 6.9 × 10 −2 )

2

2

2 1 ≅ 10 1 − ( 6.9) × 10 −6 sec −1 2

(6)

so that

ν1 = which can be written as

10 (1 − 2.40 × 10 −5 ) sec −1 2π

(7)

ν1 = ν 0 (1 − δ ) where (8)

δ = 2.40 × 10 −5

That is, ν1 is only slightly different from ν 0 .

(9)

OSCILLATIONS

81

c) The decrement of the motion is defined to be e βτ 1 where τ 1 = 1 ν 1 . Then,

e βτ 1

1.0445

3-3.

The initial kinetic energy (equal to the total energy) of the oscillator is

m = 100 g and v0 = 1 cm/sec .

1 2 mv0 , where 2

a) Maximum displacement is achieved when the total energy is equal to the potential energy. Therefore,

1 1 2 2 mv0 = kx0 2 2

x0 = or,

m 10 2 1 v0 = ×1= cm 4 k 10 10

x0 =

b) The maximum potential energy is

1 cm 10

(1)

U max = or,

1 2 1 kx0 = × 10 4 × 10 −2 2 2 U max = 50 ergs

(2)

3-4. a) Time average:

The position and velocity for a simple harmonic oscillator are given by

x = A sin ω 0 t x = ω 0 A cos ω 0 t where ω 0 = k m The time average of the kinetic energy is

T = where τ = 2π is the period of oscillation. ω0 1 t +τ

(1) (2)

τ

∫ t 1 mx 2 dt 2

(3)

82

CHAPTER 3

By inserting (2) into (3), we obtain T = or, T =

2 mA2ω 0 4

1 2 mA2ω 0 2τ

t+τ

∫ cos t 2

ω 0 t dt

(4)

(5)

In the same way, the time average of the potential energy is

U =

1

t +τ

τ

∫ t 1 2 kx dt 2 t +τ

=

1 kA2 2τ kA2 4

∫ sin t 2

ω 0 t dt

=

2 and since ω 0 = k m , (6) reduces to

(6)

U = From (5) and (7) we see that

2 mA2ω 0 4

(7)

T = U

The result stated in (8) is reasonable to expect from the conservation of the total energy.

(8)

E = T +U

(9)

This equality is valid instantaneously, as well as in the average. On the other hand, when T and U are expressed by (1) and (2), we notice that they are described by exactly the same function, displaced by a time τ 2 :

2 mA2ω 0 cos 2 ω 0 t 2 2 mA ω 0 t 2 sin ω 0 t U= 2

T=

(10)

Therefore, the time averages of T and U must be equal. Then, by taking time average of (9), we find

T = U =

b) Space average:

E 2

(11)

The space averages of the kinetic and potential energies are

OSCILLATIONS

83

A

T= and 1 1 2 ∫ 2 mx dx A0

(12)

2 mω 0 1 1 U = ∫ kx 2 dx = 2A A0 2 A

∫x

0

A

2

dx

(13)

(13) is readily integrated to give

U=

2 mω 0 A2 6

(14)

To integrate (12), we notice that from (1) and (2) we can write

2 2 x 2 = ω 0 A2 cos 2 ω 0 t = ω 0 A2 1 − sin 2 ω 0 t 2 = ω 0 A2 − x 2

(

)

(15)

(

)

∫

0 A

Then, substituting (15) into (12), we find

T=

2 mω 0 2A 2 mω 0 2A

A2 − x 2 dx

= or,

3 A3 A − 3

(16)

T=2

2 mω 0 A2 6

(17)

From the comparison of (14) and (17), we see that T = 2U To see that this result is reasonable, we plot T = T(x) and U = U(x):

1 x2 2 mω 0 A2 1 − 2 2 A 1 2 2 U = mω 0 x 2

(18)

T=

(19)

2 mA2ω 0

Energy

E = const. = 1 2 mA2ω 0 2

U = U(x) T = T(x) –A O

A

x

And the area between T(x) and the x-axis is just twice that between U(x) and the x-axis.

84 3-5.

CHAPTER 3

Differentiating the equation of motion for a simple harmonic oscillator, x = A sin ω 0 t (1)

we obtain

∆x = Aω 0 cos ω 0 t ∆t

(2)

But from (1) sin ω 0 t = Therefore, cos ω 0 t = 1 − ( x A) and substitution into (2) yields

∆t = ∆x

2

x A

(3)

(4)

ω 0 A2 − x 2

(5)

Then, the fraction of a complete period that a simple harmonic oscillator spends within a small interval ∆x at position x is given by

∆t

τ

=

∆x

ω 0τ A2 − x 2

∆t⁄τ

=

∆x

2π

A2 − x 2

(6)

x –A3 –A2 –A1 A1 A2 A3

This result implies that the harmonic oscillator spends most of its time near x = ±A, which is obviously true. On the other hand, we obtain a singularity for ∆t τ at x = ±A. This occurs because at these points x = 0, and (2) is not valid.

3-6.

k

m1 x1 m2 x2 x

Suppose the coordinates of m1 and m2 are x1 and x2 and the length of the spring at equilibrium is . Then the equations of motion for m1 and m2 are m1 x1 = − k ( x1 − x2 + m2 x2 = − k ( x2 − x1 +

) )

(1) (2)

OSCILLATIONS

85

From (2), we have x1 = 1 ( m2 x2 + kx2 − k k

)

(3)

Substituting this expression into (1), we find d2 m1 m2 x2 + ( m1 + m2 ) kx2 = 0 dt 2 from which x2 = − Therefore, x2 oscillates with the frequency m1 + m2 kx2 m1m2 (5) (4)

ω=

m1 + m2 k m1m2

(6)

We obtain the same result for x1 . If we notice that the reduced mass of the system is defined as 1

µ we can rewrite (6) as

=

1 1 + m1 m2

(7)

ω= k k

µ µ (8)

This means the system oscillates in the same way as a system consisting of a single mass µ. Inserting the given values, we obtain µ 66.7 g and ω 2.74 rad ⋅ s −1 .

3-7.

A

hb hs

Let A be the cross-sectional area of the floating body, hb its height, hs the height of its submerged part; and let ρ and ρ0 denote the mass densities of the body and the fluid, respectively. The volume of displaced fluid is therefore V = Ahs . The mass of the body is M = ρ Ahb .

86

CHAPTER 3

There are two forces acting on the body: that due to gravity (Mg), and that due to the fluid, pushing the body up ( − ρ0 gV = − ρ0 ghs A ). The equilibrium situation occurs when the total force vanishes:

0 = Mg − ρ0 gV = ρ gAhb − ρ0 ghs A which gives the relation between hs and hb : hs = hb (1)

ρ ρ0

(2)

For a small displacement about the equilibrium position ( hs → hs + x ), (1) becomes Mx = ρ Ahb x = ρ gAhb − ρ0 g ( hs + x ) A Upon substitution of (1) into (3), we have (3)

ρ Ahb x = − ρ0 gxA or, x+g

(4)

ρ0 x=0 ρ hb

(5)

Thus, the motion is oscillatory, with an angular frequency

ω2 = g

g gA ρ0 = = ρhb hs V

(6)

where use has been made of (2), and in the last step we have multiplied and divided by A. The period of the oscillations is, therefore,

τ=

Substituting the given values, τ

3-8.

y

2π

ω

= 2π

V gA

(7)

0.18 s .

2a O x m s 2a

The force responsible for the motion of the pendulum bob is the component of the gravitational force on m that acts perpendicular to the straight portion of the suspension string. This component is seen, from the figure (a) below, to be

F = ma = mv = − mg cos α

(1)

OSCILLATIONS

87

where α is the angle between the vertical and the tangent to the cycloidal path at the position of m. The cosine of α is expressed in terms of the differentials shown in the figure (b) as cos α = where ds = dx 2 + dy 2 m α F mg dy dx

dy ds

(2)

(3)

α ds S

(a)

(b)

The differentials, dx and dy, can be computed from the defining equations for x(φ) and y(φ) above: dx = a (1 − cos φ ) dφ dy = − a sin φ dφ Therefore, ds2 = dx 2 + dy 2

2 = a 2 (1 − cos φ ) + sin 2 φ dφ 2 = 2a 2 (1 − cos φ ) dφ 2

(4)

= 4 a 2 sin 2 so that

φ

2

dφ 2

(5)

ds = 2a sin Thus,

φ

2

dφ

(6)

dy − a sin φ dφ = ds 2a sin φ dφ 2 = − cos The velocity of the pendulum bob is

φ

2

= cos α

(7)

88

CHAPTER 3

v=

ds φ dφ = 2a sin dt 2 dt d dt

= −4 a from which v = −4 a Letting z ≡ cos

φ cos 2 φ cos 2

(8)

d2 dt 2

(9)

φ

2

be the new variable, and substituting (7) and (9) into (1), we have −4maz = mgz (10)

or, z+ g z=0 4a (11)

which is the standard equation for simple harmonic motion,

2 z + ω0 z = 0

(12)

If we identify

ω0 = where we have used the fact that

g

(13)

= 4a .

Thus, the motion is exactly isochronous, independent of the amplitude of the oscillations. This fact was discovered by Christian Huygene (1673).

3-9.

The equation of motion for 0 ≤ t ≤ t0 is

mx = − k ( x − x0 ) + F = − kx + ( F + kx0 ) while for t ≥ t0 , the equation is

(1)

mx = − k ( x − x0 ) = − kx + kx0

It is convenient to define

(2)

ξ = x − x0 which transforms (1) and (2) into

mξ = − kξ + F ; mξ = − kξ ;

0 ≤ t ≤ t0

(3) (4)

t ≥ t0

OSCILLATIONS

89

The homogeneous solutions for both (3) and (4) are of familiar form ξ ( t ) = Ae iω t + Be − iω t , where

ω = k m . A particular solution for (3) is ξ = F k . Then the general solutions for (3) and (4) are ξ− =

F + Ae iω t + Be − iω t ; k

0 ≤ t ≤ t0 (5) (6)

ξ+ = Ce iω t + De − iω t ;

t ≥ t0

To determine the constants, we use the initial conditions: x ( t = 0 ) = x0 and x(t = 0) = 0. Thus,

ξ− ( t = 0 ) = ξ− ( t = 0 ) = 0

The conditions give two equations for A and B:

F + A+B k 0 = iω ( A − B) 0= Then A=B=− and, from (5), we have F 2k

(7)

(8)

ξ− = x − x 0 =

F (1 − cos ω t ) ; k

0 ≤ t ≤ t0

(9)

Since for any physical motion, x and x must be continuous, the values of ξ− ( t = t0 ) and

ξ− ( t = t0 ) are the initial conditions for ξ+ ( t ) which are needed to determine C and D:

ξ+ ( t = t0 ) =

F (1 − cos ω t0 ) = Ce iω t0 + De − iω t0 k

F ξ+ ( t = t0 ) = ω sin ω t0 = iω Ce iω t0 k The equations in (10) can be rewritten as: Ce iω t0 + De − iω t0 = Ce iω t0 − De − iω t0

− De − iω t0

(10)

F (1 − cos ω t0 ) k − iF = sin ω t0 k

(11)

Then, by adding and subtracting one from the other, we obtain F − iω t0 e 1 − e iω t0 2k F iω t0 D= e 1 − e − iω t0 2k C=

(

)

(

)

(12)

90

CHAPTER 3

Substitution of (12) into (6) yields

ξ+ =

= = Thus,

F e − iω t0 − 1 e iω t + e iω t0 − 1 e − iω t 2k F iω (t − t0 ) iω t − iω t − t − e + e ( 0 ) − e − iω t e 2k F cos ω ( t − t0 ) − cos ω t k (13)

(

)

(

)

x − x0 =

F cos ω ( t − t0 ) − cos ω t ; t ≥ t0 k

(14)

3-10.

The amplitude of a damped oscillator is expressed by x ( t ) = Ae − βt cos (ω 1t + δ ) (1)

Since the amplitude decreases to 1 e after n periods, we have

β nT = β n

2π

ω1

=1

(2)

Substituting this relation into the equation connecting ω 1 and ω 0 (the frequency of undamped

2 2 oscillations), ω 1 = ω 0 − β 2 , we have 2 2 ω 0 = ω1 +

1 ω1 2 = ω 1 1 + 4π 2 n 2 2π n

2

(3)

Therefore,

ω1 1 = 1 + 2 2 ω 0 4π n so that

−1 2

(4)

ω1 1 ≅ 1− 2 2 8π n ω2

3-11.

The total energy of a damped oscillator is E (t) = 1 1 2 2 mx ( t ) + kx ( t ) 2 2 (1)

where x ( t ) = Ae − βt cos (ω 1t − δ ) x ( t ) = Ae − βt − β cos (ω 1t − δ ) − ω 1 sin (ω 1t − δ ) (2) (3)

OSCILLATIONS

91

2 ω1 = ω 0 − β 2 ,

ω0 =

k m

Substituting (2) and (3) into (1), we have

E (t) =

A2 −2 βt 2 mβ 2 + k cos 2 (ω 1t − δ ) + mω 1 sin 2 (ω 1t − δ ) e 2

(

)

+ 2mβω 1 sin (ω 1t − δ ) cos (ω 1t − δ )

(4)

Rewriting (4), we find the expression for E(t): E (t) = mA2 −2 βt 2 2 2 β cos 2 (ω 1t − δ ) + β ω 0 − β 2 sin 2 (ω 1t − δ ) + ω 0 e 2 dE : dt (5)

Taking the derivative of (5), we find the expression for

dE mA2 −2 βt 2 2βω 0 − 4β 3 cos 2 (ω 1t − δ ) e = 2 dt 2 − 4β 2 ω 0 − β 2 sin 2 (ω 1t − δ ) 0 − 2βω 2

(

)

(6)

The above formulas for E and dE dt reproduce the curves shown in Figure 3-7 of the text. To find the average rate of energy loss for a lightly damped oscillator, let us take β ω 0 . This means that the oscillator has time to complete some number of periods before its amplitude decreases considerably, i.e. the term e −2 βt does not change much in the time it takes to complete one period. The cosine and sine terms will average to nearly zero compared to the constant term in dE dt , and we obtain in this limit dE dt

3-12.

θ mg sin θ mg

2 − mβω 0 A2 e −2 βt

(7)

The equation of motion is − m θ = mg sin θ (1) (2)

θ=−

g

sin θ

If θ is sufficiently small, we can approximate sin θ ≅ θ , and (2) becomes

92

CHAPTER 3

θ=− θ which has the oscillatory solution

g

(3)

θ ( t ) = θ 0 cos ω 0 t where ω 0 = g and where θ 0 is the amplitude. If there is the retarding force 2m g θ , the equation of motion becomes − m θ = mg sin θ + 2m g θ

(4)

(5)

or setting sin θ ≅ θ and rewriting, we have

2 θ + 2ω 0θ + ω 0θ = 0

(6)

Comparing this equation with the standard equation for damped motion [Eq. (3.35)],

2 x + 2β x + ω 0 x = 0

(7)

we identify ω 0 = β . This is just the case of critical damping, so the solution for θ ( t) is [see Eq. (3.43)]

θ ( t ) = ( A + Bt ) e −ω0t

For the initial conditions θ ( 0 ) = θ 0 and θ(0) = 0, we find

(8)

θ ( t ) = θ 0 (1 + ω 0 t ) e −ω 0t

For the case of critical damping, β = ω 0 . Therefore, the equation of motion becomes x + 2β x + β 2 x = 0

3-13.

(1)

If we assume a solution of the form x (t) = y (t) e −β t

(2)

we have x = ye − β t − β ye − β t x = ye − β t − 2β ye − β t + β 2 ye − β t Substituting (3) into (1), we find ye − β t − 2β ye − β t + β 2 ye − β t + 2β ye − β t − 2β 2 ye − β t + β 2 ye − β t = 0 or, y=0 Therefore, y ( t ) = A + Bt (6) (5) (4) (3)

OSCILLATIONS

93

and x ( t ) = ( A + Bt ) e − β t which is just Eq. (3.43).

3-14.

(7)

For the case of overdamped oscillations, x(t) and x ( t ) are expressed by x ( t ) = e − β t A1eω 2t + A2 e −ω 2t x ( t ) e − β t −β A1eω 2t + + A2 e −ω 2t + A1ω 2 eω 2t − A2ω 2 e −ω 2t (1)

(

) (

)

(2)

2 where ω 2 = β 2 − ω 0 . Hyperbolic functions are defined as

cosh y = or,

ey + e− y , 2

sinh y =

ey − e−y 2

(3)

e y = cosh y + sinh y −y e = cosh y − sinh y Using (4) to rewrite (1) and (2), we have x ( t ) = ( cosh β t − sinh β t ) ( A1 + A2 ) cosh ω 2t + ( A1 − A2 ) sinh ω 2t and x ( t ) = ( cosh β t − sinh β t ) ( A1ω 2 − A1β ) ( cosh ω 2t + sinh ω 2t ) − ( A2 β + A2ω 2 ) ( cosh ω 2t − sinh ω 2t )

(4)

(5)

(6)

3-15.

We are asked to simply plot the following equations from Example 3.2: x ( t ) = Ae − β t cos (ω 1t − δ ) v(t) = − Ae − β t β cos (ω 1t − δ ) + ω 1 sin (ω 1t − δ ) (1) (2)

with the values A = 1 cm , ω 0 = 1 rad ⋅ s −1 , β = 0.1 s −1 , and δ = π rad. The position goes through x = 0 a total of 15 times before dropping to 0.01 of its initial amplitude. An exploded (or zoomed) view of figure (b), shown here as figure (B), is the best for determining this number, as is easily shown.

94

(b)

1 0.5 0 –0.5 –1 x(t) (cm) v(t) (cm/s) 0 5 10 15 20 25 30 t (s) 1 35 40 45 50

CHAPTER 3

(c)

0.5

v (cm/s)

0

–0.5

–1

–1

–0.5

0 x (cm)

0.5

1

(B) x (cm)

0.01

0

–0.01

0

5

10

15

20

25

30 t (s)

35

40

45

50

55

3-16.

If the damping resistance b is negative, the equation of motion is

2 x − 2β x + ω 0 x = 0

(1)

where β ≡ − b 2m > 0 because b < 0. The general solution is just Eq. (3.40) with β changed to –β: x ( t ) = e β t A1 exp

(

2 2 β 2 − ω 0 t + A2 exp − β 2 − ω 0 t

)

(

)

(2)

From this equation, we see that the motion is not bounded, irrespective of the relative values of 2 β 2 and ω 0 . The three cases distinguished in Section 3.5 now become:

2 2 a) If ω 0 > β 2 , the motion consists of an oscillatory solution of frequency ω 1 = ω 0 − β 2 , multiplied by an ever-increasing exponential:

OSCILLATIONS

95

x ( t ) = e β t A1e iω1t + A2 e − iω1t

2 b) If ω 0 = β 2 , the solution is

(3)

x ( t ) = ( A + Bt ) e β t which again is ever-increasing.

2 c) If ω 0 < β 2 , the solution is:

(4)

x ( t ) = e β t A1eω 2t + A2 e −ω 2t where

2 ω2 = β 2 − ω0 ≤ β

(5)

(6)

This solution also increases continuously with time. The tree cases describe motions in which the particle is either always moving away from its initial position, as in cases b) or c), or it is oscillating around its initial position, but with an amplitude that grows with the time, as in a). Because b < 0, the medium in which the particle moves continually gives energy to the particle and the motion grows without bound.

3-17.

For a damped, driven oscillator, the equation of motion is

2 x = 2β x + ω 0 x = A cos ω t

(1)

and the average kinetic energy is expressed as T = mA2 ω2 2 4 ω 0 − ω 2 2 + 4ω 2 β 2

(

)

(2)

Let the frequency n octaves above ω 0 be labeled ω 1 and let the frequency n octaves below ω 0 be labeled ω 2 ; that is

ω 1 = 2n ω 0 ω 2 = 2− n ω 0

The average kinetic energy for each case is

T

ω1

2 22 n ω 0 mA2 2 2 2 4 ω 0 − 22 n ω 0 2 + (4)22 n ω 0 β 2

(3)

=

(

)

(4)

T

ω2

=

2 2−2 n ω 0 mA2 2 2 2 4 ω 0 − 2−2 n ω 0 2 + (4)2−2 n ω 0 β 2

(

)

(5)

Multiplying the numerator and denominator of (5) by 24 n , we have

96

2 22 n ω 0 mA2 2 2 2 4 ω 0 − 22 n ω 0 2 + (4)22 n ω 0 β 2

CHAPTER 3

T

Hence, we find

ω2

=

(

)

T and the proposition is proven.

ω1

= T

ω2

(6)

3-18. Since we are near resonance and there is only light damping, we have ω 0 ω R ω , where ω is the driving frequency. This gives Q ω 0 2β . To obtain the total energy, we use the solution to the driven oscillator, neglecting the transients:

x ( t ) = D cos (ω t − δ )

We then have

(1)

E=

1 1 mD2 2 2 ω sin 2 (ω t − δ ) + ω 0 cos 2 (ω t − δ ) mx 2 + kx 2 = 2 2 2

1 2 mω 0 D2 2

(2)

The energy lost over one period is

∫ ( 2mβ x ) ⋅ ( xdt ) = 2π mωβ D

T 0

2

(3)

where T = 2π ω . Since ω

ω 0 , we have

E energy lost over one period

ω0 4πβ

Q 2π

(4)

which proves the assertion.

3-19.

The amplitude of a damped oscillator is [Eq. (3.59)]

D=

A

(ω 02 − ω 2 ) + 4ω 2β 2

2

(1)

2 At the resonance frequency, ω = ω R = ω 0 − β 2 , D becomes

DR =

A

2 2β ω 0 − β 2

(2) 1 DR : 2 A

2 2

Let us find the frequency, ω = ω ′ , at which the amplitude is 1 1 A = DR = 2 2 2 2β ω 0 − β 2 Solving this equation for ω ′ , we find

(ω

2 0

− ω′

)

+ 4ω ′ β

2

2

(3)

OSCILLATIONS

97

β2 ω ′ = ω − 2β ± 2βω 0 1 − 2 ω0

2 2 0 2

12

(4)

For a lightly damped oscillator, β is small and the terms in β 2 can be neglected. Therefore,

2 ω ′ 2 ≅ ω 0 ± 2βω 0

(5)

or,

ω ′ ≅ ω 0 1 ±

which gives

β ω0

(6)

∆ω = (ω 0 + β ) − (ω 0 − β ) = 2β We also can approximate ω R for a lightly damped oscillator:

2 ω R = ω 0 − 2β 2 ≅ ω 0

(7)

(8)

Therefore, Q for a lightly damped oscillator becomes Q≅

ω0 ω0 ≅ 2β ∆ω

(9)

3-20.

From Eq. (3.66), x= − Aω

(ω 02 − ω 2 ) + 4ω 2β 2

2

sin (ω t − δ )

(1)

Therfore, the absolute value of the velocity amplitude v is given by v0 = Aω

(ω

2 0

−ω

2 2

)

+ 4ω β

2

(2)

2

The value of ω for v0 a maximum, which is labeled ω v , is obtained from ∂ v0 ∂ω and the value is ω v = ω 0 . Since the Q of the oscillator is equal to 6, we can use Eqs. (3.63) and (3.64) to express β in terms of ω 0 : =0 ω =ω v

(3)

β2 =

2 ω0

146

(4) 2 , where vmax = v0 (ω = ω 0 ) .

We need to find two frequencies, ω 1 and ω 2 , for which v0 = vmax We find

98

CHAPTER 3

vmax A = = 2 2 2β

Aω

(ω 02 − ω 2 ) + 4ω 2β 2

2

(5)

Substituting for β in terms of ω 0 from (4), and by squaring and rearranging terms in (5), we obtain

(ω from which

2 0

2 − ω 1,2

)

2

−

2 2 2 ω 1,2 ω 0 = 0 73

(

)

(6)

2 2 ω 0 − ω 1,2 = ±

2 1 ω 1,2ω 0 ≅ ± ω 1,2ω 0 73 6

(7)

Solving for ω 1 , ω 2 we obtain ω ω 1,2 ≅ ± 0 ± ω 0 12 It is sufficient for our purposes to consider ω 1 , ω 2 positive: then (8)

ω1 ≅ so that

ω0

12

+ ω0 ;

ω2 ≅ −

ω0

12

+ ω0

(9)

∆ω = ω 1 − ω 2 = A graph of v0 vs. ω for Q = 6 is shown. v0 ω0

6

(10)

A vmax = 2β A 2 2β

∆ω

ω0 ω0 12 12 ω0 1 ω0 6

3-21.

We want to plot Equation (3.43), and its derivative: x ( t ) = ( A + Bt ) e − β t v ( t ) = [ B − β ( A + Bt ) ] e − β t (1) (2)

where A and B can be found in terms of the initial conditions A = x0 B = v0 + β x 0 (3) (4)

OSCILLATIONS

99

The initial conditions used to produce figure (a) were ( x0 , v0 ) = ( −2, 4 ) , (1, 4) , (4,−1) , (1,−4) , (−1,−4) , and (−4, 0) , where we take all x to be in cm, all v in cm ⋅ s −1 , and β = 1 s −1 . Figure (b) is a magnified view of figure (a). The dashed line is the path that all paths go to asymptotically as t → ∞. This can be found by taking the limits. lim v(t) = − β Bte − β t t →∞

(5) (6)

lim x(t ) = Bte − β t t →∞

so that in this limit, v = –βx, as required.

(a)

4 3 2 1 v (cm/s)

0 –1 –2 –3 –4

–4

–2

0 x (cm)

2

4

(b)

0.4

0.2 v (cm/s)

0

–0.2

–0.4 –0.5 –0.25 0 x (cm) 0.25 0.5

3-22.

For overdamped motion, the position is given by Equation (3.44) x ( t ) = A1e − β1t + A2 e − β2t (1)

100

CHAPTER 3

The time derivative of the above equation is, of course, the velocity: v ( t ) = − A1β1 e − β1t − A2 β 2 e − β2t

a)

(2)

At t = 0: x0 = A1 + A2 v0 = − A1β1 − A2 β 2 (3) (4)

The initial conditions x0 and v0 can now be used to solve for the integration constants A1 and A2 .

b) When A1 = 0 , we have v0 = − β 2 x0 and v ( t ) = −β 2 x ( t ) quite easily. For A1 ≠ 0 , however, we

have v ( t ) → − β1 A1 e − β1t = − β1 x as t → ∞ since β1 < β 2 .

3-23. Firstly, we note that all the δ = π solutions are just the negative of the δ = 0 solutions. The δ = π 2 solutions don’t make it all the way up to the initial “amplitude,” A , due to the

retarding force. Higher β means more damping, as one might expect. When damping is high, less oscillation is observable. In particular, β 2 = 0.9 would be much better for a kitchen door than a smaller β, e.g. the door closing (δ = 0), or the closed door being bumped by someone who then changes his/her mind and does not go through the door ( δ = π 2 ).

OSCILLATIONS

101 β2 = 0.5, δ = 0 β2 = 0.9, δ = 0

1

β2 = 0.1, δ = 0

0.5

0

–0.5

–1 β2 = 0.1, δ = π/2 β2 = 0.5, δ = π/2 β2 = 0.9, δ = π/2

1

0.5

0

–0.5

–1 β2 = 0.1, δ = π β2 = 0.5, δ = π β2 = 0.9, δ = π

1

0.5

0

0.5

–1

0

5

10

15

0

5

10

15

0

5

10

15

3-24. As requested, we use Equations (3.40), (3.57), and (3.60) with the given values to evaluate the complementary and particular solutions to the driven oscillator. The amplitude of the complementary function is constant as we vary ω, but the amplitude of the particular solution becomes larger as ω goes through the resonance near 0.96 rad ⋅ s −1 , and decreases as ω is increased further. The plot closest to resonance here has ω ω 1 = 1.1 , which shows the least distortion due to transients. These figures are shown in figure (a). In figure (b), the ω ω 1 = 6

plot from figure (a) is reproduced along with a new plot with Ap = 20 m ⋅ s −2 .

102 ω/ω1 = 1/9 1 1 ω/ω1 = 1/3 2 ω/ω1 = 1.1

CHAPTER 3

0

0

0

–1

–1 0 10 20 30 0 10 20 30 ω/ω1 = 3 ω/ω1 = 6 0.5

–2 0 10 t (s) 20 30

0.5

0

0

–0.5

–0.5 Legend: xc xp x

–1

0

10 t (s)

20

30

–1

0

10 t (s)

20

30

(a)

0.5

0.5

0

0

–0.5

–0.5

Ap = 1

–1 0 5 10 15 20 25 30

Ap = 20

–1 0 5 10 15 20 25 30

(b)

3-25. This problem is nearly identical to the previous problem, with the exception that now Equation (3.43) is used instead of (3.40) as the complementary solution. The distortion due to the transient increases as ω increases, mostly because the complementary solution has a fixed amplitude whereas the amplitude due to the particular solution only decreases as ω increases. The latter fact is because there is no resonance in this case.

OSCILLATIONS

103 ω/ω1 = 1/3 ω/ω1 = 1.1

ω/ω1 = 1/9

1

1 0.5

0

0

0

–0.5

–0.5

–1

0

5 ω/ω1 = 3

10

–1

0

5 ω/ω1 = 6

10

–1

0

5 ω/ω1 = 6, Ap = 6

10

0

0

0

–0.5

–0.5

–0.5

–1

0

Legend:

5 xc 10 xp –1

0 x 5

10

–1

0

5

10

3-26.

The equations of motion of this system are m1 x1 = − kx1 − b1 ( x1 − x2 ) + F cos ω t m2 x2 = − b2 x2 − b1 ( x2 − x1 ) (1)

The electrical analog of this system can be constructed if we substitute in (1) the following equivalent quantities: m1 → L1 ; m2 → L2 ; k→ 1 ; C b1 → R1 ; x → q b2 → R2

F → ε0 ;

Then the equations of the equivalent electrical circuit are given by 1 q1 = ε 0 cos ω t C L2 q2 + R2 q2 + R1 ( q2 − q1 ) = 0 L1q1 + R1 ( q1 − q2 ) +

(2)

Using the mathematical device of writing exp(iω t) instead of cos ω t in (2), with the understanding that in the results only the real part is to be considered, and differentiating with respect to time, we have

104

CHAPTER 3

I1 = iωε 0 e iω t C L2 I 2 + R2 I 2 + R1 I 2 − I1 = 0 L1I1 + R1 I1 − I 2 +

(

)

( )

(

)

(3)

Then, the equivalent electrical circuit is as shown in the figure:

L1 ε0 cos ωt I1(t) C 1 I2(t) R1 2 R2 L2

The impedance of the system Z is Z = iω L1 − i where Z1 is given by 1 1 1 = + Z1 R1 R2 + iω L2 Then,

Z1 = R1 R2 ( R2 + R1 ) + ω 2 L2 + iω L2 R1 2

1 + Z1 ωC

(4)

(5)

( R1 + R2 )2 + ω 2 L22

(6)

and substituting (6) into (4), we obtain

1 2 2 2 R1 R2 ( R2 + R1 ) + ω 2 L2 + i R1ω L2 + ω L1 − 2 ( R1 + R2 ) + ω L2 ω C Z= 2 2 2 ( R1 + R2 ) + ω L2

(

)

(7)

3-27.

From Eq. (3.89), F (t) =

∞ 1 a0 + ∑ ( an cos nω t + bn sin nω t ) 2 n =1

(1)

We write F (t) =

∞ 1 a0 + ∑ cn cos ( nω t − φ n ) 2 n=1

(2)

which can also be written using trigonometric relations as F (t) =

∞ 1 a0 + ∑ cn cos nω t cos φ n + sin nω t sin φ n 2 n =1

(3)

Comparing (3) with (2), we notice that if there exists a set of coefficients cn such that

OSCILLATIONS

105

cn cos φ n = an cn sin φ n = bn then (2) is equivalent to (1). In fact, from (4),

2 2 2 c n = a n + bn b tan φ n = n an

(4)

(5)

with an and bn as given by Eqs. (3.91).

3-28. Since F(t) is an odd function, F(–t) = –F(t), according to Eq. (3.91) all the coefficients an vanish identically, and the bn are given by

bn = =

ω π

∫

π ω π − ω

F ( t ′ ) sin nω t ′ dt ′

π ω 0 ω − ∫ − π sin nω t ′ dt ′ + ∫ 0 sin nω t ′ dt ′ π ω

=

ω π

0 π 1 1 cos nω t ′ + − cos nω t ′ 0ω nω − π nω ω

=

2 ( cos 0 − cos nπ ) nπ

4 for n odd = nπ 0 for n even Thus, b( 2 n + 1) = b( 2 n) Then, we have F (t) = 4 4 ( 2n + 1) π n = 0, 1, 2, … =0

(1)

(2)

π

sin ω t +

4 4 sin 3ω t + sin 5ω t + … 3π 5π

(3)

106

F(t) 1 –π ⁄ω –1 π ⁄ω t

CHAPTER 3

0.849 –π ⁄ω Terms 1 + 2 π ⁄ω –0.849 t

1.099 –π ⁄ω π ⁄ω –1.099 Terms 1 + 2 + 3 t

0.918 –π ⁄ω π ⁄ω –0.918 Terms 1 + 2 + 3 + 4 t

3-29. In order to Fourier analyze a function of arbitrary period, say τ = 2P ω instead of 2π ω , proportional change of scale is necessary. Analytically, such a change of scale can be represented by the substitution

x=

πt

P

or

t=

Px

π

(1)

for when t = 0, then x = 0, and when t = τ = 2P ω , then x = 2π ω . Thus, when the substitution t = Px π is made in a function F(t) of period 2P ω ′ , we obtain the function

Px F = f ( x) π and this, as a function of x, has a period of 2π ω . Now, f(x) can, of course, be expanded according to the standard formula, Eq. (3.91):

(2)

f ( x) = where ∞ 1 a0 + ∑ ( an cos nω x + bn sin nω x ) 2 n =1

(3)

OSCILLATIONS

107

an = bn =

ω π ω π

∫ ω f ( x ′ ) cos nω x ′ dx ′

0

2π

(4)

∫ω

0

2π

f ( x ′ ) sin nω x ′ dx ′

If, in the above expressions, we make the inverse substitutions

x= the expansion becomes

πt

P

and

dx =

π

P

dt

(5)

∞ a π t P π t nωπ t nωπ t f = F ⋅ = F ( t ) = 0 + ∑ an cos + bn sin P 2 n =1 P π P P

(6)

and the coefficients in (4) become

nωπ t ′ dt ′ 0 P ω 2ωP nωπ t ′ bn = ∫ F ( t ′ ) sin dt ′ 0 P P an =

ω

ω F t cos ( ′) P∫

2P

(7)

For the case corresponding to this problem, the period of F(t) is

4π

ω

, so that P = 2π. Then,

substituting into (7) and replacing the integral limits 0 and τ by the limits − obtain nω t ′ dt ′ − 2 ω 2ωπ nω t ′ bn = ∫ − 2ωπ F (t ′ ) sin 2 dt ′ 2π an =

τ

2

and +

τ

2

, we

ω 2π

∫ ωωπ F (t ′ ) cos

2

2π

(8)

and substituting into (6), the expansion for F(t) is F (t) = Substituting F(t) into (8) yields nω t ′ dt ′ 2 ω 2ωπ nω t ′ bn = ∫ 0 sin ω t ′ sin 2 dt ′ 2π an = a0 ∞ nω t nω t + ∑ an cos + bn sin 2 2 n =1 2 (9)

ω 2π

∫ ω sin ω t ′ cos

0

2π

(10)

Evaluation of the integrals gives

108

CHAPTER 3

b2 =

1 ; bn = 0 2

for n ≠ 2

a0 = a1 = 0

0 an ( n ≥ 2 ) = −4 2 π n −4

(

)

n even n odd

(11)

and the resulting Fourier expansion is F (t) =

ωt 4 3ω t 5ω t 7ω t 1 4 4 4 − − − +… sin ω t + cos cos cos cos 2 3π 2 5π 2 21π 2 45π 2

(12)

3-30.

The output of a full-wave rectifier is a periodic function F(t) of the form

π − sin ω t ; − ω < t ≤ 0 F (t) = π sin ω t ; 0 τ : 0 F (t) = a (t τ ) m a (t − τ ) a (t τ ) − τ t t0 2 ω0 ω1 for t < t0 x (t) = 0

(1)

x (t) =

(2)

2 where ω 1 = ω 0 − β 2 . 2 2 For the case of overdamping, ω 0 < β 2 , and consequently ω 1 = i β 2 − ω 0 is a pure imaginary

number. Hence, cos ω 1 ( t − t0 ) and sin ω 1 ( t − t0 ) are no longer oscillatory functions; instead,

2 they are transformed into hyperbolic functions. Thus, if we write ω 2 = β 2 − ω 0 (where ω 2 is real),

cos ω 1 ( t − t0 ) = cos iω 2 ( t − t0 ) = cosh ω 2 ( t − t0 ) sin ω 1 ( t − t0 ) = sin iω 2 ( t − t0 ) = i sinh ω 2 ( t − t0 ) The response is given by [see Eq. (3.105)] a β e − β (t − t0 ) − β t −t 1 − e ( 0 ) cosh ω 2 ( t − t0 ) − sinh ω 1 ( t − t0 ) for t > t0 2 ω0 ω2 for t < t0 x (t) = 0

(3)

x (t) =

(4)

For simplicity, we choose t0 = 0 , and the solution becomes

x (t) =

H ( 0) βe−β t 1 − e − β t cosh ω 2t − sinh ω 2t 2 ω0 ω2

(5)

This response is shown in (a) below for the case β = 5 ω 0 .

b) Response to an Impulse Function (in the limit τ → 0):

From Eq. (3.101) the impulse function I ( t0 , t1 ) is defined as 0 I ( t0 , t1 ) = a 0

t < t0 t0 < t < t1 t > t1

(6)

For t1 − t2 = τ → 0 in such a way that aτ is constant = b, the response function is given by Eq. (3.110):

112

CHAPTER 3

x (t) =

b

ω1

− β t −t e ( 0 ) sin ω 1 ( t − t0 ) for t > t0

(7)

Again taking the “spike” to be at t = 0 for simplicity, we have

x (t) =

b

ω1

e − βt sin ω 1 ( t )

for t > 0

(8)

2 For ω 1 = iω 2 = i β 2 − ω 0 (overdamped oscillator), the solution is

x (t) =

b

ω2

e − βt sinh ω 2t ; t > 0

(9)

This response is shown in (b) below for the case β = 5 ω 0 .

(a)

2 x H ( 0) ω 0

1

0.5

0

0

1

2

3

4

5 ω0t

6

7

8

9

10

(b) x (b ω 2 )

1

0.5

0

0

1

2

3

4

5 ω0t

6

7

8

9

10

3-33. a) In order to find the maximum amplitude of the response function shown in Fig. 3-22, we look for t1 such that x ( t ) given by Eq. (3.105) is maximum; that is,

∂ x (t) ∂t From Eq. (3.106) we have ∂ x (t) ∂t

(

) t = t1

=0

(1)

(

) = H ( 0) e ω 2 0

−β t

β2 ω 1 + sin ω 1t ω1

(2)

OSCILLATIONS

113

2 For β = 0.2ω 0 , ω 1 = ω 0 − β 2 = 0.98ω 0 . Evidently, t1 = π ω 1 makes (2) vanish. (This is the absolute maximum, as can be seen from Fig. 3-22.)

Then, substituting into Eq. (3.105), the maximum amplitude is given by x ( t ) max or, x ( t1 ) ≅ 1.53 a

2 ω0

βπ − a ω1 = x ( t1 ) = 2 1 + e ω0

(3)

(4)

b) In the same way we find the maximum amplitude of the response function shown in Fig. 3-24 by using x(t) given in Eq. (3.110); then,

∂ x (t) ∂t

(

) t = t1

β − β t −t sin ω 1 ( t − t0 ) = be ( 0 ) cos ω 1 ( t − t0 ) − ω1 t = t1

(5)

If (5) is to vanish, t1 is given by t1 − t0 = ω 1 1.37 tan −1 1 = tan −1 ( 4.9) = ω1 ω1 β ω1 1 (6)

Substituting (6) into Eq. (3.110), we obtain (for β = 0.2ω 0 ) x ( t ) max or, x ( t1 ) ≅ 0.76 aτ

× 1.37 b = x ( t1 ) = sin ( 1.37 ) e ω1 0.98ω 0 −β

(7)

ω0

(8)

The response function of an undamped (β = 0) linear oscillator for an impulse function 2π I(0,τ ), with τ = , can be obtained from Eqs. (3.105) and (3.108) if we make the following

3-34.

ω0

substitutions:

β=0; t0 = 0 ;

ω1 = ω 0

2π t1 = τ = ω0

(1)

(For convenience we have assumed that the impulse forcing function is applied at t = 0.) Hence, after substituting we have

114

CHAPTER 3

x (t) = 0 x (t) = x (t) = 1 − cos ω 0 t ω

2 0

t t0 given by Eq. (3.103) yields

A1 = x0 −

a

2 ω0

; A2 =

ω1

x0

+

β x0 βa − 2 ω1 ω 0ω1

(1)

Therefore, the response to H ( t0 ) for the initial conditions above can be expressed as

x βx − β t −t x ( t ) = e ( 0 ) x0 cos ω 1 ( t − t0 ) + 0 + 0 sin ω 1 ( t − t0 ) ω1 ω1 a β − β (t − t0 ) −β t−t + 2 1 − e ( 0 ) cos ω 1 ( t − t0 ) − sin ω 1 ( t − t0 ) e ω0 ω1 for t > t0

(2)

given by (2) for t0 < t < t1 and by a superposition of solutions for H ( t0 ) and for H ( t1 ) taken individually for t > t1 . We must be careful, however, because the solution for t > t1 must be with initial conditions x ( 0 ) = 0 , x ( 0 ) = 0 , and using t1 instead of t0 in the expression. The solution for t > t1 is then

The response to an impulse function I ( t0 , t1 ) = H ( t1 ) , for the above initial conditions will then be

equal that given by (2) for t = t1 . This can be insured by using as a solution for H ( t1 ) Eq. (3.103)

OSCILLATIONS

117

x βx − β t −t x ( t ) = e ( 0 ) x0 cos ω 1 ( t − t0 ) + 0 + 0 sin ω 1 ( t − t0 ) + x1 ( t ) ω1 ω1 where x1 ( t ) =

− β t−t ae ( 0 ) βτ β e βτ e cos ω 1 ( t − t0 − τ ) − cos ω 1 ( t − t0 ) + 2 ω0 ω1

(3)

β e βτ β + sin ω 1 ( t − t0 − τ ) − sin ω 1 ( t − t0 ) ω1 ω1

(4) for t > t1

We now allow a → ∞ as τ → 0 in such a way that aτ = b = constant; expanding (3) for this particular case, we obtain x βx b −β t−t x ( t ) = e ( 0 ) x0 cos ω 1 ( t − t0 ) + 0 + 0 + sin ω 1 ( t − t0 ) ω1 ω1 ω1 which is analogous to Eq. (3.119) but for initial conditions given above.

3-37. Any function F ( t ) m can be expanded in terms of step functions, as shown in the figure below where the curve is the sum of the various (positive and negative) step functions.

t > t0

(5)

In general, we have

2 x + 2β x + ω 0 x =

Fn ( t ) n =− ∞ m

∑

∞

∞

= where an ( t ) Hn (t) = 0

n =− ∞

∑ H (t) n (1)

t > tn = nτ t < tn = nτ

(2)

Then, since (1) is a linear equation, the solution to a superposition of functions of the form given by (2) is the superposition of the solutions for each of those functions. According to Eq. (3.105), the solution for H n ( t ) for t > tn is xn ( t ) = then, for

∞ F (t) = ∑ Hn (t) m n =− ∞

an β e − β ( t − tn ) − β t −t 1 − e ( n ) cos ω 1 ( t − tn ) − sin ω 1 ( t − tn ) 2 ω0 ω1

(3)

(4)

the solution is

118

CHAPTER 3

x (t) =

β e − β ( t − tn ) − β t −t H n ( t ) 1 − e ( n ) cos ω 1 ( t − tn ) − sin ω 1 ( t − tn ) 2 ∑ ω 0 n =− ∞ ω1 1

∞

= where

n =− ∞

∑ mH (t ) G (t ) = ∑ F ( t ) G (t ) n n n =− ∞ n n

∞

∞

(5)

1 2 Gn ( t ) = mω 0 0 or, comparing with (3)

β e − β ( t − tn ) − β ( t − tn ) cos ω 1 ( t − tn ) − sin ω 1 ( t − tn ) ; 1 − e ω1

t ≥ tn t < tn

(6)

xn ( t ) man , Gn ( t ) = 0

F (t ) m

t ≥ tn t < tn

(7)

Therefore, the Green’s function is the response to the unit step.

tn tn+1

t

Hn(t)

3-38.

The solution for x(t) according to Green’s method is x (t) = ∫ = t −∞

F ( t ′ ) G ( t , t ′ ) dt ′

F0 mω 1

∫

t

0

e − γ t ′ sin ω t ′ e − β (t − t ′) sin ω 1 ( t − t ′ ) dt ′

(1)

Using the trigonometric identity, sin ω t ′ sin ω 1 ( t − t ′ ) = we have 1 cos (ω 1 + ω ) t ′ − ω 1t − cos (ω − ω 1 ) t ′ + ω 1t 2 (2)

OSCILLATIONS

119

x (t) =

t t F0 e − β t ( β −γ ) t ′ cos (ω + ω 1 ) t ′ − ω 1t − ∫ dt ′e ( β −γ )t ′ cos (ω − ω 1 ) t ′ + ω 1t ∫ dt ′e 2mω 1 0 0

(3)

Making the change of variable, z = (ω + ω 1 ) t ′ − ω 1t , for the first integral and y = (ω − ω 1 ) t ′ + ω 1t for the second integral, we find ( β −γ )ω1t F0 e − β t e ω +ω1 x (t) = 2mω 1 ω + ω 1 ωt − ω1t

∫

( β −γ ) z

dz e

ω +ω1

cos z −

e

− ( β − γ )ω1t ωt ω −ω1

ω − ω1

ω1t

∫ dy e

(β −γ ) y ω −ω 1

cos y

(4)

After evaluating the integrals and rearranging terms, we obtain x (t) = F0 ω 2 2 m ( β − γ ) + (ω + ω 1 ) ( β − γ ) 2 + (ω − ω 1 ) 2 sin ω t 2 2 × e − γ t 2 (γ − β ) cos ω t + [ β − γ ] + ω 1 − ω 2 ω

(

)

(5)

2 2 sin ω 1t + e − β t 2 ( β − γ ) cos ω 1t + [ β − γ ] + ω 2 − ω 1 ω

(

)

3-39.

sin ω t F (t) = 0 From Equations 3.89, 3.90, and 3.91, we have F (t) = an = bn = an = a0 = a1 = 2

0 a

which is plotted in (a) below.

NONLINEAR OSCILLATIONS AND CHAOS

133

U(x) E7 E6 E5 E4 x

E3 E2 E1 –a O a

(a)

For small deviations from the equilibrium position (x = 0), the motion is just that of a harmonic oscillator. For energies E < E6 , the particle cannot reach regions with x < –a, but it can reach regions of x > a if E > E4 . For E2 < E < E4 the possibility exists that the particle can be trapped near x = a. A phase diagram for the system is shown in (b) below.

· x E2 E1 E3 E4 x

…

E7

(b)

4-10.

The system of equations that we need to solve are y x y = −0.05 y − sin x + 0.7 cos ω t (1)

The values of ω that give chaotic orbits are 0.6 and 0.7. Although we may appear to have chaos for other values, construction of a Poincaré plot that samples at the forcing frequency show that they all settle on a one period per drive cycle orbit. This occurs faster for some values of ω than others. In particular, when ω = 0.8 the plot looks chaotic until it locks on to the point (−2.50150, 0.236439) . The phase plot for ω = 0.3 shown in the figure was produced by numerical integration of the system of equations (1) with 100 points per drive cycle. The box encloses the point on the trajectory of the system at the start of a drive cycle. In addition, we also show Poincaré plot for the case ω = 0.6 in figure, integrated over 8000 drive cycles with 100 points per cycle.

134

1

CHAPTER 4

0.5

0

-0.5

–1 –1.5

–1

-0.5

0

0.5

1

1.5

3 2 1 0 –1 –2 –3 –4 –3 –2 –1 0 1 2 3 4

4-11. The three-cycle does indeed occur where indicated in the problem, and does turn chaotic near the 80th iteration. This value is approximate, however, and depends on the precision at which the calculations are performed. The behavior returns to a three-cycle near the 200th iteration, and stays that way until approximately the 270th iteration, although some may see it continue past the 300th.

1 0.8 0.6

x

0.4 0.2 0

100

200

300

400

500

iteration

NONLINEAR OSCILLATIONS AND CHAOS

135

4-12. x1 = 0.4

0.25 0.2

0.15

0.1

0.05

0

0

0.2

0.4

0.6

0.8

1

x1 = 0.75

0.25

0.2

0.15

0.1

0.05

0

0

0.2

0.4

0.6

0.8

1

These plots are created in the manner described in the text. They are created with the logistic equation xn + 1 = 0.9 ⋅ xn (1 − xn ) The first plot has the seed value x1 = 0.4 as asked for in the text. Only one additional seed has been done here ( x1 = 0.75 ) as it is assumed that the reader could easily produce more of these plots after this small amount of practice. (1)

136 4-13.

1 0.8 0.6 0.4 0.2 0 30 32 34 36 38 40 42 iteration 44 46 48 50

CHAPTER 4

x1 = 0.7 x1 = 0.700000001 1 0.8 0.6 0.4 0.2 0 30 32 34 36 38 40 42 iteration 44 46 48 50

x1 = 0.7 x1 = 0.7000000001

The plots are created by iteration on the initial values of (i) 0.7, (ii) 0.700000001, and (iii) 0.7000000001, using the equation

2 xn + 1 = 2.5 ⋅ xn 1 − xn

(

)

(1)

A subset of the iterates from (i) and (ii) are plotted together, and clearly diverge by n = 39. The plot of (i) and (iii) clearly diverge by n = 43.

4-14.

1 0.8 0.6 0.4 0.2 0

20

22

24

26

28 iteration 30

32

34

x1 = 0.9 x1 = 0.9000001 fractional difference

NONLINEAR OSCILLATIONS AND CHAOS

137

The given function with the given initial values are plotted in the figure. Here we use the notation x1 = 0.9 and y1 = 0.9000001 , with xn + 1 = f ( xn ) and yn + 1 = f ( yn ) where the function is f ( x ) = 2.5 ⋅ x 1 − x 2

(

)

(1)

The fractional difference is defined as x − y x , and clearly exceeds 30% when n = 30.

4-15. A good way to start finding the bifurcations of the function f(α,x) = α sin πx is to plot its bifurcation diagram.

1

0.8

0.6

x

0.4 0.2 0

0

0.2

0.4 α 0.6

0.8

1

One can expand regions of the diagram to give a rough estimate of the location of a bifurcation. Its accuracy is limited by the fact that the map does not converge very rapidly near the bifurcation point, or more precisely, the Lyapunov exponent approaches zero. One may continue undaunted, however, with the help of a graphical fractal generating software application, to estimate quite a few of the period doublings α n .Using Fractint for Windows, and Equation (4.47) to compute the Feigenbaum constant, we can obtain the following: n 1 2 3 4 5 6 7 8

α

0.71978 0.83324 0.85859 0.86409 0.86526 0.86551 0.865564 0.8655761

δ

4.475 4.611 4.699 4.680 4.630 4.463

One can see that although we should obtain a better value of δ as n increases, numerical precision and human error quickly degrade the quality of the calculation. This is a perfectly acceptable answer to this question. One may compute the α n to higher accuracy by other means, all of which are a great deal more complicated. See, for example, Exploring Mathematics with Mathematica, which exploits the vanishing Lyapunov exponent. Using their algorithm, one obtains the following:

138

CHAPTER 4

n 1 2 3 4 5 6 7

α

0.719962 0.833266 0.858609 0.864084 0.865259 0.865511 0.865560

δ

4.47089 4.62871 4.66198 4.65633 5.13450

Note that these are shown here only as reference, and the student may not necessarily be expected to perform to this degree of sophistication. The above values are only good to about 10 −6 , but this time only limited by machine precision. Another alternative in computing the Feigenbaum constant, which is not requested in the problem, is to use the so-called ”supercycles,” or super-stable points Rn , which are defined by

f

( 2n−1 ) R , 1 = 1 n

2 2

The values Rn obey the same scaling as the bifurcation points, and are much easier to compute since these points converge faster than for other α (the Lyapunov exponent goes to – ∞). See, for example, Deterministic Chaos: An Introduction by Heinz Georg Schuster or Chaos and Fractals: New Frontiers of Science by Peitgen, Jürgens and Saupe. As a result, the estimates for δ obtained in this way are more accurate than those obtained by calculating the bifurcation points.

4-16.

where x0 = f ( x0 ) . Now expand f(x) in a Taylor series, so that near x0 we have f ( x) where f ( x0 ) + β ( x − x0 ) = x 0 + β ( x − x0 )

The function y = f(x) intersects the line y = x at x = x0 , i.e. x0 is defined as the point

(1)

β≡

df dx

(2) x0 Now define ε n ≡ xn − x0 . If we have x1 very close to x0 , then ε 1 should be very small, and we may use the Taylor expansion. The equation of iteration xn + 1 = f ( xn ) becomes

ε n+1

βε n

(3)

If the approximation (1) remains valid from the initial value, we have ε n +1

a) The values xn − x0 = ε n form the geometric sequence ε 1 , βε 1 , β 2 ε 1 , … . b) Clearly, when β < 1 we have stability since

β nε1 .

lim ε n = 0 n →∞

Similarly we have a divergent sequence when β > 1 , although it will not really be exponentially divergent since the approximation (1) becomes invalid after some number of iterations, and normally the range of allowable xn is restricted to some subset of the real numbers.

NONLINEAR OSCILLATIONS AND CHAOS

139

4-17.

0.6 0.4 0.2 0

0

2

4

6

8

10 iteration 12

14

16

18

20

α = 0.4 α = 0.7

The first plot (with α = 0.4) converges rather rapidly to zero, but the second (with α = 0.7) does appear to be chaotic.

4-18.

1 0.8 0.6

x

0.4 0.2 0

0

0.2

0.4 α 0.6

0.8

1

The tent map always converges to zero for α < 0.5. Near α = 0.5 it takes longer to converge, and that is the artifact seen in the figure. There exists a “hole” in the region 0.5 < α < 0.7 (0.7 is approximate), where the iterations are chaotic but oscillate between an upper and lower range of values. For α > 0.7, there is only a single range of chaos, which becomes larger until it fills the range (0,1) at α = 1.

4-19.

From the definition in Equation (4.52) the Lyapunov exponent is given by

λ = lim n→∞ df 1 n −1 ∑ ln dx n i=0

(1) xi The tent map is defined as for 0 < x < 1 2 2α x f ( x) = 2α (1 − x ) for 1 2 < x < 1 This gives df dx = 2α , so we have (2)

140

CHAPTER 4

λ = lim

n − 1 ln ( 2α ) = ln ( 2α ) n→∞ n

(3)

As indicated in the discussion below Equation (4.52), chaos occurs when λ is positive: α > 1 2 for the tent map.

4-20.

0.4

0.2

y

0

–0.2

–0.4 –1.5 –1 –0.5 0 x 0.5

1

1.5

4-21.

0.4

0.2

y

0

–0.2

–0.4 –1.5 –1 –0.5 0 x 0.5 1 1.5

The shape of this plot (the attractor) is nearly identical to that obtained in the previous problem. In Problem 4-20, however, we can clearly see the first few iterations (0,0), (1,0), (–0.4,0.3),

NONLINEAR OSCILLATIONS AND CHAOS

141

whereas the next iteration (1.076,–0.12) is almost on the attractor. In this problem the initial value is taken to be on the attractor already, so we do not see any transient points.

4-22.

The following. system of differential equations were integrated numerically y x y = −0.1 y − x 3 + B cos t (1)

using different values of B in the range [9.8,13.4], and with a variety of initial conditions. The integration range is over a large number of drive cycles, throwing away the first several before starting to store the data in order to reduce the effects of the transient response. For the case B = 9.8, we have a one period per three drive cycle orbit. The phase space plot (line) and Poincaré section (boxes) for this case are overlaid and shown in figure (a). All integrations are done here with 100 points per drive cycle. One can experiment with B and determine that the system becomes chaotic somewhere between 9.8 and 9.9. The section for B = 10.0, created by integrating over 8000 drive cycles, is shown in figure (b). If one further experiments with different values of B, and one is also lucky enough to have the right initial conditions, (0,0) is one that works, then a transition will be found for B in the range (11.6,11.7). As an example of the different results one can get depending on the initial conditions, we show two plots in figure (c). One is a phase plot, overlaid with its section, for B = 12.0 and the initial condition (0,0). Examination of the time evolution reveals that it has one period per cycle. The second plot is a Poincaré section for the same B but with the initial condition (10,0), clearly showing chaotic motion. Note that the section looks quite similar to the one for B = 10.0. Another transition is in the range (13.3,13.4), where the orbits become regular again, with one period per drive cycle, regardless of initial conditions. The phase plot for B = 13.4 looks similar to the one with B = 12.0 and initial condition (0,0). To summarize, we may enumerate the above transition points by B1 , B2 , and B3 . Circumventing the actual task of computing where these transition points are, we do know that 9.8 < B1 < 9.9 , 11.6 < B2 < 11.7 , and 13.3 < B3 < 13.4 . We can then describe the behavior of the system by region. • • • • B < B1 : one period per three drive cycles B1 < B < B2 : chaotic B2 < B < B3 : mixed chaotic/one period per drive cycle (depending on initial conditions) B3 < B : one period per drive cycle

We should remind ourselves, though, that the above list only applies for B in the range we have examined here. We do not know the behavior when B < 9.8 and B > 13.4, without going beyond the scope of this problem.

142

CHAPTER 4

(a)

5

0

y

–5 –3 –2 –1 0 x 1

2

3

(b)

10

5

y

0 –5

2.4

2.6

2.8 x

3

3.2

3.4

3.6

(c)

5

y

0

–5

–3

–2

–1

0 x 1

2

3

(d)

10 5 y 0 –5 –10

2.4

2.6

2.8

3

3.2 x 3.4

3.6

3.8

NONLINEAR OSCILLATIONS AND CHAOS

143

4-23.

The Chirikov map is defined by pn + 1 = pn − K sin qn qn + 1 = qn − pn + 1 (1) (2)

The results one should get from doing this problem should be some subset of the results shown in figures (a), (b), and (c) (for K = 0.8, 3.2, and 6.4, respectively). These were actually generated using some not-so-random initial points so that a reasonably complete picture could be made. What look to be phase paths in the figures are actually just different points that come from iterating on a single initial condition. For example, in figure (a), an ellipse about the origin (just pick one) comes from iterating on any one of the points on it. Above the ellipses is chaotic orbit, then a five ellipse orbit (all five come from a single initial condition), etc. The case for K = 3.2 is similar except that there is an orbit outside of which the system is always undergoing chaotic motion. Finally, for K = 6.4 the entire space is filled with chaotic orbits, with the exception of two small lobes. Inside of these lobes are regular orbits (the ones in the left are separate from the ones in the right).

(a)

1

0.5

p⁄π

0

–0.5

–1 –1

–0.5

0 q⁄π 0.5

1

(b)

1

0.5

p⁄π

0

–0.5

–1 –1

–0.5

0 q⁄π 0.5

1

144

(c)

1

CHAPTER 4

0.5

p⁄π

0

–0.5

–1 –1

–0.5

0 q⁄π 0.5

1

4-24. a) The Van de Pol equation is

d2 x dx + ω 0 2 x = µ ( a2 − x 2 ) 2 dt dt Now look for solution in the form x(t ) = b cos ω 0 t + u(t) we have dx du = − bω 0 sin ω 0 t + dt dt and d2 x d2u = − bω 0 2 cos ω 0 t + 2 dt 2 dt Putting these into the Van de Pol equation, we obtain d 2 u(t) + ω 0 u(t) = − µ b 2 cos 2 ω 0 t + u2 (t) + 2bu(t ) cos ω 0 t − a 2 dt 2 (1)

{

}

{

− bω 0 sin ω 0 t +

du(t ) dt

}

From this one can see that u(t) is of order µ (i.e. u ~ O(µ ) ), which is assumed to be small here. Keeping only terms up to order µ , the above equation reads

d 2 u(t) + ω 0 u(t ) = − µ − b 3ω 0 sin ω 0 t cos 2 ω 0 t + a 2 bω 0 sin ω 0 t 2 dt

{

}

b2 b2 = − µbω 0 a 2 − sin ω 0 t − sin 3ω 0 t 4 4 (where we have used the identity 4 sin ω 0 t cos 2 ω 0 t = sin ω 0 t + sin 3ω 0 t ) This equation has 2 frequencies ( ω 0 and 3ω 0 ), and is complicated. However, if b = 2 a then the term sin ω 0 t disappears and the above equation becomes

NONLINEAR OSCILLATIONS AND CHAOS

145

d 2 u(t) b3 sin 3ω 0 t + ω 0 u(t) = µω 0 4 dt 2

We let b = 2 a , and the solution for this equation is

u(t ) = −

µb 3 µ a3 sin 3ω 0 t = − sin 3ω 0 t 32ω 0 4ω 0

So, finally putting this form of u(t) into (1), we obtain one of the exact solutions of Van de Pol equation:

u(t ) = 2 a cos ω 0 t −

µ a3 sin 3ω 0 t 4ω 0

b) See phase diagram below. Since µ = 0.05 is very small, then actually the second term in the expression of u(t) is negligible, and the phase diagram is very close to a circle of radius b = 2a = 2.

. x

2

1

–2

–1

1

2

x

–1

–2

4-25. We have used Mathematica to numerically solve and plot the phase diagram for the van de Pol equation. Because µ = 0.07 is a very small value, the limit cycle is very close to a circle of radius b = 2a = 2. a) In this case, see figure a), the phase diagram starts at the point (x = 1, x′ = 0) inside the limit cycle, so the phase diagram spirals outward to ultimately approach the stable solution presented by the limit cycle (see problem 4-24 for exact expression of stable solution).

146

. x 2

CHAPTER 4

1

–2

–1

1

2

x

–1

–2

b) In this case, see figure b), the phase diagram starts at the point (x = 3, x′ = 0) outside the limit cycle, so the phase diagram spirals inward to ultimately approaches the stable solution presented by the limit cycle (see problem 4-24 for exact expression of stable solution).

. x

2

1

–2

–1 –1

1

2

3

x

–2

4-26. We have used Mathematica to numerically solve and plot the phase diagram for the van de Pol equation. Because µ = 0.5 is not a small value, the limit cycle is NOT close to a circle (see problem 4-24 above). a) In this case, see figure a), the phase diagram starts at the point (x = 1, x′ = 0) inside the limit cycle, so the phase diagram spirals outward to ultimately approach the stable solution presented by the limit cycle (see problem 4-24 for exact expression of stable solution).

NONLINEAR OSCILLATIONS AND CHAOS

147

. x 2

1

–2

–1 –1

1

2

x

–2

b) In this case (see figure below), the phase diagram starts at the point (x = 3, x′ = 0) outside the limit cycle, so the phase diagram spirals inward to ultimately approaches the stable solution presented by the limit cycle (see problem 4-24 for exact expression of stable solution).

. x

2

1

–2

–1

1

2

3

x

–1

–2

148

CHAPTER 4

CHAPTER

5

Gravitation

5-1. a) Two identical masses: The lines of force (dashed lines) and the equipotential surfaces (solid lines) are as follows:

b) Two masses, +M and –M: In this case the lines of force do not continue outward to infinity, as in a), but originate on the “negative” mass and terminate on the positive mass. This situation is similar to that for two electrical charges, +q and –q; the difference is that the electrical lines of force run from +q to –q.

149

150 5-2. Inside the sphere the gravitational potential satisfies

CHAPTER 5

∇ 2 φ = 4π G ρ ( r ) Since ρ(r) is spherically symmetric, φ is also spherically symmetric. Thus,

1 ∂ 2 ∂φ = 4π G ρ ( r ) r r 2 ∂r ∂ r The field vector is independent of the radial distance. This fact implies

∂φ = constant ≡ C ∂r

(1)

(2)

(3)

Therefore, (2) becomes

2C = 4π Gρ r or, (4)

ρ=

C 2π Gr

(5)

5-3. In order to remove a particle from the surface of the Earth and transport it infinitely far away, the initial kinetic energy must equal the work required to move the particle from r = Re to r = ∞ against the attractive gravitational force:

∫

∞ Re

G

Me m 1 2 dr = mv0 2 r 2

(1)

where Me and Re are the mass and the radius of the Earth, respectively, and v0 is the initial velocity of the particle at r = Re . Solving (1), we have the expression for v0 :

v0 =

2G Me Re

(2)

Substituting G = 6.67 × 10 −11 m 3/kg ⋅ s 2 , Me = 5.98 × 10 24 kg , Re = 6.38 × 10 6 m , we have v0 ≅ 11.2 km/sec (3)

5-4.

The potential energy corresponding to the force is U = − ∫ F dx = mk 2 dx mk 2 =− 2 ∫ x 3 2x (1)

The central force is conservative and so the total energy is constant and equal to the potential energy at the initial position, x = d:

GRAVITATION

151

E = constant =

1 1 k2 1 k2 mx 2 − m 2 = − m 2 2 2 x 2 d

0

(2)

Rewriting this equation in integrable form,

dt = − ∫ d 0

dx 1 1 k 2 − 2 d x

2

=−

d x dx k ∫ d2 − x2 d

(3)

where the choice of the negative sign for the radical insures that x decreases as t increases. Using Eq. (E.9), Appendix E, we find t= or t= d2 k d d2 − x2 k

0

(4) d 5-5.

The equation of motion is mx = −G Mm x2 (1)

Using conservation of energy, we find 1 2 1 1 x − G M = E = −G M 2 x x∞

1 1 dx = − 2GM − dt x x∞

(2)

(3)

where x∞ is some fixed large distance. Therefore, the time for the particle to travel from x∞ to x is t =−∫ x x∞

1 =− GM 1 1 2GM − x x∞ dx

x∞

∫

x

xx∞ dx 2 ( x∞ − x )

Making the change of variable, x → y 2 , and using Eq. (E.7), Appendix E, we obtain t= x∞ x −1 x ( x∞ − x ) − x∞ sin 2GM x∞ x∞ x (4)

If we set x = 0 and x = x∞ 2 in (4), we can obtain the time for the particle to travel the total distance and the first half of the distance.

152

0

CHAPTER 5

T0 = x∞ 2

x∞

∫

dt =

1 x∞ GM 2

x∞ 2

32

32

(5)

T1 2 = Hence,

x∞

∫

dt =

1 GM

π 1 + 2

(6)

T1 2 T0 Evaluating the expression, T1 2 T0 or

=

1+

π

2

π

= 0.818

(7)

T1 2 T0

≅

9 11

(8)

5-6. z P α θ φ x s r2drd(cos θ)dφ r y

Since the problem has symmetry around the z-axis, the force at the point P has only a z-component. The contribution to the force from a small volume element is dg z = −G where ρ is the density. Using cos α = a ρ s2 r 2 dr d ( cos θ ) dφ cos α

(1)

z − r cos θ and integrating over the entire sphere, we have s

+1 −1

2π

g z = −Gρ ∫ r 2 dr ∫ d ( cos θ ) ∫ dφ

0 0

(r

z − r cos θ

2

+ z 2 − 2rz cos θ

)

32

(2)

Now, we can obtain the integral of cos θ as follows:

GRAVITATION

153

+1 −1

I=

∫

(

z − r cos θ r + z 2 − 2rz cos θ

2

)

32

d ( cos θ )

=−

∂ ∂z

+1 −1

∫ (r

2

+ z 2 − 2rz cos θ

)

12

d ( cos θ )

Using Eq. (E.5), Appendix E, we find

I=−

∂ ∂z

1 2 r + z 2 − 2rz cos θ − rz

(

)

12

+1

−1 (3)

=−

∂ 2 2 = ∂z z z 2

Therefore, substituting (3) into (2) and performing the integral with respect to r and φ, we have

g z = −Gρ = −G But

a3 2 ⋅ 2π 3 z2 (4)

4π 3 1 a ρ 2 z 3

4π 3 a ρ is equal to the mass of the sphere. Thus, 3 g z = −GM 1 z2 (5)

Thus, as we expect, the force is the same as that due to a point mass M located at the center of the sphere.

5-7.

dx x s R P

The contribution to the potential at P from a small line element is dΦ = −G where ρ = M

ρ s dx

(1)

is the linear mass density. Integrating over the whole rod, we find the potential Φ = −G M

∫

2

− 2

1 x 2 + R2

dx

(2)

154

CHAPTER 5

Using Eq. (E.6), Appendix E, we have M

2 2 2 + R2 + GM 2 4 =− ln 2 + R2 − + 4 2 2

Φ = −G

ln x + x 2 + R2 −

(3)

Φ=−

GM

ln

+ 4 R2 + 2 + 4 R2 −

5-8. z z0 α a z rdrdθdz y r 2 + ( z0 − z )

2

x

θ r

Since the system is symmetric about the z-axis, the x and y components of the force vanish and we need to consider only the z-component of the force. The contribution to the force from a small element of volume at the point (r,θ,z) for a unit mass at (0,0, z0 ) is dg z = −Gρ r + ( z0 − z )

2

rdrdθ dz

2

cos α

= −Gρ

( z0 − z) rdrdθ dz 32 r 2 + ( z0 − z ) 2

(1)

where ρ is the density of the cylinder and where we have used cos α =

( z0 − z ) 2 r 2 + ( z0 − z )

. We can

find the net gravitational force by integrating (1) over the entire volume of the cylinder. We find g z = −Gρ ∫ rdr

0

a

2π

∫ dθ ∫ dz

0 0 a z0 −

z0 − z

32

2 r 2 + ( z0 − z )

Changing the variable to x = z0 − z , we have g z = 2π Gρ ∫ rdr

0

z0

∫

xdx r + x2

2 32

(2)

Using the standard integral,

GRAVITATION

155

∫ we obtain

xdx

(

a2 ± x

2 3

)

=

−1 a ± x2

2

(3)

a r g z = −2π Gρ ∫ dr r2 + z − 0 (0 Next, using Eq. (E.9), Appendix E, we obtain

)

2

−

2 r 2 + z0 r

(4)

g z = −2π Gρ a 2 + ( z0 −

)2 −

2 a 2 + z0 +

(5)

Now, let us find the force by first computing the potential. The contribution from a small element of volume is dΦ = −Gρ Integrating over the entire volume, we have dΦ = −Gρ ∫ dz

0 2π

rdrdθ dz r + ( z0 − z )

2 2

(6)

∫ dθ ∫ dr

0 0

a

r r 2 + ( z0 − z )

2

(7)

Using Eq. (E.9), Appendix E, again, we find

2 dΦ = −2π Gρ ∫ dz a 2 + ( z0 − z ) − ( z0 − z ) 0

(8)

Now, we use Eqs. (E.11) and (E.8a), Appendix E, and obtain

(z − Φ = −2π Gρ − 0 2 + Thus, the force is z0 2

)

a 2 + ( z0 −

)2 +

a2 ln −2 ( z0 − 2

) + 2 ( z0 − ) 2 + a 2

2

2 a 2 + z0 −

a2 1 2 ln −2z0 + 2 z0 + a 2 − z0 + 2 2

156

CHAPTER 5

∂Φ 1 2 = −2π Gρ gz = − a + ( z0 − ∂z 0 2

)

2

1 + 2

( z0 − ) 2 2 a 2 + ( z0 − )

1− z0

1− a2 + 2 −(z − 0

z0 −

( z0 − ) 2 + a 2 ) + ( z0 − ) 2 + a 2

−

1 2 1 2 a + z0 − 2 2

z +a z a − + 2 2 2 − z0 + z0 + a 2 a 2 + z0

2 0 2 2 0 2

(9)

or, g z = −2π Gρ a 2 + ( z0 − and we obtain the same result as in (5). In this case, it is clear that it is considerably easier to compute the force directly. (See the remarks in Section 5.4.)

5-9.

a θ R r P

)2 −

2 a 2 + z0 +

(10

The contribution to the potential at the point P from a small line element d is Φ = −G ∫

ρd r M . Using 2π a

(1)

where ρ is the linear mass density which is expressed as ρ = r = R2 + a 2 − 2aR cos θ and d = adθ, we can write (1) as Φ=− GM 2π

2π

∫

0

dθ R + a − 2aR cos θ

2 2

(2)

This is the general expression for the potential. If R is much greater than a, we can expand the integrand in (2) using the binomial expansion:

−1 2

a2 1 a = 1 − 2 cos θ − 2 R R2 + a 2 − 2aR cos θ R R 1

2 1 a a2 3 a a2 1 = 1 + 2 cos θ − 2 + 2 cos θ − 2 + … R 8 R R R 2 R

(3)

GRAVITATION

157

3

a If we neglect terms of order and higher in (3), the potential becomes R

Φ=− GM 2π GM 2π

2π

∫

0

3 a2 a a2 1 + cos θ − + cos 2 θ dθ R 2 2 2R 2R (4)

=− or,

a2 3 a2 2π − π 2 + π 2 2 R R

Φ ( R) ≅ −

GM 1 a 2 1+ R 4 R2

(5)

We notice that the first term in (5) is the potential when mass M is concentrated in the center of GMa 2 . the ring. Of course this is a very rough approximation and the first correction term is − 4 R3

5-10.

P φ θ a x R r R cos θ

R sin θ

Using the relations

x=

( R sin θ ) 2 + a2 − 2aR sin θ cos φ

(1) (2) (3)

r = x 2 + R2 cos 2 θ = R 2 + a 2 − 2 aR sin θ cos φ

ρ = the potential is expressed by

Φ = −G ∫

M (the linear mass density), 2π a

−GM 2π R

2π

ρd r =

∫

0

dφ a a2 1 − 2 sin θ cos φ − 2 R R

3

(4)

If we expand the integrand and neglect terms of order ( a R) and higher, we have a a2 1 − 2 sin θ cos φ − 2 R R Then, (4) becomes

−1 2

≅ 1+

a 1 a2 3 a2 sin θ cos φ − sin 2 θ cos 2 φ + R 2 R2 2 R2

(5)

158

CHAPTER 5

Φ≅− Thus,

GM 1 a2 3 a2 2π − 2π + π sin 2 θ 2 2 2π R 2R 2R

Φ ( R) ≅ −

GM R

1 a2 3 2 1 − 2 R 2 1 − 2 sin θ

(6)

5-11.

P r dm z a θ

The potential at P due to a small mass element dm inside the body is dΦ = −G dm dm = −G 2 2 r z + a − 2 za cos θ (1)

Integrating (1) over the entire volume and dividing the result by the surface area of the sphere, we can find the average field on the surface of the sphere due to dm: dΦ ave = 1 4π a 2 π 2π a 2 sin θ dθ −G dm ∫ z 2 + a 2 − 2za cos θ 0

(2)

Making the variable change cos θ = x, we have dΦ ave Using Eq. (E.5), Appendix E, we find G = − dm ∫ 2 −1

+1

(z

)

dx

2

+ a 2 − 2 zax

)

(3)

dΦ ave = − =− =−

G 1 dm − 2 za

(z

2

+ a 2 − 2za +

1 za

(z

2

+ a 2 + 2 za

)

G − ( z − a) + ( z + a) dm 2 za G dm z (4)

This is the same potential as at the center of the sphere. Since the average value of the potential is equal to the value at the center of the sphere at any arbitrary element dm, we have the same relation even if we integrate over the entire body.

GRAVITATION

159

5-12.

dm

r'

O

r

R

P

Let P be a point on the spherical surface. The potential dΦ due to a small amount of mass dm inside the surface at P is dΦ = − Gdm r (1)

The average value over the entire surface due to dm is the integral of (1) over dΩ divided by 4π. Writing this out with the help of the figure, we have dΦ ave = − 2π sin θ dθ Gdm π ∫0 r ′ 2 + R2 − 2r ′R cos θ 4π (2)

Making the obvious change of variable and performing the integration, we obtain dΦ ave = − Gdm 1 du Gdm ∫−1 r ′ 2 + R2 − 2r ′Ru = − R 4π (3)

We can now integrate over all of the mass and get Φ ave = − Gm R . This is a mathematical statement equivalent to the problem’s assertion.

5-13.

R1 R0 R2 ρ1 ρ2

R0 = position of particle. For R1 < R0 < R2 , we calculate the force by assuming that all mass for which r < R0 is at r = 0, and neglect mass for which r > R0 . The force is in the radially inward direction ( −er ). The magnitude of the force is F= where M = mass for which r < R0 GMm 2 R0

160

CHAPTER 5

M= So F = − 4π Gm 3 3 3 ρ1R1 + ρ2 R0 − ρ2 R1 er 2 3R0

4 4 3 3 3 π R1 ρ1 + π R0 − R1 ρ2 3 3

(

)

(

)

( ρ − ρ ) R3 4 F = − π Gm 1 22 1 + ρ2 R0 er 3 R0

5-14.

Think of assembling the sphere a shell at a time (r = 0 to r = R).

For a shell of radius r, the incremental energy is dU = dm φ where φ is the potential due to the mass already assembled, and dm is the mass of the shell. So 3 Mr 2 dr 3M dm = ρ 4π r 2 dr = 4π r 2 dr = 3 R3 4π R

φ=−

So

Gm r3 where m = M 3 r R

U = ∫ du = 3 Mr 2 dr GMr 2 ∫ R3 − R3 r =0

R

=−

3GM 2 4 r dr R6 ∫ 0 U=− 3 GM 2 5 R

R

5-15. When the mass is at a distance r from the center of the Earth, the force is in the inward radial direction and has magnitude Fr : m r

Fr =

Gm 4 3 π r ρ where ρ is the mass density of the Earth. The equation of motion is r2 3

GRAVITATION

161

Fr = mr = − or

Gm 4 3 π r ρ r2 3 4π Gρ 3

r + ω 2 r = 0 where ω 2 =

This is the equation for simple harmonic motion. The period is T= 2π

ω

=

3π Gρ

Substituting in values gives a period of about 84 minutes.

5-16.

z y x h r M θ r 2 + h2

For points external to the sphere, we may consider the sphere to be a point mass of mass M. Put the sheet in the x-y plane. Consider force on M due to the sheet. By symmetry, Fx = Fy = 0 Fz = ∫ dFz = With dm = ρs 2π rdr and cos θ = we have Fz = 2πρsGMh h r + h2

2

∞

r =0

∫ (r

∞

GMdm cos θ 2 + h2

)

r =0

∫

(r

rdr

2

+ h2

)

32

1 Fz = −2πρsGMh 2 r + h2

(

)

12 0

∞

Fz = 2πρsGM The sphere attracts the sheet in the z -direction with a force of magnitude 2πρsGM

162

5-17.

y

CHAPTER 5

water

x

moon (not to scale) Earth

Start with the hint given to us. The expression for g x and g y are given by gx = GMm y GMe y 2GMm x GMe x − − ; gy = − 3 3 D3 R3 D R (1)

where the first terms come from Equations (5.54) and the second terms come from the standard assumption of an Earth of uniform density. The origin of the coordinate system is at the center of the Earth. Evaluating the integrals:

∫

xmax 0

2 2GMm GMe xmax ; g x dx = − 3 3 R 2 D

∫

ymax 0

2 GMm GMe ymax − 3 g y dy = − 3 R 2 D

(2)

To connect this result with Example 5.5, let us write (1) in the following way

2 ymax GMe 2 GMm 2 2 xmax − ymax xmax + = 3 3 2 2R D

(

)

(3)

The right-hand side can be factored as GMe GMe x + ymax )( xmax − ymax ) = ( 2 R ) ( h ) = gh 3 ( max 2R 2R 3

2 If we make the approximation on the left-hand side of (3) that xmax Equation (5.55). Turning to the exact solution of (3), we obtain 2 ymax

(4) R 2 , we get exactly

h = 2R

Me M m Me 2 M m + 3 − − 3 R D R3 D3 Me Mm Me 2 M m + 3 + − 3 R D R3 D3

(5)

Upon substitution of the proper values, the answer is 0.54 m, the same as for Example 5.5. Inclusion of the centrifugal term in g x does not change this answer significantly.

GRAVITATION

163

5-18.

From Equation (5.55), we have with the appropriate substitutions

hmoon hsun Substitution of the known values gives

3GMm r 2 3 2 gD3 M R = = m es 3GMs r 2 Ms D 3 2 gRes

(1)

hmoon 7.350 × 10 22 kg 1.495 × 1011 m = hsun 1.993 × 10 30 kg 3.84 × 10 8 m

3

2. 2

(2)

5-19.

ωearth

ωmoon

Because the moon’s orbit about the Earth is in the same sense as the Earth’s rotation, the difference of their frequencies will be half the observed frequency at which we see high tides. Thus 1 1 1 = − 2Ttides Tearth Tmoon which gives Ttides 12 hours, 27 minutes. (1)

5-20.

The differential potential created by a thin loop of thickness dr at the point (0,0,z) is 2π rdrM −GM d ( r 2 ) −2GM ( z 2 + R2 − z ) = ⇒ Φ( z) = ∫ dΦ( z) = 2 2 2 2 2 2 R R2 πR z +r z +r −G

dΦ( z) =

Then one can find the gravity acceleration,

2 2 ˆ dΦ = − k 2GM z + R − z ˆ g( z ) = − k dz R2 z 2 + R2

ˆ where k is the unit vector in the z-direction.

164

5-21.

CHAPTER 5

(We assume the convention that D > 0 means m is not sitting on the rod.)

The differential force dF acting on point mass m from the element of thickness dx of the rod, which is situated at a distance x from m, is dF = G ( M L) mdx x

2

GMm ⇒ F = ∫ dF = L

L+ D

D

∫

dx GMm = 2 x D(L + D)

And that is the total gravitational force acting on m by the rod.

CHAPTER

6

(1)

Some Methods in the Calculus of Variations

6-1. If we use the varied function

y (α , x ) = x + α sin π (1 − x ) Then

dy = 1 − απ cos π (1 − x ) dx Thus, the total length of the path is

(2)

S=∫

0 1

1

dy 1 + dx dx

12

2

= ∫ 2 − 2απ cos π (1 − x ) + α 2π 2 cos 2 π (1 − x )

0

dx

(3)

Setting π (1 − x ) ≡ u , the expression for S becomes

S= 1

π

∫

0

π

1 2 1 − απ cos u + α 2π 2 cos 2 2

u

12

du

(4)

The integral cannot be performed directly since it is, in fact, an elliptic integral. Because α is a small quantity, we can expand the integrand and obtain S= 2

π

∫

0

π

2 1 1 2 2 1 1 2 2 2 2 1 − απ cos u − α π cos u − απ cos u − α π cos u + … du 8 2 2 2

(5)

If we keep the terms up to cos 2 u and perform the integration, we find

S= 2+ which gives

2 2 2 π α 16

(6)

165

166

CHAPTER 6

∂S 2 2 π α = ∂α 8 Therefore

∂S ∂α and S is a minimum when α = 0.

6-2.

(7)

=0 a=0 (8)

The element of length on a plane is dS = dx 2 + dy 2 (1)

from which the total length is

( x2 , y2 )

S=

( x1 , y1 )

∫

dx 2 + dy 2 =

x2

x1

∫

dy 1 + dx dx

2

(2)

If S is to be minimum, f is identified as dy f = 1+ dx Then, the Euler equation becomes d d 1 + y′2 = 0 dx dy ′ where y ′ = dy . (4) becomes dx d y′ dx 1 + y ′ 2 or, y′ 1 + y′2 from which we have = constant ≡ C (6) =0 (5) (4)

2

(3)

y′ = Then,

C2 = constant ≡ a 1 − C2

(7)

y = ax + b This is the equation of a straight line.

(8)

SOME METHODS IN THE CALCULUS OF VARIATIONS

167

6-3.

The element of distance in three-dimensional space is dS = dx 2 + dy 2 + dz 2 (1)

Suppose x, y, z depends on the parameter t and that the end points are expressed by ( x1 (t1 ) , y1 (t1 ) , z1 (t1 )) , ( x2 (t2 ) , y2 (t2 ) , z2 (t2 )) . Then the total distance is

S=∫

t1 t2

dx dy dz dt + dt + dt dt

2

2

2

(2)

The function f is identified as f = x2 + y2 + z2 Since (3)

∂ f ∂f ∂ f = = = 0 , the Euler equations become ∂x ∂y ∂z

d ∂f =0 dt ∂x d ∂f =0 dt ∂y d ∂f =0 dt ∂z

(4)

from which we have

= constant ≡ C1 x2 + y2 + z2 y = constant ≡ C2 x2 + y2 + z2 z = constant ≡ C3 x2 + y2 + z2 x From the combination of these equations, we have

(5)

y x = C1 C2 y z = C2 C3

If we integrate (6) from t1 to the arbitrary t, we have

(6)

168

CHAPTER 6

x − x1 y − y1 = C1 C2 y − y1 z − z1 = C2 C3

On the other hand, the integration of (6) from t1 to t2 gives

(7)

x2 − x1 y2 − y1 = C1 C2 y2 − y1 z2 − z1 = C2 C3

(8)

from which we find the constants C1 , C2 , and C3 . Substituting these constants into (7), we find

y − y1 x − x1 z − z1 = = x2 − x1 y2 − y1 z2 − z1

This is the equation expressing a straight line in three-dimensional space passing through the two points ( x1 , y1 , z1 ) , ( x2 , y2 , z2 ) .

(9)

6-4. z 1 dS y φ

ρ 2 x

The element of distance along the surface is dS = dx 2 + dy 2 + dz 2 In cylindrical coordinates (x,y,z) are related to (ρ,φ,z) by x = ρ cos φ y = ρ sin φ z=z from which dx = − ρ sin φ dφ dy = ρ cos φ dφ dz = dz

(1)

(2)

(3)

SOME METHODS IN THE CALCULUS OF VARIATIONS

169

Substituting (3) into (1) and integrating along the entire path, we find S = ∫ ρ dφ + dz =

2 2 2 1 2

φ2 φ1

∫

ρ 2 + z 2 dφ

(4)

where z =

dz . If S is to be minimum, f ≡ ρ 2 + z 2 must satisfy the Euler equation: dφ ∂f ∂ ∂f − =0 ∂ z ∂φ ∂ z (5)

Since

∂f = 0 , the Euler equation becomes ∂z ∂ ∂φ z

ρ + z2

2

=0

(6)

from which z

ρ + z2

2

= constant ≡ C

(7)

or, z= Since ρ is constant, (8) means dz = constant dφ and for any point along the path, z and φ change at the same rate. The curve described by this condition is a helix.

6-5.

z

(x2,y2) ds (x1,y1) y

C2 ρ 1 − C2

(8)

x

The area of a strip of a surface of revolution is dA = 2π × ds = 2π × dx 2 + dy 2 Thus, the total area is (1)

170 x2 CHAPTER 6

A = 2π where y =

x1

∫x

1 + y 2 dx

(2)

dy . In order to make A a minimum, f ≡ x 1 + y 2 must satisfy equation (6.39). Now dx ∂f = 1 + y2 ∂x xy ∂f = ∂y 1 + y2

Substituting into equation (6.39) gives

1 + y2 = xy 2 d x 1 + y2 − dx 1 + y2 d x = dx 1 + y 2

= Multiplying by

1 + y 2 − xy ( dy dx ) 1 + y 2 1 + y2

(

)

−1 2

1 + y 2 and rearranging gives − dy dx = x y 1 + y2

(

)

Integration gives − ln x + ln a = y2 1 ln 2 1 + y2

where ln a is a constant of integration. Rearranging gives y2 = Integrating gives y = b + a cosh −1 or x = a cosh which is the equation of a catenary. y−b a x a

(

1 x a2 − 1

2

)

SOME METHODS IN THE CALCULUS OF VARIATIONS

171

6-6.

0 θ=π y

θ=0 2a (x2,y2) x

(x1,y1)

If we use coordinates with the same orientation as in Example 6.2 and if we place the minimum point of the cycloid at (2a,0) the parametric equations are x = a (1 + cos θ ) y = a (θ + sin θ ) Since the particle starts from rest at the point ( x1 , y1 ) , the velocity at any elevation x is [cf. Eq. 6.19] v = 2 g ( x − x1 ) Then, the time required to reach the point ( x2 , y2 ) is [cf. Eq. 6.20] t= x2 (1)

(2)

x1

∫

1 + y′2 2 g ( x − x1 )

12

dx

(3)

Using (1) and the derivatives obtained therefrom, (3) can be written as t= a g 1 + cos θ ∫ cos θ − cosθ1 θ2 = 0 θ1 12

dθ

(4)

Now, using the trigonometric identity, 1 + cos θ = 2 cos 2 θ 2 , we have a g θ1 t=

∫

0

cos cos 2

θ

2

dθ

θ

2

− cos 2

θ1

2

=

a g

θ1

∫

0

cos sin

2

θ

2

dθ

2

θ1

2

− sin

θ

2

(5)

Making the change of variable, z = sin θ 2 , the expression for t becomes sin θ1

2

t=2

a g

∫

0

dz sin 2

θ1

2

(6) − z2

The integral is now in standard form:

172

CHAPTER 6

∫

Evaluating, we find

x = sin −1 a a =x dx

2 2

(7)

t=π

a g

(8)

Thus, the time of transit from ( x1 , y1 ) to the minimum point does not depend on the position of the starting point.

6-7.

a θ1 θ1 n1 n2 b θ2 (n2 > n1) c n1 c v2 = n2 v1 =

x d

The time to travel the path shown is (cf. Example 6.2) 1 + y′2 ds t=∫ =∫ dx v v Although we have v = v(y), we only have dv dy ≠ 0 when y = 0. The Euler equation tells us y′ d dx v 1 + y ′ 2 Now use v = c n and y′ = –tan θ to obtain n sin θ = const. This proves the assertion. Alternatively, Fermat’s principle can be proven by the method introduced in the solution of Problem 6-8.

6-8.

(1)

=0

(2)

(3)

To find the extremum of the following integral (cf. Equation 6.1)

J = ∫ f ( y , x ) dx we know that we must have from Euler’s equation ∂f =0 ∂y This implies that we also have

SOME METHODS IN THE CALCULUS OF VARIATIONS

173

∂f ∂J =∫ dx = 0 ∂y ∂y giving us a modified form of Euler’s equation. This may be extended to several variables and to include the imposition of auxiliary conditions similar to the derivation in Sections 6.5 and 6.6. The result is ∂g j ∂J + ∑ λ j ( x) =0 ∂y i ∂y i j when there are constraint equations of the form g j ( yi , x ) = 0

a) The volume of a parallelepiped with sides of lengths a1 , b1 , c1 is given by

V = a1b1c1 We wish to maximize such a volume under the condition that the parallelepiped is circumscribed by a sphere of radius R; that is,

2 2 2 a1 + b1 + c1 = 4 R 2

(1)

(2)

We consider a1 , b1 , c1 as variables and V is the function that we want to maximize; (2) is the constraint condition: g { a1 , b1 , c1 } = 0 Then, the equations for the solution are

∂g ∂V +λ =0 ∂a1 ∂a1 ∂g ∂V +λ =0 ∂b1 ∂b1 ∂g ∂V +λ =0 ∂c1 ∂c1 from which we obtain b1c1 + 2λ a1 = 0 a1c1 + 2λ b1 = 0 a1b1 + 2λ c1 = 0 Together with (2), these equations yield a1 = b1 = c1 = 2 R 3 2 R. 3 (6)

(3)

(4)

(5)

Thus, the inscribed parallelepiped is a cube with side

174

CHAPTER 6

b) In the same way, if the parallelepiped is now circumscribed by an ellipsoid with semiaxes a, b, c, the constraint condition is given by

2 a1 b2 c2 = 12 = 1 2 = 1 4 a 2 4b 4z

(7)

where a1 , b1 , c1 are the lengths of the sides of the parallelepiped. Combining (7) with (1) and (4) gives

2 2 2 a1 b1 c1 = 2 = 2 a2 b c

(8)

Then, a1 = a 2 2 2 , b1 = b , c1 = c 3 3 3 (9)

6-9.

The average value of the square of the gradient of φ ( x1 , x2 , x3 ) within a certain volume V

is expressed as

I=

1 V

∫∫∫ ( ∇φ ) ∫∫∫

2

dx1 dx2 dx3

1 v= V In order to make I a minimum,

∂φ 2 ∂ φ 2 ∂ φ 2 dx1dx2 dx3 + + ∂x ∂x ∂x 1 2 3

(1)

∂φ ∂φ ∂φ f = + + ∂x1 ∂x2 ∂x3 must satisfy the Euler equation:

3 ∂f ∂ ∂f −∑ =0 ∂ φ i = 1 ∂x i ∂ φ ∂ ∂x i

2

2

2

(2)

If we substitute f into (2), we have

∑ ∂x i =1

3

∂ ∂φ =0 i ∂x i

(3)

which is just Laplace’sequation:

∇2 φ = 0 Therefore, φ must satisfy Laplace’s equation in order that I have a minimum value. (4)

SOME METHODS IN THE CALCULUS OF VARIATIONS

175

6-10. This problem lends itself to the method of solution suggested in the solution of Problem 6-8. The volume of a right cylinder is given by

V = π R2 H

The total surface area A of the cylinder is given by

(1)

A = Abases + Aside = 2π R 2 + 2π RH = 2π R ( R + H )

We wish A to be a minimum. (1) is the constraint condition, and the other equations are

(2)

∂g ∂A +λ =0 ∂R ∂R ∂g ∂A +λ =0 ∂H ∂H where g = V − π R 2 H = 0 . The solution of these equations is R= 1 H 2

(3)

(4)

6-11. y 1 2a R θ ds

}

The constraint condition can be found from the relation ds = Rdθ (see the diagram), where ds is the differential arc length of the path: ds = dx 2 + dy 2 which, using y = ax 2 , yields 1 + 4a 2 x 2 dx = Rdθ If we want the equation of constraint in other than a differential form, (2) can be integrated to yield A + Rθ = x 2 4 ax 2 + 1 + 1 ln 2ax + 4 a 2 x 2 + 1 4a (2)

(

)

12

= Rdθ

(1)

(

)

(3)

where A is a constant obtained from the initial conditions. The radius of curvature of a parabola, y = ax 2 , is given at any point (x,y) by r0 ≥ 1 2a . The condition for the disk to roll with one and only one point of contact with the parabola is R < r0 ; that is,

176

CHAPTER 6

R<

1 2a

(4)

6-12.

The path length is given by s = ∫ ds = ∫ 1 + y ′ 2 + z ′ 2 dx

(1)

and our equation of constraint is g ( x, y, z) = x2 + y 2 + z2 − ρ2 = 0

(2)

The Euler equations with undetermined multipliers (6.69) tell us that y′ d dx 1 + y ′ 2 + z ′ 2

dg =λ = 2λ y dy

(3)

with a similar equation for z. Eliminating the factor λ, we obtain y′ 1 d y dx 1 + y ′ 2 + z ′ 2

1 d z′ − z dx 1 + y ′ 2 + z ′ 2

=0

(4)

This simplifies to z y ′′ 1 + y ′ 2 + z ′ 2 − y ′ ( y ′y ′′ + z ′z ′′ ) − y z ′′ 1 + y ′ 2 + z ′ 2 − z ′ ( y ′y ′′ + z ′z ′′ ) = 0 zy ′′ + ( yy ′ + zz ′ ) z ′y ′′ − yz ′′ − ( yy ′ + zz ′ ) y ′z ′′ = 0

(

)

(

)

(5) (6)

and using the derivative of (2),

( z − xz ′ ) y ′′ = ( y − xy ′ ) z ′′

This looks to be in the simplest form we can make it, but is it a plane? Take the equation of a plane passing through the origin:

Ax + By = z

(7)

(8)

and make it a differential equation by taking derivatives (giving A + By′ = z′ and By″ = z″) and eliminating the constants. The substitution yields (7) exactly. This confirms that the path must be the intersection of the sphere with a plane passing through the origin, as required.

6-13. For the reason of convenience, without lost of generality, suppose that the closed curve passes through fixed points A(-a,0) and B(a,0) (which have been chosen to be on axis Ox). We denote the part of the closed curve above and below the Ox axis as y1 ( x ) and y2 ( x ) respectively. (note that y1 > 0 and y2 < 0 )

The enclosed area is

J ( y1 , y2 ) =

−a

∫

a

y1 ( x)dx −

−a

∫

a

y2 ( x)dx =

−a

∫ ( y1 (x) − y2 (x)) dx =

a

−a

∫

a

f ( y1 , y2 )dx

SOME METHODS IN THE CALCULUS OF VARIATIONS

177

The total length of closed curve is

K ( y1 , y2 ) = ′ ′

−a

∫

a

1 + ( y1 ) dx + ′

2

−a

∫

a

1 + ( y2 ) dx = ′

2

−a

∫{ a 1 + ( y1 ) + 1 + ( y1 ) ′ ′

2

2

}

dx =

−a

∫ g ( y1′ , y2′ ) dx

a

Then the generalized versions of Eq. (6.78) (see textbook) for this case are ∂g y1 ∂f ′ d ∂f d ∂g d − +λ − =0 = 0 ⇒ 1− λ 2 ∂y1 dx ∂y1 dx 1 + ( y1 ) ′ ′ ′ ∂y1 dx ∂y1 ∂g y2 ∂f ′ d ∂f d ∂g d − +λ − =0 = 0 ⇒ 1− λ 2 ∂y2 dx ∂y2 dx 1 + ( y2 ) ′ ′ ′ ∂y2 dx ∂y2 Analogously to Eq. (6.85); from (1) we obtain from (2) we obtain (1)

(2)

( x − A1 ) 2 + ( y1 − A2 ) 2 = λ 2 ( x − B1 ) 2 + ( y2 − B2 ) 2 = λ 2

(3) (4)

where constants A’s, B’s can be determined from 4 initial conditions

( x = ± a, y1 = 0)

and

( x = ± a, y2 = 0)

We note that y1 < 0 and y2 > 0 , so actually (3) and (4) altogether describe a circular path of radius λ . And this is the sought configuration that renders maximum enclosed area for a given path length.

6-14. It is more convenient to work with cylindrical coordinates (r, φ ,z) in this problem. The constraint here is z = 1 – r , then dz = –dr

ds2 = dr 2 + r 2 dφ 2 + dz 2 = 2 ( dr 2 + r 2 dβ 2 )

where we have introduced a new angular coordinate β =

φ

2

In this form of ds2 , we clearly see that the space is 2-dimensional Euclidean flat, so the shortest line connecting two given points is a straight line given by: r= cos ( β − β 0 )

r0

=

φ − φ0 cos 2

r0

this line passes through the endpoints (r = 1, φ = ± the shortest path equation

π

2

) , then we can determine unambiguously

r(φ ) =

2 2 φ cos 2

cos

π and z = 1 – r

178

CHAPTER 6

Accordingly, the shortest connecting length is l=

π dr 2 ∫ dφ 2 dφ + r = 2 2 sin 2 2 −π 2

2

π 2

6-15.

dy 2 I[ y] = ∫ − y 2 dx dx 0

1

a) Treating I[y] as a mechanical action, we find the corresponding Euler-Lagrange equation

y( x ) = −

d2 y dx 2

Combining with the boundary conditions (x = 0, y = 0) and (x = 1, y = 1), we can determine unambiguously the functional form of y( x ) = (sin x) (sin 1) .

b) The corresponding minimum value of the integral is

1 dy 2 1 I[ y] = ∫ − y 2 dx = dx cos 2 x = cot (1) = 0.642 sin 2 1 ∫ 0 0 dx 1

c) If x = y then I[y] = ( 2 3 ) = 0.667.

6-16. a) S is arc length

9 dy dz dy S = ∫ dx 2 + dy 2 + dz 2 = ∫ dx 1 + + = ∫ dx 1 + + x = ∫ L dx dx dx dx 4 Treating S and L like a mechanical action and Lagrangian respectively, we find the canonical momentum associated with coordinate y p= 2

2

2

δL

dy δ dx

= 1+

dy dx 9 dy x+ dx 4

2

Because L does not depend on y explicitly, then E-L equation implies that p is constant (i.e. dp dx = 0 ), then the above equation becomes

9 32 1+ x dy 9 9 4 ⇒y = p =p +B dx 1 + x = A 1 + x dx 1 − p2 1 − p2 ∫ 4 4 where A and B are constants. Using boundary conditions (x = 0, y = 0) and (x = 1, y = 1) one can determine the arc equation unambiguously

SOME METHODS IN THE CALCULUS OF VARIATIONS

179

32 9 y( x ) = 3 2 1 + x − 1 13 − 8 4

8

and

z = x3 2

b) z 1 0.75 0.5 0.25 y 0 1 0.75 0.5 0.25 x 1 0. 0.75 0.5 0.25 00

6-17. a) Equation of a ellipse

x2 y2 + =1 a2 b2 which implies xy ≤ ab 2 because 2xy x 2 y 2 ≤ 2 + 2 ab a b

so the maximal area of the rectangle, whose corners lie on that ellipse, is Max[A] = Max[4xy] = 2 iab. This happens when x= a 2 and y= b 2

b) The area of the ellipse is A0 = π ab ; so the fraction of rectangle area to ellipse area is then

Max[ A] 2 = A0 π

6-18. One can see that the surface xy = z is “locally” symmetric with respect to the line x = − y = −z where x > 0, y < 0, z < 0. This line is a parabola. This implies that if the particle starts from point (1,-1,-1) (which belongs to the symmetry line) under gravity ideally will move downward along this line. Its velocity at altitude z (z < –1) can be found from the conservation of energy.

v( z) = −2 g( z + 1)

180

CHAPTER 6

CHAPTER

7

Hamilton’s Principle— Lagrangian and Hamiltonian Dynamics

7-1. Four coordinates are necessary to completely describe the disk. These are the x and y coordinates, the angle θ that measures the rolling, and the angle φ that describes the spinning (see figure). φ θ y

x

Since the disk may only roll in one direction, we must have the following conditions:

dx cos φ + dy sin φ = R dθ

(1)

dy = tan φ dx

(2)

These equations are not integrable, and because we cannot obtain an equation relating the coordinates, the constraints are nonholonomic. This means that although the constraints relate the infinitesimal displacements, they do not dictate the relations between the coordinates themselves, e.g. the values of x and y (position) in no way determine θ or φ (pitch and yaw), and vice versa.

7-2.

Start with the Lagrangian L= = m v + at + θ cos θ 0 2

(

) + ( θ sin θ ) + mg

2 2

cos θ

(1) (2)

m ( v0 + at ) 2 + 2 ( v0 + at ) θ cos θ + 2θ 2 + mg cos θ 2

181

182

CHAPTER 7

Now let us just compute

d ∂L d 2 = m ( v0 + at ) cos θ + m θ dt ∂θ dt

(3)

= ma cos θ − m ( v0 − at ) θ sin θ + m 2θ ∂L = − m ( v0 + at ) θ sin θ − mg sin θ ∂θ According to Lagrange’s equations, (4) is equal to (5). This gives Equation (7.36)

(4) (5)

θ=

g

sin θ +

a

cos θ = 0

(6)

To get Equation (7.41), start with Equation (7.40)

η=− and use Equation (7.38)

g cos θ e − a sin θ e

η

(7)

tan θ e = −

a g

(8)

to obtain, either through a trigonometric identity or a figure such as the one shown here, g 2 + a2 θe

g

a

cos θ e = Inserting this into (7), we obtain

g g +a

2 2

sin θ e =

a g + a2

2

(9)

η=− as desired.

a2 + g 2

η

(10)

We know intuitively that the period of the pendulum cannot depend on whether the train is accelerating to the left or to the right, which implies that the sign of a cannot affect the frequency. From a Newtonian point of view, the pendulum will be in equilibrium when it is in line with the effective acceleration. Since the acceleration is sideways and gravity is down, and the period can only depend on the magnitude of the effective acceleration, the correct form is clearly

a2 + g 2 .

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

183

7-3. θ R ρ φ θ

If we take angles θ and φ as our generalized coordinates, the kinetic energy and the potential energy of the system are

T=

2 1 1 m ( R − ρ ) θ + I φ 2 2 2

(1)

(2)

U = R − ( R − ρ ) cos θ mg

where m is the mass of the sphere and where U = 0 at the lowest position of the sphere. I is the moment of inertia of sphere with respect to any diameter. Since I = ( 2 5) mρ 2 , the Lagrangian

becomes L = T −U = 1 1 2 m ( R − ρ ) θ 2 + mρ 2φ 2 − R − ( R − ρ ) cos θ mg 2 5 (3)

When the sphere is at its lowest position, the points A and B coincide. The condition A0 = B0 gives the equation of constraint: f (θ , φ ) = ( R − ρ ) θ − ρφ = 0 Therefore, we have two Lagrange’s equations with one undetermined multiplier:

∂L d − ∂θ dt ∂f ∂L ∂θ + λ ∂ θ = 0 ∂f ∂ L d ∂L − +λ =0 ∂φ dt ∂φ ∂φ

(4)

(5)

After substituting (3) and ∂f ∂θ = R − ρ and ∂f ∂φ = − ρ into (5), we find − ( R − ρ ) mg sin θ − m ( R − ρ ) θ + λ ( R − ρ ) = 0

2

(6) (7)

− From (7) we find λ:

2 mρ 2φ − λρ = 0 5

λ = − mρφ or, if we use (4), we have

2 5

(8)

λ = − m ( R − ρ) θ

Substituting (9) into (6), we find the equation of motion with respect to θ :

2 5

(9)

184

CHAPTER 7

θ = −ω 2 sin θ where ω is the frequency of small oscillations, defined by

(10)

ω=

5g 7 ( R − ρ)

(11)

7-4. y m r θ x

If we choose (r,θ ) as the generalized coordinates, the kinetic energy of the particle is T= 1 1 m x 2 + y 2 = m r 2 + r 2θ 2 2 2 ∂U ∂r

(

)

(

)

(1)

Since the force is related to the potential by f =− we find U= A (2)

α

rα

(3)

where we let U(r = 0) = 0. Therefore, the Lagrangian becomes L= 1 A m r 2 + r 2θ 2 − r α 2 α

(

)

(4)

Lagrange’s equation for the coordinate r leads to mr − mrθ 2 + Ar α −1 = 0 Lagrange’s equation for the coordinate θ leads to d mr 2θ = 0 dt Since mr 2θ = is identified as the angular momentum, (6) implies that angular momentum is conserved. Now, if we use , we can write (5) as mr − Multiplying (7) by r , we have mrr − r 2 + Ar r α −1 = 0 mr 3 (8)

2 3

(5)

(

)

(6)

mr

+ Ar α −1 = 0

(7)

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

185

which is equivalent to

2 d A α d 1 d mr 2 + r =0 + 2 dt 2 dt 2mr dt α

(9)

Therefore, d (T + U ) = 0 dt

(10)

and the total energy is conserved.

7-5.

y r m φ x

Let us choose the coordinate system so that the x-y plane lies on the vertical plane in a gravitational field and let the gravitational potential be zero along the x axis. Then the kinetic energy and the potential energy are expressed in terms of the generalized coordinates (r,φ) as

T= U=

1 m r 2 + r 2φ 2 2

(

)

(1) (2)

A

α

r α + mgr sin φ

from which the Lagrangian is

L = T −U =

1 A m r 2 + r 2φ 2 − r α − mgr sin φ α 2

(

)

(3)

Therefore, Lagrange’s equation for the coordinate r is mr − mrφ 2 + Ar α −1 + mg sin φ = 0 Lagrange’s equation for the coordinate φ is d mr 2φ + mgr cos φ = 0 dt (4)

(

)

(5)

Since mr 2φ is the angular momentum along the z axis, (5) shows that the angular momentum is not conserved. The reason, of course, is that the particle is subject to a torque due to the gravitational force.

186 7-6. y M S α ξ φ m α x CHAPTER 7

Let us choose ξ,S as our generalized coordinates. The x,y coordinates of the center of the hoop are expressed by x = ξ + S cos α + r sin α y = r cos α + ( Therefore, the kinetic energy of the hoop is − S) sin α (1)

Thoop = =

1 1 m x 2 + y 2 + Iφ 2 2 2 1 m ξ + S cos α 2

(

)

(

) + ( −S sin α ) + 1 Iφ 2

2 2

2

(2)

Using I = mr 2 and S = rφ , (2) becomes Thoop = 1 m 2S 2 + ξ 2 + 2ξS cos α 2 (3)

In order to find the total kinetic energy, we need to add the kinetic energy of the translational motion of the plane along the x-axis which is Tplane = Therefore, the total kinetic energy becomes T = mS2 + The potential energy is U = mgy = mg r cos α + ( − S) sin α Hence, the Lagrangian is l = mS 2 + 1 ( m + M ) ξ 2 + mξS cos α − mg r cos α + ( − S) sin α 2 (7) (6) 1 ( m + M ) ξ 2 + mξS cos α 2 (5) 1 Mξ2 2 (4)

from which the Lagrange equations for ξ and S are easily found to be 2mS + mξ cos α − mg sin α = 0 (8) (9)

( m + M ) ξ + mS cos α = 0

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

187

or, if we rewrite these equations in the form of uncoupled equations by substituting for ξ and

S , we have

m cos 2 α 2 − S − g sin α = 0 m+ M mg sin α cos α ξ=− 2 2 ( m + M ) − m cos α Now, we can rewrite (9) as d ( m + M ) ξ + mS cos α = 0 dt where we can interpret ( m + M ) ξ as the x component of the linear momentum of the total system and mS cos α as the x component of the linear momentum of the hoop with respect to the plane. Therefore, (11) means that the x component of the total linear momentum is a constant of motion. This is the expected result because no external force is applied along the x-axis.

7-7.

y1 φ1 x1 m φ2 x2 m x y2 y

(10)

(11)

If we take (φ1 , φ 2 ) as our generalized coordinates, the x,y coordinates of the two masses are x1 = cos φ1 y1 = sin φ1 x2 = cos φ1 + cos φ 2 y2 = sin φ1 + sin φ2 Using (1) and (2), we find the kinetic energy of the system to be T= = = m 2 m 2 2 2 x1 + y1 + x2 + y2 2 2 m 2 m 2

2

(1)

(2)

(

)

(

)

2 2 2 φ1 + φ1 + φ 2 + 2φ1φ2 ( sin φ1 sin φ 2 + cos φ1 cos φ 2 ) 2 2 2φ1 + φ2 + 2φ1φ 2 cos (φ1 − φ 2 )

2

(3)

188

CHAPTER 7

The potential energy is U = − mgx1 − mgx2 = − mg 2 cos φ1 + cos φ2 Therefore, the Lagrangian is (4)

2 1 2 L = m 2 φ1 + φ2 + φ1φ2 cos (φ1 − φ2 ) + mg 2 cos φ1 + cos φ 2 2 from which

∂L = m 2φ1φ 2 sin (φ1 − φ 2 ) − 2mg sin φ1 ∂φ1 ∂L 2 2 = 2m φ1 + m φ 2 cos (φ1 − φ 2 ) ∂φ1 ∂L = − m 2φ1φ 2 sin (φ1 − φ2 ) − mg sin φ 2 ∂φ 2 ∂L 2 2 = m φ2 + m φ1 cos (φ1 − φ2 ) ∂φ 2 The Lagrange equations for φ1 and φ 2 are

2 2φ1 + φ 2 cos (φ1 − φ 2 ) + φ 2 sin (φ1 − φ 2 ) + 2

(5)

(6)

g

sin φ1 = 0

(7)

2 φ2 + φ1 cos (φ1 − φ 2 ) − φ1 sin (φ1 − φ2 ) +

g

sin φ 2 = 0

(8)

7-8.

U1 y v2 θ2 θ1 v1 U2 x

Let us choose the x,y coordinates so that the two regions are divided by the y axis: U1 U ( x) = U2 x0

If we consider the potential energy as a function of x as above, the Lagrangian of the particle is L= 1 m x2 + y2 − U ( x) 2

(

)

(1)

Therefore, Lagrange’s equations for the coordinates x and y are

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

189

mx +

dU ( x ) =0 dx

(2) (3)

my = 0 Using the relation mx = dP dP dx Px dpx d mx = x = x = dt dt dx dt m dx Px dPx dU ( x ) + =0 m dx dx

(4)

(2) becomes (5)

Integrating (5) from any point in the region 1 to any point in the region 2, we find

∫

1

2

dU ( x ) Px dPx dx + ∫ dx = 0 m dx dx 1

2

(6)

Px22

2m or, equivalently,

−

Px21

2m

+ U 2 − U1 = 0

(7)

1 1 2 2 mx1 + U1 = mx2 + U 2 2 2 Now, from (3) we have d my = 0 dt

(8)

and my is constant. Therefore, my1 = my2

(9)

From (9) we have 1 1 2 2 my1 = my2 2 2 Adding (8) and (10), we have 1 1 2 2 mv1 + U1 = mv2 + U 2 2 2 From (9) we also have mv1 sin θ 1 = mv2 sin θ 2

(10)

(11)

(12)

Substituting (11) into (12), we find sin θ 1 v2 U1 − U 2 = = 1 + T1 sin θ 2 v1

12

(13)

190

CHAPTER 7

This problem is the mechanical analog of the refraction of light upon passing from a medium of a certain optical density into a medium with a different optical density.

7-9.

y O x ξ φ m M

α

Using the generalized coordinates given in the figure, the Cartesian coordinates for the disk are (ξ cos α, –ξ sin α), and for the bob they are ( sin φ + ξ cos α, – cos φ – ξ sin α). The kinetic energy is given by 1 1 1 2 2 T = Tdisk + Tbob = Mξ 2 + Iθ 2 + m x bob + y bob 2 2 2

(

)

(1)

Substituting the coordinates for the bob, we obtain

T=

1 1 1 ( M + m) ξ 2 + Iθ 2 + m 2φ 2 + m φξ cos (φ + a) 2 2 2

(2)

The potential energy is given by

U = Udisk + U bob = Mgydisk + mgy bob = − ( M + m) gξ sin α − mg cos φ

(3)

Now let us use the relation ξ = Rθ to reduce the degrees of freedom to two, and in addition substitute I = MR2 2 for the disk. The Lagrangian becomes 1 1 3 L = T − U = M + m ξ 2 + m 2φ 2 + m φξ cos ( φ + a) + ( M + m) gξ sin α + mg cos φ 4 2 2 The resulting equations of motion for our two generalized coordinates are 3 2 M + m ξ − ( M + m) g sin α + m φ cos ( φ + α ) − φ sin ( φ + α ) = 0 2 (5) (6) (4)

φ + ξ cos ( φ + α ) +

1

g

sin φ = 0

7-10. y x M S –y x

M

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

191

Let the length of the string be so that

(S − x) − y =

Then,

x = −y

a) The Lagrangian of the system is

(1)

(2)

L=

1 1 Mx 2 + My 2 − Mgy = My 2 − Mgy 2 2 d ∂L ∂L − = 2 My + Mg = 0 dt ∂y ∂y

(3)

Therefore, Lagrange’s equation for y is (4)

from which y=− g 2

(5)

Then, the general solution for y becomes y (t) = − g 2 t + C1t + C2 4

(6)

If we assign the initial conditions y ( t = 0 ) = 0 and y ( t = 0 ) = 0 , we find y (t) = − g 2 t 4

(7)

b) If the string has a mass m, we must consider its kinetic energy and potential energy. These are

Tstring = U string = − m

1 my 2 2 y mg 2 y =− 2 2 mg 2 1 my 2 + y 2 2

(8) (9)

yg

Adding (8) and (9) to (3), the total Lagrangian becomes

L = My 2 − Mgy +

(10)

Therefore, Lagrange’s equation for y now becomes

( 2 M + m) y −

mg

y + Mg = 0

(11)

In order to solve (11), we arrange this equation into the form

( 2 M + m) y =

mg M y − m

(12)

192

CHAPTER 7

Since

d2 dt 2

M d2 = y− y , (12) is equivalent to m dt 2 d2 dt 2 M y − m = mg ( 2 M + m) M y − m

(13)

which is solved to give y− M = Aeγ t + Be − γ t m

(14)

where

γ=

mg ( 2 M + m)

(15)

If we assign the initial condition y ( t = 0 ) = 0 ; y ( t = 0 ) = 0 , we have

A = +B = − M 2m

Then, y (t) = M (1 − cosh γ t ) m

(16)

7-11. x y′ er′ m φ x′

y

The x,y coordinates of the particle are x = R cos ω t + R cos ( φ + ω t ) y = R sin ω t + R sin ( φ + ω t )

(1)

Then, x = − Rω sin ω t − R φ + ω sin ( φ + ω t ) y = Rω cos ω t + R φ + ω cos ( φ + ω t )

(

)

(

)

(2)

Since there is no external force, the potential energy is constant and can be set equal to zero. The Lagrangian becomes

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

193

L=

1 m x2 + y2 2

(

)

(3)

= from which

m 2 2 2 2 2 R ω + R ( φ + ω ) + 2R ω ( φ + ω ) cos φ 2

∂L = − mR2ω φ + ω sin φ ∂φ

(

)

(4)

d ∂L d = mR 2 φ + ω + ω cos φ dt ∂φ dt

(

)

(5)

Therefore, Lagrange’s equation for φ becomes

φ + ω 2 sin φ = 0

(6)

which is also the equation of motion for a simple pendulum. To make the result appear reasonable, note that we may write the acceleration felt by the particle in the rotating frame as a = ω 2 R ( i ′ + er ) ′

(7)

where the primed unit vectors are as indicated in the figure. The part proportional to e′ does r not affect the motion since it has no contribution to the torque, and the part proportional to i′ is constant and does not contribute to the torque in the same way a constant gravitational field provides a torque to the simple pendulum.

7-12.

m r θ

Put the origin at the bottom of the plane

L = T −U =

1 m r 2 + r 2 θ 2 − mgr sin θ 2

(

)

θ = αt; θ = α

L=

1 m r 2 + α 2 r 2 − mgr sin α t 2 mr = mα 2 r − mg sin α t

(

)

Lagrange’s equation for r gives

or r − α 2 r = − g sin α t

(1)

194

CHAPTER 7

The general solution is of the form r = rp + rh where rh is the general solution of the homogeneous equation r − α 2 r = 0 and rp is a particular solution of Eq. (1). So rh = Aeα t + Be −α t

For rp , try a solution of the form rp = C sin α t . Then rp = −C α 2 sin α t . Substituting into (1) gives

−C α 2 sin α t − C α 2 sin α t = − g sin α t

C=

g 2α 2 g

So r ( t ) = Aeα t + Be −α t +

2α 2

sin α t

We can determine A and B from the initial conditions: r ( 0 ) = r0 r ( 0) = 0

(2) (3)

(2) implies r0 = A + B (3) implies 0 = A − B + g 2α 2 g 1 r0 − 2α 2 2 g 1 r0 + 2α 2 2

Solving for A and B gives:

A= r (t) = B=

g αt 1 g −α t g 1 r0 − 2α 2 e + 2 r0 + 2α 2 e + 2α 2 sin α t 2

or r ( t ) = r0 cosh α t + g

2α 2

( sin α t − sinh α t )

7-13. a) a b θ m

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

195

x=

1 2 at − b sin θ 2

y = − b cos θ x = at − bθ cos θ y = bθ sin θ L= = 1 m x 2 + y 2 − mgy 2 1 m a 2t 2 − 2 at bθ cos θ + b 2θ 2 + mgb cos θ 2 d ∂L ∂L = dt ∂θ ∂θ gives

(

)

(

)

d − mat b cos θ + mb 2θ = mat bθ sin θ − mgb sin θ dt This gives the equation of motion

θ+

g a sin θ − cos θ = 0 b b

b) To find the period for small oscillations, we must expand sin θ and cos θ about the equilibrium point θ 0 . We find θ 0 by setting θ = 0 . For equilibrium,

g sin θ 0 = a cos θ 0 or tan θ 0 = a2 + g 2 θ0 g

a g a Using the first two terms in a Taylor series expansion for sin θ and cos θ gives

f (θ ) sin θ cos θ tan θ 0 =

f (θ 0 ) + f ′ (θ ) θ =θ (θ − θ 0 )

0

sin θ 0 + (θ − θ 0 ) cos θ 0 cos θ 0 − (θ − θ 0 ) sin θ 0

a implies sin θ 0 = g cos θ 0 =

a a + g2

2

,

g a2 + g 2

196

Thus sin θ cos θ 1

CHAPTER 7

a + g2

2

( a + gθ − gθ 0 ) ( g − aθ + aθ 0 )

1

a2 + g 2

Substituting into the equation of motion gives 0 =θ + This reduces to

g b a2 + g

( a + gθ − gθ 0 ) − 2 g 2 + a2 b

a b a2 + g 2

( g − aθ + aθ 0 )

θ+

θ=

g 2 + a2 b

θ0

The solution to this inhomogeneous differential equation is

θ = θ 0 + A cos ωθ + B sin ωθ where (g ω=

Thus 2π

2

+ a2 b1 2

)

14

T=

ω

=

(g

2π b1 2

2

+ a2

)

14

7-14. a θ b m

x = b sin θ y= 1 2 at − b cos θ 2

x = bθ cos θ y = at + bθ sin θ T= 1 1 m x 2 + y 2 = m b 2θ 2 + a 2t 2 + 2abtθ sin θ 2 2

(

)

(

)

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

197

1 U = mgy = mg at 2 − b cos θ 2 L = T −U = Lagrange’s equation for θ gives d mb 2θ + mabt sin θ = mabtθ cos θ − mgb sin θ dt b 2θ + ab sin θ + abtθ cos θ = abtθ cos θ − gb sin θ 1 1 m b 2θ 2 + a 2t 2 + 2 abtθ sin θ + mg b cos θ − at 2 2 2

(

)

θ+

For small oscillations, sin θ

a+ g sin θ = 0 b

θ θ+ a+ g θ =0. b

Comparing with θ + ω 2θ = 0 gives T= 2π

ω

= 2π

b a+ g

7-15. θ k

m

b = unextended length of spring = variable length of spring T= U= 1 m 2

(

2

+

2

θ2)

1 1 2 2 k ( − b ) + mgy = k ( − b ) − mg cos θ 2 2 1 m 2

L = T −U =

(

2

+

2

θ2) −

1 ( − b ) 2 + mg cos θ 2

Taking Lagrange’s equations for and θ gives : d m = m θ 2 − k ( − b ) + mg cos θ dt

198

CHAPTER 7

θ:

This reduces to

d m 2θ = − mg sin θ dt

− θ2 +

k ( − b ) − g cos θ = 0 m g sin θ = 0

θ+

2

θ+

7-16. x = a sin ωt θ b m

For mass m: x = a sin ω t + b sin θ y = − b cos θ x = aω cos ω t + bθ cos θ y = bθ sin θ Substitute into T= 1 m x2 + y2 2 U = mgy and the result is L = T −U = 1 m a 2ω 2 cos 2 ω t + 2abωθ cos ω t cos θ + b 2θ 2 + mgb cos θ 2

(

)

(

)

Lagrange’s equation for θ gives d mabω cos ω t cos θ + mb 2θ = − mabwθ cos ω t sin θ − mgb sin θ dt − abω 2 sin ω t cos θ − abωθ cos ω t sin θ + b 2θ = − abωθ cos ω t sin θ − gb sin θ or

(

)

θ+

g a sin θ − ω 2 sin ω t cos θ = 0 b b

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

199

7-17. y A h θ C qθ mg B

Using q and θ (= ω t since θ (0) = 0), the x,y coordinates of the particle are expressed as x = h cos θ + q sin θ = h cos ω t + q ( t ) sin ω t y = h sin θ − q cos θ = h sin ω t − q ( t ) cos ω t from which x = − hω sin ω t + qω cos ω t + q sin ω t y = hω cos ω t + qω sin ω t − q cos ω t Therefore, the kinetic energy of the particle is T= = The potential energy is U = mgy = mg ( h sin ω t − q cos ω t ) Then, the Lagrangian for the particle is L= 1 1 1 mh 2ω 2 + mq2ω 2 + mq2 − mgh sin ω t + mgq cos ω t − mhω q 2 2 2 q − ω 2 q = g cos ω t The complementary solution and the particular solution for (6) are written as (5) (4) 1 m x2 + y2 2 (2) (1)

(

) )

(3)

1 m h 2ω 2 + q2ω 2 + q2 − mhω q 2

(

Lagrange’s equation for the coordinate is (6)

qc ( t ) = A cos ( iω t + δ ) g qP ( t ) = − 2 cos ω t 2ω so that the general solution is q ( t ) = A cos ( iω t + δ ) − Using the initial conditions, we have g 2ω 2 cos ω t

(7)

(8)

200

=0 q ( 0 ) = − iω A sin δ = 0 q ( 0 ) = A cos δ − g 2ω 2 Therefore,

CHAPTER 7

(9)

δ = 0, A = and q (t) = or, q (t) = g 2ω 2 q(t) g 2ω 2

(10)

g 2ω 2

( cos iωt − cos ωt ) ( cosh ωt − cos ωt )

(11)

(12)

g 2ω 2

t

In order to compute the Hamiltonian, we first find the canonical momentum of q. This is obtained by p= Therefore, the Hamiltonian becomes ∂L = mq − mω h ∂q (13)

H = pq − L = mq2 − mω hq − so that H= 1 2 1 1 mq − mω 2 h 2 − mω 2 q2 + mgh sin ω t − mgq cos ω t 2 2 2 (14) 1 1 1 mω 2 h 2 − mω 2 q2 − mq2 + mgh sin ω t − mgq cos ω t + mω qh 2 2 2

Solving (13) for q and substituting gives p2 1 H= + ω hp − mω 2 q2 + mgh sin ω t − mgq cos ω t 2m 2 (15)

The Hamiltonian is therefore different from the total energy, T + U. The energy is not conserved in this problem since the Hamiltonian contains time explicitly. (The particle gains energy from the gravitational field.)

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

201

7-18. y S R θ O A B θ x –y m C x

From the figure, we have the following relation: AC = − s = − Rθ where θ is the generalized coordinate. In terms of θ, the x,y coordinates of the mass are (1)

x = AC cos θ + R sin θ = ( − Rθ ) cos θ + R sin θ y = R cos θ − AC sin θ = R cos θ − ( − Rθ ) sin θ from which x = Rθθ sin θ − θ sin θ y = Rθθ cos θ − θ cos θ Therefore, the kinetic energy becomes T= The potential energy is U = mgy = mg R cos θ − ( − Rθ ) sin θ Then, the Lagrangian is L = T −U = 1 m 2θ 2 + R 2θ 2θ 2 − 2R θθ 2 − mg R cos θ − ( − Rθ ) sin θ 2 1 1 m x 2 + y 2 = m 2θ 2 + R 2θ 2θ 2 − 2R θθ 2 2 2

(2)

(3)

(

)

(4)

(5)

(6)

Lagrange’s equation for θ is

( − Rθ ) θ − Rθ 2 − g cos θ = 0

Now let us expand about some angle θ 0 , and assume the deviations are small. Defining ε ≡ θ − θ 0 , we obtain

(7)

ε+

g sin θ 0 g cos θ 0 ε= − Rθ 0 − Rθ 0 cos θ 0 sin θ 0

(8)

The solution to this differential equation is

ε = A sin (ω t + δ ) + where A and δ are constants of integration and

(9)

202

CHAPTER 7

ω≡

g sin θ 0 − Rθ 0

(10)

is the frequency of small oscillations. It is clear from (9) that θ extends equally about θ 0 when θ0 = π 2 .

7-19.

P θ d m1 m1g φ m2 m2 g

Because of the various constraints, only one generalized coordinate is needed to describe the system. We will use φ, the angle between a plane through P perpendicular to the direction of the gravitational force vector, and one of the extensionless strings, e.g., 2 , as our generalized coordinate. The, the kinetic energy of the system is T= The potential energy is given by U = − m1 g

1

1 m1 2

( φ)

1

2

+

1 m2 2

( φ)

2

2

(1)

sin (π − ( φ + θ ) ) − m2 g

2

sin φ

(2)

from which the Lagrangian has the form L = T −U = 1 m1 2

(

2 1

+ m2

2 2

)φ

2

+ m1 g

1

sin ( φ + θ ) + m2 g

2

sin φ

(3)

The Lagrangian equation for φ is m2 g

2

cos φ + m1 g

1

cos ( φ + θ ) − m1

(

2 1

+ m2

2 2

)φ = 0

(4)

This is the equation which describes the motion in the plane m1 , m2 , P . To find the frequency of small oscillations around the equilibrium position (defined by φ = φ0 ), we expand the potential energy U about φ0 : U ( φ ) = U ( φ0 ) + U ′ ( φ 0 ) φ + = 1 U ′′ (φ0 ) φ 2 2 1 U ′′ (φ0 ) φ 2 + … 2 (5)

where the last equality follows because we can take U (φ0 ) = 0 and because U ′ (φ0 ) = 0 . From (4) and (5), the frequency of small oscillations around the equilibrium position is

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

203

ω2 =

The condition U ′ (φ0 ) = 0 gives tan φ0 = or, sin φ0 =

U ′′ (φ0 ) m1

2 1

+ m2

2 2

(6)

m2

+ m1 1 cos θ m1 1 sin θ

2

(7)

(m

m2

2 + m2

2 2 2

+ m1

1

cos θ

1 2

2 2 1 1

+ 2m1m2

cos θ

)

12

(8)

Then from (2), (7), and (8), U ′′ (φ0 ) is found to be U ′′ (φ0 ) = g sin φ0 ( m2 =

2

+ m1 + m1

1

cos θ + m1 cos θ )

2 1

1

sin θ cot φ0 ) m2

2 + m1 1 cos θ + 2 m1 2 sin 2 θ 1 m2 2 + m1 1 cos θ

(m (

g ( m2

2 + m2 2 1 2 2

2

1

2 2 1 1

+ 2m2 m1

2 2

cos θ

)

12

2 = g m1

2 + m2

+ 2m2 m1

2 1

cos θ

)

12

(9)

Finally, from (6) and (9), we have

ω which, using the relation,

2

(m =g

2 2 1 1

2 + m2

(m

2 2

+ 2m1m2 + m2

1 2 2 2

2 1 1

)

cos θ

)

12

(10)

cos θ = can be written as

2 1

+ 2

2 2

− d2

(11)

1 2

ω =

2

g ( m1 + m2 ) m1

( (m

2 1

+ m2 + m2

2 2 2 2

2 1 1

)−d m m )

2 1 2

12

(12)

Notice that ω 2 degenerates to the value g (so that 1 = 2 ).

appropriate for a simple pendulum when d → 0

204

CHAPTER 7

7-20. The x-y plane is horizontal, and A, B, C are the fixed points lying in a plane above the hoop. The hoop rotates about the vertical through its center. z A C

B

y θ′

x R

A′ C′

B′ θ

The kinetic energy of the system is given by

T=

1 2 1 MR2 2 1 ∂z 2 θ + M θ Iω + Mz 2 = 2 2 2 2 ∂θ θ =0

2

(1)

For small θ, the second term can be neglected since ( ∂z ∂θ ) The potential energy is given by

U = Mgz

=0

(2)

where we take U = 0 at z = – . Since the system has only one degree of freedom we can write z in terms of θ. When θ = 0, z = − . When the hoop is rotated thorough an angle θ, then

z2 = so that

2

− ( R − R cos θ ) − ( R sin θ )

2

2

(3)

z = − and the potential energy is given by

2

+ 2R 2 ( cos θ − 1)

12

(4)

U = − Mg for small θ, cos θ − 1 ≅ −θ 2 2 ; then,

2

+ 2R2 ( cos θ − 1)

12

(5)

R 2θ 2 U ≅ − Mg 1 − 2 R 2θ 2 ≅ − Mg 1 − 2 2 From (1) and (6), the Lagrangian is L = T −U =

12

(6)

R 2θ 2 1 MR 2θ 2 + Mg 1 − , 2 2 2

(7)

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

205

for small θ. The Lagrange equation for θ gives

θ+ where g

θ =0

(8)

ω=

g

(9)

which is the frequency of small rotational oscillations about the vertical through the center of the hoop and is the same as that for a simple pendulum of length .

7-21. ω θ

From the figure, we can easily write down the Lagrangian for this system.

T=

mR 2 2 θ + ω 2 sin 2 θ 2 U = − mgR cos θ

(

)

(1) (2)

The resulting equation of motion for θ is

θ − ω 2 sin θ cos θ +

g sin θ = 0 R

(3)

The equilibrium positions are found by finding the values of θ for which 0 =θ θ =θ 0

g = ω 2 cos θ 0 − sin θ 0 R

(4)

Note first that 0 and π are equilibrium, and a third is defined by the condition cos θ 0 =

g ω 2R

(5)

To investigate the stability of each of these, expand using ε = θ − θ 0

ε = ω 2 cos θ 0 −

g − ε sin θ 0 ( sin θ 0 + ε cos θ 0 ) ω 2R

(6)

206

For θ 0 = π , we have

CHAPTER 7

ε = ω2 1 + indicating that it is unstable. For θ 0 = 0 , we have

g ε ω 2R

(7)

ε = ω2 1 −

g ε ω 2R

(8)

which is stable if ω 2 < g R and unstable if ω 2 > g R . When stable, the frequency of small oscillations is

ω 2 − g R . For the final candidate, ε = −ω 2 sin 2 θ 0 ε

(9)

with a frequency of oscillations of

ω 2 − ( g ω R) , when it exists. Defining a critical frequency

2 2

ω c2 ≡ g R , we have a stable equilibrium at θ 0 = 0 when ω < ω c , and a stable equilibrium at θ 0 = cos −1 (ω c2 ω 2 ) when ω ≥ ω c . The frequencies of small oscillations are then ω 1 − (ω c ω ) and ω 1 − (ω c ω ) , respectively.

4

To construct the phase diagram, we need the Hamiltonian

H ≡θ

∂L −L ∂θ

(10)

which is not the total energy in this case. A convenient parameter that describes the trajectory for a particular value of H is 1 K≡ = 2 2 mω c R 2 so that we’ll end up plotting

H

θ 2 ω 2 − sin 2 θ − cos θ ω c ω c

(11)

θ ω 2 ω = 2 ( K + cos θ ) + ω sin θ c c

2

2

(12)

for a particular value of ω and for various values of K. The results for ω < ω c are shown in figure (b), and those for ω > ω c are shown in figure (c). Note how the origin turns from an attractor into a separatrix as ω increases through ω c . As such, the system could exhibit chaotic behavior in the presence of damping.

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

207

2

1.5

K

1

0.5

0

3

2

1

0

1

2

3

θ

(b)

2

1.5

K

1

0.5

0

3

2

1

0

1

2

3

θ

(c)

7-22.

The potential energy U which gives the force F ( x, t) = k −( t τ ) e x2 ∂U ∂x (1)

must satisfy the relation F=− we find U= k −t τ e x (3) (2)

208

CHAPTER 7

Therefore, the Lagrangian is L = T −U = The Hamiltonian is given by H = px x − L = x so that H=

2 px k − t τ + e 2m x

1 k mx 2 − e − t τ x 2 ∂L −L ∂x

(4)

(5)

(6)

The Hamiltonian is equal to the total energy, T + U, because the potential does not depend on velocity, but the total energy of the system is not conserved because H contains the time explicitly.

7-23.

The Hamiltonian function can be written as [see Eq. (7.153)]

H = ∑ p j qj − L j (1)

For a particle which moves freely in a conservative field with potential U, the Lagrangian in rectangular coordinates is

L=

1 m x2 + y2 + z2 − U 2

(

)

and the linear momentum components in rectangular coordinates are

px =

∂L = mx ∂x py = my pz = mz

(2)

1 H = mx 2 + my 2 + mz 2 − m x 2 + y 2 + z 2 − U 2

(

)

=

1 1 2 2 2 m x2 + y2 + z2 + U = px + py + pz 2 2m

(

)

(

)

(3)

which is just the total energy of the particle. The canonical equations are [from Eqs. (7.160) and (7.161)]

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

209

px = mx = − py = my = − pz = mz = − These are simply Newton’s equations.

7-24.

∂U = Fx ∂x ∂U = Fy ∂y ∂U = Fz ∂z (4)

θ

m

The kinetic energy and the potential energy of the system are expressed as T= 1 m 2

(

2

+ 2θ 2 =

)

U = − mg cos θ so that the Lagrangian is L = T −U = The Hamiltonian is H = pθ θ − L = =

1 m α 2 + 2θ 2 2

(

)

(1)

1 m α 2 + 2θ 2 + mg cos θ 2 ∂L θ−L ∂θ

(

)

(2)

2 pθ 1 − mα 2 − mg cos θ 2 2m 2

(3)

which is different from the total energy, T + U. The total energy is not conserved in this system because work is done on the system and we have d (T + U ) ≠ 0 dt (4)

210

7-25.

z z r

CHAPTER 7

m y

θ x

In cylindrical coordinates the kinetic energy and the potential energy of the spiraling particle are expressed by 1 m r 2 + r 2θ 2 + z 2 2 U = mgz T= Therefore, if we use the relations, i.e., z = kθ r = const. z = kθ the Lagrangian becomes L= Then the canonical momentum is pz = or, z= pz r2 m 2 + 1 k (5) r2 ∂L = m 2 + 1 z ∂z k (4) 1 r2 2 m 2 z + z 2 − mgz 2 k (3) (2)

(1)

The Hamiltonian is

H = pz z − L = p z pz r2 m 2 + 1 k −

2 pz

r2 2m 2 + 1 k

+ mgz

(6)

or, H= 1 2

2 pz

r2 m 2 + 1 k

+ mgz

(7)

Now, Hamilton’s equations of motion are

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

211

− so that

∂H = pz ; ∂z

∂H =z ∂p z

(8)

−

∂H = − mg = pz ∂z pz r2 m 2 + 1 k =z

(9) (10)

∂H = ∂p z

Taking the time derivative of (10) and substituting (9) into that equation, we find the equation of motion of the particle:

z=

g r k 2 + 1

2

(11)

It can be easily shown that Lagrange’s equation, computed from (3), gives the same result as (11).

7-26. a) θ m

L = T −U = L=

1 2 2 m θ − mgy 2

1 2 2 m θ + mg cos θ 2 ∂L = m 2θ ∂θ

pθ = so

θ=

pθ m 2

Since U is velocity-independent and the coordinate transformations are time-independent, the Hamiltonian is the total energy H = T +U = The equations of motion are

2 pθ − mg cos θ 2m 2

212 p ∂H ∂H = θ 2 and pθ = − = − mg sin θ ∂pθ m ∂θ

CHAPTER 7

θ=

b)

a

x m2 m1

T=

1 1 1 x2 m1 x 2 + m2 x 2 + I 2 2 2 2 a

where I = moment of inertia of the pulley U = − m1 gx − m2 g ( − x ) px = So x= px I m1 + m2 + a 2 H=T+U H=

2 px

∂L ∂T I = = m1 + m2 + 2 x ∂x ∂x a

I 2 m1 + m2 + 2 a

− m1 gx − m2 g ( − x )

The equations of motion are x= px ∂H = I ∂p x 2 m1 + m2 + 2 a ∂H = m1 g − m2 g = g ( m1 − m2 ) ∂p x

px = −

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

213

7-27. a)

k,b θ m1

m2

xi , yi = coordinates of mi Using , θ as polar coordinates x2 = x1 + cos θ y2 = y1 + sin θ x2 = x1 + cos θ − θ sin θ y2 = y1 + sin θ + θ cos θ If we substitute (1) and (2) into L = T −U = the result is L= 1 1 ( m1 + m2 ) x12 + y12 + m2 2 2 1 1 1 2 2 2 2 2 m1 x1 + y1 + m2 x2 + y2 − k ( − b ) 2 2 2 (1) (2)

(

)

(

)

(

)

(

2

+ 2θ 2

)

1 2 + m2 ( x1 cos θ + y1 sin θ ) + m2 θ ( y1 cos θ − x1 sin θ ) − k ( − b ) 2 The equations of motion are x1 : d ( m1 + m2 ) x1 + m2 cos θ − m2 θ sin θ = 0 dt = m1 x1 + m2 x2 = px So px = constant y1 : d ( m1 + m2 ) y1 + m2 sin θ + m2 θ cos θ = 0 dt = m1 y1 + m2 y2 = py So py = constant : d m2 + m2 ( x1 cos θ + y1 sin θ ) = m2 θ 2 − k ( − b ) + m2θ ( y1 cos θ − x1 sin θ ) dt

which reduces to

214

CHAPTER 7

− θ 2 + x1 cos θ + y1 sin θ +

k ( − b) = 0 m2

θ:

d m2 2θ + m2 ( y1 cos θ − x1 sin θ ) dt = − m2 ( x1 sin θ − y1 cos θ ) + m2 θ ( − x1 cos θ − y1 sin θ )

which reduces to

θ+

b)

2

θ+

cos θ

y1 −

sin θ

x1 = 0

As was shown in (a) ∂L = px = constant ∂x1 ∂L = py = constant ∂y1 (total linear momentum)

c)

Using L from part (a) px1 = py1 = p = ∂L = ( m1 + m2 ) x1 + m2 cos θ − m2 θ sin θ ∂x1 ∂L = ( m1 + m2 ) y1 + m2 sin θ − m2 θ cos θ ∂y1

∂L = m2 x1 cos θ + m2 y1 sin θ + m2 ∂

pθ = − m2 x1 sin θ + m2 y1 cos θ + m2 2θ Inverting these equations gives (after much algebra) x1 = y1 = = sin θ 1 pθ px1 − p cos θ + m1 cos θ 1 pθ py1 − p sin θ − m1

1 m1 + m2 p − px1 cos θ − py1 sin θ + m1 m2 1 m1 + m2 pθ px1 sin θ − py1 cos θ + m1 m2 H = T +U

θ=

Since the coordinate transformations are time independent, and U is velocity independent,

Substituting using the above equations for qi in terms of pi gives

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

215

H=

m1 + m2 1 2 2 px1 + py1 + m2 2m1

2 2 pθ p + 2 − 2 p px1 cos θ + py1 sin θ pθ 1 2 +2 px1 sin θ − py1 cos θ + k ( − b ) 2

(

)

(

)

The equations of motion are x1 = y1 = = sin θ 1 ∂H = pθ px1 − p cos θ + ∂px1 m1 cos θ 1 ∂H = pθ py1 − p sin θ − ∂py1 m1

1 m1 + m2 ∂H p − px1 cos θ − py1 sin θ = m1 m2 ∂p 1 m1 + m2 ∂H pθ + px1 sin θ − py1 cos θ = ∂pθ m1 m2 ∂H =0 ∂x1 py1 = − ∂H =0 ∂y1

θ=

px1 = −

2 p ∂H ( m1 + m2 ) pθ p =− = + θ 2 px1 sin θ − py1 cos θ − k ( − b ) 3 m1m2 m1 ∂

(

)

pθ = −

p ∂H p = − px1 sin θ + py1 cos θ − θ px1 cos θ + py1 sin θ m1 ∂θ m1

(

)

Note: This solution chooses as its generalized coordinates what the student would most likely choose at this point in the text. If one looks ahead to Section 8.2 and 8.3, however, it would show another choice of generalized coordinates that lead to three cyclic coordinates ( xCM , yCM , and θ ), as shown in those sections. F = − kr −2 U = − kr −1 L = T −U = pr = pθ = 1 k m r 2 + r 2θ 2 + 2 r

7-28.

so

(

)

p ∂L = mr so r = r m ∂r so θ = pθ mr 2

∂L = mr 2θ ∂θ

Since the coordinate transformations are independent of t, and the potential energy is velocityindependent, the Hamiltonian is the total energy.

216

1 k m r 2 + r 2θ 2 − 2 r

CHAPTER 7

H = T +U = =

(

)

2 p2 k 1 pr m 2 + r 2 2θ 4 − mr r 2 m 2 pr p2 k + θ2− r 2m 2mr

H= Hamilton’s equations of motion are

r=

∂H p r = ∂p r m p2 k ∂H = θ3 − 2 r ∂r mr ∂H =0 ∂θ

θ=

p ∂H = θ ∂pθ mr 2

pr = − pθ = −

7-29. a k θ m

b = unextended length of spring = variable length of spring

a)

x = sin θ y= 1 2 at − cos θ 2 1 m x2 + y2 2

x = sin θ + θ cos θ y = at − cos θ + θ sin θ

Substituting into T =

(

)

U = mgy + gives L = T −U = 1 m 2

2

1 2 k ( − b) 2

at 2 k 2 + 2θ 2 + a 2t 2 + 2at θ sin θ − cos θ + mg cos θ − − 2 ( − b) 2

(

)

Lagrange’s equations give:

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

217

:

d m − amt cos θ = m θ 2 + matθ sin θ + mg cos θ − k ( − b ) dt d m 2θ + mat sin θ = mat sin θ − mg sin θ + mat θ cos θ dt

θ:

Upon simplifying, the equations of motion reduce to: − θ 2 − ( a + g ) cos θ + k ( − b) = 0 m sin θ = 0 = p + at cos θ m (1)

θ+

b)

2

θ+

a+ g

p = pθ = or

∂L = m − mat cos θ ∂

or

∂L = m 2θ + mat sin θ ∂θ

θ=

pθ at sin θ − 2 m

(2)

Since the transformation equations relating the generalized coordinates to rectangular coordinates are not time-independent, the Hamiltonian is not the total energy. H = ∑ pi qi − L = p + pθ θ − L Substituting (1) and (2) for H= and θ and simplifying gives

p2 p2 1 1 at 2 + θ 2 − pθ sin θ + atp cos θ + k ( − b ) + mgat 2 − mg cos θ 2 m 2m 2 2 are p at ∂H = θ 2 − sin θ ∂pθ m ∂H p = + at cos θ m ∂p agreeing with (1) and (2)

The equations for θ and

θ=

=

The equations for p and pθ are

2 pθ ∂H at p =− = − 2 pθ sin θ − k ( − b ) + mg cos θ + ∂ m 3

or p + pθ = − at

2

pθ sin θ + k ( − b ) − mg cos θ +

2 pθ =0 m 3

at ∂H = − pθ cos θ + at p sin θ − mg sin θ ∂θ

218

CHAPTER 7

or pθ −

c)

at

pθ cos θ − at p sin θ + mg sin θ = 0

sin θ

θ , cos θ 1 −

θ2

2 θ2 k − θ 2 − ( a + g ) 1 − + ( − b ) = 0 2 m

Substitute into Lagrange’s equations of motion

θ+

For small oscillations, θ 1, θ 1,

2

θ+

a+ g

θ−

at θ

=0

1 . Dropping all second-order terms gives + k k = a+ g+ b m m

θ+

For θ,

a+ g

θ =0

Tθ = The solution to the equation for is

2π

ω

= 2π

a+ g

=

homogeneous

+

particular

= A cos So for ,

k k m t + B sin t + (a + g) + b m m k

T =

2π

ω

= 2π

m k

7-30. a) From the definition of a total derivative, we can write

∂g ∂qk ∂g ∂pk dg ∂g = + ∑ + dt ∂t ∂ p k ∂t k ∂qk ∂t Using the canonical equations

(1)

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

219

∂qk ∂H = qk = ∂t ∂p k

∂pk ∂H = pk = − ∂t ∂qk

(2)

we can write (1) as

∂ g ∂H ∂ g ∂ H dg ∂g = + ∑ − dt ∂t ∂ p k ∂ qk k ∂qk ∂pk

(3)

or dg ∂g = + g , H dt ∂t

b)

(4)

qj =

∂q j ∂t

=

∂H ∂p j

(5)

According to the definition of the Poisson brackets,

∂q ∂ H ∂g j ∂H qj , H = ∑ j − ∂pk ∂qk k ∂ qk ∂ p k

(6)

but

∂q j ∂qk = δ jk and ∂q j ∂pk = 0 for any j,k

(7)

then (6) can be expressed as

∂H q j , H = ∂p = q j j In the same way, from the canonical equations, pj = − so that ∂p ∂H ∂p j ∂ H pj , H = ∑ j − ∂ p k ∂ qk k ∂qk ∂pk but ∂p j ∂pk then, = δ jk and ∂p j ∂ qk = 0 for any j,k (11) (10) ∂H ∂q j (9) (8)

220

∂H = pj , H ∂q j

CHAPTER 7

pj = −

(12)

c)

∂p ∂ p j ∂ p k ∂ p j pk , p j = ∑ k − ∂p ∂ q ∂ q ∂p ∂pk = 0 for any k, ∂q

(13)

since, (14)

the right-hand side of (13) vanishes, and

pk , p j = 0 In the same way, ∂ q ∂ q j ∂ qk ∂ q j qk , q j = ∑ k − ∂ q ∂p ∂p ∂ q since ∂q j ∂p the right-hand side of (16) vanishes and qk , q j = 0

d)

(15)

(16)

= 0 for any j,

(17)

(18)

∂q ∂p j ∂qk ∂p j qk , p j = ∑ k − ∂ q ∂p ∂p ∂q = ∑δk δ j or, qk , p j = δ kj

e) Let g ( pk , qk ) be a quantity that does not depend explicitly on the time. If g ( pk , qk ) commutes with the Hamiltonian, i.e., if

(19)

(20)

g , H = 0 then, according to the result in a) above, dg =0 dt and g is a constant of motion.

(21)

(22)

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

221

7-31. A spherical pendulum can be described in terms of the motion of a point mass m on the surface of a sphere of radius , where corresponds to the length of the pendulum support rod. The coordinates are as indicated below. z θ mg φ x

y

The kinetic energy of the pendulum is T= and the potential energy is U = mg cos θ The Lagrangian is L= so that the momenta are pθ = pφ = The Hamiltonian then becomes H = pθ θ + pφ φ − 1 m 2

2

1 1 1 I1φ 2 + I 2θ 2 = m 2 2 2

2

(φ

2

sin 2 θ + θ 2

)

(1)

(2)

1 m 2

2

(φ

2

sin 2 θ + θ 2 − mg cos θ

)

(3)

∂L = m 2θ ∂θ

(4) (5)

∂L = m 2φ sin 2 θ ∂φ

(φ

2

sin 2 θ + θ 2 + mg cos θ (6)

)

p2 = θ 2 + V θ , pφ 2m

(

)

which is just the total energy of the system and where the effective potential is

2 pθ V θ , pφ = + mg cos θ 2m 2 sin 2 θ

(

)

(7)

When pφ = 0 , V(θ,0) is finite for all θ, with a maximum at θ = 0 (top of the sphere) and a minimum at θ = π (bottom of the sphere); this is just the case of the ordinary pendulum. For different values of pφ , the V–θ diagram has the appearance below:

222

CHAPTER 7

V

0

π 2

π θ Pφ = 0

When pφ > 0 , the pendulum never reaches θ = 0 or θ = π because V is infinite at these points. The V–θ curve has a single minimum and the motion is oscillatory about this point. If the total energy (and therefore V) is a minimum for a given pφ ,θ is a constant, and we have the case of a conical pendulum. For further details, see J. C. Slater and N. H. Frank, Mechanics, McGraw-Hill, New York, 1947, pp. 79–86.

7-32.

The Lagrangian for this case is L = T −U = 1 k m r 2 + r 2θ 2 + r 2 sin 2 θφ 2 + 2 r

(

)

(1)

where spherical coordinates have been used due to the symmetry of U. The generalized coordinates are r, θ, and φ, and the generalized momenta are pr = pθ = pφ = ∂L = mr ∂r ∂L = mr 2θ ∂θ ∂L = mr 2φ sin 2 θ ∂φ (2) (3) (4)

The Hamiltonian can be constructed as in Eq. (7.155): H = pr r + pθ θ + pφ φ − L = = 1 k m r 2 + r 2θ 2 + r 2φ 2 sin 2 θ − 2 r

2 2 2 k pφ pθ 1 pr + + − 2 2 2 mr sin θ r 2 m mr

(

)

(5)

Eqs. (7.160) applied to H as given in (5) reproduce equations (2), (3), and (4). The canonical equations of motion are obtained applying Eq. (7.161) to H: pr = − pφ p2 ∂H k = − 2 + θ3 + 3 ∂r r mr mr sin 2 θ

2

(6)

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

223

2 pφ cot θ ∂H pθ = − = ∂θ mr 2 sin 2 θ

(7) (8)

pφ = −

∂H =0 ∂φ

The last equation implies that pφ = const , which reduces the number of variables on which H depends to four: r , θ , pr , pθ : H= 1 2m

2 2 pθ const k pr + 2 + 2 − r r sin 2 θ r

(9)

For motion with constant energy, (9) fixes the value of any of the four variables when the other three are given. From (9), for a given constant value of H = E, we obtain p 2 sin 2 θ + const 2mk + pr = 2 mE − θ 2 r sin 2 θ r

12

(10)

and so the projection of the phase path on the r − pr plane are as shown below. pr 2mE θ3 θ2 θ1 θ1 < θ2 < θ3

7-33.

x

m1 x′ m2 m3

Neglect the masses of the pulleys T= 1 1 1 2 2 m1 x 2 + m2 ( x ′ − x ) + m3 ( − x − x ′ ) 2 2 2

U = − m1 gx − m2 g ( − x + x ′ ) − m3 g ( − x + ′ − x ′ )

224

1 1 ( m1 + m2 + m3 ) x 2 + 2 ( m2 + m3 ) x ′ 2 + xx ′ ( m3 − m2 ) 2 + g ( m1 − m2 − m3 ) x + g ( m2 − m3 ) x ′ + constant ∂L = ( m1 + m2 + m3 ) x + ( m3 − m2 ) x ′ ∂x ∂L = ( m3 − m2 ) x + ( m2 + m3 ) x ′ ∂x ′

CHAPTER 7

L=

We redefine the zero in U such that the constant in L = 0. px = px ′ = (1) (2)

Solving (1) and (2) for px and px ′ gives x = D−1 ( m2 + m3 ) px + ( m2 − m3 ) px ′ x ′ = D−1 ( m2 + m3 ) px + ( m1 + m2 + m3 ) px ′ where D = m1m3 + m1m2 + 4 m2 m3

H =T +U = 1 1 ( m1 + m2 + m3 ) x 2 + 2 ( m2 + m3 ) x ′ 2 + ( m3 − m2 ) xx 2 − g ( m1 − m2 − m3 ) x − g ( m2 − m3 ) x ′

Substituting for x and x′ and simplifying gives H= 1 1 ( m2 + m3 ) D−1 px2 + 2 ( m1 + m2 + m3 ) D−1 px2′ 2 + ( m2 − m3 ) D−1 px px ′ − g ( m1 − m2 + m3 ) x − g ( m2 − m3 ) x ′

where D = m1m3 + m1 m2 + 4 m2 m3 The equations of motion are x= x′ = ∂H = ( m2 + m3 ) D−1 px + ( m2 − m3 ) D−1 px ′ ∂p x ∂H = ( m2 − m3 ) D−1 px ′ + ( m1 + m2 + m3 ) D−1 px ′ ∂p x ′ ∂H = g ( m1 − m2 − m3 ) ∂x ∂H = g ( m2 − m3 ) ∂x ′

px = − px ′ = −

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

225

7-34. x r R θ m M

The coordinates of the wedge and the particle are xM = x yM = 0 The Lagrangian is then L= M+m 2 m 2 x + r + r 2θ 2 + 2 xr cos θ − 2 xrθ sin θ + mgr sin θ 2 r xm = r cos θ + x ym = − r sin θ (1)

(

)

(2)

Note that we do not take r to be constant since we want the reaction of the wedge on the particle. The constraint equation is f ( x , θ , r ) = r − R = 0 .

a) Right now, however, we may take r = R and r = r = 0 to get the equations of motion for x and θ. Using Lagrange’s equations,

x = aR θ sin θ + θ 2 cos θ

(

)

(3) (4)

θ= where a ≡ m ( M + m) .

x sin θ + g cos θ R

b) We can get the reaction of the wedge from the Lagrange equation for r

λ = mx cos θ − mRθ 2 − mg sin θ

(5)

We can use equations (3) and (4) to express x in terms of θ and θ , and substitute the resulting expression into (5) to obtain

λ=

a−1 2 Rθ + g sin θ 2 1 − a sin θ

(

)

(6)

To get an expression for θ , let us use the conservation of energy

H= M+m 2 m 2 2 x + R θ − 2xRθ sin θ − mgR sin θ = − mgR sin θ 0 2 2

(

)

(7)

where θ 0 is defined by the initial position of the particle, and − mgR sin θ 0 is the total energy of the system (assuming we start at rest). We may integrate the expression (3) to obtain x = aRθ sin θ , and substitute this into the energy equation to obtain an expression for θ

θ2 =

2 g ( sin θ − sin θ 0 )

R 1 − a sin 2 θ

(

)

(8)

226

CHAPTER 7

Finally, we can solve for the reaction in terms of only θ and θ 0

λ=−

mMg 3 sin θ − a sin 3 θ − 2 sin θ 0

(

( M + m) ( 1 − a sin θ )

2

)

2

(9)

7-35. We use zi and pi as our generalized coordinates, the subscript i corresponding to the ith particle. For a uniform field in the z direction the trajectories z = z(t) and momenta p = p(t) are given by

z i = z i 0 + vi 0 t −

1 2 gt 2 pi = pi 0 − mgt

(1)

where zi 0 , pi 0 , and vi 0 = pi 0 m are the initial displacement, momentum, and velocity of the ith particle. Using the initial conditions given, we have z1 = z0 + p0 t 1 2 − gt m 2

(2a) (2b)

p1 = p0 − mgt z2 = z0 + ∆z0 + p2 = p0 − mgt z 3 = z0 + p0 t 1 2 − gt m 2

(2c) (2d) (2e) (2f) (2g) (2h)

( p0 + ∆p0 ) t − 1 gt 2 m 2

p3 = p0 + ∆p0 − mgt z4 = z0 + ∆Z0 +

( p0 + ∆p0 ) t − 1 gt 2 m 2

p4 = p0 + ∆p0 − mgt

The Hamiltonian function corresponding to the ith particle is

H i = Ti + Vi = p2 mzi2 + mgzi = i + mgzi = const. 2 2m

(3)

From (3) the phase space diagram for any of the four particles is a parabola as shown below.

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

227

p0 + ∆p0 p0

Area at t = 0

1 p

2

3

4

Area at t = t1 ∆p0

∆z0 z0 z0 + ∆z0 z

From this diagram (as well as from 2b, 2d, 2f, and 2h) it can be seen that for any time t, p1 = p2 p3 = p 4

(4) (5)

Then for a certain time t the shape of the area described by the representative points will be of the general form p (p3,z3) 3 (p4,z4) 4

1 (p1,z1)

2 (p2,z2) z

where the base 1 2 must parallel to the top 3 4 . At time t = 0 the area is given by ∆z0 ∆p0 , since it corresponds to a rectangle of base ∆z0 and height ∆p0 . At any other time the area will be given by

A = base of parallelogram t = t = ( z2 − z1 ) t = t

1

{

1

= ( z 4 − z 3 ) t = t = ∆z 0

1

}

x height of parallelogram t = t = ( p3 − p1 )

1

{

t = t1

= ( p 4 − p2 ) =∆p0 ∆z0 Thus, the area occupied in the phase plane is constant in time.

7-36.

t = t1

= ∆p0

}

(6)

The initial volume of phase space accessible to the beam is

2 2 V0 = π R0 π p0

(1)

After focusing, the volume in phase space is

2 2 V1 = π R1 π p1

(2)

228

CHAPTER 7

where now p1 is the resulting radius of the distribution of transverse momentum components of the beam with a circular cross section of radius R1 . From Liouville’s theorem the phase space accessible to the ensemble is invariant; hence,

2 2 2 2 V0 = π R0 π p0 = V1 = π R1 π p1

(3)

from which p1 = R0 p0 R1 (4)

If R1 < R0 , then p1 > p0 , which means that the resulting spread in the momentum distribution has increased. This result means that when the beam is better focused, the transverse momentum components are increased and there is a subsequent divergence of the beam past the point of focus.

7-37.

Let’s choose the coordinate system as shown:

x1

x2

x3 m1

m3 m2

The Lagrangian of the system is

2 2 2 1 dx dx dx L = T − U = m1 1 + m2 2 + m3 3 + g ( m1 x1 + m2 x2 + m3 x3 ) dt dt 2 dt

with the constraints x1 + y = l1 and x2 − y + x3 − y = l2 (1)

which imply 2x1 + x2 + x3 − (2l1 + l2 ) = 0 ⇒ 2

d 2 x1 d 2 x2 d 2 x3 + 2 + 2 =0 dt 2 dt dt

The motion equations (with Lagrange multiplier λ ) are

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

229

m1 g − m1 m2 g − m2 m3 g − m3 Combining (1)–(4) we find

d 2 x1 + 2λ = 0 dt 2 d 2 x2 +λ =0 dt 2 d 2 x3 +λ =0 dt 2

(2) (3) (4)

λ=

−4 g 4 1 1 + + m1 m2 m3

Finally, the string tension that acts on m1 is (see Eq. (2)) T1 = m1 g − m1 8g d 2 x1 = −2λ = 2 4 1 1 dt + + m1 m2 m3

7-38.

The Hamiltonian of the system is H =T +U = kx 2 bx 4 p 2 kx 2 bx 4 1 dx m + + = + + 2 dt 2 4 2m 2 4 dx ∂H p = = dt ∂p m dp ∂H =− = −(kx + bx 3 ) dt ∂x

2

The Hamiltonian motion equations that follow this Hamiltonian are

7-39.

z

The Lagrangian of the rope is mgz 2 1 dz mz z 1 dz L = T − U = m − − g = m + 2 dt b 2 2 dt 2b

2 2

230

CHAPTER 7

from which follows the equation of motion mgz d2 z ∂L d ∂L = ⇒ =m 2 ∂z dt ∂z b dt

7-40.

θ1 m θ2 m

2m

x

We choose the coordinates for the system as shown in the figure. The kinetic energy is

2 2 2 dθ 1 dx 1 1 2 dθ 1 dx dx T = 2m + m b cos θ 1 + + 2b dt dt dt 2 2 dt dt 2 2 dθ 1 dθ 2 dθ 2 1 dx dθ 1 cos θ 1 + b cos θ 2 + b sin θ 1 + b sin θ 2 + m +b dt dt dt dt 2 dt

The potential energy is U = − mgb cos θ 1 − mg(b cos θ 1 + b cos θ 2 ) And the Lagrangian is 1 dx dθ 1 dx dθ dθ cos θ 1 + mb 2 2 L = T − U = 2m + mb 2 1 + 2mb dt dt dt 2 dt dt

2 2 2

+ mb

dx dθ 2 dθ dθ cos θ 2 + mb 2 1 2 cos (θ 1 − θ 2 ) + 2mgb cos θ 1 + mgb cos θ 2 dt dt dt dt d 2θ d2 x d 2θ + b 2 21 cos θ 1 + 22 cos θ 2 2 dt dt dt

From this follow 3 equations of motion ∂L d ∂L = ∂x dt ∂x ⇒ 0=4

2 dθ 2 dθ − b 2 1 sin θ 1 + 2 sin θ 2 dt dt

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS

231

∂L d ∂L = ∂θ 1 dt ∂θ 1

⇒ − 2 g sin θ 1 = 2b

2

d 2θ 1 d 2θ d2 x + 2 2 cos θ 1 + b 22 cos(θ 1 − θ 2 ) dt 2 dt dt

dθ + b 2 sin(θ 1 − θ 2 ) dt ∂L d ∂L = ∂θ 2 dt ∂θ 2 − g sin θ 2 = b ⇒ d 2θ 2 d 2 x d 2θ dθ + 2 cos θ 2 + b 21 cos (θ 1 − θ 2 ) − b 1 sin (θ 1 − θ 2 ) 2 dt dt dt dt

2

7-41.

For small angle of oscillation θ we have

T=

So the Lagrangian reads

1 2 dθ 1 db mb + m dt 2 2 dt

2

2

and

U = − mgb cos θ

L = T −U =

1 2 dθ 1 db mb + m + mgb cos θ dt 2 2 dt

2 2

from which follow 2 equations of motion

∂L d ∂L = ∂b dt ∂b ∂L d ∂ L = ∂θ dt ∂θ

d2b dα dθ ⇒ b + g cos θ = 2 = − dt dt dt

2

⇒ − mgb sin θ = 2mb

db dθ d 2θ dθ d 2θ + mb 2 2 = −2mbα + mb 2 2 dt dt dt dt dt

232

CHAPTER 7

CHAPTER

8

Central-Force Motion

8-1. x3 m1 r1 r2 x2 m2

x1

In a uniform gravitational field, the gravitational acceleration is everywhere constant. Suppose the gravitational field vector is in the x1 direction; then the masses m1 and m2 have the gravitational potential energies:

( ( ( U g1) = − F (1) x11) = − m1α x11) ( 2) ( 2) ( 2 ) ( 2) U g = − F x1 = − m2α x1

(1)

( ( ( where r1 = x11) , x21) , x31) and where α is the constant gravitational acceleration. Therefore,

(

)

introducting the relative coordinate r and the center of mass coordinate R according to r = r1 − r2 m1r1 + m2 r2 = ( m1 + m2 ) R

(2)

we can express r1 and r2 in terms of r and R by

m1 r2 = − r+R m1 + m2 r1 = m2 r+R m1 + m2

(3)

233

234

CHAPTER 8

which differ from Eq. (8.3) in the text by R. The Lagrangian of the two-particle system can now be expressed in terms of r and R:

L= = 1 1 2 2 ( ( m1 r1 + m2 r2 − U ( r ) − U g1) − U g2) 2 2 m2 m1 1 1 r + R + m2 − r + R − U (r) m1 2 2 m1 + m2 m1 + m2 m2 m1 x + X + m2α − x + X + m1α m1 + m2 m1 + m2

2 2

(4)

where x and X are the x1 components of r and R, respectively. Then, (4) becomes

2 m2 2 1 m1 2 1 1 L = m1 r + m2 r + ( m1 + m2 ) R 2 2 2 m1 + m2 m1 + m2 m1m2 − m1m2 x + ( m1 + m2 ) α X − U (r) + α m1 + m2 2

(5)

Hence, we can write the Lagrangian in the form

L=

2 1 1 2 µ r − U ( r ) + ( m1 + m2 ) R + ( m1 + m2 ) α X 2 2

(6)

where µ is the reduced mass:

µ=

m1m2 m1 + m2

(7)

Therefore, this case is reducible to an equivalent one-body problem.

8-2.

Setting u = 1 r , Eq. (8.38) can be rewritten as

θ = −∫

2µE

2

+

du 2µ k

2

(1) u−u

2

where we have used the relation du = − 1 r 2 dr . Using the standard form of the integral [see

Eq. (E.8c), Appendix E]: 2ax + b −1 sin −1 + const. 2 −a b − 4 ac

(

)

∫ we have

dx ax 2 + bx + c

=

(2)

2 2µ k − + 2 −1 r θ + const. = sin 2 2µ k + 8 µ E 2 2

(3)

CENTRAL-FORCE MOTION

235

or, equivalently,

sin (θ + const.) =

2 2µ k − + 2 r 2 µE 2µ k 2 +8 2

(4)

We can choose the point from which θ is measured so that the constant in (4) is − π 2 . Then,

cos θ =

1 −1 µk r 2E 2 1+ µk 2

2

(5)

which is the desired expression.

8-3. When k → k 2 , the potential energy will decrease to half its former value; but the kinetic energy will remain the same. Since the original orbit is circular, the instantaneous values of T and U are equal to the average values, T and U . For a 1 r 2 force, the virial theorem states

T =− Hence,

1 U 2

(1)

1 1 E = T +U = − U +U = U 2 2 Now, consider the energy diagram

E D

(2)

C

r –k/r

B

A

where CB = E CA = U CD = UC original total energy original potential energy original centrifugal energy

The point B is obtained from CB = CA − CD . According to the virial theorem, E = (1 2) U or

CB = (1 2) CA. Therefore, CD = CB = BA

236

CHAPTER 8

Hence, if U suddenly is halved, the total energy is raised from B by an amount equal to (1 2) CA or by CB . Thus, the total energy point is raised from B to C; i.e., E(final) = 0 and the orbit is parabolic.

8-4. Since the particle moves in a central, inverse-square law force field, the potential energy is

U=− so that the time average is

U =−

k r

(1)

τ∫

0

1

τ

k dt r

(2)

Since this motion is a central motion, the angular momentum is a constant of motion. Then,

µ r 2 θ ≡ = const. from which dt = Therefore, (2) becomes U =− 1

2π

(3)

µr 2

dθ

(4)

τ

∫

0

k µr 2 kµ dθ = − τ r

2π

∫ rdθ

0

(5)

Now, substituting α r = 1 + ε cos θ and τ = ( 2µ U =− k α 2π a 3 2

)π

2π 0

α a 3 2 , (5) becomes

1 (6)

∫ 1 + ε cos θ dθ

where a is the semimajor axis of the ellipse. Using the standard integral [see Eq. (E.15), Appendix E],

2π

∫ 1 + ε cos θ dθ =

0

1

2π 1− ε2

(7)

and the relation,

α = a (1 − ε 2 )

(6) becomes U =− The kinetic energy is k a

(8)

(9)

CENTRAL-FORCE MOTION

237

2 1 T = − µr 2 + 2 2µ r 2

(10)

and the time average is

T =−

Part of this integral is trivial,

1

τ

∫ T dt =

0

τ

1 2π

kµ a3

2π

∫ Tr

0

2

dθ

(11)

T =

1 2π

2π kµ µ π 2 ( rr ) 2 dθ + a3 2 ∫ µ 0

(12)

To evaluate the integral above, substitute the expression for r and make a change of variable 1 2

2π

∫ ( rr )

0

2

1 ε dθ = 2 µ

2 2π

∫ (1 + ε cos θ )

0

sin 2 θ dθ

2

ε = µ

2 1

−1

∫ (1 + ε x )

1 − x 2 dx

2

(13)

The reader is invited to evaluate this integral in either form. The solution presented here is to integrate by parts twice, which gives a third integral that can be looked up in a table:

−1

∫ (1 + ε x )

1

1 − x 2 dx

2

1 − x2 =− ε (1 + ε x )

1

−

−1

ε ∫

1

1

x dx 1 − x 2 (1 + ε x )

(14)

−1

1 x sin −1 x =− ε 1 + εx

1

−1

−∫

sin −1 x dx 2 −1 ( 1 + ε x )

1

(15)

1

1 2 tan −1 = − 2 sin −1 x + 2 ε 1− ε

(1 − ε )(1 − x ) (1 + ε )(1 + x ) −1

(16)

=

π 1 −1 + 2 ε 1 − ε2 k 2a

(17)

Substituting this into (13) and then into (12), we obtain the desired answer,

T =

(18)

This explicitly verifies the virial theorem, which states that for an inverse-square law force,

T =−

1 U 2

(19)

238

8-5.

a m1 m2

CHAPTER 8

Suppose two particles with masses m1 and m2 move around one another in a circular orbit with radius a. We can consider this motion as the motion of one particle with the reduced mass µ moving under the influence of a central force G m1m2 a 2 . Therefore, the equation of motion before the particles are stopped is

µ aω 2 = G where 1

m1m2 a2

2π

(1)

µ

The radius of circular motion is

=

1 1 + , m1 m2

ω=

τ

(2)

G m1m2 τ 2 a= 2 4π µ

13

(3)

After the circular motion is stopped, the particle with reduced mass µ starts to move toward the force center. We can find the equation of motion from the conservation of energy: −G or, 2G m1m2 x= µ Therefore, the time elapsed before the collision is t = ∫ dt = − ∫ a m1m2 1 2 mm = µx − G 1 2 a 2 x

12

(4)

1 1 − x a

(5)

0

dx 2G m1m2 1 1 x − a µ

(6)

where the negative sign is due to the fact that the time increases as the distance decreases. Rearranging the integrand, we can write t=− aµ 2G m1m2

∫ a 0

x dx a−x

(7)

Setting x ≡ y 2 ( dx = 2 y dy ) , the integral in (7) becomes I=∫ a 0

x dx = 2 a−x

∫ a 0

y2 a − y2

dy

CENTRAL-FORCE MOTION

239

Using Eq. (E.7), Appendix E, we find y a − y2 a y + sin −1 I = 2 − 2 2 a Therefore, t= or, t= aµ aπ 2G m1m2 2

0

=− a πa

2

(8)

τ

4 2

(9)

8-6. m1 x1 O r r = x2 − x1

m2 x2

When two particles are initially at rest separated by a distance r0 , the system has the total energy E0 = −G m1m2 r0 (1)

The coordinates of the particles, x1 and x2 , are measured from the position of the center of mass. At any time the total energy is E= 1 1 mm 2 2 m1 x1 + m2 x2 − G 1 2 2 2 r p = m1 x1 + m2 x2 = 0 From the conservation of energy we have E = E0 , or −G Using (3) in (4), we find x1 = v1 = m2 2G 1 1 − M r r0 (5) x1 = v2 = − m1 2G M 1 1 − r r0 m1m2 1 1 mm 2 2 = m1 x1 + m2 x2 − G 1 2 r0 2 2 r (4) (2)

and the linear momentum, at any time, is (3)

240

CHAPTER 8

8-7. Since F ( r ) = − kr is a central force, angular momentum is conserved and the areal velocity, dA dt = 2µ , is trivially constant (see Section 8.3). In order to compute U , we start with

dt =

dr

2 2 E−U − µ 2µ r 2

(1)

and U= kr 2 2 (2)

The time average of the potential energy becomes

U = 1

τ

2

∫ U dt

0 rmax rmin

τ

=

τ∫

kr 3 2

dr

2 2 2 kr 4 Er − − 2 2µ µ

(3)

Substituting r2 = x (4) becomes dr = 1 dx 2r (4)

U =

k 2τ

µ

2 ∫

2 rmax 2 rmin

x dx k − + Ex − x 2 2µ 2

2

(5)

Using the integrals in Eqs. (E.9) and (E.8c), Appendix E,

∫

(5) becomes k U = 2τ

x dx ax 2 + bx + c

=

2ax + b b 1 ax 2 + bx + c + sin −1 2 a 2a − a b − 4 ac

(6)

12 µ E 2

− 2 k k

2 E − kr 2 2 k 4 2 − − r + Er − sin 2 12 2 2µ E2 − k k µ rmin

−1

rmax

(7)

But rmax and rmin were originally defined as the roots of vanishes at both limits of integration. On the other hand,

E−U −

2

2µ r 2

. Hence, the second term

CENTRAL-FORCE MOTION

241 rmax rmin

τ =2∫

dr

2 2 E−U − 2 µ 2µ r rmax rmin

= 2µ

∫

rdr

2 k − r 4 + Er 2 − 2 2µ

(8)

or, using (5),

τ=

µ

2 ∫

2 rmax

dx −

12 2 k 2 x + Ex − 2 2µ

2 rmin

µ 2 −1 =− sin

2 k

2 12 E2 − k µ E − kr 2

rmax

(9) rmin Using (9) to substitute for τ in (7), we have U = Now, T =E− U = The virial theorem states: T = In our case n = 1, therefore, T = U = E 2 (13) n+1 U 2 when U = kr n +1 (12) E 2 (11) E 2 (10)

8-8.

The general expression for θ (r) is [see Eq. (8.17)]

θ (r) = ∫

(

r 2 dr

)

2 2µ E − U − 2 2µ r

(1)

where U = − ∫ kr dr = − kr 2 2 in the present case. Substituting x = r 2 and dx = 2r dr into (1), we have

242

CHAPTER 8

θ (r) =

1 2∫

dx µk 2µE x 2 x2 + 2 x − 1

(2)

Using Eq. (E.10b), Appendix E,

∫x

dx ax + bx + c

2

=

bx + 2c 1 sin −1 −c x b 2 − 4 ac

(3)

and expressing again in terms of r, we find µE 2 2 r − 1 1 −1 θ ( r ) = sin 2 2 2 µ E 2 µk + 2 r 4 or, sin 2 (θ − θ 0 ) = 1 k 1+ µE2

2

+ θ0

(4)

−

1 r2

2

µE

2

k 1+ µE2

(5)

In order to interpret this result, we set 1+ k ≡ ε′ 2 µE 2 ≡ α′ µE

2

(6)

and specifying θ 0 = π 4 , (5) becomes

α′ r2 or,

= 1 + ε ′ cos 2θ

(7)

α ′ = r 2 + ε ′r 2 ( cos 2 θ − sin 2 θ )

Rewriting (8) in x-y coordinates, we find

(8)

α ′ = x2 + y2 + ε ′ ( x2 − y2 ) or, y2 x2 1= + α′ α′ 1 + ε′ 1 − ε′ Since a′ > 0, ε′ > 1 from the definition, (10) is equivalent to

(9)

(10)

CENTRAL-FORCE MOTION

243

1=

y2 x2 + α′ α′ 1 + ε′ 1− ε′

(11)

which is the equation of a hyperbola.

8-9. (a) By the virial theorem, T = − U 2 for a circular orbit.

The firing of the rocket doesn’t change U, so U f = U i But Tf = So E f = 2Ti + U i = −U i + U i = 0

Ef Ei =0

1 m v 2 + v 2 = 2Ti 2

(

)

The firing of the rocket doesn’t change the angular momentum since it fires in a radial direction. f i

=1

(b) E = 0 means the orbit is parabolic. The satellite will be lost.

E(r) = 0 T (r) = E − U = V (r) = U (r) + Behavior of V(r) is determined by 2 2µ r 2 − GM m r e s

U (r) = − GMe ms r

2

GMe ms r

2µ r 2

=−

2 GMe ms + 2µ r 2 r

for small r for large r

244

CHAPTER 8

Energy V(r) T(r) r0 E(r) = 0

U(r)

Minimum in V(r) is found by setting

dV = 0 at r = r0 dr

2 GMe ms 0=− + 3 µ r0 r02

r0 = −

2

µGMe ms

8-10.

For circular motion T= 1 meω 2 re2 2 GMs me re

U=−

We can get ω 2 by equating the gravitational force to the centripetal force GMs me = meω 2 re re2 or

ω2 =

So T=

GMs re3

GM GMs me 1 1 me re2 ⋅ 3 s = =− U re 2 2re 2 E=T +U = 1 U 2

If the sun’s mass suddenly goes to

1 its original value, T remains unchanged but U is halved. 2

CENTRAL-FORCE MOTION

245

1 1 1 U=− U+ U=0 2 2 2

E′ = T ′ + U ′ = T +

The energy is 0, so the orbit is a parabola. For a parabolic orbic, the earth will escape the solar system.

8-11.

For central-force motion the equation of orbit is [Eq. (8.21)] d2 1 1 µr 2 + = − 2 F (r) 2 dθ r r

(1)

r force center θ

a

In our case the equation of orbit is r = 2 a cos θ

(2)

Therefore, (1) becomes 1 d2 1 4 a2 µ ( cos θ )−1 + 2a ( cos θ )−1 = − 2 F ( r ) cos2 θ 2 a dθ 2 But we have d2 d sin θ −1 cos θ ) = 2 ( dθ cos 2 θ dθ = Therefore, we have 2 sin 2 θ 8 a3 µ 1 1 + + = − 2 F ( r ) cos 2 θ cos θ cos 3 θ cos θ or, (5) 2 sin 2 θ 1 + cos θ cos 3 θ (4)

(3)

F (r) = − so that

2 8 a2 =− µ 8 a 3 µ cos 5 θ

2

2

1 r5

(6)

F (r) = −

k r5

(7)

246

8-12.

CHAPTER 8

re

βre

The orbit of the comet is a parabola (ε = 1), so that the equation of the orbit is

α r = 1 + cos θ

(1)

We choose to measure θ from perihelion; hence

r (θ = 0 ) = β rE

Therefore,

(2)

α=

2

µk

= 2β rE

(3)

Since the total energy is zero (the orbit is parabolic) and the potential energy is U = − k r , the time spent within the orbit of the Earth is

T = 2∫

rE

dr

2 2 k r − 2µ r 2 µ rE rE

ΒrE

=

2µ k

∫β

r dr r − β rE rE (4)

2µ 2 ( −2β rE − r ) = r − β rE − 3 k β rE from which

T=

2µ k

2 3 2 3 rE ( 2β + 1) 1 − β

4π 2 µE 3 rE k′

(5)

Now, the period and the radius of the Earth are related by

2 τE =

(6)

or,

CENTRAL-FORCE MOTION

247

3 rE 2 =

k′ τE µE 2π

(7)

Substituting (7) into (5), we find

T=

2µ 2 k 3

k′ τE ( 2β + 1) 1 − β µE 2π

(8)

where k = GMs µ and k ′ = GMs µE . Therefore,

T=

1 3π

2 (1 − β ) (1 + 2β ) τ E

(9)

where τ E = 1 year . Now, β = rMercury rE = 0.387 . Therefore,

T= so that

1 3π

2 (1 − 0.387 ) (1 + 2 × 0.387 ) × 365 days

T = 76 days

(10)

8-13.

Setting u ≡ 1 r we can write the force as

F=−

k λ − = − ku 2 − λ u3 r2 r3

(1)

Then, the equation of orbit becomes [cf. Eq. (8.20)]

d2u µ 1 + u = − 2 2 − ku2 − λ u3 2 dθ u

(

)

(2)

from which

d2u µλ µ k + u 1 − 2 = 2 2 dθ or, (3)

d 2 u µλ µk 1 + 1 − 2 u − 2 =0 2 µλ dθ 1− 2

If we make the change of variable,

(4)

v=u−

µk

2

1 1−

µλ

2

(5)

we have

d 2 v µλ + 1− 2 v = 0 dθ 2

(6)

248 or, CHAPTER 8

d2v + β 2v = 0 dθ 2

(7)

where β 2 = 1 − µλ 2 . This equation gives different solutions according to the value of λ. Let us consider the following three cases:

i)

λ<

2

µ:

For this case β 2 > 0 and the solution of (7) is

v = A cos ( βθ − δ )

By proper choice of the position θ = 0, the integration constant δ can be made to equal zero. Therefore, we can write 1 = A cos βθ + r

2

µk − µλ

(9)

When β = 1 (λ = 0), this equation describes a conic section. Since we do not know the value of the constant A, we need to use what we have learned from Kepler’s problem to describe the motion. We know that for λ = 0, 1 µk = 2 (1 + ε cos θ ) r and that we have an ellipse or circle (0 ≤ ε < 1) when E < 1, a parabola (ε = 1) when E = 0, and a hyperbola otherwise. It is clear that for this problem, if E ≥ 0, we will have some sort of parabolic or hyperbolic orbit. An ellipse should result when E < 0, this being the only bound orbit. When β ≠ 1, the orbit, whatever it is, precesses. This is most easily seen in the case of the ellipse, where the two turning points do not have an angular separation of π. One may obtain most constants of integration (in particular A) by using Equation (8.17) as a starting point, a more formal approach that confirms the statements made here. ii) λ=

2

µ

For this case β 2 = 0 and (3) becomes

d2u µk = 2 dθ 2 so that 1 µk 2 = θ + Aθ + B r 2 2

(10)

u=

(11)

from which we see that r continuously decreases as θ increases; that is, the particle spirals in toward the force center.

CENTRAL-FORCE MOTION

249

iii)

λ>

2

µ

For this case β 2 < 0 and the solution (7) is v = A cosh

(

−β 2 θ − δ

)

(12)

δ may be set equal to zero by the proper choice of the position at which θ = 0. Then,

1 = A cosh r

(

−β 2 θ +

)

2

µk − µλ

(13)

Again, the particle spirals in toward the force center.

8-14.

The orbit equation for the central-force field is [see Eq. (8.17)] 2µ r 4 dr = 2 dθ

2 2 E−U − 2 2µ r

(1)

But we are given the orbit equation: r = kθ 2 from which dr 2 2 dθ = 4 k θ Substituting (2) into (3), we have dr 2 r dθ = 4 k k = 4 kr From (1) and (4), we find the equation for the potential U: 4 kr = from which U = E− and F ( r ) = − ∂U ∂r . Therefore, F (r) = − 6k 1 + µ r4 r3

2

2 2µ r 4 E−U − 2 2 2µ r 2 2

(2)

(3)

(4)

(5)

2k

2

µ

2 1 1 − 3 r 2µ r 2

(6)

(7)

250

8-15.

A m r θ P b 2

CHAPTER 8

B

Let us denote by v the velocity of the particle when it is infinitely far from P and traveling along the line AB. The angular momentum is

= where we have used m = 1. Therefore,

k = mvb 2 = vb 2 b

(1)

v=

k 2 b2

(2)

The total energy E of the particle is equal to the initial kinetic energy: E= 1 2 k v = 4 2 4b (3)

The general orbit equation for a force, F ( r ) = − k r 5 , is dθ = dr r

2 2 k 2 E + 4 − 2 4r 2r

(4)

Substituting for and E from (1) and (3), we have dθ = k 1 b r2 dr k k k + 4− 2 2 4 b r 2b 2r dr r − 2b 2 r 2 + b 4

4

=b 2 =b 2 or,

dr

(r

2

− b2

)

2

dθ = − b 2

dr r − b2

2

(5)

where we have taken the negative square root because r decreases as θ increases (see the diagram). We can now use the integral [see Eq. (E.4b), Appendix E]

CENTRAL-FORCE MOTION

251

∫a x

2

dx 1 = − coth −1 2 2 −b ab

ax b

(6)

from which we obtain r θ = 2 coth −1 + θ 0 b or, θ − θ0 r = b coth 2 Now, coth φ → ∞ as φ → 0, since r → ∞ as θ → 0, we must have θ 0 = 0 . Thus, r = b coth θ

(7)

(8)

(

2

)

(9)

Notice that r is always greater than b (because coth φ → 1 as φ → ∞), so that the denominator in (5) never equals zero nor changes sign. Thus, r always decreases as θ increases. This is, the particle spirals in toward P but never approaches closer than a distance b.

8-16.

The total energy of the particle is

E = T +U

(1)

a principle that by no means pushes the philosophical envelope of physical interpretation. The impulse that causes v → v + δv changes the kinetic energy, not the potential energy. We therefore have

δ E = δ T = δ mv 2 = mv δ v

1 2 By the virial theorem, for a nearly circular orbit we have 1 E = − mv 2 2 so that

(2)

(3)

δE

−E

=

2δ v v

(4)

where we have written –E since E < 0. The major and minor axes of the orbit are given by

a=−

k 2E

b=

−2µE

(5)

Now let us compute the changes in these quantities. For a we have

δ a = −δ and for b we have

k kδ E δE = 2 = a 2E 2E −E

(6)

252

CHAPTER 8

δb = δ

δ + = −2µE −2µE3 −2µE

δE δ 1 − 2 δ E = b − 2E

(7)

Easily enough, we can show that δ

= δ v v and therefore

δa a 8-17.

=

δb b =

δE

−E

=

2δ v v

(8)

The equation of the orbit is

α r from which

= 1 + ε cos θ

(1)

r= where α =

α

1 + ε cos θ

(2)

2E 2 . Therefore, the radial distance r can vary from the mk 2 maximum value α (1 − ε ) to the minimum value α (1 + ε ) . Now, the angular velocity of the particle is given by

2

µ k and ε = 1 +

ω=

µr 2

(3)

Thus, the maximum and minimum values of ω become

ω max =

µr

2 min

=

µ

α 1 + ε α 1 − ε

2

2

ω min =

2 µ rmax

=

µ

2

(4)

Thus,

ω max 1 + ε = =n ω min 1 − ε from which we find

(5)

ε=

n −1 n +1

(6)

8-18. Kepler’s second law states that the areal velocity is constant, and this implies that the angular momentum L is conserved. If a body is acted upon by a force and if the angular momentum of the body is not altered, then the force has imparted no torque to the body; thus,

CENTRAL-FORCE MOTION

253

the force must have acted only along the line connecting the force center and the body. That is, the force is central. Kepler’s first law states that planets move in elliptical orbits with the sun at one focus. This means the orbit can be described by Eq. (8.41):

α r = 1 + ε cos θ

with 0 < ε < 1

(1)

On the other hand, for central forces, Eq. (8.21) holds:

d2 dθ 2

µr 1 1 r + r = − 2 F (r)

2

(2)

Substituting 1 r from (1) into the left-hand side of (2), we find 1 µ = − 2 r2 F (r) a which implies, that (3)

F (r) = −

2

αµ r 2

(4)

8-19. The semimajor axis of an orbit is defined as one-half the sum of the two apsidal distances, rmax and rmin [see Eq. (8.44)], so

1 1 α α α [ rmax + rmin ] = 2 1 + ε + 1 − ε = 1 − ε 2 2

(1)

This is the same as the semimajor axis defined by Eq. (8.42). Therefore, by using Kepler’s Third Law, we can find the semimajor axis of Ceres in astronomical units:

kC 2 τC aC 4π 2 µC = aE kE τ 2 4π 2 µE E where kc = γ Ms mc , and 1

(2)

µc

=

1 1 + Ms mc

Here, Ms and mc are the masses of the sun and Ceres, respectively. Therefore, (2) becomes

aC Ms + mc = aE Ms + me from which

τc τE

2 13

(3)

254

1 333, 480 + aC 8, 000 = ( 4.6035) 2 aE 333, 480 + 1 so that

13

CHAPTER 8

(4)

aC ≅ 2.767 aE

The period of Jupiter can also be calculated using Kepler’s Third Law:

(5)

4π 2 µ J 3 aJ τ J kJ = τ E 4π 2 µE 3 aE kE from which

12

M + m aJ 3 E = s Ms + mJ aE

12

(6)

12 τ J 333, 480 + 1 = ( 5.2028 ) 3 τ E 333, 480 + 318.35

(7)

Therefore,

τJ ≅ 11.862 τE

The mass of Saturn can also be calculated from Kepler’s Third law, with the result

(8)

ms ≅ 95.3 me

(9)

8-20.

Using Eqs. (8.42) and (8.41) for a and r, we have 1 1 + ε cos θ a cos θ cos θ = ∫ dt r τ 0 1− ε2

4 4

τ

(1)

From Kepler’s Second Law, we can find the relation between t and θ:

dt =

τ τ 1 α2 dA = dθ π ab π ab 2 (1 + ε cos θ ) 2

(2)

since dA = (1 2) r 2 dθ . Therefore, (1) becomes 1 1 a cos θ = r τ 1− ε2

4

(

)

4

τ a2 π ab 2

2π

∫ cos θ (1 + ε cos θ )

0

2

dθ

(3)

It is easily shown that the value of the integral is 2πε. Therefore,

CENTRAL-FORCE MOTION

255

1 a cos θ = r 1 − ε2

4

(

)

4

1 2 α ε ab

(4)

After substituting a and b in terms of ε and α [see Eqs. (8.42) and (8.43)], we obtain

ε a cos θ = r 1 − ε2

4

(

)

52

(5)

8-21. If we denote the total energy and the potential of the family of orbits by E and U(r), we have the relation

2 1 2 µr + + U (r) = E 2 2µ r 2

(1)

from which

2

1 = 2µ r 2 E − U ( r ) − µ r 2 2

(2)

Here, E and U(r) are same for all orbits, and the different values of result from different values of (1 2) µ r 2 . For stable circular motion, r = 0 , but for all other motions, r ≠ 0 . Therefore, for noncircular motions, r 2 > 0 and is smaller than for the circular case. That is, the angular momentum of the circular orbit is the largest among the family.

8-22.

For the given force, F ( r ) = − k r 3 , the potential is U (r) = − k 2r 2 (1)

and the effective potential is V (r) = The equation of the orbit is [cf. Eq. (8.20)] d2u µ + u = − 2 2 − ku3 2 dθ u 1 2 2 1 µ − k r2 (2)

(

)

(3)

or, d2u µk + 1− 2 u = 0 dθ 2 Let us consider the motion for various values of . (4)

256

i)

2

CHAPTER 8

= µk :

In this case the effective potential V(r) vanishes and the orbit equation is d2u =0 dθ 2 with the solution u= 1 = Aθ + B r (6) (5)

and the particle spirals toward the force center. ii) 2

> µk :

In this case the effective potential is positive and decreases monotonically with increasing r. For any value of the total energy E, the particle will approach the force center and will undergo a reversal of its motion at r = r0 ; the particle will then proceed again to an infinite distance.

V(r) E

r0

r

Setting 1 − µ k

2

≡ β 2 > 0 , (4) becomes d2u + β 2u = 0 dθ 2 (7)

with the solution u= 1 = A cos ( βθ − δ ) r (8)

Since the minimum value of u is zero, this solution corresponds to unbounded motion, as expected from the form of the effective potential V(r). iii) 2

< µk :

2

For this case we set µ k

− 1 ≡ G 2 > 0 , and the orbit equation becomes d2u − G2u = 0 2 dθ (9)

with the solution u= 1 = A cosh ( βθ − δ ) r (10)

so that the particle spirals in toward the force center.

CENTRAL-FORCE MOTION

257

In order to investigate the stability of a circular orbit in a 1 r 3 force field, we return to Eq. (8.83) and use g ( r ) = k µ r 3 . Then, we have x− or,

2 1 =0 x + k − ⋅ 3 µ µρ 1 + ( x ρ ) 3 2

µ ρ 1 + ( x ρ )

2 3

3

=−

k

µρ 1 + ( x ρ )

3

3

(11)

(12)

Since r r = p = 0 , Eq. (8.87) shows that k =

2

µ . Therefore, (12) reduces to x=0 (13)

so that the perturbation x increases uniformly with the time. The circular orbit is therefore not stable. We can also reach the same conclusion by examining the basic criterion for stability, namely, that ∂V ∂r = 0 and r=ρ ∂ 2V ∂r 2

>0 r=ρ The first of these relations requires k = 2 µ while the second requires 2 µ > k . Since these requirements cannot be met simultaneously, no stable circular orbits are allowed.

8-23.

Start with the equation of the orbit:

α r and take its time derivative

= 1 + ε cos θ

(1)

r ε ε = θ sin θ = sin θ 2 r α αµ r 2 Now from Equation (8.45) and (8.43) we have 2µ 2πµ aα 1− ε2

(2)

τ= so that from (2)

⋅ π ab =

(3)

r max = as desired.

ε 2π aε ⋅ = µ α τ 1 − ε2

(4)

258

8-24.

(b) b (a) r θ

CHAPTER 8

ra

rp

a) With the center of the earth as the origin, the equation for the orbit is

α r Also we know

= 1 + ε cox θ

(1)

rmin = a (1 − ε ) rmax = a (1 + ε ) rmin = rp = 300 km + re = 6.67 × 10 6 m rmax = ra = 3500 km + re = 9.87 × 10 6 m a= 1 ra + rp = 8.27 × 106 m 2

(2)

(

)

Substituting (2) gives ε = 0.193. When θ = 0, a = 1+ ε rmin which gives α = 7.96 × 106 m . So the equation of the orbit is 7.96 × 10 6 m = 1 + 0.193 cos θ r When θ = 90°, r = α = 7.96 × 10 6 m The satellite is 1590 km above the earth.

b)

b θ β a – rmin

CENTRAL-FORCE MOTION

259

θ =π −β

= π − tan −1 Using b = α a b a − rmin

θ = π − tan −1

Substituting into (1) gives

αa a − rmin

101°

r = 8.27 × 10 6 m ; which is 1900 km above the earth

8-25. Let us obtain the major axis a by exploiting its relationship to the total energy. In the following, let M be the mass of the Earth and m be the mass of the satellite.

E=−

GMm 1 GMm 2 = mvp = 2a 2 rp

(1)

where rp and vp are the radius and velocity of the satellite’s orbit at perigee. We can solve for a and use it to determine the radius at apogee by 2GM ra = 2 a − rp = rp − 1 2 rp vp Inserting the values G = 6.67 × 10 −11 N ⋅ m 2 ⋅ kg −2 M = 5.976 × 10 24 kg rp = 6.59 × 106 m vp = 7.797 × 10 3 m ⋅ s −1 we obtain ra 1.010 rp = 6.658 × 106 m , or 288 km above the earth’s surface. We may get the mra va = mrp vp giving va = 27,780 km ⋅ hr −1 . The period can be found from Kepler’s third law t2 = 4π 2 a 3 GM (8) (7) (3) (4) (5) (6)

−1

(2)

speed at apogee from the conservation of angular momentum,

Substitution of the value of a found from (1) gives τ = 1.49 hours.

260

8-26.

r ra rp

CHAPTER 8

First, consider a velocity kick ∆v applied along the direction of travel at an arbitrary place in the orbit. We seek the optimum location to apply the kick.

E1 = initial energy = 1 GMm mv 2 − 2 r

E2 = final energy = 1 GMm 2 m ( v + ∆v ) − 2 r

We seek to maximize the energy gain E2 − E1 : E2 − E1 = 1 m 2v ∆v + ∆ v 2 2

(

)

For a given ∆v , this quantity is clearly a maximum when v is a maximum; i.e., at perigee. Now consider a velocity kick ∆V applied at perigee in an arbitrary direction:

∆v v2

v1

The final energy is 1 GMm 2 mv2 = 2 rp This will be a maximum for a maximum v 2 ; which clearly occurs when v 1 and ∆v are along the same direction. Thus, the most efficient way to change the energy of an elliptical orbit (for a single engine thrust) is by firing along the direction of travel at perigee.

CENTRAL-FORCE MOTION

261

8-27.

By conservation of angular momentum

mra va = mrp vp or

Substituting gives

va =

rp vp ra

va = 1608 m/s

8-28. Use the conservation of energy for a spacecraft leaving the surface of the moon with just enough velocity vesc to reach r = ∞:

Ti + U i = Tf + U f

1 GM m m 2 =0+0 mvesc − 2 rm

vesc = where 2GMm rm

Mm = mass of the moon = 7.36 × 10 22 kg rm = radius of the moon = 1.74 × 10 6 m

Substituting gives

vesc = 2380 m/s vmax = v + v0 , vmin = v − v0 mva ra = mvb rb or 8-29.

From conservation of angular momentum we know

vmax rmin = vmin rmax ;

Also we know

rmax vmax = rmin vmin

(1)

rmin = a (1 − e ) rmax = a (1 + e )

Dividing (3) by (2) and setting the result equal to (1) gives

(2) (3)

262

CHAPTER 8

rmax 1 + e vmax = = rmin 1 − e vmin vmin (1 + e ) = vmax (1 − e ) e ( vmin + vmax ) = vmax − vmin e ( 2v ) = 2v0 e= v0 v

8-30. To just escape from Earth, a velocity kick must be applied such that the total energy E is zero. Thus

GMe m 1 2 =0 mv2 − 2 r where (1)

v2 = velocity after kick Me = 5.98 × 10 24 kg G = 6.67 × 10 −11 Nm 2/kg 2 r = 200 km + re = 200 km + 6.37 × 10 6 m = 6.57 × 106 m Substituting into (1) gives v2 = 11.02 km/sec . For a circular orbit, the initial velocity v1 is given by Eq. (8.51) v1 = GMe = 7.79 km/sec r

Thus, to escape from the earth, a velocity kick of 3.23 km/sec must be applied.

CENTRAL-FORCE MOTION

263

Since E = 0, the trajectory is a parabola.

parabolic escape orbit

Earth

circular orbit

8-31.

From the given force, we find dF ( r ) 2k 4 k ′ = F′ (r) = 3 + 5 dr r r (1)

Therefore, the condition of stability becomes [see Eq. (8.93)] k ρ 2 + 2k ′ F ′ ( ρ) 3 ρ 5 3 + = + >0 1 F ( ρ) ρ − kρ 2 + k ′ ρ 4 2

(

)

ρ

(

)

(2)

or, kρ2 − k ′ >0 ρ kρ2 + k ′

(

)

(3)

Therefore, if ρ 2 k > k ′ , the orbit is stable.

8-32.

For this force, we have dF ( r ) k 2k = F′ ( r ) = 3 e−r a + 2 e−r a dr r ar k = 3 e−r a r r 2 + a

(1)

Therefore, the condition of stability [see Eq. (8.93)] becomes

F′ (r) 3 + = F (r) r This condition is satisfied if r < a. r − 2 + + 3 a >0 r

(2)

264

CHAPTER 8

8-33. The Lagrangian of the particle subject to a gravitational force is written in terms of the cylindrical coordinates as

L = T −U = From the constraint r 2 = 4 az , we have

1 m r 2 + r 2θ 2 + z 2 − mgz 2

(

)

(1)

z= Therefore, (1) becomes L= Lagrange’s equation for θ is

rr 2a

(2)

mg 2 r2 1 m 1 + 2 r 2 + r 2θ 2 − r 2 4a 4a d ∂L d ∂L − =− mr 2θ = 0 ∂θ dt ∂θ dt

(3)

(

)

(4)

This equation shows that the angular momentum of the system is constant (as expected): mr 2θ = = const. Lagrange’s equation for r is mg m d r2 ∂ L d ∂L r − m 1 + 2 r = 0 − = 2 rr 2 + mrθ 2 − dt 4 a 2a ∂r dt ∂r 4 a from which mg m 2 r2 m rr + mrθ 2 − r − m 1 + 2 r − 2 rr 2 = 0 2 4a 2a 2a 4a After rearranging, this equation becomes

2 mg r2 m 1 m 1 + 2 r + 2 rr 2 + r− =0 m r3 4a 2a 4a

(5)

(6)

(7)

(8)

For a circular orbit, we must have r = r = 0 or, r = ρ = constant. Then,

2 mg ρ = 2a mρ 3

(9)

or,

2

=

m2 g 4 ρ 2a

(10)

Equating this with

2

= m2 ρ 4θ 2 , we have m2 ρ 4θ 2 = m2 g 4 ρ 2a (11)

CENTRAL-FORCE MOTION

265

or,

θ2 =

g 2a

(12)

Applying a perturbation to the circular orbit, we can write r → ρ + x where This causes the following changes: r 2 → ρ 2 + 2ρ x x 1− 3 1 ρ → ρ3 r3 r→x r→x from which, we have rr 2 → ( ρ + x ) x 2 ≅ 0, in lowest order 2 2 2 r r → ρ + 2ρ x x ≅ ρ x , in lowest order (15) x

ρ

1

(13)

(14)

(

)

Thus, (8) becomes

2 mg 1 ρ + x) − m 1 + 2 ρ 2 x + ( mρ 3 2a 4a

x 1 − 3 ρ = 0

(16)

But

2 mg ρ = 2a mρ 3

(17)

so that (16) becomes mg ρ2 3 2 m 1 + 2 x + x+ x=0 2a mρ 4 4a Substituting (17) into (18), we find 2mg ρ2 m 1 + 2 x + x=0 a 4a or, x+ 2g a+ (19) (18)

ρ2

4a

x=0

(20)

266

Therefore, the frequency of small oscillations is

CHAPTER 8

ω= where 2g a + z0

(21)

z0 =

ρ2

4a

8-34.

The total energy of the system is E= 1 m r 2 + r 2θ 2 + r 2 cot 2 α + mgr cot α 2

(

)

(1)

or, E= Substituting 1 1 m 1 + cot 2 α r 2 + mr 2θ 2 + mgr cot α 2 2

(

)

(2)

= mr 2θ , we have E=

2 1 + mgr cot α m 1 + cot 2 α r 2 + 2 2mr 2

(

)

(3)

Therefore, the effective potential is V (r) =

2

2mr 2

+ mgr cot α

(4)

At the turning point we have r = 0 , and (3) becomes a cubic equation in r: mgr 3 cot α − Er 2 +

2

2m

=0

(5)

V(r

)

E

Energy

mg

ta r co

r1

r2

r

This cubic equation has three roots. If we attempt to find these roots graphically from the intersections of E = const. and V ( r ) = 2 2mr 2 + mgr cot α , we discover that only two of the roots are real. (The third root is imaginary.) These two roots specify the planes between which the motion takes place.

CENTRAL-FORCE MOTION

267

8-35.

If we write the radial distance r as r = ρ + x,

ρ = const.

(1)

then x obeys the oscillatory equation [see Eqs. (8.88) and (8.89)]

2 x + ω0 x = 0

(2)

where

ω0 =

3g ( ρ )

ρ

+ g ′ ( ρ)

(3)

The time required for the radius vector to go from any maximum value to the succeeding minimum value is ∆t = where τ 0 = 2π , the period of x. Thus, ∆t =

τ0

2

(4)

ω0

π ω0 πω ω0

(5)

The angle through which the particle moves during this time interval is

φ = ω∆t =

(6)

where ω is the angular velocity of the orbital motion which we approximate by a circular motion. Now, under the force F ( r ) = − µ g ( r ) , ω satisfies the equation

µρω 2 = − F ( r ) = µ g ( ρ )

Substituting (3) and (7) into (6), we find for the apsidal angle g ( ρ)

(7)

φ=

πω = ω0

π ρ

ρ

+ g ′ ( ρ)

3 g ( ρ)

=

π g ′ ( ρ) 3+ρ g ( ρ)

(8)

Using g ( r ) =

k 1 , we have µ rn g ′ ( ρ) n =− g ( ρ) ρ (9)

Therefore, (8) becomes

φ =π

3−n

(10)

268

CHAPTER 8

In order to have the closed orbits, the apsidal angle must be a rational fraction of 2π. Thus, n must be n = 2, − 1, − 6, … n = 2 corresponds to the inverse-square-force and n = –1 corresponds to the harmonic oscillator force.

8-36.

The radius of a circular orbit in a force field described by F (r) = − k −r a e r2 (1)

is determined by equating F(r) to the centrifugal force:

2 k −ρ a = e r2 mρ 3

(2)

Hence, the radius ρ of the circular orbit must satisfy the relation

ρe − ρ a =

2

mk

(3)

Since the orbit in which we are interested is almost circular, we write r (θ ) = ρ [1 + δ (θ ) ] where δ (θ ) 1 for all values of θ. (With this description, the apsides correspond to the maximum and minimum values of δ.) We can express the following quantities in terms of δ by using (4): u= 1 1 = (1 − δ ) r ρ (5a) (4)

d2 1 1 d 2δ =− dθ 2 r ρ dθ 2

(5b)

F ( u) = − ku2 e −1 au ≅− Then, substitution into Eq. (8.20) yields − 1 d 2δ 1 mke − ρ a + (1 − δ ) = (1 − pδ a) 2 ρ dθ 2 ρ (6) k e−ρ a

ρ 2 (1 + δ )

2

(1 − ρδ a)

(5c)

Multiplying by ρ, using (3) and simplifying, (6) reduces to d 2δ + (1 − ρ a ) δ = 0 dθ 2 (7)

CENTRAL-FORCE MOTION

269

This equation obviously has two types of solution depending on whether ρ a is larger than or smaller than 1; we consider only ρ < a. (In fact, there is no stable circular orbit for ρ > a.) For the initial condition, we choose δ = δ 0 to be a maximum (i.e., an apside) at θ = 0. Then, we have

δ = δ 0 cos (1 − ρ a) θ , for ρ < a

12

(8)

This solution describes an orbit with well-defined apsides. The advance of the apsides can be found from (8) by computing for what value of θ is δ again a maximum. Thus,

θ=

The advance of the apside is given by

2π 1− ρ a

(9)

∆ = θ − 2π = 2π 1 − (1 − ρ a ) In the particular case in which ρ

−1 2

(10)

a we obtain, by extending (10),

ρ ∆ ≅ 2π − 2π 1 + 2a so that ∆≅

(11)

πρ a (12)

8-37.

From the equations in Section 8.8 regarding Hohmann transfers: ∆v = ∆v1 + ∆v2 ∆v = vt1 − v1 + v2 − vt2 ∆v = 2k r2 k k 2k r1 + − − mr1 r1 + r2 mr1 mr2 mr2 r1 + r2 (1)

Substituting k = GMe = 6.67 × 10 −11 Nm 2/kg 2 5.98 × 10 24 kg m r1 = initial height above center of Earth = 2re r2 = final height above center of Earth = 3re re = radius of the Earth = 6.37 × 10 6 m gives ∆v 1020 m/s

(

)(

)

270

8-38.

CHAPTER 8

Substitute the following into Eq. (1) of problem 8-37: k = GMs = 6.67 × 10 −11 Nm 2/kg 2 1.99 × 10 30 kg m r1 = mean Earth-sun distance 1.50 × 1011 m r2 = mean Venus-sun distance 1.08 × 1011 m

(

)(

)

The result is ∆v = −5275 m/s . The answer is negative because r2 < r1 ; so the rocket must be fired in the direction opposite to the motion (the satellite must be slowed down). ∆v = 5275 m/s; opposite to direction of motion. From Eq. (8.58), the time is given by T =π Substituting gives m 32 at = π k m k r1 + r2 2

32

(1)

τ

148 days

8-39. We must calculate the quantity ∆v1 for transfers to Venus and Mars. From Eqs. (8.54), (8.53), and (8.51):

∆v1 = vt1 − v1 = where k = GMs = 6.67 × 10 −11 m 3/s 2 -kg 1.99 × 10 30 kg m r1 = mean Earth-sun distance = 150 × 10 9 m Venus 108 − sun distance = × 109 m r2 = mean Mars 228 Substituting gives ∆vVenus = −2.53 km/sec ∆vMars = 2.92 km/sec where the negative sign for Venus means the velocity kick is opposite to the Earth’s orbital motion. Thus, a Mars flyby requires a larger ∆v than a Venus flyby.

2k r2 k − mr1 r1 + r2 mr1

(

)(

)

CENTRAL-FORCE MOTION

271

8-40. To crash into the sun, we calculate ∆v1 from Eq. (8.54) with r1 = mean distance from sun to Earth, and r2 = radius of the sun. Using Eqs. (8.54), (8.53), and (8.51) we have

( ∆v1 )sun =

Substituting

2GMs r2 GMs − r1 r1 + r2 r1

G = 6.67 × 10 −11 Nm 2/kg 2 Ms = 1.99 × 10 30 kg r1 = rse = 1.5 × 1011 m r2 = rsun = 6.96 × 10 8 m gives

( ∆v1 )sun = −26.9 km/sec

To escape from the solar system, we must overcome the gravitational pull of both the sun and Earth. From conservation of energy ( Efinal = 0 ) we have: − Substituting values gives v = 43500 m/s Now

∆v = v − v i =v− GMs rse

GMs m GMe m 1 − + mv 2 = 0 rse re 2

= ( 43500 − 29700 ) m/s

( ∆v) escape = 13.8 km/s

To send the waste out of the solar system requires less energy than crashing it into the sun.

8-41.

From the equations in Section 8.8 regarding Hohmann transfers ∆v = ∆v1 + ∆v2 = vt1 − v1 + v2 − vt2

where

272

2k r2 ; v1 = mr1 r1 + r2 2k mr2 r1 ; v1 = r1 + r2

CHAPTER 8

vt1 = vt2 = Substituting

k mr1 k mr2

k = GMe = 6.67 × 10 −11 Nm 2/kg 2 5.98 × 10 24 kg m r1 = 200 km + re = 6.37 × 10 6 m + 2 × 10 5 m r2 = mean Earth-moon distance = 3.84 × 10 8 m gives ∆v = 3966 m/s From Eq. (8.58), the time of transfer is given by T =τ Substituting gives m 32 at = π k m r1 + r2 k 2

32

(

)(

)

τ = 429, 000 sec. 5 days

8-42.

G = 6.67 × 10 −11 Nm 2/kg 2 Me = 5.98 × 10 24 kg r1 = 2 × 10 5 m + 6.37 × 10 6 m r2 = ? r3 = mean Earth-moon distance = 3.84 × 10 8 m We can get r2 from Kepler’s Third Law (with τ = 1 day) GMe τ 2 r2 = 2 4π We know E = − GMm 2r

13

= 4.225 × 107 m

CENTRAL-FORCE MOTION

273

So

E ( r1 ) = −

GMe m = −3.04 × 1011 J 2r1

E ( r2 ) = −4.72 × 1010 J E ( r3 ) = −5.19 × 10 9 J To place the satellite in a synchronous orbit would require a minimum energy of E ( r2 ) − E ( r1 ) = 2.57 × 1011 J

8-43.

In a circular orbit, the velocity v0 of satellite is given by

2 mv0 GMm GM = ⇒ v0 = 2 R R R

where M is the Earth’s mass. Conservation of energy implies

2 2 mv1 GMm mv2 GMm − = − 2 2 2R R

Conservation of angular momentum gives mRv1 = m2Rv2 From these equations, we find v1 = so the velocity need to be increased by a factor 4GM 3R 4 3 to change the orbit.

8-44.

The bound motion means that E = k −r a e . r

mv 2 +V 0 region, with the base in the x-y plane. By symmetry, x = y = 0 .

∫φ ∫θ ∫ z= π π ∫φ ∫θ ∫

=0 2 =0 =0

2π

π 2

r2

r = r1 2 r2

ρ zr 2 dr sin θ dθ dφ ρ r 2 dr sin θ dθ dφ

= 0 r = r1

Using z = r cos θ and doing the integrals gives z= ( 8(r

3 r24 − r14

3 2

−r

3 1

) )

9-2. z z=−

h ρ+h a

h y

a x

By symmetry, x = y = 0 . Use cylindrical coordinates ρ, φ, z.

ρ0 = mass density

277

278 h − ρ+ h a 0 φ =0 ρ=0 z=0 h a − ρ+h 2π 4 0 φ =0 ρ =0 z =0 2π a

CHAPTER 9

z=

∫ ∫ ∫

ρ z ρ dρ dφ dz ρ ρ dρ dφ dz

=

∫ ∫ ∫

h 4

The center of mass is on the axis 3 of the cone h from the vertex. 4

9-3. z h y

a x

By symmetry, x = y = 0 . From problem 9-2, the center of mass of the cone is at z =

1 h. 4

From problem 9-1, the center of mass of the hemisphere is at z=− So the problem reduces to i z1 = 1 1 h ; m1 = ρ1 π a 2 h 4 3 3 2 a ; m2 = ρ2 π a 3 8 3 3 a ( r2 = a, r1 = 0 ) 8

i z2 = −

z= for ρ1 = ρ2

m1 z1 + m2 z2 ρ1 h 2 − 3ρ2 a 2 = m1 + m2 4 ( ρ1 h + 2ρ2 a )

z=

h 2 − 3a3 4 ( 2a + h)

DYNAMICS OF A SYSTEM OF PARTICLES

279

9-4. y a θ/2 θ′ θ/2 a

x

By symmetry, y = 0 . If σ = mass length then M = σ aθ So x= x= Using M = σ aθ and x = a cos θ ′ , 1 M

∫θ

θ 2

′ =−

θ

2

xdm

1 M

∫θ

θ 2

′ =−

θ

2

xσ adθ ′

x= =

θ∫ θ a 1

θ 2

− 2

a cos θ ′dθ ′ =

θ a θ sin 2 − sin − 2 θ

θ

2 sin

θ

2

x= 2a

θ

sin

θ

2

y=0

9-5. mi ri

ri – r0 r0

280

CHAPTER 9

ri = position of the i th particle m1 = mass of the i th particle M = ∑ mi = total mass g = constant gravitational field Calculate the torque about r0

τ = ∑τi

= ∑ ( ri − r0 ) × Fi = ∑ ( ri − r0 ) × mi g = ∑ r1 × mi g − ∑ r0 × mi g = =

(∑ m r ) × g − (∑ m ) r i i i

0

×g

( ∑ m r ) × g − Mr i i

0

×g

Now if the total torque is zero, we must have

∑m r or r0 =

i i

= Mr0

1 M

∑m r

i i

which is the definition of the center of mass. So or center of gravity = center of mass.

τ = 0 about r0 = rCM

9-6.

Since particle 1 has F = 0, r0 = v 0 = 0 , then r1 = 0 . For particle 2 ˆ F2 = F0 x then r =

F0 ˆ x m

Integrating twice with r0 = v 0 = 0 gives r2 = F0 2 ˆ t x 2m

rCM =

F m1r1 + m2 r2 ˆ = 0 t2 x m1 + m2 4m

DYNAMICS OF A SYSTEM OF PARTICLES

281

rCM = v CM = a CM =

F0 2 ˆ t x 4m F0 ˆ tx 2m F0 ˆ x 2m

9-7. y H a O 52˚ 52˚ a H x

By symmetry y = 0 m0 = 16 mH Let mH = m, m0 = 16 m Then x= x= 1 M

∑m

1

3

i

xi

a cos 52° 1 ( 2ma cos 52°) = 9 18 m x = 0.068 a

9-8. By symmetry, x = 0 . Also, by symmetry, we may integrate over the x > 0 half of the triangle to get y . σ = mass/area

y=

∫

∫

a 2 x=0 a 2 x=0

∫

∫

a −x 2 y =0 a −x 2 y =0

σ y dy dx σ dy dx a 3 2

=

a 3 2

y=

282

9-9.

z vz POW! m1 v1 45˚ y m 2 vy

CHAPTER 9

Let the axes be as shown with the projectile in the y-z plane. At the top just before the explosion, v the velocity is in the y direction and has magnitude v0 y = 0 . 2 v0 = 2 2E0 m1 + m2 2 E0 m1 + m2

v0 y =

=

where m1 and m2 are the masses of the fragments. The initial momentum is E0 , 0 pi = ( m1 + m2 ) 0, m1 + m2 The final momentum is p f = p1 + p2 p1 = m1 ( 0, 0, v1 )

p2 = m2 vx , vy , vz The conservation of momentum equations are px : py : pz : The energy equation is 0 = m2 vx or vx = 0 or

(

)

E0 ( m1 + m2 ) = m2 vy

vy =

1 E0 ( m1 + m2 ) m2

0 = m1v1 + m2 vz or v1 = −

m2 vz m1

E0 1 1 1 2 2 2 + E0 = m1v1 + m2 vy + vz ( m1 + m2 ) 2 m1 + m2 2 2 or

2 2 2 3E0 = m1v1 + m2 vy + vz

(

)

(

)

Substituting for vy and v1 gives

DYNAMICS OF A SYSTEM OF PARTICLES

283

E0 m1 ( 2m2 − m1 ) 2 m2 ( m1 + m2 ) m2 vz gives m1 E0 ( 2m2 − m1 ) m1 ( m1 + m2 )

vz =

v1 = −

v1 = − So

m1 travels straight down with speed = v1 m2 travels in the y-z plane

2 2 v2 = v y + v z

(

)

12

=

E0 ( 4 m1 + m2 ) m2 ( m1 + m2 ) m1 ( 2m2 − m1 ) ( m1 + m2 )

θ = tan −1

vz = tan −1 vy

The mass m1 is the largest it can be when v1 = 0 , meaning 2m2 = m1 and the mass ratio is m1 1 = m2 2

9-10. y B x θ A x1 x2

First, we find the time required to go from A to B by examining the motion. The equation for the y-component of velocity is vy = v0 sin θ − gt

(1)

At B, vy = 0 ; thus tB = v0 sin θ g . The shell explodes giving m1 and m2 horizontal velocities v1 and v2 (in the original direction). We solve for v1 and v2 using conservation of momentum and energy.

px : E:

( m1 + m2 ) v0 cos θ = m1v1 + m2v2

1 1 1 ( m1 + m2 ) v02 cos2 θ + E = m1v12 + m2v22 2 2 2

(2) (3)

Solving for v2 in (2) and substituting into (3) gives an equation quadratic in v1 . The solution is

284

CHAPTER 9

v1 = v0 cos θ ± and therefore we also must have

2 m2 E m1 ( m1 + m2 )

(4)

v2 = v0 cos θ ∓

2 m1E m2 ( m1 + m2 )

(5)

Now we need the positions where m1 and m2 land. The time to fall to the ocean is the same as the time it took to go from A to B. Calling the location where the shell explodes x = 0 gives for the positions of m1 and m2 upon landing:

x1 = v1tB ;

Thus

x2 = v2tB v0 sin θ v1 − v2 g

(6)

x1 − x2 =

Using (4) and (5) and simplifying gives

(7)

x1 − x2 =

v0 sin θ g

m2 2E m1 + m1 + m2 m2 m1

(8)

9-11.

The term in question is

∑∑ fαβ α β α ≠b

For n = 3, this becomes

f12 + f13 + f 21 + f 23 + f 31 + f 32 = f12 + f 21 + f13 + f 31 + f 23 + f 32

But by Eq. (9.1), each quantity in parentheses is zero. Thus

(

) (

) (

)

α =1 β =1 α ≠β

∑∑ fαβ = 0

3

3

9-12. a)

v = v0 + u ln

m0 m

Assuming v0 = 0 , we have

m 100 v = 100 ln s 98 v = 2.02 m/s; yes, he runs out of gas.

DYNAMICS OF A SYSTEM OF PARTICLES

285

b) Relative to Stumblebum’s original frame of reference we have:

Before throwing tank 98 kg → 2.02 m/s After throwing the tank we want Stumblebum’s velocity to be slightly greater than 3 m/s (so that he will catch up to the orbiter). 8 kg V← Conservation of momentum gives 90 kg → 3m/s

( 98 kg ) ( 2.02 m/s) = ( 90 kg ) ( 3 m/s) − ( 8 kg ) v v = 9 m/s (This velocity is relative to Stumblebum’s original reference frame; i.e., before he fires his pressurized tank.) Since Stumblebum is traveling towards the orbiter at 2.02 m/s, he must throw the tank at v = 9 m/s + 2.02 m/s

v = 11 m/s

9-13.

From Eq. (9.9), the total force is given by

∑ Fα α (e)

+ ∑∑ f αβ α β α ≠β

As shown in Section 9.3, the second term is zero. So the total force is

∑ Fα α (e)

It is given that this quantity is zero. Now consider two coordinate systems with origins at 0 and 0′ mα rα rα′

r0 O

O′

where

r 0 is a vector from 0 to 0′ rα is the position vector of mα in 0

286 rα is the position vector of mα in 0′ ′

CHAPTER 9

We see that ra = r0 + rα ′ The torque in 0 is given by

( τ = ∑ rα × Fαe )

α

The torque in 0′ is

( τ ′ = ∑ rα × Fαe ) ′

α

( = ∑ rα − r 0 × Fαe )

α

(

)

( ( = ∑ rα × Fαe ) − ∑ r 0 × Fαe )

α

α

( = τ − r 0 × ∑ Fαe )

α

But it is given that

Thus

∑ Fα α (e)

=0

τ = τ′

9-14. Neither Eq. (9.11) or Eq. (9.31) is valid for a system of particles interacting by magnetic forces. The derivations leading to both of these equations assumes the weak statement of Newton’s Third Law [Eq. (9.31) assumes the strong statement of the Third Law also], which is

fαβ = − fβα

That this is not valid for a system with magnetic interactions can be seen by considering two particles of charge q1 and q2 moving with velocities v1 and v2 :

v2 f21 q2

f12 q1

v1

Now

f ij = qi v i × Bij where Bij is the magnetic field at qi due to the motion of q j .

DYNAMICS OF A SYSTEM OF PARTICLES

287

Since f ij is perpendicular to both v i and Bij (which is either in or out of the paper), f ij can only be parallel to f ji if v i and v j are parallel, which is not true in general. Thus, equations (9.11) and (9.31) are not valid for a system of particles with magnetic interactions.

9-15.

σ = mass/length

F= dp becomes dt

mg = mv + mv where m is the mass of length x of the rope. So

m = σ x; m = σ x

σ x g =σ x xg=x dv +σ x v dt

dv dx + v2 dx dt dv + v2 dx

x g = xv

Try a power law solution:

v = ax n ;

Substituting,

dv = nax n −1 dx

x g = x ax n nax n −1 + a 2 x 2 n or ( )(

)

x g = a 2 ( n + 1) x 2 n

Since this must be true for all x, the exponent and coefficient of x must be the same on both sides of the equation. Thus we have: 1 = 2n or n = 1 2

g = a 2 ( n + 1) or a =

So

2g 3

v=

2 gx 3

288

CHAPTER 9

dv dv dx dv 2 gx = =v = a= dt dx dt dx 3 a= Ti = 0 Tf = Ui = 0 g 3

12

g 2 gx 3 3

−1 2

( y = 0 on table)

1 1 2 gL mgL mv 2 = m = 2 2 3 3 L 2

U f = mgh = − mg So Ei = 0; E f = − mgL 6

Energy lost =

mgL 6

9-16.

T

x

T1

T2

The equation of motion for the falling side of the chain is, from the figure,

ρ (b − x)

2

x=

ρ (b − x)

2

g + T2

(1)

From Example 9.2, we have for the energy conservation case x=g− Substitution gives us T2 = g 2bx − x 2 2 (b − x)

2

(

) =g+

x2 2 (b − x)

(2)

ρx 2

4

(3)

To find the tension on the other side of the bend, change to a moving coordinate system in which the bend is instantaneously at rest. This frame moves downward at a speed u = x 2 with respect to the fixed frame. The change in momentum at the bend is

DYNAMICS OF A SYSTEM OF PARTICLES

289

∆p = ( ρ∆x ) ⋅ ( 2u) = 2ρu2 ∆t = Equating this with the net force gives T1 + T2 = Using equation (3), we obtain T1 =

ρx 2

2

∆t

(4)

ρx 2

2

(5)

ρx 2

4

(6)

as required. Note that equation (5) holds true for both the free fall and energy conservation cases.

9-17.

As the problem states, we need to perform the following integral

12

τ=

∫ ε 1 − 2α dα 2α (1 − α )

(1)

Our choice of ε is 10 −4 for this calculation, and the results are shown in the figure. We plot the natural velocity dα d τ = x 2 gb vs. the natural time τ.

2

1.5 dα dτ

1

0.5

0

0

0.1

0.2

0.3

0.4 τ 0.5

0.6

0.7

0.8

9-18. Once we have solved Problem 9-17, it becomes an easy matter to write the expression for the tension (Equation 9.18):

T 1 + 2α − 6α 2 = mg 2 (1 − 2α ) This is plotted vs. the natural time using the solution of Problem 9-17.

(1)

290

20

CHAPTER 9

15

T mg

10

5

0

0

0.1

0.2

0.3

0.4 τ 0.5

0.6

0.7

0.8

9-19. ceiling x

released at t=0 b

at time t

table

The force that the tabletop exerts on the chain counteracts the force due to gravity, so that we may write the change in momentum of the center of the mass of the chain as dp = ρbg − F dt We can write out what the momentum is, though: p = ρ (b − x) x which has a time derivative dp = ρ − x 2 + ( b − x ) x = ρ ( bg − 3 gx ) dt where we have used x = g and x = 2 gx . Setting this last expression equal to (1) gives us F = 3ρ gx (4) (3) (2) (1)

Although M. G. Calkin (personal communication) has found that experimentally the time of fall for this problem is consistently less than the value one would obtain in the above treatment by about 1.5%, he also finds evidence that suggests the free fall treatment is more valid if the table is energy absorbing.

DYNAMICS OF A SYSTEM OF PARTICLES

291

9-20. a–x a+x

Let ρ = mass/length The force on the rope is due to gravity F = ( a + x) ρ g − ( a − x) ρ g = 2 xρ g dp dv dv =m = 2 aρ dt dt dt So F = dp becomes dt xg = a Now dv dv dx dv = ⋅ =v dt dx dt dx So xg = av or vdv = Integrating yields 1 2 g 2 v = x +c 2 2a Since v = 0 when x = 0, c = 0. Thus v2 = When the rope clears the nail, x = a. Thus v = ga g 2 x a g xdx a dv dx dv dt

292

CHAPTER 9

9-21. Let us call x the length of rope hanging over the edge of the table, and L the total length of the rope. The equation of motion is

mgx gx dx 2 =m 2 ⇒ x= L dt L Let us look for solution of the form x = Aeωt + Be −ωt Putting this into equation of motion, we find

ω=

g L

Initial conditions are x(t = 0) = x0 = 0.3 m ; v(t = 0) = 0 m/s . From these we find A = B = x0 2 . Finally x = x0 cosh (ω t) . When x = L , the corresponding time is t= L cosh −1 = 0.59 s. ω x0 1

9-22.

Let us denote (see figure) mass of neutron and deuteron respectively velocity of deuteron before collision velocity of neutron and deuteron, respectively after collision ν2 ν0 2m 2m m ψ α ν1

m and 2m v0 v1 and v2

a) Conservation of energy:

2 2 2 2mv0 mv1 2mv2 v2 v2 2 = + ⇒ v0 = 1 + 2 2 2 2 2 2

Conservation of momentum is 2mv0 = 2mv2 + v1

2 2 2 ⇒ 4v2 + 4v0 = v1 + 8v0 v2 cos ψ

Solving these equations, we obtain 2 sets of solutions v1 = 2v0 6 − 4 cos 2 ψ + 2 cos ψ 4 cos 2 ψ − 3 3 v2 =

2 2 2v0 cos ψ + 4v0 cos 2 ψ − 3v0

3

DYNAMICS OF A SYSTEM OF PARTICLES

293

or numerically v1 = 5.18 km/s v1 = 19.79 km/s v2 = 14.44 km/s and v2 = 5.12 km/s

b) Let us call α the lab scattering angle of the neutron, then from the sine theorem we have

mv1 2mv2 = sin ψ sin α ⇒ α = 74.84° and α = 5.16°

c) From a) we see that cos ψ =

⇒ sin α = 2

v2 sin ψ v1

2 2 2 4v0 + 4v2 − v1 8v0 v2

=

2 2 2v0 + 6v2 3 ≥ 8 v0 v 2 2

⇒

ψ ≤ 300 ⇒ ψ max = 30°

9-23.

Conservation of momentum requires v f to be in the same direction as u1 (component pi = m1u1 p f = ( m1 + m2 ) v f pi = p f → v f = m1 u1 m1 + m2

of v f ⊥ to u1 must be zero).

The fraction of original kinetic energy lost is

2 2 m1 u1 1 1 2 m1u1 − ( m1 + m2 ) 2 Ki − K f 2 ( m1 + m2 )2 = 1 Ki 2 m1u1 2

=

m1 −

2 m1 m1 + m2 m1

=

m2 m1 + m2

294

9-24.

v

CHAPTER 9

x

O aθ

θ ω0 b

ω

v0 m

The energy of the system is, of course, conserved, and so we have the following relation involving the instantaneous velocity of the particle: 1 1 2 mv 2 = mv0 2 2 (1)

The angular momentum about the center of the cylinder is not conserved since the tension in the string causes a torque. Note that although the velocity of the particle has both radial and angular components, there is only one independent variable, which we chose to be θ. Here ω = θ is the angular velocity of the particle about the point of contact, which also happens to be the rate at which the point of contact is rotating about the center of the cylinder. Hence we may write v0 = ω 0 b ; v = ω 0 ( b − aθ ) (2)

From (1) and (2), we can solve for the angular velocity after turning through an angle θ

ω=

ω0 a 1− θ b

(3)

The tension will then be (look at the point of contact) T = mω 2 ( b − aθ ) = mω 0ω b (4)

9-25. The best elements are those that will slow down the neutrons as much as possible. In a collision between m1 (the neutron) and m2 (moderator atom), we would thus want to minimize T1 (kinetic energy of the neutron after the collision); or alternatively, maximize T2 (kinetic energy of the moderator atom after the collision). From Eq. (9.88)

T2 4 m1 m2 cos 2 ζ = T0 ( m1 + m2 ) 2 Since one cannot control the angle ζ, we want to maximize the function f=

( m1 + m2 ) 2

m1m2

DYNAMICS OF A SYSTEM OF PARTICLES

295

with respect to m2 . ( m1 = constant)

2 2 m1 m1 − m2 df = = 0 when m1 = m2 dm2 ( m1 + m2 ) 4

(

)

By evaluating

d2 f one can show that the equilibrium point is a maximum. Thus, T2 is a 2 dm2 m = m

1 2

maximum for m1 = m2 . Back to reactors, one would want elements whose mass is as close as possible to the neutron mass (thus, as light as possible). Naturally, there are many other factors to consider besides mass, but in general, the lower the mass of the moderator, the more energy is lost per collision by the neutrons.

9-26.

The internal torque for the system is

N = r1 × f 12 + r 2 × f 21

(1)

where f12 is the force acting on the first particle due to the second particle. Now f 21= −f12

(2)

Then,

N = r1 − r 2 × f12

(

)

r r 2 − r1 = k r1 − r 2 × r 2 − r1 − v0

(

) (

)

(

)

=

kr r1 − r 2 × r1 − r 2 v0

(

) (

)

(3)

This is not zero in general because r1 − r 2 and r1 − r 2 are not necessarily parallel. The internal torque vanishes only if the internal force is directed along the line joining two particles. The system is not conservative.

9-27.

(

)

(

)

The equation for conservation of py in the lab system is (see fig. 9-10c): 0 = m1v1 sin ψ − m2 v2 sin ζ

Thus sin ζ = or sin ζ = m1T1 sin ψ m2T2 m1v1 sin ψ m2 v2

296

9-28.

CHAPTER 9

Using the notation from the chapter: m1 : Ti = T0 , T f = T1 m2 : Ti = 0; Tf = T2

Thus T0 = T1 + T2 or 1 = T1 T2 + T0 T0 T1 or, T0 (1)

If we want the kinetic energy loss for m1 to be a maximum, we must minimize equivalently, maximize From Eq. (9.88): T2 4 m1 m2 = cos 2 ζ T0 ( m1 + m2 ) 2 To maximize this, ζ = 0 (it can’t = 180°). T2 4 m1 m2 = T0 ( m1 + m2 ) 2 T2 . T0

The kinetic energy loss for m1 is T0 − T1 . The fraction of kinetic energy loss is thus T0 − T1 T T = 1 − 1 = 2 (from (1)) T0 T0 T0 T0 − T1 T0 = max ( m1 + m2 )2 v1 4 m1 m2

ζ = 0 implies ψ = 0, 180° (conservation of pv ). So the reaction is as follows

Before:

m1 m2 v2 m2

After: m1 v1

px : m1v = m1v1 + m2 v2 E: Solving for v1 gives v1 = So m2 travels in + x direction m1 − m2 v m1 + m2 1 1 1 2 2 m1v 2 = m1v1 + m2 v2 2 2 2

DYNAMICS OF A SYSTEM OF PARTICLES

297

+ x direction if m1 > m2 m1 travels in − x direction if m1 < m2

9-29.

From Eq. (9.69) tan ψ = sin θ cos θ + ( m1 m2 )

From Eq. (9.74)

θ = π − 2ζ

Substituting gives tan ψ = or tan ψ = sin ( 2ζ ) ( m1 m2 ) − cos ( 2ζ ) sin ( π − 2ζ ) ( m1 m2 ) + cos (π − 2ζ )

9-30.

Before: m1 After: m1 v1 m2 v1 m2 v2

a)

∆py = ( 0.06 kg ) 16 m/s cos 15° − ( −8 m/s ) cos 45° = 1.27 N ⋅ sec ∆px = ( 0.06 kg ) 16 m/s cos 15° ( −8 m/s ) sin 45° = −0.09 N ⋅ sec

The impulse P is the change in momentum. So

P = ( −0.09x + 1.27 y ) N ⋅ sec

b)

P = ∫ F dt = F ∆t F = − ( 9x + 127 y ) N

So

298

9-31.

CHAPTER 9

From Eq. (9.69) tan ψ = sin θ

m1 − cos θ m2

From Eq. (9.74)

θ =π −φ

Substituting gives

tan ψ =

sin φ m1 − cos φ m2

9-32.

pi = mu1 p f = mv1 + 2mv2

Conservation of momentum gives u1 = v1 + 2v2 ∆T = = or v1 = u1 − 2v2

1 1 2 2 2 mu1 − mv1 − mv2 2 2 1 1 2 2 2 2 mu1 − m u1 − 4u1v2 + 4v2 − mv2 2 2

(

)

2 = 2mu1v2 − 3mv2

d ( ∆T ) = 0 implies 2u1 = 6v2 dv2

or v2 =

u1 3

d 2 ( ∆T ) < 0, so this is a maximum 2 dv2 v1 = u1 − 2v2 = v1 = v2 = u1 3 u1 3

DYNAMICS OF A SYSTEM OF PARTICLES

299

9-33.

From Eq. (9.87b) in the text, we have T1 m = T0 ( m1 + m2 ) 2 1

2 1 2 cos ψ + m2 − sin 2 ψ m1 2

2 2 cos 2 ψ + m2 − sin 2 ψ + 2 cos ψ m2 − sin 2 ψ = 2 m2 m1 m1 1+ m1

Substituting m2 m1 ≡ α and cos ψ ≡ y we have T1 −2 = (1 + α ) 2 y 2 + α 2 − 1 + 2 y α 2 + y 2 − 1 T0

1 α = 12 α=4 0 α=1 α=2 π/2 ψ π

(1)

T1 T0

9-34.

Before After v1 u1 m m 45˚ m m v2 θ x

Cons. of pz : Cons. of py :

mu1 = mv1 cos 45° + mv2 cos θ 0 = mv1 sin 45° − mv2 sin θ

(1) (2)

Cons. of energy (elastic collision) 1 1 1 2 2 2 mu1 = mv1 − mv2 2 2 2 Solve (1) for cos θ : cos θ = Solve (2) for sin θ : sin θ = v1 2 v2 u1 − v1 v2 2 (3)

300

CHAPTER 9

Substitute into cos 2 θ + sin 2 θ = 1 , simplify, and the result is

2 2 2 u1 = v2 − v1 + 2 u1v1

Combining this with (3) gives

2 2v1 = 2 u1v1

We are told v1 ≠ 0 , hence v1 = u1

2

Substitute into (3) and the result is v2 = u1

2

Since v1 = v2 , (2) implies 0 = 45°

9-35.

From the following two expressions for T1 T0 ,

2 T1 v1 = 2 T0 u1

Eq. (9.82)

2

T1 m = T ( m1 + m2 ) 2

2 1

2 cos ψ ± m2 − sin 2 ψ m1

Eq. (9.87b)

we can find the expression for the final velocity v1 of m1 in the lab system in terms of the scattering angle ψ : v1 =

2 m m1u1 cos ψ ± 2 − sin 2 ψ m1 + m2 m1

(1)

If time is to be constant on a certain surface that is a distance r from the point of collision, we have r = v1t0

(2)

Thus,

2 m2 m1u1t0 cos ψ ± − sin 2 ψ r= m1 + m2 m1

(3)

This is the equation of the required surface. Let us consider the following cases:

DYNAMICS OF A SYSTEM OF PARTICLES

301

i)

m2 = m1 : r= u1t0 cos ψ ± 1 − sin 2 ψ = u1t0 cos ψ 2

(4)

(The possibility r = 0 is uninteresting.) iI) m2 = 2m1 : r= u1t0 cos ψ ± 4 − sin 2 ψ 3

(5)

iII)

m2 = ∞ : Rewriting (3) as

mut r= 1 10 m 1+ 1 m2 and taking the limit m2 → ∞ , we find

2 cos ψ 1 sin 2 ψ ± − 2 m2 m2 m1

(6)

r = u1t0 All three cases yield spherical surfaces, but with the centers displaced: m2 = ∞ m2 = 2m1 m1 = m2 O

(7)

m1

–v1t0

−

v1t0 3

m2

v1t0

This result is useful in the design of a certain type of nuclear detector. If a hydrogenous material is placed at 0 then for neutrons incident on the material, we have the case m1 = m2 . Therefore, neutrons scattered from the hydrogenous target will arrive on the surface A with the same time delay between scattering and arrival, independent of the scattering angle. Therefore, a coincidence experiment in which the time delay is measured can determine the energies of the incident neutrons. Since the entire surface A can be used, a very efficient detector can be constructed.

9-36.

Since the initial kinetic energies of the two particles are equal, we have 1 1 1 2 2 2 m1u1 = m2 u2 = α 2 m2 u1 2 2 2 (1)

or, m1 = α2 m2 (2)

302

CHAPTER 9

Now, the kinetic energy of the system is conserved because the collision is elastic. Therefore, 1 1 1 2 2 2 2 m1u1 = m2 u2 = m1u1 = m2 v2 2 2 2 since v1 = 0 . Momentum is also conserved, so we can write m1u1 + m2 u2 = ( m1 + α m2 ) u1 = m2 v2 Substituting the second equality in (4) into (3), we find m + α m2 2 1 m u = m2 1 u1 2 m2

2 1 1 2

(3)

(4)

(5)

or, 1 m1 = m2 2 Using m1 m2 = α 2 , (6) becomes 2α 2 = α 2 + α solving for α, we obtain m1 + α m2

2

(6)

(

)

2

(7)

α = −1 ∓ 2;

This gives us m1 = 3 ± 2 2; m2

α2 = 3 ∓ 2 2

(8)

u2 + :α < 0 = − 1 ± 2 with u1 − :α > 0

(

)

(9)

9-37.

Impulse = ∫ F dt =∫

6 × 10 −3 t=0

( 360 − 10 t ) dt

7 2

3 107 = 360 ⋅ 6 × 10 −3 − ⋅ 6 × 10 −3 N ⋅ s 3

(

)

Impulse = 1.44 Since the initial velocity is zero, v f = ∆v

kg m s

Impulse = ∆P = m∆v So

DYNAMICS OF A SYSTEM OF PARTICLES

303

kg m s vf = 0.003 kg 1.44 vmuzzle = 480 m s

9-38. v1 ψ

v1′ θ

θ−ψ

V

1 2 m1u1 2 1 2 T1 = m1v1 2 T0 = Thus,

2 T1 v1 = 2 T0 u1

(1)

(2)

Now, from the diagram above, we have v1 = V cos ψ + v1 cos (θ − ψ ) ′ Using Eq. (9.68) in the text, this becomes m v1 = V cos ψ + 2 cos (θ − ψ ) m1 Thus,

2 v1 V 2 m = 2 cos ψ + 2 (θ − ψ ) 2 u1 u1 m1 2

(3)

(4)

(5)

Using Eq. (9.84) in the text, V m1 = u1 m1 + m2 Therefore, we find

2 T1 m1 = T0 ( m1 + m2 ) 2

(6)

cos (θ − ψ ) cos ψ + ( m1 m2 )

2

(7)

If we define

304

CHAPTER 9

S ≡ cos ψ + we have

cos (θ − ψ ) ( m1 m2 )

(8)

2 T1 m1 = × S2 T0 ( m1 + m2 ) 2

(9)

as desired.

9-39.

θ u θ′ v y x

As explained in Section 9.8, the component of velocity parallel to the wall is unchanged. So vx = u sin θ vy is given by

ε= or vy uy

=

vy u cos θ

vy = ε u cos θ Thus v = u2 sin 2 θ + ε 2 u2 cos 2 θ v = u sin 2 θ + ε 2 cos 2 θ tan θ ′ = or u sin θ ε u cos θ

12 12

θ ′ = tan −1 tan θ ε

1

DYNAMICS OF A SYSTEM OF PARTICLES

305

9-40. Because of the string, m2 is constrained to move in a circle of radius a. Thus, initially, m2 will move straight up (taken to be the y direction). Newton’s rule applies to the velocity component along u1 . The perpendicular component of velocity (which is zero) is unchanged. Thus m1 will move in the original direction after the collision.

From conservation of py we have m1u1 sin α = m1v1 sin α + m2 v2 From Newton’s rule we have v2 cos ( 90° − α ) − v1 u1 (1)

ε= or v1 = v2 sin α − ε u1 Substituting (2) into (1) and solving for v2 gives v2 = (2) then gives v1 = u1 m1 sin 2 α − ε m2 m1 sin α + m2

2

(2)

( ε + 1) m1u1 sin α m1 sin 2 α + m2

straight up

(

)

along u1

9-41.

Using y = v0 t −

1 2 gt and v = v0 − gt , we can get the velocities before and after the 2 where h1 = 1 2 gt1 2 2 h1 = − 2 gh1 g

collision: Before: u1 = − gt1 So After: 0 = v0 − gt2 h2 = v0 t2 − = So

u1 = − g or t2 = v0 g

1 2 gt2 2 or v0 = 2 gh2 v1 = 2 gh2

2 2 v0 1 v0 − g 2 g

306

CHAPTER 9

Thus

ε=

v2 − v1 = u2 − u1

2 gh2 2 gh1

ε=

h2 h1

Tlost = Ti − Tf

Fraction lost = = Ti − T f Ti

2 2 u1 − v1 h1 − h2 h = = 1− 2 2 u1 h1 h1

Ti − Tf Ti

= 1− ε2

9-42. θ 30˚ 5 m/s y x

As explained in Section 9.8, the velocity component in the y-direction is unchanged. m vy = uy = 5 sin 30° = 2.5 m/s s For the x component we have 0.8 = vx ux = vx vx = m 3 m 5 5 s cos 30° 2 s vx = 2 3 vf = 1 73 2 m s

4.3 m/s 36°

θ = tan −1

2.5 2 3

DYNAMICS OF A SYSTEM OF PARTICLES

307

9-43. v2 4m T0 m 4m α 45˚ m v1

Conservation of px gives 2mT0 = 4 mv2 cos α + or cos α = Conservation of py gives 0= or sin α = Substituting into sin 2 α + cos 2 α = 1 gives v2 1= 1 2 + 32v2 Simplifying gives

2 v2 = 2 v1 T0 v1 T0 m + − 16 8 m 8m

1 mv1 2

2mT0 −

1 mv1 2

4 mv2

1 mv1 − 4 mv2 sin α 2

v1 4 2 v2

2mT0 +

1 2 2 m v1 − 2 mT0 mv1 2 2 16 m2 v2

(1)

The equation for conservation of energy is T0 − or

2 2 5T0 = 3mv1 + 12mv2

T0 1 1 2 2 = mv1 + ( 4 m) v2 6 2 2

(2)

Substituting (1) into (2) gives a quadratic in v1 :

2 15mv1 − 6 T0 m v1 − 14T0 = 0

308

CHAPTER 9

Using the quadratic formula (taking the positive sign since v1 > 0 ) gives: v1 = 1.19 T0 m

Substituting this into the previous expressions for cos α and sin α and dividing gives tan α = sin α = 1.47 cos α

Thus α , the recoil angle of the helium, is 55.8°.

9-44. v0 x

Fgrav = mg = µ xg where µ = mass/length Fimpulse = mv + mv + mv0 , since v = v0 , v = 0. We have m= So the total force is

2 F ( x ) = µ xg + µv0

d dx ( µ x ) = µ = µv0 dt dt

We want F(x = a)

2 F ( a ) = µ ag + µv0

or v2 F = µ ag 1 + 0 ag

9-45. Since the total number of particles scattered into a unit solid angle must be the same in the lab system as in the CM system [cf. Eq. (9.124) in the text],

σ (θ ) 2π sin θ dθ = σ (ψ ) ⋅ 2π sin ψ dψ

Thus, sin ψ dψ sin θ dθ

(1)

σ (θ ) = σ (ψ )

(2)

DYNAMICS OF A SYSTEM OF PARTICLES

309

The relation between θ and ψ is given by Eq. (9.69), which is tan ψ = sin θ cos θ + x (3)

where x = m1 m2 . Using this relation, we can eliminate ψ from (2):

sin ψ =

1 1 1+ tan 2 ψ

=

1

( cos θ + x ) 2 1+ sin 2 θ

=

sin θ 1 + 2 x cos θ + x 2

(4)

d ( tan ψ ) cos θ ( cos θ + x ) + sin 2 θ dψ dψ 2 = = cos ψ dθ d ( tan ψ ) dθ ( cos θ + x )2 Since cos 2 ψ = 1 , (5) becomes 1 + tan 2 ψ dψ = dθ 1 + x cos θ 1 + x cos θ 1 = 2 2 sin θ ( cos θ + x ) 1 + 2x cos θ + x 2 1+ ( cos θ + x ) 2

(5)

(6)

Substituting (4) and (6) into (2), we find

σ (θ ) = σ (ψ )

(1 + 2x cos θ + x )

1 + x cos θ

2 32

(7)

9-46. The change in angle for a particle of mass µ moving in a central-force field is [cf. Eq. (9.121)]. Let ψ = capital θ here.

∆ψ = ∫

rmax rmin

2µ E − U − θ (

(

r 2 dr

2

)

2µ r 2

)

(1)

b ψ a

rmin

In the scattering from an impenetrable sphere, rmin is the radius of that sphere. Also, we can see from the figure that θ = π − 2ψ . For r > rmin , U = 0. Thus (1) becomes ∆ψ = ∫ Substituting b 2µT0′ = ; E = T0′ (3)

∞

a

(

r 2 dr

2

)

2µE −

r2

(2)

310

CHAPTER 9

(2) becomes ∆ψ = ∫

∞

dr r2 r 2 −1 b

∞

α

(4)

This integral can be solved by using Eq. (E. 10b), Appendix E: −2 ψ = sin 4 r b2 a

−1

(5)

Thus, sin ψ = b a (6)

Therefore, we can find the relation between θ and b by substituting (θ = π − 2ψ ) into (6). We have b = a cos

θ

2

(7)

Now, the differential cross section is given by Eq. (9.120):

σ (θ ) =

From (7), we have

b db sin θ dθ

(8)

σ (θ ) =

Total cross section is given by

1 θ a θ a2 a cos × sin = sin θ 2 2 2 4

(9)

σ t = ∫ σ (θ ) dΩ = so that

a2 ⋅ 4π 4

(10)

σ t = π a2

(11)

9-47. The number of recoil particles scattered into unit solid angle in each of the two systems, lab and CM, are the same. Therefore,

σ ( φ ) sin φ dφ = σ (ζ ) sin ζ dζ

(1)

where φ and ζ are the CM and lab angles, respectively, of the recoil particle. From (1) we can write [cf. Eq. (9.125) in the text]

σ ( φ ) sin ζ dζ = σ (ζ ) sin φ dφ

(2)

DYNAMICS OF A SYSTEM OF PARTICLES

311

Now, in general, φ = 2ζ [see Eq. (9.74)]. Hence, sin ζ sin ζ 1 = = sin φ sin 2ζ 2 cos ζ and dζ 1 = dφ 2 Using (3) and (4) in (2), we have (4) (3)

σ (φ ) 1 = σ (ζ ) 4 cos ζ

For m1 = m2 , the Rutherford scattering cross section is [Eq. (9.141)]

(5)

σ (θ ) =

k2 1 × 2 4 4T0 sin (θ 2)

(6)

Also for this case, we have [Eqs. (9.71) and (9.75)] π ψ = −ζ 2

ψ=

θ

2

(7)

Hence, sin

θ

π = sin ψ = sin − ζ = cos ζ 2 2

(8)

and since the CM recoil cross section σ (φ) is the same as the CM scattering cross section σ (θ ) , (6) becomes

σ (φ ) =

Using (5) to express σ (ζ ), we obtain

k2 1 × 2 4T0 cos 4 ζ

(9)

σ (ζ ) = σ ( φ ) × 4 cos ζ or, (10)

σ (ζ ) =

k2 1 × 2 T0 cos 3 ζ

(11)

312

CHAPTER 9

9-48. In the case m1 m2 , the scattering angle ψ for the incident particle measured in the lab system is very small for all energies. We can then anticipate that σ (ψ) will rapidly approach zero as ψ increases.

Eq. (9.140) gives the Rutherford cross section in terms of the scattering angle in the CM system:

σ CM (θ ) =

From Eq. (9.79) we see that for m1 m2 , T0′ = Furthermore, from Eq. (9.69), tan ψ =

( 4T0′)

k2

2

sin (θ 2)

4

1

(1)

m2 m T0 ≅ 2 T0 m1 + m2 m1 sin θ m1 + cos θ m2

(2)

≅

m2 sin θ m1

(3)

and therefore, since ψ is expected to be small for all cases of interest, sin θ ≅ Then, m cos θ = 1 − 1 ψ m2 and

2 m1 1 sin (θ 2) = 1 − 1 − ψ 2 m2 2 2

m1 m tan ψ ≅ 1 ψ m2 m2

(4)

(5)

(6)

(Notice that ψ to unity.)

1 , but since m1

m2 , the quantity m1ψ m2 is not necessarily small compared

With the help of (2) and (6), we can rewrite the CM cross section in terms of ψ as

mk 1 σ CM (ψ ) = 1 2 2 2m2T0 1 − 1 − m1 ψ m 2 According to Eq. (9.129),

2

(7)

DYNAMICS OF A SYSTEM OF PARTICLES

313

2

2 m1 cos ψ + 1 − m1 sin ψ m2 m2 σ LAB (ψ ) = σ CM (θ ) 2 m 1 − 1 sin ψ m2

(8)

We can compute σ LAB (ψ ) with the help of (7) and the simplifications introduced in the righthand side of (8) by the fact that ψ 1 :

2 m1 + 1 − m1 ψ 2 m2 m2 m1k σ LAB (ψ ) ≅ 2 2 2 2m2T0 m1 m1 1 − 1 − ψ 1− ψ m2 m2 2

(9)

and so,

σ LAB (ψ ) ≅

(m k

2 1

2 2m2 T0 2

)

2

2 1 − 1 − m1 ψ m2

(10)

2

m 1− 1 ψ m2

This expression shows that the cross section has a second-order divergence at ψ = 0. For values of ψ > m2 m1 , (9) gives complex values for σ lab . This result is due to the approximations involved in its derivation, making our result invalid for angles larger than m2 m1 .

9-49. The differential cross section for Rutherford scattering in the CM system is [cf. Eq. (9.140) in the text]

σ (θ ) = where [cf. Eq. (9.79)] T0′ =

1 k2 2 16T0′ sin 4 θ 2 m2 T0 m1 + m2

(1)

(2)

314

CHAPTER 9

Thus, 1 m1 + m2 k2 σ (θ ) = 2 16T0 sin 4 θ m2 2 1 k2 m = 1+ 1 2 θ m2 16T0 sin 4 2 Since m1 m2 1 , we expand m1 m1 +… 1 + ≅ 1+ 2 m2 m2 Thus, to the first order in m1 m2 , we have

2 2 2

(3)

(4)

σ (θ ) =

m k2 1 1+ 2 1 2 θ 16T0 sin 4 m2 2

(5)

This result is the same as Eq. (9.140) except for the correction term proportional to m1 m2 .

9-50.

The potential for the given force law is U (r) = k 2r 2 (1)

First, we make a change of variable, z = 1 r . Then, from Eq. (9.123), we can write

θ =∫

zmax

b dz k 2 z 1 − b 2 + 2 mu0 b z zmax

k zmax = b 2 + 2 mu0

0

−1 2

0

=

k 2 b + 2 mu0

2

sin −1

(2)

=

πb

2 b2 + k 2 mu0

Solving (2) for b = b(θ ), bθ = k 2 mu0 2θ

π − 4θ 2

2

(3)

DYNAMICS OF A SYSTEM OF PARTICLES

315

According to Fig. 9-22 and Eq. (9.122),

θ= so that b(θ ) can be rewritten as (θ): b (θ ) =

1 (π − θ ) 2

(4)

k 2 mu0

π −θ θ ( 2π − θ )

(5)

The differential cross section can now be computed from Eq. (9.120):

σ (θ ) = with the result

b db sin θ dθ

(6)

σ (θ ) =

2 mu0θ 2 ( 2π − θ ) sin θ 2

kπ 2 (π − θ )

(7)

9-51. n θ φ=π−θ p

In the CM system, whenever the neutron is scattered through the angle θ, the proton recoils at the angle φ = π – θ. Thus, the neutron scattering cross section is equal to the recoil cross section at the corresponding angles: dN p dN n = dΩ (θ ) dΩ ( φ ) Thus, dN p dTp dN n = dΩ (θ ) dTp dΩ ( φ ) where dN p dTp is the energy distribution of the recoil protons. According to experiment, dN p dTp = const. Since mp ≅ mn , Tp is expressed in terms of the angle ψ by using Eq. (9.89b): Tp = T0 sin 2 ψ We also have ψ = (3) (2) (1)

θ

2

for the case mp ≅ mn . Thus, dΩ ( φ ) dTp = d 1 T0 sin 2 ψ 2π sin φ dφ

(

)

(4)

316

CHAPTER 9

or,

dΩ ( φ )

dTp

=

T0 d π − φ sin 2 2π sin φ dφ 2 T0 φ φ T sin cos = 0 2π sin φ 2 2 4π (5)

=

Therefore, we find for the angular distribution of the scattered neutron, dN p T0 dN n = ⋅ dΩ (θ ) dTp 4π (6)

Since dN p dTp = const., dN n dΩ is also constant. That is, the scattering of neutrons by protons is isotropic in the CM system.

9-52. Defining the differential cross section σ (θ ) in the CM system as in Eq. (9.116), the number of particles scattered into the interval from θ to θ + dθ is proportional to

dN ∝ σ (θ ) sin θ dθ = −σ (θ ) d ( cos θ ) From Eq. (9.87a) and the assumption of elastic collisions (i.e., T0 = T1 + T2 ), we obtain T2 2m1 m2 = (1 − cos θ ) T0 ( m1 + m2 ) 2 or, solving for cos θ, cos θ = where Tm = 4 m1 m2 Tm − 2T2 Tm

(1)

(2)

(3)

( m1 + m2 )2

T0 is the maximum energy attainable by the recoil particle in the lab

system. Then, (1) can be rewritten as dN ∝ 2σ (θ ) dT2 Tm (4)

and consequently, we obtain the desired result for the energy distribution: dN ∝ σ (θ ) dT2 (5)

DYNAMICS OF A SYSTEM OF PARTICLES

317

9-53.

m1 = mass of particle α 1 ,

m2 = mass of

238

U

u1 , u′ : velocity of particle α in LAB and CM before collision 1 v1 , v′ : velocity of particle ∝ in LAB and CM after collision 1 u2 , u′ : velocity of 2 v2 , v′ ” velocity of 2 u2 = 0 , T1 =

238

U in LAB and CM before collision U in LAB and CM after collision

238

2 m1u1 = 7.7MeV 2

ψ = 90° is angle through which particle α is deflected in LAB θ ζ is angle through which particle α and is recoil angle of

238 238

U are deflected in CM

U in LAB v1 m1

u1 m2

m1 ζ θ v2′

–vCM

v2

a) Conservation of momentum in LAB:

m1 u1 = m2 v2 cos ζ ⇒ v1 = u1 tan ζ m1 v1 = m2 v2 sin ζ Conservation of energy in LAB:

2 2 2 m1u1 m1v1 m2 v2 = + 2 2 2

From these equations we obtain the recoil scattering angle of tan ζ =

b) The velocity of CM of system is

238

U

m2 − m1 m2 + m1

⇒ ζ = 44.52°

vCM = The velocity of

238

m1u1 m1 + m2

' U in CM after collision is v2 = v2 − vCM . From the above figure we can obtain 238

the scattering angle of particle

U in CM to be

318

CHAPTER 9

tan θ =

v2 sin ζ = v2 cos ζ − vcm

2 2 m2 − m1 2 m1

⇒ θ = 89.04°

238

In CM, clearly after collision, particle α moves in opposite direction of that of

c) The kinetic energy of particle

238

U.

U after collision in LAB is

2

2 2 2 m2 v2 m2 m1v0 m1 u1 2m1 T1 = 0.25 MeV = = = m cos ζ 2 2 2 m1 + m2 m1 + m2

Evidently, conservation of energy is satisfied.

d) The impact parameter in CM is given in Section 9.10.

b= where k = so

k θ cot 2 2T0′

q1q2 1 and T0′ = m1u1 2 + m2 u2 2 is the total energy of system in CM, ′ ′ 2 9πε 0 b= q1 q2 m1 + m2 θ cot = 1.8 × 10 −14 m 2 2 4πε 0 m1 m2 u1

(

)

We note that b is the impact parameter of particle α with respect to CM, so the impact ( m1 + m2 ) b = 1.83 × 10−14 m. parameter of particle α with respect to 238 U is m2

e) In CM system, the orbit equation of particle α is

α r′ = (1 + ε ) cos θ

where θ = 0

corresponds to r = rmin

⇒ rmin = ′

α

1+ ε

is closest distance from particle α to the center of mass, and ′ 4πε 0 ( m1u1b ) α= = m1k q1q2 m1

2 2

and

ε = 1+ 2

′ 4πε 0 ( m1u1b ) E 2 = 1 + 2E m1k q1 q2 m1 ′ 4πε 0 ( m1u1b ) q1q2 m1

2

2

= 1 + m1u1 2 ′

But the actual minimum distance between particles is rmin = m1 + m2 rmin = 0.93 × 10 −14 m. ′ m2

DYNAMICS OF A SYSTEM OF PARTICLES

319

f)

Using formula

σ LAB (ψ ) = σ CM (θ ) where x = m1 , ψ = 90° , m2

(

x cos ψ + 1 − x 2 sin 2 ψ 1 − x sin ψ

2 2

)

2

σ CM (θ ) =

( 4T0′)

k2

2

1 θ sin 4 2

We find this differential cross section in LAB at ψ = 90°:

σ LAB (ψ = 90°) = 3.16 × 10 −28 m 2

g) Since

dN = σ (θ ) sin ψ dψ dφ we see that the ratio of probability is N

σ (ψ ) sin ψ = 11.1 σ (ψ ') sin ψ ′

9-54.

Equation 9.152 gives the velocity of the rocket as a function of mass: v = v0 + u ln m0 m = µ ln 0 m m m0 m

( v0 = 0 )

p = mv = mu ln To maximize p, set dp =0 dm 0= ln m0 =1 m

dp m = u ln 0 − 1 dm m m0 =e m or m = e −1 m0

To check that we have a maximum, examine d2 p dm2 m = m e−1

0

d2 p u =− 2 dm m

d2 p u e < 0 , so we have a maximum. =− 2 dm m = m e−1 m0

0

m = e −1 m0

320

9-55.

CHAPTER 9

The velocity equation (9.165) gives us: m v ( t ) = − gt + u ln 0 m (t) (1)

where m ( t ) = m0 − α t , the burn rate α = 9m0 10τ , the burn time τ = 300 s , and the exhaust velocity u = 4500 m ⋅ s -1 . These equations are good only from t = 0 to t = τ. First, let us check that the rocket does indeed lift off at t = 0: the thrust α u = 9um0 10τ = 13.5 m ⋅ s −2 ⋅ m0 > m0 g , as required. To find the maximum velocity of the rocket, we need to check it at the times t = 0 and t = τ, and also check for the presence of any extrema in the region 0 < t < τ. We have v(0) = 0, v(τ ) = − gτ + u ln 10 = 7400 m ⋅ s −1 , and calculate αu dv αu = −g + =g − 1 > 0 dt m (t) m (t) g (2)

The inequality follows since α u > m0 g > m ( t ) g . Therefore the maximum velocity occurs at t = τ, where v = − gτ + u ln 10 = 7400 m ⋅ s −1 . A similar single-stage rocket cannot reach the moon since v ( t ) < u ln m0 m final = u ln 10 10.4 m ⋅ s −1 , which is less than escape velocity and independent of fuel burn rate.

9-56. a) Since the rate of change of mass of the droplet is proportional to its cross-sectional area, we have

(

)

dm = kπ r 2 dt If the density of the droplet is ρ, m= so that dm dm dr dr = = 4πρr 2 = π kr 2 dt dr dt dt Therefore, dr k = dt 4 ρ or, r = r0 + as required. k t 4ρ 4π ρ r3 3

(1)

(2)

(3)

(4)

(5)

DYNAMICS OF A SYSTEM OF PARTICLES

321

b) The mass changes with time, so the equation of motion is

F= Using (1) and (2) this becomes

d dv dm = mg ( mv) = m + v dt dt dt

(6)

4π 3 dv 4π 3 + π kr 2 v = ρr ρr g 3 3 dt or, dv 3k v=g + dt 4 ρr Using (5) this becomes dv 3k v + =g dt 4 ρ r + k t 0 4ρ If we set A = 3k k and B = , this equation becomes 4ρ 4ρ dv A + v=g dt r0 + Bt and we recognize a standard form for a first-order differential equation: dv + P (t) v = Q (t) dt in which we identify P (t) = The solution of (11) is

− P ( t ) dt ∫ P (t) dt Q dt + constant v (t) = e ∫ ∫ e

(7)

(8)

(9)

(10)

(11)

A ; Q(t) = g r0 + Bt

(12)

(13)

Now,

∫ P (t ) dt = ∫ r

A = 3 . Therefore, B e∫

0

A A dt = ln ( r0 + Bt ) B + Bt

3

= ln ( r0 + Bt ) since

(14)

Pdt

= ( r0 + Bt )

3

(15)

322

CHAPTER 9

Thus,

−3 3 v ( t ) = ( r0 + Bt ) ∫ ( r0 + Bt ) g dt + constant −3 g 4 = ( r0 + Bt ) ( r0 + Bt ) + C 4B

(16)

The constant C can be evaluated by setting v ( t = 0 ) = v0 : v0 = so that C = v0 r03 − We then have v (t) = or, v (t) = 1 ( Bt ) 3 g 4 3 4B ( Bt ) + 0 r0 g 4 r0 4B (18) 1 g 4 r0 + C r03 4B

(

)

(17)

( r0 + Bt )

1

3

g 4 4 g 3 4B ( r0 + Bt ) + v0 r0 − 4B r0

(19)

( )

(20)

where 0 r03 means “terms of order r03 and higher.” If r0 is sufficiently small so that we can neglect these terms, we have

( )

v (t) ∝ t as required.

9-57.

(21)

Start from our definition of work:

W = ∫ F dx = ∫

dp dx = ∫ v dp dt

(1)

We know that for constant acceleration we must have v = at (zero initial velocity). From Equation (9.152) this means

m = m0 e − at u

We can then compute dp:

(2)

at dp = d ( mv ) = d ( mat ) = ma dt + at dm = m0 ae − at u 1 − dt u

This makes our expression for the work done on the rocket

(3)

Wr =

m0 a t ( at )( u − at ) e − at u dt u ∫0

(4)

DYNAMICS OF A SYSTEM OF PARTICLES

323

The work done on the exhaust, on the other hand, is given with v → (v – u) and dp → dmexhaust (v − u) , so that

We =

m0 a t ( at − u) 2 e − at u dt u ∫0

(5)

The upper limit on the integrals is the burnout time, which we can take to be the final velocity divided by the acceleration. The total work done by the rocket engines is the sum of these two quantities, so that

W=

m0 a v a 2 − at u 2 ∫ 0 u − uat e dt = m0 u u

(

)

− ∫ 0 (1 − x ) e v u

x

dx

(6)

where the obvious substitution was made in the last expression. Upon evaluating the integral we find

W = m0 uve − v u = muv where m is the mass of the rocket after its engines have turned off and v is its final velocity.

9-58.

(7)

From Eq. (9.165) the velocity is

v=

dy m = − gt + u ln 0 dt m m0 − gt + u ln m dt

∫ dy = ∫

Since

m = −α , dt = − dm α dt y+C =− a

1 2 u gt − 2 α

∫ ln

m0 dm m

∫ ln x dx = x 1 + ln x , so we have y+C =−

1 2 u gt − 2 α

a

m0 m + m ln m

Evaluate C using y = 0 when t = 0, m = m0

C=− y= u ( m0 − m) −

um0

α

α

1 2 mu m0 ln gt − ; m0 − m = α t m 2 α 1 2 mu m0 ln gt − 2 α m

y = ut −

At burnout, y = yB , t = tB

324

CHAPTER 9

yB = utB −

After burnout, the equations are

1 2 mu m0 gtB − ln m 2 α

y = y 0 + v0 t −

Calling the top of the path the final point

1 2 gt and v = v0 − gt 2 or

v f = 0 = vB − gt f y − y0 =

t f = vB g y0 = 0

2 2 2 vB vB vB − = ; g 2g 2g 2 vB 2g

y=

9-59. In order to immediately lift off, the thrust must be equal in magnitude to the weight of the rocket. From Eq. (9.157):

Thrust = v0α So

v0 = velocity of fuel v0α = mg

or

v0 = mg α

9-60.

The rocket will lift off when the thrust just exceeds the weight of the rocket. Thrust = − u

dm = uα dt

Weight = mg = ( m0 − α t ) g Set thrust = weight and solve for t:

uα = ( m0 − α t ) g ; t =

m0

α

−

u g

With m0 = 70000 kg , α = 250 kg/s , u = 2500 m/s , g = 9.8 m/s 2 t 25 sec

The design problem is that there is too much fuel on board. The rocket sits on the ground burning off fuel until the thrust equals the weight. This is not what happens in an actual launch. A real rocket will lift off as soon as the engines reach full thrust. The time the rocket sits on the ground with the engines on is spent building up to full thrust, not burning off excess fuel.

DYNAMICS OF A SYSTEM OF PARTICLES

325

9-61.

From Eq. (9.153), the velocity after the first state is: v1 = v0 + u ln k

After the second stage: v2 = v1 + u ln k = v0 + 2u ln k After the third stage: v3 = v2 + u ln k = v0 + 3u ln k After the n stages: vn = vn −1 + u ln k = v0 + nu ln k vn = v0 + nu ln k

9-62. To hover above the surface requires the thrust to counteract the gravitational force of the moon. Thus:

−u −

dm 1 = mg dt 6

6u dm = dt g m

Integrate from m = m0 to 0.8 m0 and t = 0 to T: T=− 6 ( 2000 m/s ) 6u ln 0.8 = − ln 0.8 9.8 m/s 2 g T = 273 sec

9-63. a) With no air resistance and constant gravity, the problem is simple:

1 2 mv0 = mgh 2

2 giving the maximum height of the object as h = v0 2 g v0 g 610 s .

(1) 1800 km . The time it takes to do this is

b) When we add the expression for air resistance, the differential equation that describes the projectile’s ascent is

F=m

dv 1 v = − mg − cW ρ Av 2 = − mg 1 + v dt 2 t

2

(2)

326

CHAPTER 9

where vt = 2mg cW ρ A

2.5 km ⋅ s −1 would be the terminal velocity if the object were falling

from a sufficient height (using 1.3 kg ⋅ m −3 as the density of air). Solution of this differential equation gives v ( t ) = vt tan tan −1

v0 gt − v vt t

(3)

This gives v = 0 at time τ = ( vt g ) tan −1 ( v0 vt )

300 s . The velocity can in turn be integrated to

give the y-coordinate of the projectile on the ascent. The height it reaches is the y-coordinate at time τ:

v2 v h = t ln 1 + 0 2 g vt

2

(4)

which is

600 km.

c) Changing the acceleration due to gravity from –g to − GMe

( Re + y )

2

= − g Re ( Re + y )

2

changes our differential equation for y to y 2 R 2 e y = − g + v t Re + y Using the usual numerical techniques, we find that the projectile reaches a height of a flight time of 330 s. (5) 630 km in

d) Now we must replace the ρ in the air resistance equation with ρ ( y ) . Given the dependence

of vt on ρ, we may write the differential equation ρ ( y) y 2 R 2 e y = −g + ρ0 vt Re + y (6)

where we use log 10 ρ ( y ) = 0.11 − (5 × 10 −5 )y and ρ0 = 1.3 , with the ρ ’s in kg ⋅ m −3 and y in meters. The projectile then reaches a height of 2500 km in a flight time of 940 s. This is close to the 2600 km. height to which the projectile rises when there is no air resistance, which is

DYNAMICS OF A SYSTEM OF PARTICLES

327

2.5 2

height (1000 km)

1.5 1 0.5 0

0

0.1

0.2

0.3

0.4

0.5 t (1000 s)

0.6

0.7

0.8

0.9

1

(a)

(b)

(c)

(d)

9-64. We start with the equation of motion for a rocket influenced by an external force, Eq. (9.160), with Fext including gravity, and later, air resistance. a) There is only constant acceleration due to gravity to worry about, so the problem can be solved analytically. From Eq. (9.166), we can obtain the rocket’s height at burnout

yb = utb −

1 2 mu m0 gtb − ln 2 α mb

(1)

where mb is the mass of the rocket at burnout and α = ( m0 − mb ) tb . Substitution of the given values gives yb

2 250 km . After burnout, the rocket travels an additional vb 2 g , where vb is the

rocket velocity at burnout. The final height the rocket ends up being is taken into account.

3700 km, after everything

b) The situation, and hence the differential equation, becomes more complicated when air resistance is added. Substituting Fext = − mg − cW ρ Av 2 2 (with ρ = 1.3 kg ⋅ m −3 ) into Equation (9.160), we obtain

c ρ Av 2 dv uα = −g− W dt m 2m

(2)

We must remember that the mass m is also a function of time, and we must therefore include it also in the system of equations. To be specific, the system of equations we must use to do this by computer are v y 2 cW ρ Av uα v = m − g − 2m m −α These must be integrated from the beginning until the burnout time, and therefore must be integrated with the substitution α = 0. Firstly, we get the velocity and height at burnout to be vb 7000 m ⋅ s −1 and yb 230 km . We can numerically integrate to get the second part of the

(3)

328

CHAPTER 9

journey, or use the results of Problem 9-63(b) to help us get the additional distance travelled with air resistance, analytically. The total height to which the rocket rises is 890 km in a total flight time of 410 s.

2

c) The variation in the acceleration of gravity is taken into account by substituting

GMe vb

( Re + y )

= g Re ( Re + y ) for g in the differential equation in part (b). This gives

2

6900 m ⋅ s −1 , yb

230 km , with total height

950 km and time-of-flight

460 s.

d) Now one simply substitutes the given expression for the air density, ρ (y) for ρ, into the differential equation from part (c). This gives vb 8200 m ⋅ s −1 , yb 250 km , and total height

8900 km with time-of-flight

10 8

2900 s.

height (km)

6 4 2 0

0

0.5

1

1.5 t (1000 s)

2

2.5

3

(a)

(b)

(c)

(d)

9-65.

Total impulse Total mass Burn time Rocket cross section area Drag coefficient Drag force

P = 8.5 N ⋅ s m0 = 0.054 kg t f = 1.5 s S=

π d2

4

= 4.5 × 10 −4 m 2

cw = 0.75 D= 1 ρcw Sv 2 = Kv 2 = 2 × 10 −4 v 2 N 2

where ρ = 1.2 kg/m 3 is density of air and v is rocket’s speed Rocket exhaust speed u = 800 m/s

a) The total mass of propellant is

∆m =

P = 0.0106 kg u

DYNAMICS OF A SYSTEM OF PARTICLES

329

Since m ~ 20% m0 , we will assume that the rocket’s mass is approximately constant in this problem. The equation of motion of rocket is m0 (where α = dv = − mo g + uα − Kv 2 dt

∆m = 7.1 × 10 −3 kg/s is fuel burn rate) tf ⇒ m0 dv = dt ( uα − m0 g ) − Kv2

Using the initial condition at t = 0, v = 0, we find v(t) = At burn-out, t = t f = 1.5 s , we find v f = v(t f ) = 114.3 m/s The height accordingly is given by h(t) = ∫ v(t) dt =

0

( uα − m0 g ) tanh

K

K ( uα − m0 g ) m0 2

k ( uα − m g ) m0 0 ln cosh t 2 k m0

At burn out, t = t f , we find the burn-out height h f = h(t f ) = 95.53 m

b)

After the burn-out, the equation of motion is m0 dv = − m0 g − Kv 2 dt ⇒ m0 dv = − dt m0 g + Kv 2

with solution v(t) = − Kg m0 g tan t − C K m0

Using the initial condition at t = t f , v(t) = v f , we find the constant C = –1.43 rad, so v(t) = − and the corresponding height h(t) = h f + ∫ v(t) dt = h f + tf t

Kg m0 g tan t − 1.43 K m0

m0 K

Kg − 1.43 0.88 + ln cos t m0

330

CHAPTER 9

When the rocket reaches its maximum height (at t = tmax ) the time tmax can be found by setting v(tmax ) = 0. We then find tmax = 7.52 s. And the maximum height the rocket can reach is hmax = h(tmax ) = 334 m

c)

Acceleration in burn-out process is (see v(t) in a)) a(t ) = dv uα − m0 g = dt m0 1 k ( uα − m g ) 0 cosh 2 t 2 m0

Evidently, the acceleration is maximum when t = 0 and amax = a(t = 0) = uα − m0 g = 95.4 m/s 2 m0

d) In the fall-down process, the equation of motion is

m0 dv = m0 g − Kv 2 , dt With the initial condition t = tmax , v = 0, we find ( t ≥ tmax ) v(t) = − m0 g Kg tanh ( t − tmax ) K m0

(v(t) is negative for t ≥ tmax , because then the rocket falls downward) The height of the rocket is h = hmax + Kg m0 ln cosh ( t − tmax ) K m0

tmax

∫

t

v(t) dt = hmax −

To find the total flight-time, we set h = 0 and solve for t. We find ttotal = 17.56 s

e) Putting t = ttotal into the expression of V(t) in part d), we find the speed at ground impact to

be vg = − m0 g Kg tanh ( ttotal − tmax ) = −49.2 m/s K m0

9-66. If we take into account the change of the rocket’s mass with time m = m0 − α t , where α is the fuel burn rate,

α = 7.1 × 10 −3 kg/s as calculated in problem 9-65. The equation of motion for the rocket during boost phase is

DYNAMICS OF A SYSTEM OF PARTICLES

331

( m0 − α t ) dt = uα − Kv2

Integrating both sides we obtain finally

dv

⇒

dv dt = Kv − uα α t − m0

2

1 + C αt − m 2 ( 0) v(t) = 1 − C (α t − m ) 2 0

uα Ku K α

Ku

α

where C is a constant. Using the initial condition v(t ) = 0 at t = 0 , we can find C and the velocity is 2 1 − 1 − αt m0 uα v(t) = K 2 1 + 1 − αt m0

a) The rocket speed at burn-out is (note t f = 1.5 s ).

Ku ∝

Ku ∝

v f = v(t f ) = 131.3 m/s

b) The distance the rocket has traveled to the burn-out is

h f = ∫ v(t) dt = 108.5 m θ tf

9-67.

From Equation (9.167) we have H bo = − g m0 − m2 f 2α

2

(

) + u m

α

f

mf ln + m0 − m f m0

Using numerical values from Example 9.12

α = −1.42 × 10 4 kg/s m0 = 2.8 × 106 kg m f = 0.7 × 106 kg u = 2.600 m/s we find H bo = 97.47 km . From Equation (9.168) we find vbo = − g(m0 − m f ) m + u ln o mf

α

vbo = 2125 m/s

332

CHAPTER 9

CHAPTER

10

Motion in a Noninertial Reference Frame

10-1. (1) (2)

The accelerations which we feel at the surface of the Earth are the following: :

Gravitational

980 cm/sec 2

Due to the Earth’s rotation on its own axis:

2π rad/day rω = 6.4 × 10 cm × 86400 sec/day

2

( (

8

)

2

= 6.4 × 10 8 × 7.3 × 10 −5 (3) Due to the rotation about the sun:

) (

)

2

= 3.4 cm/sec 2

rω = 1.5 × 10

2

( (

13

2π rad/year cm × 86400 × 365 sec/day

)

2

7.3 × 10 −5 2 = 1.5 × 1013 × = 0.6 cm/sec 365

)

2

10-2.

The fixed frame is the ground. y a θ x

The rotating frame has the origin at the center of the tire and is the frame in which the tire is at rest. From Eqs. (10.24), (10.25):

a f = R f + a r + ω × r + ω × ( ω × r ) + 2ω × v r

333

334 Now we have

R f = − a cos θ i + a sin θ j r = r0 i ω= Substituting gives V k r0 vr = ar = 0 ω= a k r0

CHAPTER 10

a f = − a cos θ i + a sin θ j + a j −

v2 i r0

(1)

v2 a f = −i + a cos θ + j ( sin θ + 1) a r0 We want to maximize a f , or alternatively, we maximize a f :

2

af

2

=

v4 2 av 2 cos θ + a 2 + 2 a 2 sin θ + a 2 sin 2 θ + a 2 cos 2 θ + 2 r0 r0 v4 2av 2 + 2a2 + cos θ + a 2 sin 2 θ r02 r0

=

d af dθ

2

=−

2av 2 cos θ + 2a 2 cos θ r0 ar0 v2 v2 a 2 r02 + v 4

= 0 when tan θ =

(Taking a second derivative shows this point to be a maximum.) tan θ = and sin θ = Substituting into (1) ar0 a r + v4

2 2 0

ar0 implies cos θ = v2

v2 av 2 a f = −i + r0 a 2 r02 + v 4 This may be written as

ar0 + j + 1 a a 2 r02 + v 4

a f = a + a 2 + v 4 r02

MOTION IN A NONINERTIAL REFERENCE FRAME

335

A θ

This is the maximum acceleration. The point which experiences this acceleration is at A: where tan θ = ar0 v2

10-3.

We desire Feff = 0 . From Eq. (10.25) we have

Feff = F − mR f − mω × r − mω × ( ω × r ) − 2mω × v r ω r

0

The only forces acting are centrifugal and friction, thus µs mg = mω 2 r , or

r=

µs g ω2

10-4. Given an initial position of (–0.5R,0) the initial velocity (0,0.5ωR) will make the puck motionless in the fixed system. In the rotating system, the puck will appear to travel clockwise in a circle of radius 0.5R. Although a numerical calculation of the trajectory in the rotating system is a great aid in understanding the problem, we will forgo such a solution here. 10-5.

The effective acceleration in the merry-go-round is given by Equation 10.27:

x = ω 2 x + 2ω y y = ω 2 y − 2ω x

These coupled differential equations must be solved with the initial conditions x0 ≡ x ( 0 ) = −0.5 m , y0 ≡ y ( 0 ) = 0 m , and x ( 0 ) = y ( 0 ) = v0 2 m ⋅ s −1 , since we are given in the problem that the initial velocity is at an angle of 45° to the x-axis. We will vary v0 over some range that we know satisfies the condition that the path cross over ( x0 , y0 ) . We can start by

(1) (2)

looking at Figures 10-4e and 10-4f, which indicate that we want v0 > 0.47 m ⋅ s −1 . Trial and error can find a trajectory that does loop but doesn’t cross its path at all, such as v0 = 0.53 m ⋅ s −1 . From here, one may continue to solve for different values of v0 until the wanted crossing is eyeball-suitable. This may be an entirely satisfactory answer, depending on the inclinations of the instructor. An interpolation over several trajectories would show that an accurate answer to the problem is v0 = 0.512 m ⋅ s −1 , which exits the merry-go-round at 3.746 s. The figure shows this solution, which was numerically integrated with 200 steps over the time interval.

336

1

CHAPTER 10

0.5

y (m)

0

0.5

–1 –1

–0.5

0 x (m)

0.5

1

10-6. z z = f(r) m r

Consider a small mass m on the surface of the water. From Eq. (10.25)

Feff = F − mR f − mω × r − mω × ( ω × r ) − 2mω × v r

In the rotating frame, the mass is at rest; thus, Feff = 0 . The force F will consist of gravity and the force due to the pressure gradient, which is normal to the surface in equilibrium. Since R f = ω = v r = 0 , we now have 0 = mg + Fp − mω × ( ω × r ) where Fp is due to the pressure gradient.

Fp θ θ′ mω2r

mg

Since Feff = 0 , the sum of the gravitational and centrifugal forces must also be normal to the surface. Thus θ ′ = θ. tan θ ′ = tan θ =

ω 2r g MOTION IN A NONINERTIAL REFERENCE FRAME

337

but tan θ = Thus dz dr

z=

ω2

2g

r 2 + constant

The shape is a circular paraboloid.

10-7. For a spherical Earth, the difference in the gravitational field strength between the poles and the equator is only the centrifugal term:

g poles − gequator = ω 2 R For ω = 7.3 × 10 −5 rad ⋅ s −1 and R = 6370 km, this difference is only 34 mm ⋅ s −2 . The disagreement with the true result can be explained by the fact that the Earth is really an oblate spheroid, another consequence of rotation. To qualitatively describe this effect, approximate the real Earth as a somewhat smaller sphere with a massive belt about the equator. It can be shown with more detailed analysis that the belt pulls inward at the poles more than it does at the equator. The next level of analysis for the undaunted is the “quadrupole” correction to the gravitational potential of the Earth, which is beyond the scope of the text.

10-8.

ω z y λ

x

Choose the coordinates x, y, z as in the diagram. Then, the velocity of the particle and the rotation frequency of the Earth are expressed as ω = ( −ω cos λ , 0, ω sin λ ) so that the acceleration due to the Coriolis force is a = −2ω × r = 2ω ( 0, − z cos λ , 0 ) v = ( 0, 0, z )

(1)

(2)

338

CHAPTER 10

This acceleration is directed along the y axis. Hence, as the particle moves along the z axis, it will be accelerated along the y axis: y = −2ω z cos λ Now, the equation of motion for the particle along the z axis is z = v0 − gt z = v0 t − where v0 is the initial velocity and is equal to h: 1 2 gt 2 (4) (5) (3)

2gh if the highest point the particle can reach is (6)

v0 = 2 gh From (3), we have y = −2ω z cos λ + c but the initial condition y ( z = 0 ) = 0 implies c = 0. Substituting (5) into (7) we find

1 y = −2ω cos λ v0 t − gt 2 2 = ω cos λ gt 2 − 2v0 t 2

(7)

(

)

(8)

Integrating (8) and using the initial condition y(t = 0) = 0, we find

1 y = ω cos λ gt 2 − v0 t 2 3 From (5), the time the particle strikes the ground (z = 0) is 1 0 = v0 − gt t 2 so that t= Substituting this value into (9), we have 2v0 g

(9)

(10)

1 8v 3 4v 2 y = ω cos λ g 30 − v0 20 g 3 g v3 4 = − ω cos λ 0 3 g2 If we use (6), (11) becomes (11)

MOTION IN A NONINERTIAL REFERENCE FRAME

339

4 8h3 y = − ω cos λ 3 g The negative sign of the displacement shows that the particle is displaced to the west.

(12)

10-9. Choosing the same coordinate system as in Example 10.3 (see Fig. 10-9), we see that the lateral deflection of the projectile is in the x direction and that the acceleration is

ax = x = 2 ω z vy = 2 (ω sin λ )(V0 cos α ) Integrating this expression twice and using the initial conditions, x ( 0 ) = 0 and x ( 0 ) = 0 , we obtain x ( t ) = ω V0 t 2 cos α sin λ Now, we treat the z motion of the projectile as if it were undisturbed by the Coriolis force. In this approximation, we have z ( t ) = V0 t sin α − 1 2 gt 2

(1)

(2)

(3)

from which the time T of impact is obtained by setting z = 0: T= 2V0 sin α g (4)

Substituting this value for T into (2), we find the lateral deflection at impact to be x (T ) = 4ωV03 sin λ cos α sin 2 α 2 g (5)

10-10. In the previous problem we assumed the z motion to be unaffected by the Coriolis force. Actually, of course, there is an upward acceleration given by −2ω x vy so that

z = 2ω V0 cos α cos λ − g

(1)

from which the time of flight is obtained by integrating twice, using the initial conditions, and then setting z = 0: T′ = 2V0 sin α g − 2ωV0 cos α cos λ (2)

Now, the acceleration in the y direction is ay = y = 2ω x vz = 2 ( −ω cos λ )(V0 sin α − gt ) Integrating twice and using the initial conditions, y ( 0 ) = V0 cos α and y ( 0 ) = 0 , we have (3)

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CHAPTER 10

y (t) =

1 ω gt 3 cos λ − ωV0 t 2 cos λ sin α + V0 t cos α 3

(4)

Substituting (2) into (4), the range R′ is

R′ =

4ωV03 sin 3 α cos λ 2V02 cos α cos λ 8 ωV03 g sin 3 α cos λ − + 3 ( g − 2ωV0 cos α cos λ ) 3 ( g − 2ωV0 cos α cos λ ) 2 g − 2ωV0 cos α cos λ

(5)

We now expand each of these three terms, retaining quantities up to order ω but neglecting all quantities proportional to ω 2 and higher powers of ω. In the first two terms, this amounts to neglecting 2ωV0 cos α cos λ compared to g in the denominator. But in the third term we must use

2V02 cos α sin α 2ωV0 cos α cos λ g 1 − g ≅ 2ωV0 2V02 cos α sin α 1 + cos α cos λ g g 4ωV03 sin α cos 2 α cos λ 2 g

= R0 + ′

(6)

where R′ is the range when Coriolis effects are neglected [see Example 2.7]: 0 R0 = ′ 2V02 cos α sin α g (7)

The range difference, ∆R ′ = R ′ − R0 , now becomes ′ ∆R ′ = 4ωV03 1 cos λ sin α cos 2 α − sin 3 α 2 g 3 (8)

Substituting for V0 in terms of R′ from (7), we have, finally, 0 ∆R′ = 2R0 1 ′ ω cos λ cot1 2 α − tan 3 2 α g 3 (9)

MOTION IN A NONINERTIAL REFERENCE FRAME

341

10-11. d= Rθ

R sin θ

θ

This problem is most easily done in the fixed frame, not the rotating frame. Here we take the Earth to be fixed in space but rotating about its axis. The missile is fired from the North Pole at some point on the Earth’s surface, a direction that will always be due south. As the missile travels towards its intended destination, the Earth will rotate underneath it, thus causing it to miss. This distance is: ∆ = (transverse velocity of Earth at current latitude) × (missile’s time of flight) = ω R sin θ × T = dω R d sin R v (1) (2)

Note that the actual distance d traveled by the missile (that distance measured in the fixed frame) is less than the flight distance one would measure from the Earth. The error this causes in ∆ will be small as long as the miss distance is small. Using R = 6370 km, ω = 7.27 × 10 −5 rad ⋅ s −1 , we obtain for the 4800 km, T = 600 s flight a miss distance of 190 km. For a 19300 km flight the missile misses by only 125 km because there isn’t enough Earth to get around, or rather there is less of the Earth to miss. For a fixed velocity, the miss distance actually peaks somewhere around d = 12900 km. Doing this problem in the rotating frame is tricky because the missile is constrained to be in a path that lies close to the Earth. Although a perturbative treatment would yield an order of magnitude estimate on the first part, it is entirely wrong on the second part. Correct treatment in the rotating frame would at minimum require numerical methods.

10-12.

ε r0 λ Fs z

x

342

CHAPTER 10

Using the formula

Feff = ma f − mω × ( ω × r ) − 2mω × v r

(1)

we try to find the direction of Feff when ma f (which is the true force) is in the direction of the z axis. Choosing the coordinate system as in the diagram, we can express each of the quantities in (1) as

ω = (−ω cos λ , 0, ω sin λ ) r = (0, 0, R) ma f = (0, 0, − mg0 ) Hence, we have ω × r = Rω cos λ e y and (1) becomes (3) vr = 0

(2)

ex Feff = − mg0 ez − m −ω cos λ 0 from which, we have

ey

ez (4)

0 ω sin λ Rω cos λ 0

Feff = − mg0 ez + mRω 2 sin λ cos λ ex + mRω 2 cos 2 λ e z Therefore, ( Ff )x = mRω 2 sin λ cos λ 2 2 ( Ff )z = − mg 0 + mRω cos λ

(5)

(6)

The angular deviation is given by tan ε = ( Ff )x ( Ff )z = Rω 2 sin λ cos λ g 0 − Rω 2 cos 2 λ (7)

Since ε is very small, we can put ε ≅ ε . Then, we have

ε=

Rω 2 sin λ cos λ g 0 − Rω 2 cos 2 λ

45° .

(8)

It is easily shown that ε is a maximum for λ

Using R = 6.4 × 10 8 cm , ω = 7.3 × 10 −5 sec −1 , g = 980 cm/sec 2 , the maximum deviation is

ε≅

1.7 ≅ 0.002 rad 980

(9)

MOTION IN A NONINERTIAL REFERENCE FRAME

343

10-13. ω z z′ ε

Earth x x′ λ

The small parameters which govern the approximations that need to be made to find the southerly deflection of a falling particle are:

δ≡ and height of fall h = R radius of Earth

(1)

α≡

centrifugal force Rω 2 = g0 purely gravitational force

(2)

The purely gravitational component is defined the same as in Problem 10-12. Note that although both δ and α are small, the product δα = hω 2 g 0 is still of order ω 2 and therefore expected to contribute to the final answer. Since the plumb line, which defines our vertical direction, is not in the same direction as the outward radial from the Earth, we will use two coordinate systems to facilitate our analysis. The unprimed coordinates for the Northern Hemisphere-centric will have its x-axis towards the south, its y-axis towards the east, and its z-axis in the direction of the plumb line. The primed coordinates will share both its origin and its y′-axis with its unprimed counterpart, with the z′and x′-axes rotated to make the z′-axis an outward radial (see figure). The rotation can be described mathematically by the transformation

x = x ′ cos ε + z ′ sin ε y = y′ z = − x ′ sin ε + z ′ cos ε where (3) (4) (5)

ε≡ as found from Problem 10-12.

Rω 2 sin λ cos λ g

(6)

a) The acceleration due to the Coriolis force is given by

a X ≡ −2ω × v ′

Since the angle between ω and the z′-axis is π – λ, (7) is most appropriately calculated in the primed coordinates:

(7)

344 x ′ = 2ω y ′ sin λ y ′ = −2ω ( z ′ cos λ + x ′ sin λ ) z ′ = 2ω y ′ cos λ

In the unprimed coordinates, the interesting component is

CHAPTER 10

(8) (9) (10)

x = 2ω y ( sin λ cos ε + cos λ sin ε )

At our level approximation this becomes 2ω y sin λ

(11)

x

(12)

Using the results for y and z , which is correct to order ω (also found from Example 10.3),

x

2ω 2 gt 2 sin λ cos λ

(13)

Integrating twice and using the zeroth order result for the time-of-fall, t = 2 h g , we obtain for the deflection

dX =

2 h2 2 ω sin λ cos λ 3 g

(14)

b) The centrifugal force gives us an acceleration of

a c ≡ −ω × ( ω × r ′ )

The component equations are then

(15)

x ′ = ω 2 sin λ x ′ sin λ + ( R + z ′ ) cos λ y′ = ω 2 y′ z ′ = ω 2 cos λ x ′ sin λ + ( R + z ′ ) cos λ − g0

(16) (17) (18)

where we have included the pure gravitational component of force as well. Now transform to the unprimed coordinates and approximate

x ω 2 ( R + z ) sin λ cos λ − g0 sin ε

We can use Problem 10-12 to obtain sin ε to our level of approximation sin ε

(19)

ε

Rω 2 sin λ cos λ g0 x ω 2 z sin λ cos λ

(20)

The prompts a cancellation in equation (19), which becomes simply (21)

Using the zeroth order result for the height, z = h − gt 2 2 , and for the time-of-fall estimates the deflection due to the centrifugal force

dc

5 h2 2 ω sin λ cos λ 6 g

(22)

MOTION IN A NONINERTIAL REFERENCE FRAME

345

c) Variation in gravity causes the acceleration

ag ≡ −

GM r + g0 k r3

(23)

where r = x ′i + y ′j + ( R + z ′ ) k is the vector pointing to the particle from the center of the spherical Earth. Near the surface

r 2 = x ′ 2 + y ′ 2 + ( z ′ + R)

2

R2 + 2Rz ′

(24)

so that (23) becomes, with the help of the binomial theorem,

ag

Transform and get the x component

−

g0 ( x ′i + y ′j − 2 z ′k ) R

(25)

x

=

g0 ( − x ′ cos ε + 2z ′ sin ε ) R g0 − ( x cos ε − z sin ε ) cos ε + 2 ( x sin ε + z cos ε ) sin ε R g0 ( − x + 3z sin ε ) R

(26) (27) (28)

Using (20),

x

3ω 2 z sin λ cos λ

(29)

where we have neglected the x R term. This is just thrice the part (b) result,

dg

5 h2 2 ω sin λ cos λ 2 g

(30)

Thus the total deflection, correct to order ω 2 , is

d

4

h2 2 ω sin λ cos λ g

(31)

(The solution to this and the next problem follow a personal communication of Paul Stevenson, Rice University.)

10-14. The solution to part (c) of the Problem 10-13 is modified when the particle is dropped down a mineshaft. The force due to the variation of gravity is now

ag ≡ −

g0 r + g0 k R

(1)

As before, we approximate r for near the surface and (1) becomes

ag

In the unprimed coordinates,

−

g0 ( x ′i + y ′j + z ′k ) R

(2)

346

CHAPTER 10

x

− g0

x R

(3)

To estimate the order of this term, as we probably should have done in part (c) of Problem 10-13, we can take x ~ h 2ω 2 g , so that

x ~ ω 2h ×

h R

(4)

which is reduced by a factor h R from the accelerations obtained previously. We therefore have no southerly deflection in this order due to the variation of gravity. The Coriolis and centrifugal forces still deflect the particle, however, so that the total deflection in this approximation is

d

3 h2 2 ω sin λ cos λ 2 g

(5)

10-15.

The Lagrangian in the fixed frame is

L=

1 mv 2 − U rf f 2

( )

(1)

where v f and rf are the velocity and the position, respectively, in the fixed frame. Assuming we have common origins, we have the following relation

v f = v r + ω × rr where vr and rr are measured in the rotating frame. The Lagrangian becomes

(2)

L=

The canonical momentum is

m 2 2 v + 2v r ⋅ ( ω × rr ) + ( ω × rr ) − U ( rr ) r 2

∂L = mv r + m ( ω × rr ) ∂v r

(3)

pr ≡

The Hamiltonian is then

(4)

H ≡ v r ⋅ pr − L =

1 1 2 mvr2 − U ( rr ) − m ( ω × rr ) 2 2

(5)

H is a constant of the motion since ∂L ∂t = 0 , but H ≠ E since the coordinate transformation equations depend on time (see Section 7.9). We can identify Uc = −

1 2 m ( ω × rr ) 2 (6)

as the centrifugal potential energy because we may find, with the use of some vector identities, −∇U c =

m 2 2 2 ∇ ω rr − ( ω ⋅ rr ) 2

(7) (8)

= m ω 2 rr − ( ω ⋅ rr ) ω

MOTION IN A NONINERTIAL REFERENCE FRAME

347

(9)

= − mω × ( ω ⋅ rr ) which is the centrifugal force. Computing the derivatives of (3) required in Lagrange’s equations

d ∂L = ma r + mω × v r dt ∂v r ∂L = m∇ ( v r × ω ) ⋅ rr − ∇ (U c + U ) ∂rr = − m ( ω × v r ) − mω × ( ω × rr ) − ∇U

The equation of motion we obtain is then

(10) (11) (12)

ma r = −∇U − mω × ( ω × rr ) − 2m ( ω × v r )

(13)

If we identify Feff = ma r and F = −∇U , then we do indeed reproduce the equations of motion given in Equation 10.25, without the second and third terms.

10-16. The details of the forces involved, save the Coriolis force, and numerical integrations in the solution of this problem are best explained in the solution to Problem 9-63. The only thing we do here is add an acceleration caused by the Coriolis force, and re-work every part of the problem over again. This is conceptually simple but in practice makes the computation three times more difficult, since we now also must include the transverse coordinates in our integrations. The acceleration we add is

a c = 2ω vy sin λi − ( vx sin λ + vz cos λ ) j + vy cos λ k where we have chosen the usual coordinates as shown in Figure 10-9 of the text.

a) Our acceleration is

(1)

a = − gk + a C

As a check, we find that the height reached is 1800 km, in good agreement with the result of 77 km, to the west.

(2)

Problem 9-63(a). The deflection at this height is found to be

b) This is mildly tricky. The correct treatment says that the equation of motion with air resistance is (cf. equation (2) of Problem 9-63 solution)

v a = − g k + 2 vt The deflection is calculated to be

c) d)

v + aC

(3)

8.9 km. 10 km. 160 km.

Adding the vaiation due to gravity gives us a deflection of Adding the variation of air density gives us a deflection of

348

CHAPTER 10

Of general note is that the deflection in all cases was essentially westward. The usual small deflection to the north did not contribute significantly to the total transverse deflection at this precision. All of the heights obtained agreed well with the answers from Problem 9-63. Inclusion of the centrifugal force also does not change the deflections to a significant degree at our precision.

10-17. Due to the centrifugal force, the water surface of the lake is not exactly perpendicular to the Earth’s radius (see figure).

B

m

β

Ta n

g

ge

nt

Wa ter sur β

to

fac

e

Ea

rth

A

α

su r

C

fa ce

The length BC is (using cosine theorem)

BC =

AC 2 + (mg ) 2 − 2 ACmg cos α

where AC is the centrifugal force AC = mω 2 R cos α with α = 47° and Earth’s radius R ≅ 6400 km , The angle β that the water surface is deviated from the direction tangential to the Earth’s surface is

BC AC = sin α sin β

⇒ sin β =

AC sin α = 4.3 × 10−5 BC

So the distance the lake falls at its center is h = r sin β where r = 162 km is the lake’s radius. So finally we find h = 7 m.

MOTION IN A NONINERTIAL REFERENCE FRAME

349

10-18.

Let us choose the coordinate system Oxyz as shown in the figure.

α

O β νy

νx ν

y

x

The projectile’s velocity is

vx v0 cos β v = v y = v0 sin β − gt 0 0

The Earth’s angular velocity is

where β = 37°

−ω cos α ω = −ω sin α 0

So the Coriolis acceleration is

where α = 50°

ac = 2v × ω = −2v0 ω cos β sin α + 2 ( v0 sin β − gt ) ω cos α e z

The velocity generated by Coriolis force is

(

)

vc = ∫ ac dt = 2v0 ω t ( cos β sin α − sin β cos α ) − gt 2ω cos α

0

t

And the distance of deviation due to the Coriolis force is

gt 3ω cos α zc = ∫ vc dt = −v0 ω t sin (α − β ) − 3 0 t 2

The flight time of the projectile is t = distance due to Coriolis force to be

2v0 sin β . If we put this into zc , we find the deviation 2

zc ~ 260 m

350

10-19.

CHAPTER 10

The Coriolis force acting on the car is

Fc = 2m v × ω ⇒ Fc = 2mvω sin α where α = 65°, m = 1300 kg, v = 100 km/hr. So Fc = 4.76 N.

10-20. Given the Earth’s mass, M = 5.976 × 1024 kg , the magnitude of the gravitational field vector at the poles is

g pole =

GM = 9.866 m/s 2 2 R pole

The magnitude of the gravitational field vector at the equator is

g eq =

GM − ω 2 R e q = 9.768 m/s 2 2 Req

where ω is the angular velocity of the Earth about itself. If one use the book’s formula, we have

g (λ = 90°) = 9.832 m/s 2 at the poles and g (λ = 0°) = 9.780 m/s 2 at the equator

10-21.

The Coriolis acceleration acting on flowing water is

ac = 2v × ω ⇒

ac = 2vω sin α

Due to this force, the water is higher on the west bank. As in problem 10-17, the angle β that the water surface is deviated from the direction tangential to Earth’s surface is

sin β =

ac g +a

2 2 c

=

2vω sin α g + 4v ω sin α ∝

2 2 2 2

= 2.5 × 10−5

The difference in heights of the two banks is

∆h = sin β = 1.2 × 10−3 m where = 47 m is the river’s width.

The Coriolis acceleration is ac = 2v × ω . This acceleration ac pushes lead bullets

10-22.

eastward with the magnitude ac = 2vω cos α = 2 gt ω cos α , where α = 42°. The velocity generated by the Coriolis force is

MOTION IN A NONINERTIAL REFERENCE FRAME

351

vc (t ) = ∫ a dt = gt 2ω cos α and the deviation distance is

∆xc = ∫ vc (t ) dt =

The falling time of the bullet is t =

gt 3 ω cos α 3

2h g . So finally

∆xc =

ω 8h3

3 g

cos α = 2.26 × 10−3 m

352

CHAPTER 10

CHAPTER

11

Dynamics of Rigid Bodies

11-1.

The calculation will be simplified if we use spherical coordinates:

x = r sin θ cos φ y = r sin θ sin φ z = r cos θ z (1)

y x

Using the definition of the moment of inertia,

I ij = ∫ ρ ( r ) δ ij we have

∑x k 2 k

− xi x j dv

(2)

I 33 = ρ ∫ r 2 − z 2 dv = ρ ∫ r − r cos θ r dr d ( cos θ ) dφ

2 2 2 2

( (

)

)

(3)

or,

I 33 = ρ ∫ r 4 dr

0 R +1

−1

∫ (1 − cos θ ) d ( cos θ ) ∫ dφ

2 0

2π

= 2πρ

R5 4 ⋅ 5 3

(4)

353

354

CHAPTER 11

The mass of the sphere is

M= Therefore, I 33 = 2 MR 2 5 (6) 4π ρ R3 3 (5)

Since the sphere is symmetrical around the origin, the diagonal elements of {I} are equal:

I11 = I 22 = I 33 =

2 MR2 5

(7)

A typical off-diagonal element is

I12 = ρ ∫ ( − xy ) dv

= − ρ ∫ r sin θ sin φ cos φ r dr d ( cos θ ) dφ

2 2 2

(8)

This vanishes because the integral with respect to φ is zero. In the same way, we can show that all terms except the diagonal terms vanish. Therefore, the secular equation is

I11 − I

0

I 22 − I

0 0

I 33 − I

0 0 From (9) and (7), we have

=0

(9)

0

I1 = I 2 = I 3 =

2 MR2 5

(10)

11-2. a) Moments of inertia with respect to the xi axes: x3 = x3′ R h CM x1′ x1

x2′ x2

It is easily seen that I ij = 0 for i ≠ j. Then the diagonal elements I ii become the principal moments I i , which we now calculate. The computation can be simplified by noting that because of the symmetry, I1 = I 2 ≠ I 3 . Then, I1 = I 2 = I1 + I 2 ρ = 2 2

∫ ( 2x

2 3

2 2 + x1 + x2 dv

)

(1)

DYNAMICS OF RIGID BODIES

355

which, in cylindrical coordinates, can be written as I1 = I 2 = where

ρ

2∫

2π 0

dφ ∫ dz ∫

0

h

Rz h

0

(r

2

+ 2z 2 rdr

)

(2)

ρ=

M 3M = V π R2 h

(3)

Performing the integration and substituting for ρ, we find I1 = I 2 = I 3 is given by

2 2 I 3 = ρ ∫ x1 + x2 dv = ρ ∫ r 2 ⋅ rdr dφ dz

3 M R2 + 4 h 2 20

(

)

(4)

(

)

(5)

from which

I3 =

3 MR 2 10

(6)

b) Moments of inertia with respect to the xi′ axes:

Because of the symmetry of the body, the center of mass lies on the x′ axis. The coordinates of 3 the center of mass are (0 , 0, z0 ) , where

z0 =

Then, using Eq. (11.49),

∫ x ′ dv = 3 h ∫ dv 4

3

(7)

I ij = I ij − M a 2δ ij − ai a j ′

In the present case, a1 = a2 = 0 and a3 = ( 3 4 ) h , so that

(8)

I1 = I1 − ′ I2 = I2 − ′ I3 = I3 − ′

9 3 1 Mh 2 = M R2 + h2 16 20 4 9 3 1 Mh 2 = M R2 + h 2 16 20 4 3 MR2 10

11-3.

The equation of an ellipsoid is

2 2 2 x1 x2 x3 + 2 + 2 =1 a2 b c

(1)

356

CHAPTER 11

which can be written in normalized form if we make the following substitutions: x1 = aξ , x2 = bη , x3 = cζ Then, Eq. (1) reduces to (2)

ξ2 +η2 +ζ 2 = 1

This is the equation of a sphere in the (ξ,η,ζ) system. If we denote by dv the volume element in the xi system and by dτ the volume element in the (ξ,η,ζ ) system, we notice that the volume of the ellipsoid is

(3)

V = ∫ dv = ∫ dx1 dx2 dx3 = abc ∫ dξ dη dζ 4 = abc ∫ dτ = π abc 3 because ∫ dτ is just the volume of a sphere of unit radius. The rotational inertia with respect to the x3 -axis passing through the center of mass of the ellipsoid (we assume the ellipsoid to be homogeneous), is given by

I3 = = M V

(4)

∫(x

2 1

2 + x2 dv

)

M abc ∫ a 2ξ 2 + b 2η 2 dτ V

(

)

(5)

In order to evaluate this integral, consider the following equivalent integral in which z = r cos θ :

∫a

2

z 2 dv = ∫ a 2 z 2 ( r dr r sin θ dθ dφ ) = a2

∫

2π 0

dφ ∫ cos 2 θ sin θ dθ ∫

0

π

R =1 0

r 4 dr

= a 2 × 2π × = Therefore, 4π a 2 15

2 1 × 3 5 (6)

∫ (a ξ

2

2

+ b 2η 2 dτ =

)

4π 2 a + b2 15

(

)

(7)

and I3 = 1 M a2 + b 2 5

(

)

(8)

Since the same analysis can be applied for any axis, the other moments of inertia are

DYNAMICS OF RIGID BODIES

357

I1 =

1 M b2 + c2 5

(

) )

(9)

1 I 2 = M a2 + c2 5

(

11-4.

The linear density of the rod is

ρ =

m

(1)

For the origin at one end of the rod, the moment of inertia is

I = ∫ ρ x 2 dx =

0

m

3

3

=

m 3

2

(2)

If all of the mass were concentrated at the point which is at a distance a from the origin, the moment of inertia would be I = ma 2 Equating (2) and (3), we find a= This is the radius of gyration.

11-5.

z J a Q M z–a

(3)

3

(4)

a) The solid ball receives an impulse J; that is, a force F(t) is applied during a short interval of time τ so that

J = ∫ F ( t ′ ) dt ′

The equations of motion are dp =F dt dL =r×F dt

(1)

(2) (3)

358

CHAPTER 11

which, for this case, yield ∆p = ∫ F ( t ′ ) dt ′ = J ∆L = ∫ r × F ( t ′ ) dt ′ = r × J Since p(t = 0) = 0 and L(t = 0) = 0, after the application of the impulse, we have p = MVCM = J; L = I 0 ω = r × J = ( z − a ) J

(4) (5)

ω

ω

(6)

so that

VCM = J M

(7)

and ω= J ω ( z − a) ω I0

(8)

where I 0 = ( 2 5) Ma 2 . The velocity of any point a on the ball is given by Eq. (11.1):

vα = VCM + ω × rα

For the point of contact Q, this becomes

(9)

v Q = VCM − ω a

=

J J

(10)

J 5 ( z − a) 1− 2a M

Then, for rolling without slipping, v Q = 0 , and we have

2a = 5 ( z − a) so that z= 7 a 5

(11)

(12)

b) Many billiard tricks are performed by striking the ball at different heights and at different angles in order to impart slipping and spinning motion (“English”). For the table not to introduce spurious effects, the rail must be at such a height that the ball will be “reflected” upon collision.

Consider the case in which the ball is incident normally on the rail, as in the diagram. We have the following relationships:

DYNAMICS OF RIGID BODIES

359 y VCM x

Before Collision Linear Momentum px = − MVCM py = 0 Angular Momentum Lx = 0 Ly = * Lz = 0

After Collision px = + MVCM ′ py = 0 ′ Lx = 0 ′ Ly = − Ly ′ Lz = 0 ′

* The relation between Ly and VCM depends on whether or not slipping occurs.

Then, we have

∆p = −2 px = J = 2 MVCM ∆L = −2Ly = 2I 0 ω = J ( z − a) so that 2I 0 ω = 2 MVCM ( z − a ) from which

(13) (14)

(15)

z−a=

I0 ω 2 Ma 2ω 2 a2ω = = MVCM 5 MVCM 5 VCM

(16)

If we assume that the ball rolls without slipping before it contacts the rail, then VCM = ω a , and we obtain the same result as before, namely,

z−a= or, 2 a 5

(17)

z=

7 a 5

(18)

Thus, the height of the rail must be at a height of ( 2 5) a above the center of the ball.

11-6. Let us compare the moments of inertia for the two spheres for axes through the centers of each. For the solid sphere, we have

Is =

2 MR 2 5

(see Problem 11-1)

(1)

360

CHAPTER 11

For the hollow sphere,

R sin θ

θ

I h = σ ∫ dφ ∫ ( R sin θ ) R2 sin θ dθ

2 0 0

2π

π

= 2πσ R 4 ∫ sin 3 θ dθ

0

π

8 = πσ R 4 3 or, using 4πσ R 2 = M , we have

Ih =

2 MR 2 3

(2)

Let us now roll each ball down an inclined plane. [Refer to Example 7.9.] The kinetic energy is

T=

1 1 M y 2 + Iθ 2 2 2

(3)

where y is the measure of the distance along the plane. The potential energy is

U = Mg ( − y ) sin α

(4)

where is the length of the plane and α is the angle of inclination of the plane. Now, y = Rθ, so that the Lagrangian can be expressed as

L=

1 1 I 2 M y2 + y + Mgy sin α 2 2 R2

(5)

where the constant term in U has been suppressed. The equation of motion for y is obtained in the usual way and we find

y=

gMR 2 sin α MR2 + I

(6)

Therefore, the sphere with the smaller moment of inertia (the solid sphere) will have the greater acceleration down the plane.

DYNAMICS OF RIGID BODIES

361

11-7. x R θ

d φ

r

The force between the force center and the disk is, from the figure

F = − kr

(1)

Only the component along x does any work, so that the effective force is Fx = − kr sin φ = − kx . This corresponds to a potential U = kx 2 2 . The kinetic energy of the disk is

T=

1 1 3 Mx 2 + Iθ 2 = Mx 2 2 2 4

(2)

where we use the result I = MR2 2 for a disk and dx = R dθ. Lagrange’s equations give us 3 Mx + kx = 0 2 This is simple harmonic motion about x = 0 with an angular frequency of oscillations (3)

ω=

2k 3m

(4)

11-8. x3′ h

w x1′

x2′

d

We let x′ be the vertical axis in the fixed system. This would be the axis (i.e., the hinge line) of 3 the door if it were properly hung (no self-rotation), as indicated in the diagram. The mass of the door is M=ρwhd. The moment of inertia of the door around the x3 axis is ′ m 1 2 2 I3 = ∫ dh′ ∫ w ′ d dw ′ = 3 Mw whd 0 0 h w

(1)

where the door is considered to be a thin plate, i.e., d

w,h.

362

CHAPTER 11

The initial position of the self-closing door can be expressed as a two-step transformation, starting with the position in the diagram above. The first rotation is around the x′ -axis through 1 an angle θ and the second rotation is around the x′′-axis through an angle ψ: 1 x3′ w h

x3′ θ

x3″

x3′

x3 = x3″ w3 = w

x2′ x1′ x2′

x1′ = x1″

θ x2″

x1″

ψ

x1

x2′ ψ x2 x2″

The x′ -axes are the fixed-system axes and the xi -axes are the body system (or rotating) axes 1 which are attached to the door. Here, the Euler angle φ is zero. The rotation matrix that transforms the fixed axes into the body axes ( xi′ → xi ) is just Eq. (11.99) with φ = 0 and θ → – θ since this rotation is performed clockwise rather than counterclockwise as in the derivation of Eq. (11.99):

cos ψ λ = − sin ψ 0

cos θ sin ψ cos θ cos ψ sin θ

− sin θ sin ψ − sin θ cos ψ cos θ

(2)

The procedure is to find the torque acting on the door expressed in the fixed coordinate system and then to obtain the x3 component, i.e., the component in the body system. Notice that when the door is released from rest at some initial angle ψ 0 , the rotation is in the direction to decrease ψ. According to Eq. (11.119),

I 3ω 3 = N 3 = I 3 ψ

(3)

where ω 1 = ω 2 = 0 since φ = θ = 0 . In the body ( xi ) system the coordinates of the center of mass of the door are

0 1 R = w 2 h

(4)

where we have set the thickness equal to zero. In the fixed ( x′) system, these coordinates are i obtained by applying the inverse transformation λ −1 to R; but λ −1 = λ t , so that

− w sin ψ 1 R ′ = λ R = w cos θ cos ψ + h sin θ 2 − w sin θ cos ψ + h cos θ t (5)

Now, the gravitational force acting on the door is downward, and in the x′ coordinate system i is

F ′ = − Mg e′ 3

(6)

DYNAMICS OF RIGID BODIES

363

There the torque on the door, expressed in the fixed system, is

N′ = R ′ × F ′ e1 ′ 1 = − Mg − w sin ψ 2 0 e2 ′ w cos θ cos ψ + h sin θ 0 e′ 3 − w sin θ cos ψ + h cos θ 1

w cos θ cos ψ + h sin θ 1 = − Mg w sin ψ 2 0

(7)

so that in the body system we have

w cos θ cos 2 ψ + h sin θ cos ψ + w cos θ sin 2 ψ 1 N ′ = λN ′ = − Mg − h sin θ sin ψ 2 w sin θ sin ψ

(8)

Thus,

−N3 =

1 Mgw sin θ sin ψ 2

(9)

and substituting this expression into Eq. (3), we have

−

1 1 Mgw sin θ sin ψ = I 3 ψ = Mw 2 ψ 2 3

(10)

where we have used Eq. (1) for I 3 . Solving for ψ ,

ψ =−

3 g sin θ sin ψ 2w

(11)

This equation can be integrated by first multiplying by ψ :

∫ ψ ψ dt = 2 ψ

=

1

2

=−

3 g sin θ ∫ sin ψ ψ dt 2w (12)

3 g sin θ cos ψ 2w

where the integration constant is zero since cos ψ = 0 when ψ = 0 . Thus,

ψ =±

3g sin θ cos ψ w

(13)

We must choose the negative sign for the radical since ψ < 0 when cos ψ > 0 . Integrating again, from ψ = 90° to ψ = 0°,

0

∫ π

2

dψ =− cos ψ

T 3g sin θ ∫ dt w 0

(14)

364 where T = 2 sec. Rewriting Eq. (14), π CHAPTER 11

∫

0

2

dψ =T cos ψ

3g sin θ w

(15)

Using Eq. (E. 27a), Appendix E, we find 1 Γ π 4 −1 2 ∫ cos ψ dψ = 2 3 0 Γ 4 π 2

(16)

From Eqs. (E.20) and (E.23), 1 1 1 Γ = Γ 1 = 0.906 4 4 4 1 Γ = 3.624 4 And from Eqs. (E.20) and (E.24), 3 3 3 Γ = Γ 1 = 0.919 4 4 4 3 Γ = 1.225 4 Therefore, π (17)

(18)

∫

0

2

dψ π 3.624 = = 2.62 2 1.225 cos ψ

(19)

Returning to Eq. (15) and solving for sin θ, sin θ = w 2 × ( 2.62) 3 gT 2 (20)

Inserting the values for g, w(= 1m), and T(= 2 sec), we find

θ = sin −1 ( 0.058 ) or, θ ≅ 3.33°

(21)

DYNAMICS OF RIGID BODIES

365

11-9. y C Q

C′ P θ R a O

x

The diagram shows the slab rotated through an angle θ from its equilibrium position. At equilibrium the contact point is Q and after rotation the contact point is P. At equilibrium the position of the center of mass of the slab is C and after rotation the position is C′. Because we are considering only small departures from θ = 0, we can write QP ≅ Rθ Therefore, the coordinates of C′ are (see enlarged diagram below) (1)

r = OA + AC ′ so that a sin θ − Rθ cos θ 2 a y = R + cos θ + Rθ sin θ 2

(2)

x = R +

(3)

C′ C θ

Rθ A

Q

Rθ

P

θ

O

Consequently,

366

x = R + a cos θ − R cos θ + Rθ sin θ θ 2

CHAPTER 11

a = cos θ + Rθ sin θ θ 2 a y = − R + sin θ + Rθ cos θ + R sin θ θ 2 a = − sin θ + Rθ cos θ θ 2 from which a2 x 2 + y 2 = + R 2θ 2 θ 2 4

(4)

The kinetic energy is

T=

1 1 M x 2 + y 2 + Iθ 2 2 2

(

)

(5)

where I is the moment of inertia of the slab with respect to an axis passing through the center of mass and parallel to the z-axis:

I=

1 M 12

(

2

+ a2

)

(6)

Therefore,

T=

1 f1 (θ ) θ 2 2

(7)

where

a2 f1 (θ ) = M + R 2θ 2 + I 4

(8)

The potential energy is

U = Mgy = − f 2 (θ )

(9)

where a f 2 (θ ) = − Mg R + cos θ + Rθ sin θ 2

(10)

and where Eq. (3) has been used for y. The Lagrangian is

L=

1 f1 (θ ) θ 2 + f 2 (θ ) 2

(11)

The Lagrange equation for θ is

DYNAMICS OF RIGID BODIES

367 d ∂ L ∂L − =0 dt ∂θ ∂θ

(12)

Now,

∂L = f1 (θ ) θ ∂θ d ∂L = f1 (θ ) θ + f1 (θ ) θ dt ∂θ a2 = M + R 2θ 2 + I θ + 2 MR 2 θ θ 2 4 ∂L 1 = f1′(θ ) θ 2 + f 2′ (θ ) ∂θ 2 a = MR 2 θ θ 2 + Mg R + sin θ − Rθ cos θ − R sin θ 2

(13)

(14)

Combining, we find

a2 2 2 2 2 M + R θ + I θ + MR θ θ − Mg R + 4 a sin θ − Rθ cos θ − R sin θ = 0 2

(15)

For the case of small oscillations, θ 2

θ and θ 2

θ , so that Eq. (15) reduces to

a Mg R − 2 θ+ θ =0 2 Ma +I 4 The system is stable for oscillations around θ = 0 only if a Mg R − 2 = ω2 > 0 2 Ma +I 4 This condition is satisfied if R − a 2 > 0 , i.e., R> Then, the frequency is a Mg R − 2 2 Ma 1 + M 2 + a2 4 12 a 2

(16)

(17)

(18)

ω=

(

)

(19)

Simplifying, we have

368

CHAPTER 11

ω=

a 12 g R − 2

(

2

+ 4a2

)

(20)

According to Eqs. (9) and (10), the potential energy is a U (θ ) = Mg R + cos θ + Rθ sin θ 2 This function has the following forms for R > a 2 and R < a 2 :

R> a 2

U(θ)

(21)

Mg R +

a 2

U(θ)

R<

a 2

−π/2

π/2 θ

θ

To verify that a stable condition exists only for R > a 2 , we need to evaluate ∂ 2U ∂θ 2 at θ = 0: ∂U a = Mg − sin θ + Rθ cos θ ∂θ 2 ∂ 2U a = Mg − cos θ + R cos θ − Rθ sin θ ∂θ 2 2 and a ∂ 2U = Mg R − 2 ∂θ θ = 0 2 so that ∂ 2U a > 0 if R > 2 ∂θ 2

11-10.

z θ m R

(22) (23)

(24)

(25)

When the mass m is at one pole, the z component of the angular momentum of the system is Lz = Iω = 2 MR 2ω 5 (1)

After the mass has moved a distance vt = Rθ along a great circle on the surface of the sphere, the z component of the angular momentum of the system is

DYNAMICS OF RIGID BODIES

369

2 Lz = MR 2 + mR 2 sin 2 θ φ 5

(2)

where φ is the new angular velocity. Since there is no external force acting on the system, angular momentum must be conserved. Therefore, equating (1) and (2), we have

φ=

2 MR 2 + mR2 sin 2 θ 5

2 MR 2ω 5

(3)

Substituting θ = vt R and integrating over the time interval during which the mass travels from one pole to the other, we have t= πR

V

φ=

t=0

∫

2 MR 2 + mR 2 sin 2 ( vt R) 5

2 MR2ω 5

dt

(4)

Making the substitutions, vt R ≡ u, dt = ( R v ) du we can rewrite (4) as (5)

φ=∫

0

π

R du 2 v 2 2 2 MR + mR sin u 5 π 2

2 MR 2ω 5

2Rω = v

∫ 1 + β sin

0

du

2

u

(6)

where β ≡ 5m 2 M and where we have used the fact that the integrand is symmetric around u = π 2 to write φ as twice the value of the integral over half the range. Using the identity sin 2 u = we express (6) as 2Rω φ= v or, changing the variable to x = 2u, π 2

1 (1 − cos 2u) 2

(7)

∫

0

du 1 1 1 + β − β cos 2u 2 2

(8)

φ=

Rω v

∫

π

0

dx 1 1 1 + β − β cos x 2 2

(9)

Now, we can use Eq. (E.15), Appendix E, to obtain

370

(1 + β ) tan ( x 2) 2Rω φ= tan −1 v 1+ β 1+ β = π CHAPTER 11

0

π Rω v 2M 2 M + 5m 2M 2 M + 5m (10)

= ωT

where T = π R v is the time required for the particle to move from one pole to the other. If m = 0, (10) becomes

φ ( m = 0) = ωT

Therefore, the angle of retardation is

(11)

α = φ ( m = 0 ) − φ ( m) or, (12)

α = ω T 1 −

2M 2 M + 5m

(13)

11-11. a) No sliding:

2 2

P

P

From energy conservation, we have mg 2 = mg 2 + 1 1 2 mvC.M. + Iω 2 2 2 (1)

where vCM is the velocity of the center of mass when one face strikes the plane; vC.M. is related to ω by vCM = 2

ω

(2)

I is the moment of inertia of the cube with respect to the axis which is perpendicular to one face and passes the center: I= Then, (1) becomes 1 m 6

2

(3)

DYNAMICS OF RIGID BODIES

371

mg 2 from which, we have

(

1 ω 1 m 2 2 1 ω = m 2ω 2 2 −1 = m + 2 2 2 6 3

)

2

(4)

ω2 =

b) Sliding without friction:

3 g 2

(

2 −1

)

(5)

In this case there is no external force along the horizontal direction; therefore, the cube slides so that the center of mass falls directly downward along a vertical line.

P

θ

P

While the cube is falling, the distance between the center of mass and the plane is given by y= 2 cos θ (6)

Therefore, the velocity of center of mass when one face strikes the plane is

=−

0 =π 4

y

2

sin θ θ θ =π 4

=−

1 1 θ=− ω 2 2

(7)

From conservation of energy, we have mg from which we have 1 1 11 = mg + m − ω + m 2 ω 2 2 6 2 2 2 2

2

(8)

ω2 =

12 g 5

(

2 −1

)

(9)

11-12.

According to the definition of the principal moments of inertia,

2 I j + I k = ∫ ρ xi2 + xk dv + ∫ ρ xi2 + x 2 dv j 2 = ∫ ρ x 2 + xk dv + 2 ∫ ρ xi2 dv j

(

)

(

)

(

)

= I i + 2 ∫ ρ xi2 dv since

(1)

∫ρ x we have

2 i

dv > 0

372

I j + Ik ≥ Ii

CHAPTER 11

(2)

11-13.

We get the elements of the inertia tensor from Eq. 11.13a:

2 2 I11 = ∑ mα xα ,2 + xα ,3

α

(

) ( )

= 3m b 2 + 4 m 2b 2 + 2m b 2 = 13mb 2 Likewise I 22 = 16 mb 2 and I 33 = 15mb 2 I12 = I 21 = − ∑ mα xα ,1 xα ,2 α ( )

( )

= −4 m b 2 − 2m − b 2 = −2mb 2 Likewise I13 = I 31 = mb 2 and I 23 = I 32 = 4mb 2 Thus the inertia tensor is 13 −2 1 {I } = mb −2 16 4 1 4 15

2

( )

( )

The principal moments of inertia are gotten by solving −2 1 13 − λ mb −2 16 − λ 4 =0 1 4 15 − λ

2

Expanding the determinant gives a cubic equation in λ:

λ 3 − 44λ 2 + 622λ − 2820 = 0

Solving numerically gives

λ1 = 10.00 λ 2 = 14.35 λ 3 = 19.65

Thus the principal moments of inertia are I1 = 10 mb 2 I 2 = 14.35 mb 2 I 3 = 19.65 mb 2 To find the principal axes, we substitute into (see example 11.3):

DYNAMICS OF RIGID BODIES

373

(13 − λi ) ω1i − 2ω 2i + ω 3i = 0

−2 ω 1i + (16 − λ i ) ω 2 i + 4ω 3 i = 0

ω 1i + 4ω 2 i + (15 − λi ) ω 3 i = 0

For i = 1, we have ( λ1 = 10 ) 3ω 11 − 2ω 21 + ω 31 = 0 −2ω 11 − 6ω 21 + 4ω 31 = 0

ω 11 − 4ω 21 + 5ω 31 = 0

Solving the first for ω 31 and substituting into the second gives

ω 11 = ω 21

Substituting into the third now gives

ω 31 = −ω 21 or ω 11 : ω 21 : ω 31 = 1 : 1 : − 1

So, the principal axis associated with I1 is 1 ( x + y − z) 3 Proceeding in the same way gives the other two principal axes: i = 2: i = 3: −.81x + .29y − .52z −.14 x + .77 y + .63z

We note that the principal axes are mutually orthogonal, as they must be.

11-14.

z

y x

Let the surface of the hemisphere lie in the x-y plane as shown. The mass density is given by

ρ=

M M 3M = = V 2 π b 3 2π b 3 3

First, we calculate the center of mass of the hemisphere. By symmetry

374 xCM = yCM = 0 zCM = 1 M

CHAPTER 11

∫ ρz dv v b

Using spherical coordinates ( z = r cos θ , dv = r 2 sin θ dr dθ dφ ) we have

zCM =

ρ

M

2π

φ =0

∫

π 2

dφ

θ =0

∫

sin θ cos θ dθ

r =0

∫r

3

dr

3 1 1 3 2π ) b 4 = b = 3( 2π b 2 4 8 We now calculate the inertia tensor with respect to axes passing through the center of mass: z′ = z

y′

3 b 8

x′

By symmetry, I12 = I 21 = I13 = I 31 = I 23 = I 32 = 0 . Thus the axes shown are the principal axes. Also, by symmetry I11 = I 22 . We calculate I11 using Eq. 11.49: 3 I11 = J11 − M v 8

2

(1)

where J11 = the moment of inertia with respect to the original axes

J11 = ρ ∫ y 2 + z 2 dv v (

)

= ∫ r 2 sin 2 θ sin 2 φ + r 2 cos 2 θ r 2 sin θ dr dθ dφ v (

)

=

3M 2π b 3

r =0

∫

b

r 4 dr

2π 2 2 2 ∫ ∫ sin θ sin φ + cos θ dφ sin θ dθ θ = 0 φ = 0 π 2

(

)

3 Mb 2 = 10π = Thus, from (1) 2 Mb 2 5

π 2

θ =0

∫ (π sin

3

θ + 2π cos 2 θ sin θ ) dθ

I11 = I 22 = Also, from Eq. 11.49

2 9 83 Mb 2 − Mb 2 = Mb 2 5 64 320

DYNAMICS OF RIGID BODIES

375

I 33 = J 33 − M ( 0 ) = J 33 ( I 33 = J 33 should be obvious physically) So

I 33 = ρ ∫ x 2 + y 2 dv v (

)

= ρ ∫ r 4 sin 3 θ dr dθ dφ = v 2 Mb 2 5

Thus, the principal axes are the primed axes shown in the figure. The principal moments of inertia are I11 = I 22 = I 33 = 83 Mb 2 320

2 Mb 2 5

11-15.

P θ g

We suspend the pendulum from a point P which is a distance from the center of mass. The rotational inertia with respect to an axis through P is

2 I = MR0 + M 2

(1)

where R0 is the radius of gyration about the center of mass. Then, the Lagrangian of the system is L = T −U = Lagrange’s equation for θ gives Iθ + Mg sin θ = 0 For small oscillation, sin θ ≅ θ . Then, (3) Iθ 2 − Mg (1 − cos θ ) 2 (2)

θ+ or, Mg θ =0 I

(4)

θ+

g R +

2 0

2

θ =0

(5)

376 from which the period of oscillation is

CHAPTER 11

τ=

2π

ω

= 2π

2 R0 + g

2

(6)

If we locate another point P′ which is a distance ′ from the center of mass such that the period of oscillation is also τ, we can write

2 R0 + g 2 from which R0 = 2 2 R0 + ′ 2 g ′

=

(7)

′ . Then, the period must be

τ = 2π or, ′+ g

2

(8)

τ = 2π

+ ′ g

(9)

This is the same as the period of a simple pendulum of the length + ′. Using this method, one does not have to measure the rotational inertia of the pendulum used; nor is one faced with the problem of approximating a simple pendulum physically. On the other hand, it is necessary to locate the two points for which τ is the same.

11-16.

The rotation matrix is cos θ ( λ ) = − sin θ 0 sin θ cos θ 0 0 0 1 (1)

The moment of inertia tensor transforms according to

(Ι ′) = (λ ) (Ι ) (λ t )

That is

cos θ ( I ′ ) = − sin θ 0 sin θ cos θ 0 1 ( A + B) 0 2 1 0 ( A − B) 2 1 0 1 ( A − B) 0 2 cos θ 1 ( A + B) 0 sin θ 2 C 0 0 − sin θ cos θ 0 0 0 1

(2)

DYNAMICS OF RIGID BODIES

377

cos θ = − sin θ 0

sin θ cos θ 0

1 1 2 ( A + B) cos θ + 2 ( A − B) sin θ 0 1 1 0 ( A − B) cos θ + ( A + B) sin θ 2 2 1 0

−

1 1 ( A + B) sin θ + ( A − B) cos θ 2 2 1 1 − ( A − B) sin θ + ( A + B) cos θ 2 2 0

0 0 C

1 1 2 2 2 ( A + B) cos θ + ( A − B) cos θ sin θ + 2 ( A + B) sin θ 1 1 = − ( A − B) sin 2 θ + ( A − B) cos 2 θ 2 2 0 1 1 0 ( A − B) cos 2 θ − ( A − B) sin 2 θ 2 2 1 1 ( A + B) sin 2 θ − ( A − B) sin θ cos θ + ( A + B) cos 2 θ 0 2 2 C 0 or 1 2 ( A + B) + ( A − B) cos θ sin θ 1 1 ( I′ ) = − ( A − B) sin 2 θ + ( A − B) cos 2 θ 2 2 0 If θ = π 4 , sin θ = cos θ = 1 2 . Then, A 0 0 ( I′ ) = 0 B 0 0 0 C 1 1 ( A − B) cos 2 θ − ( A − B) sin 2 θ 0 2 2 1 0 ( A + B) − ( A − B) cos θ sin θ 2 C 0

(3)

(4)

11-17. x3 x2

x1

378

CHAPTER 11

The plate is assumed to have negligible thickness and the mass per unit area is ρs . Then, the inertia tensor elements are I11 = ρs

∫ (r

2

2 − x1 dx1 dx2

)

2 2 2 = ρs ∫ x2 + x3 dx1dx2 = ρs ∫ x2 dx1 dx2 ≡ A

(

)

(1) (2) (3)

I 22 = ρs I 33 = ρs

∫ (r ∫ (r

2

2 2 − x2 dx1 dx2 = ρs ∫ x1 dx1 dx2 ≡ B 2 2 2 − x3 dx1 dx2 = ρs ∫ x1 + x2 dx1 dx2

) )

2

(

)

Defining A and B as above, I 33 becomes I 33 = A + B Also, I12 = ρs I 21 = ρs I13 = ρs I 23 = ρs Therefore, the inertia tensor has the form A −C {I} = −C B 0 0 0 0 A + B (4)

∫ ( − x x ) dx

1 2

1

dx2 ≡ −C dx2 = −C dx2 = 0 = I 31 dx2 = 0 = I 32

(5) (6) (7) (8)

∫ ( − x x ) dx

2 1 1 3

1

∫ ( − x x ) dx

2 3

1

∫ ( − x x ) dx

1

(9)

11-18. The new inertia tensor {I′} is obtained from {I} by a similarity transformation [see Eq. (11.63)]. Since we are concerned only with a rotation around the x3 -axis , the transformation matrix is just λ φ , as defined in Eq. (11.91). Then,

− I′ = λ φ I λ φ 1

(1)

where

− t λ φ1 = λ φ

(2)

Therefore, the similarity transformation is cos θ I′ = − sin θ 0 sin θ cos θ 0 0 A −C 0 −C B 1 0 0 0 cos θ 0 sin θ A + B 0 − sin θ cos θ 0 0 0 1

Carrying out the operations and simplifying, we find

DYNAMICS OF RIGID BODIES

379

1 ( B − A) sin 2θ 2 0 0 A + B

2 2 A cos θ − C sin 2θ + B sin θ 1 {I′} = −C cos 2θ + ( B − A) sin 2θ 2 0

−C cos 2θ +

A sin 2 θ + C sin 2θ + B cos 2 θ 0

(3)

Making the identifications stipulated in the statement of the problem, we see that I11 = A′ , I 22 = B′ ′ ′ I12 = I 21 = −C ′ ′ ′ and I 33 = A + B = A′ + B′ ′ Therefore 0 A ′ −C ′ 0 {I ′ } = − C ′ B ′ 0 0 A′ + B′ In order that x1 and x2 be principal axes, we require C′ = 0: C cos 2θ − or, tan 2θ = from which 2C B− A (8) 1 ( B − A) sin 2θ = 0 2 (7) (5) (4)

(6)

θ=

1 2C tan −1 2 B − A

(9)

Notice that this result is still valid if A = B. Why? (What does A = B mean?)

11-19.

x2 θ = π/2 θ=π η θ θ=0

x1

The boundary of the plate is given by r = keαθ . Any point (η,θ ) has the components x1 = η cos θ x2 = η sin θ (1)

380

The moments of inertia are

CHAPTER 11

I1 = A = ρ ∫ π π

0

∫

keαθ

0

2 x2 η dη dθ

= ρ ∫ sin 2 θ dθ ∫

0

keαθ

0

η 3 dη

The integral over θ can be performed by using Eq. (E.18a), Appendix E, with the result

I1 = A = where ρk 4 P 2α

(2)

P=

In the same way,

e 4πα − 1 16 1 + 4α 2

(

)

(3)

I2 = B = ρ ∫ π π

0

∫

keαθ

0

2 x1 η dη dθ

= ρ ∫ cos 2 θ dθ ∫

0

keαθ

0

η 3 dη

(4)

Again, we use Eq. (E.18a) by writing cos 2 θ = 1 − sin 2 θ , and we find

I2 = B =

Also

ρk 4 P ( 1 + 8α 2 ) 2α π 0

(5)

I12 = −C = − ρ ∫ π ∫

keαθ

0

x1 x2 η dη dθ keαθ 0

(6)

= − ρ ∫ cos θ sin θ dθ ∫

0

η dη

3

In order to evaluate the integral over θ in this case we write cos θ sin θ = (1 2) sin 2θ and use Eq. (E.18), Appendix E. We find

I12 = −C = ρk 4 P

Using the results of problem 11-17, the entire inertia tensor is now known. According to the result of Problem 11-18, the angle through which the coordinates must be rotated in order to make {I} diagonal is

(7)

θ=

1 2C tan −1 2 B − A

(8)

Using Eqs. (2), (5), and (7) for A, B, C, we find 2C 1 = B − A 2α (9)

DYNAMICS OF RIGID BODIES

381

so that tan 2θ = Therefore, we also have

1 + 4α 2 2θ 2α 1

1 2α

(10)

sin 2θ = cos 2θ =

1 1 + 4α 2α 1 + 4α 2

2

(11)

Then, according to the relations specified in Problem 11-18,

I1 = A′ = A cos 2 θ − C sin 2θ + B sin 2 θ ′

Using cos 2 θ = (1 2)(1 + cos 2θ ) and sin 2 θ = (1 2)(1 − cos 2θ ) , we have

(12)

I1 = A′ = ′

Now,

1 1 ( A + B) + ( A − B) cos 2θ − C sin 2θ 2 2 ρk 4 P (1 + 4α 2 ) α

(13)

A+B=

(14)

A − B = −4αρk 4 P

Thus,

I1 = A′ = ′ or, ρk 4 P (1 + 4α 2 ) − 2αρk 4 P × 2α 2 − ρk 4 P × 1 2 2α 1 + 4α 1 + 4α

I1 = A′ = ρk 4 P ( Q − R ) ′

(15)

(16)

where

Q=

1 + 4α 2 2α R = 1 + 4a2

(17)

Similarly,

I 2 = B′ = ρk 4 P ( Q + R ) ′

(18)

and, of course,

382

I 3 = A′ + B′ = I1 + I 2 ′ ′ ′

CHAPTER 11

(19)

We can also easily verify, for example, that I12 = −C ′ = 0 . ′

11-20. We use conservation of energy. When standing upright, the kinetic energy is zero. Thus, the total energy is the potential energy

E = U1 = mg

b 2

(

b is the height of the center of mass above the floor.) 2

When the rod hits the floor, the potential energy is zero. Thus

E = T2 =

1 2 Iω 2

where I is the rotational inertia of a uniform rod about an end. For a rod of length b, mass/length σ:

I end = ∫ σ x 2 dx =

0

b

1 3 1 σ b = mb 2 3 3

Thus

T2 =

1 mb 2 ω 2 6

By conservation of energy

U1 = T2 mg b 1 = mb 2 ω 2 2 6

ω=

3g b

11-21.

Using I to denote the matrix whose elements are those of {I} , we can write

L = Iω L ′ = I′ω ′

We also have x ′ = λ x and x ′ = λ t x ′ and therefore we can express L and ω as

(11.54) (11.54a)

L = λ t L′ ω = λ tω ′

(11.55a) (11.55b)

substituting these expressions into Eq. (11.54), we have

DYNAMICS OF RIGID BODIES

383 λ t L ′ = Iλ t ω ′

and multiplying on the left by λ, λλ t L ′ = λ Iλ t ω ′

or

L ′ = λ Iλ t ω ′

(

)

by virture of Eq. (11.54a), we identify

I′ = λ Iλ t

(11.61)

11-22.

According to Eq. (11.61),

I ij = ∑ λ ik I k λ −j1 ′ k ,l

(1)

Then, tr {I′} = ∑ I ii = ∑∑ λ ik I k λ −i1 ′ i i k,

= ∑ Ik k, ∑λ i −1 i

λ ik

(2)

= ∑ I k δ k = ∑ I kk k, k

so that tr {I′} = tr {I}

(3)

This relation can be verified for the examples in the text by straightforward calculations. Note: A translational transformation is not a similarity transformation and, in general, tr {I} is not invariant under translation. (For example, tr {I} will be different for inertia tensors expressed in coordinate system with different origins.)

11-23.

We have

I′ = λIλ −1

(1)

Then,

384

CHAPTER 11

I′ = λ Iλ −1

= λ × I × λ −1 = λλ −1 × I

so that,

I′ = I

This result is easy to verify for the various examples involving the cube.

11-24.

x2 –a/2 a/2 x1

(2)

−

3 a 2

The area of the triangle is A = 3 a 2 4 , so that the density is

ρ=

M 4M = A 3 a2

(1)

a) The rotational inertia with respect to an axis through the point of suspension (the origin) is x2 x1 θ 2 2 I 3 = ρ ∫ x1 + x2 dx1 dx2 a2 0

(

)

= 2ρ

∫

0

dx1

−

∫

3 ( a − 2 x1 ) 2

(x

2 1

2 + x2 ) dx2

=

3 4 1 ρ a = Ma 2 24 6

(2)

When the triangle is suspended as shown and when θ = 0, the coordinates of the center of mass are (0, x2 , 0) , where

DYNAMICS OF RIGID BODIES

385

1 ρ x2 dx1 dx2 M∫ a2 x2 =

2ρ = M

=−

∫

0

dx1

− 3 ( a − 2 x1 )

∫

0

x2 dx2

a

2 3

(3)

The kinetic energy is

T=

1 1 I 3θ 2 = Ma 2θ 2 2 12

Mga (1 − cos θ ) 2 3

(4)

and the potential energy is

U=

(5)

Therefore,

L= Mga 1 cos θ Ma 2θ 2 + 12 2 3

(6)

where the constant term has been suppressed. The Lagrange equation for θ is

θ+ 3

g sin θ = 0 a

(7)

and for oscillations with small amplitude, the frequency is

ω=

3

g a

(8)

b) The rotational inertia for an axis through the point of suspension for this case is x2 x1

−

0 − x2

3 a 2

I 3 = 2ρ ′

∫

3a − 2

dx2

∫ (x

0

3

2 1

2 + x2 dx1

)

=

5 Ma 2 12

Mga 5 cos θ Ma 2θ 2 + 24 3

(9)

The Lagrangian is now

L=

(10)

386 and the equation of motion is

CHAPTER 11

θ+

12 g sin θ = 0 5 3 a

(11)

so that the frequency of small oscillations is

ω=

12 g 5 3 a

(12)

which is slightly smaller than the previous result.

11-25.

x2 R ρ

r

θ

x1

2ρ

The center of mass of the disk is (0, x2 ) , where

x2 = 2 ∫ x2 dx1 dx2 + ∫ x2 dx1 dx2 M lower upper semicircle semicircle

ρ

=

M ∫ 0

ρ

R

∫ ( r sin θ ) ⋅ rdrdθ + 2 ∫ ∫ π ( r sin θ ) ⋅ rdrdθ π R 2π 0 0

=−

2 ρR 3 3 M 1 1 M = ρ ⋅ π R 2 + 2ρ ⋅ π R 2 2 2

=

(1)

Now, the mass of the disk is

3 ρπ R2 2

(2)

so that x2 = −

4 R 9π

(3)

The direct calculation of the rotational inertia with respect to an axis through the center of mass is tedious, so we first compute I with respect to the x3 -axis and then use Steiner’s theorem.

DYNAMICS OF RIGID BODIES

387

I 3 = ρ ∫ 0 =

R

∫

π

0

r 2 ⋅ rdr dθ + 2 ∫

R 0

∫π

2π

r 2 ⋅ rdr dθ

3 1 πρR 4 = MR2 4 2

2 I 0 = I 3 − Mx2

(4)

Then,

= =

1 16 MR2 − M ⋅ R2 2 81π 2 1 32 MR2 1 − 2 81π 2 (5)

When the disk rolls without slipping, the velocity of the center of mass can be obtained as follows: Thus ρ 2ρ x2 θ

CM

θ

R

xCM = Rθ − x2 sin θ yCM = R − x2 cos θ xCM = Rθ − x2 θ cos θ yCM = x2 θ sin θ

(x where 2 CM

2 2 + yCM = V 2 = R 2θ 2 + x2 θ 2 − 2θ 2 R x2 cos θ

)

V 2 = a 2θ 2

(6)

2 a = R 2 + x2 − 2R x2 cos θ

(7)

Using (3), a can be written as a = R 1+ The kinetic energy is 16 8 cos θ − 2 81π 9π (8)

388

T = Ttrans + Trot = Substituting and simplifying yields T= The potential energy is 1 U = Mg R + x2 cos θ 2 = Thus the Lagrangian is

L= 1 8 3 8 MR Rθ 2 − cos θ − g 1 − cos θ 2 2 9π 9π

CHAPTER 11

1 1 Mv 2 + I 0θ 2 2 2

(9)

1 3 8 MR2θ 2 − cos θ 2 2 9π

(10)

1 8 MgR 1 − cos θ 2 9π

(11)

(12)

11-26.

Since ω φ = φ lies along the fixed x′ -axis , the components of ω φ along the body axes 3 ωφ ω φ ω φ

( xi ) are given by the application of the transformation matrix λ [Eqs. (11.98) and (11.99)]:

( ) φ 0 ( ) = φ = λ 0 φ ( ) φ

1 2 3 1 2 3 1 2 3

(1)

Carrying out the matrix multiplication, we find ωφ ω φ ω φ which is just Eq. (11.101a). The direction of ω θ = θ coincides with the line of nodes and lies along the x′′′ axis. The 1 components of ωθ along the body axes are therefore obtained by the application of the transformation matrix λ ψ which carries the xi′′′ system into the xi system: (ω θ ) 1 (ω θ ) 2 = λ ψ (ω θ ) 3 θ 0 = θ 0 cos ψ − sin ψ 0

( ) sin ψ sin θ ( ) = φ cos ψ sin θ cos θ ( )

(2)

(3)

DYNAMICS OF RIGID BODIES

389

which is just Eq. (11.101b). Finally, since ω ψ lies along the body x3 -axis , no transformation is required:

ω ψ ω ψ ω ψ which is just Eq. (11.101c). Combining these results, we obtain ωφ ω = ωφ ω φ

( ) 0 ( ) = ψ 0 ( ) 1

1 2 3

(4)

( ) + (ω ) + (ω ) ( ) + (ω ) + ( ω ) ( ) + (ω ) + ( ω )

1

θ 1

ψ 1

2 3

θ 2 θ 3

ψ 2 ψ 3

φ sin ψ sin θ + θ cos ψ = φ φ cos ψ sin θ − θ sin ψ θ cos θ + ψ which is just Eq. (11.02).

11-27.

x3′ x3

(5)

L α θ

ω x2

Initially:

L1 = 0 = I1ω 1 L2 = L sin θ = I1ω 2 = I1ω sin α L3 = L xos θ = I 3ω 3 = I 3ω cos α

Thus tan θ = From Eq. (11.102)

L2 I1 = tan α L3 I 3

(1)

ω 3 = φ cos θ + ψ

Since ω 3 = ω cos α , we have

390

CHAPTER 11

φ cos θ = ω cos α − ψ

From Eq. (11.131)

I 3 − I1 ω3 I1

(2)

ψ = −Ω = −

(2) becomes

φ cos θ =

I3 ω cos α I1

(3)

From (1), we may construct the following triangle

I3 tan α

θ I3

from which cos θ =

I3

I + I tan 2 α

2 3 2 1

12

Substituting into (3) gives

φ=

ω

I1

2 2 I1 sin 2 α + I 3 cos 2 α

11-28.

From Fig. 11-7c we see that ω φ = φ is along the x3 -axis , ω θ = θ is along the line of ′ ω φ = φ e′ ′ 3

nodes, and ω ψ = ψ is along the x3 -axis . Then, (1)

whee e′ is the unit vector in the x3 direction. ′ 3 Projecting the lines of nodes into the x1 - and x2 -axes , we obtain ′ ′ ω θ = θ ( e1 cos φ + e′ sin φ ) ′ ′ 2 along the x3 -axis and a component normal to this axis: ′ ω ψ = ψ ( e12 sin θ + e′ cos θ ) ′ ′ 3 where (3) (2)

ωψ has components along all three of the xi′ axes. First, we write ω ψ in terms of a component ′ ′

e12 = e1 sin φ − e′ cos φ ′ ′ 2

Then, ω ψ = ψ ( e1 sin θ sin φ − e′ sin θ cos φ + e′ cos θ ) ′ ′ 2 3

(4)

(5)

DYNAMICS OF RIGID BODIES

391

Collecting the various components, we have

ω 1 = θ cos φ + ψ sin θ sin φ ′ ω 2 = θ sin φ − ψ sin θ cos φ ′ ω 3 = ψ cos θ + φ ′

When the motion is vertical θ = 0. Then, according to Eqs. (11.153) and (11.154),

Pφ = I 3 φ + ψ = Pψ

(6)

11-29.

(

)

(1)

and using Eq. (11.159), we see that

Pψ = Pφ = I 3 ω 3

(2)

Also, when θ = 0 (and θ = 0 ), the energy is [see Eq. (11.158)]

E=

1 2 I 3 ω 3 + Mgh 2

(3)

Furthermore, referring to Eq. (11.160),

E′ = E −

1 2 I 3 ω 3 = Mgh 2

(3)

If we wish to examine the behavior of the system near θ = 0 in order to determine the conditions for stability, we can use the values of Pψ , Pφ , and E′ for θ = 0 in Eq. (11.161). Thus,

2 2 I 3 ω 3 (1 − cos θ ) 1 + Mgh cos θ Mgh = I12θ 2 + 2 2I12 sin 2 θ 2

(5)

Changing the variable to z = cos θ and rearranging, Eq. (5) becomes z2 =

(1 − z ) 2 2 2 2 2 Mgh I12 (1 + z ) − I 3 w3

I12

(6)

The questions concerning stability can be answered by examining this expression. First, we note that for physically real motion we must have z 2 ≥ 0 . Now, suppose that the top is spinning very rapidly, i.e., that ω 3 is large. Then, the term in the square brackets will be negative. In such a case, the only way to maintain the condition z 2 ≥ 0 is to have z = 1, i.e., θ = 0. Thus, the motion at θ = 0 will be stable as long as

2 2 4 Mgh I12 − I 3 ω 3 < 0

(7)

or, 4 Mgh I12 ω c the motion is stable and for ω < ω c there is nutation:

ωc =

2 Mgh I12

I3

(9)

If the top is set spinning with ω 3 > ω c and θ = 0, the motion will be stable. But as friction slows the top, the critical angular velocity will eventually be reached and nutation will set in. This is the case of the “sleeping top.” If we set θ = 0 , Eq. (1.162) becomes

11-30.

(P − P E′ = V (θ ) = φ 2 I12 1 − cos 2 θ

(

ψ

cos θ

)

2

)

+ Mgh cos θ

(1)

Re-arranging, this equation can be written as

( 2 Mgh I12 ) cos3 θ − ( 2E′ I12 + Pψ2 ) cos2 θ + 2 ( Pφ Pψ − Mgh I12 ) cos θ + ( 2E′ I12 − Pφ2 ) = 0 which is cubic in cos θ.

(2)

V(θ ) has the form shown in the diagram. Two of the roots occur in the region −1 ≤ cos θ ≤ 1 , and one root lies outside this range and is therefore imaginary.

V(θ)

+1 –1 cos θ

DYNAMICS OF RIGID BODIES

393

11-31.

The moments of inertia of the plate are

I2 = I 3 = I1 + I 2 = I 2 (1 + cos 2α ) = 2I 2 cos 2 α I1 = I 2 cos 2α

(1)

We also note that I1 − I 2 = − I 2 (1 − cos 2α ) = −2I 2 sin α

2

(2)

Since the plate moves in a force-free manner, the Euler equations are [see Eq. (11.114)]

( I1 − I 2 ) ω1ω 2 − I 3ω 3 = 0

( I 2 − I 3 ) ω 2ω 3 − I1ω1 = 0 ( I 3 − I1 ) ω 3ω1 − I 2ω 2 = 0 Substituting (1) and (2) into (3), we find (3)

( −2I

sin 2 α ω 1ω 2 − 2I 2 cos 2 α ω 3 = 0 ( − I 2 cos 2α ) ω 2ω 3 − ( I 2 cos 2α ) ω1 = 0 I 2 ω 3ω 1 − I 2ω 2 = 0

2

)

(

)

(4)

These equations simplify to

ω 3 = −ω 1ω 2 tan 2 α ω 1 = −ω 2ω 3 ω 2 = ω 3ω 1

From which we can write

(5)

ω 1ω 2ω 3 = ω 2ω 2 = −ω 1ω 1 = −ω 3ω 3 cot 2 α

Integrating, we find

2 2 2 2 2 2 ω 2 − ω 2 ( 0 ) = −ω 1 + ω 1 ( 0 ) = −ω 3 cot 2 α + ω 3 ( 0 ) cot 2 α

(6)

(7)

Now, the initial conditions are

394

CHAPTER 11

ω 2 ( 0) = 0 ω 3 ( 0 ) = Ω sin α Therefore, the equations in (7) become

2 2 2 ω 2 = −ω 1 + Ω 2 cos 2 α = −ω 3 cot 2 α + Ω 2 cos 2 α

ω 1 ( 0 ) = Ω cos α

(8)

(9)

From (5), we can write

2 2 2 ω 2 = ω 3 ω1 2 2 and from (9), we have ω 1 = ω 3 cot 2 α . Therefore, (10) becomes 2 ω 2 = ω 3 cot α 2 2 and using ω 3 = Ω 2 sin 2 α − ω 2 tan 2 α from (9), we can write (11) as

(10)

(11)

ω2 = − cot α ω tan α − Ω 2 sin 2 α

2 2 2

(12)

Since ω 2 = dω 2 dt , we can express this equation in terms of integrals as

∫ω

− Solving for ω 2 ,

2 2

dω 2 = − cot α ∫ dt tan 2 α − Ω 2 sin 2 α

(13)

Using Eq. (E.4c), Appendix E, we find ω tan α 1 tanh −1 2 = −t cot α ( tan α )( Ω sin α ) Ω sin α (14)

ω 2 ( t ) = Ω cos α tanh ( Ω t sin α )

11-32. a) The exact equation of motion of the physical pendulum is

(15)

Iθ + MgL sin θ = 0 where I = Mk 2 , so we have

θ=− or gL sin θ k2

gL d ( cos θ ) d θ = 2 dt k dθ

()

or

DYNAMICS OF RIGID BODIES

395

θ d (θ ) = so gL d ( cos θ ) k2

θ2 =

2 gL cos θ + a k2

where a is a constant determined by the initial conditions. Suppose that at t = 0 , θ = θ 0 and at −2 gl that initial position the angular velocity of the pendulum is zero, we find a = 2 cos θ 0 . So k finally

θ=

2 gl ( cos θ − cos θ 0 ) k2

b) One could use the conservation of energy to find the angular velocity of the pendulum at any angle θ , but it is exactly the result we obtained in a), so at θ = 10 , we have

ω =θ =

2 gL ( cos θ − cos θ 0 ) = 53.7 s−1 k2

11-33. Cats are known to have a very flexible body that they can manage to twist around to a feet-first descent while falling with conserved zero angular momentum. First they thrust their back legs straight out behind their body and at the same time they tuck their front legs in. Extending their back legs helps to resist spinning, since rotation velocity evidently is inversely proportional to inertia momentum. This allows the cat to twist their body differently to preserve zero angular momentum: the front part of the body twisting more than the back. Tucking the front legs encourages spinning to a downward direction preparing for touchdown and as this happens, cats can easily twist the rear half of their body around to catch up with the front.

However, whether or not cats land on their feet depends on several factors, notably the distance they fall, because the twist maneuver takes a certain time, apparently around 0.3 sec. Thus the minimum height required for cats falling is about 0.5m.

11-34. The Euler equation, which describes the rotation of an object about its symmetry axis, say 0x, is

I x ω x − I y − I z ω yω z = N x where N x = − b ω x is the component of torque along Ox. Because the object is symmetric about Ox, we have I y = I z , and the above equation becomes Ix dω x = −b ω x dt ⇒ ωx = e

− b t Ix

(

)

ω x0

396

CHAPTER 11

CHAPTER

12

Coupled Oscillations

12-1. m1 = M k1 x1 k12 x2 m2 = M k2

The equations of motion are

Mx1 + (κ 1 + κ 12 ) x1 − κ 12 x2 = 0 Mx2 + (κ 2 + κ 12 ) x2 − κ 12 x1 = 0 We attempt a solution of the form x1 ( t ) = B1 e iωt x2 ( t ) = B2 e iωt Substitution of (2) into (1) yields

(1)

(2)

(κ

+ κ 12 − Mω 2 B1 − κ 12 B2 = 0 −κ 12 B1 + κ 2 + κ 12 − Mω 2 B2 = 0

1

)

(

)

(3)

In order for a non-trivial solution to exist, the determinant of coefficients of B1 and B2 must vanish. This yields

2 κ 1 + κ 12 − Mω 2 κ 2 + κ 12 − Mω 2 = κ 12

(

)

(

)

(4)

from which we obtain

ω2 =

κ 1 + κ 2 + 2κ 12

2M

±

1 2M

2 (κ 1 − κ 2 )2 + 4κ 12

(5)

This result reduces to ω 2 = (κ + κ 12 ± κ 12 ) M for the case κ 1 = κ 2 = κ (compare Eq. (12.7)].

397

398

CHAPTER 12

If m2 were held fixed, the frequency of oscillation of m1 would be

2 ω 01 =

1 (κ 1 + κ 12 ) M

(6)

while in the reverse case, m2 would oscillate with the frequency

2 ω 02 =

1 (κ 2 + κ 12 ) M

(7)

Comparing (6) and (7) with the two frequencies, ω + and ω − , given by (5), we find

2 ω+ =

1 κ + κ 2 + 2κ 12 + 1 2M

2 (κ 1 − κ 2 )2 + 4κ 12

(8)

> so that

1 1 2 κ 1 + κ 2 + 2κ 12 + (κ 1 − κ 2 ) = M (κ 1 + κ 12 ) = ω 01 2M

ω + > ω 01

Similarly,

2 ω− =

(9)

1 2M

κ + κ + 2κ − 2 12 1

2 (κ 1 − κ 2 )2 + 4κ 12

(10)

< so that

1 1 2 κ 1 + κ 2 + 2κ 12 − (κ 1 − κ 2 ) = M (κ 2 + κ 12 ) = ω 02 2M

ω − < ω 02

If κ 1 > κ 2 , then the ordering of the frequencies is

(11)

ω + > ω 01 > ω 02 > ω −

12-2.

(12)

From the preceding problem we find that for κ 12

κ1,κ2

M (1)

ω1 ≅

If we use

κ 1 + κ 12

M

; ω2 ≅

κ 2 + κ 12

ω 01 =

κ1

M

; ω 02 =

κ2

M

(2)

then the frequencies in (1) can be expressed as

COUPLED OSCILLATIONS

399

ω 1 = ω 01 1 + ω 2 = ω 02 where κ 12 ≅ ω 01 (1 + ε 1 ) κ1

κ 1 + 12 ≅ ω 02 (1 + ε 2 ) κ2

(3)

ε1 =

For the initial conditions [Eq. 12.22)],

κ 12 κ ; ε 2 = 12 2κ 1 2κ 2

(4)

x1 ( 0 ) = D, x2 ( 0 ) = 0, x1 ( 0 ) = 0, x2 ( 0 ) = 0 , the solution for x1 ( t ) is just Eq. (12.24):

(5)

ω + ω2 x1 ( t ) = D cos 1 2 Using (3), we can write

ω − ω2 t cos 1 2

t

(6)

ω 1 + ω 2 = (ω 01 + ω 02 ) + ( ε1ω 01 + ε 2ω 02 )

≡ 2Ω + + 2 ε +

(7)

ω 1 − ω 2 = (ω 01 − ω 02 ) + ( ε 1ω 01 − ε 2ω 02 )

≡ 2Ω − + 2ε − Then, x1 ( t ) = D cos ( Ω + t + ε + t ) cos ( Ω − t + ε − t ) Similarly, ω + ω2 x2 ( t ) = D sin 1 2 ω − ω2 t sin 1 2 t

(8)

(9)

= D sin ( Ω + t + ε + t ) sin ( Ω − t + ε − t ) Expanding the cosine and sine functions in (9) and (10) and taking account of the fact that ε + and ε − are small quantities, we find, to first order in the ε’s, x1 ( t ) ≅ D cos Ω + t cos Ω − t − ε + t sin Ω + t cos Ω − t − ε − t cos Ω + t sin Ω − t x2 ( t ) ≅ D sin Ω + t sin Ω − t + ε + t cos Ω + t sin Ω − t + ε − t sin Ω + t cos Ω − t When either x1 ( t ) or x2 ( t ) reaches a maximum, the other is at a minimum which is greater than zero. Thus, the energy is never transferred completely to one of the oscillators.

(10)

(11) (12)

400 12-3.

CHAPTER 12

The equations of motion are m 2 x2 + ω 0 x1 = 0 M m 2 x2 + x2 + ω 0 x2 = 0 M x1 +

(1)

We try solutions of the form x1 ( t ) = B1 e iωt ; x2 ( t ) = B2 e iωt (2)

We require a non-trivial solution (i.e., the determinant of the coefficients of B1 and B2 equal to zero), and obtain

2 (ω 02 − ω 2 ) − ω 4 m = 0 2

M

(3)

so that

2 ω 0 − ω 2 = ±ω 2

m M

(4)

and then

ω2 =

2 ω0

m 1± M

(5)

Therefore, the frequencies of the normal modes are

ω1 =

2 ω0

1+

m M

2 0

(6)

ω2 =

ω

1−

m M

where ω 1 corresponds to the symmetric mode and ω 2 to the antisymmetric mode. By inspection, one can see that the normal coordinates for this problem are the same as those for the example of Section 12.2 [i.e., Eq. (12.11)].

12-4.

The total energy of the system is given by

E = T +U = Therefore, 1 1 1 2 2 2 2 2 M x1 + x2 + κ x1 + x2 + κ 12 ( x2 − x1 ) 2 2 2

(

)

(

)

(1)

COUPLED OSCILLATIONS

401

dE = M ( x1 x1 + x2 x2 ) + κ ( x1 x1 + x2 x2 ) + κ 12 ( x2 − x1 )( x2 − x1 ) dt = Mx1 + κ x1 − κ 12 ( x2 − x1 ) x1 + Mx2 + κ x2 + κ 12 ( x2 − x1 ) x2 = Mx1 + (κ + κ 12 ) x1 − κ 12 x2 x1 + Mx2 − κ 12 x1 + (κ + κ 12 ) x2 x2

(2)

which exactly vanishes because the coefficients of x1 and x2 are the left-hand sides of Eqs. (12.1a) and (12.1b). An analogous result is obtained when T and U are expressed in terms of the generalized coordinates η1 and η2 defined by Eq. (12.11):

T= U=

1 2 2 M η1 + η2 4

(

)

(3) (4)

1 1 2 2 2 κ η1 + η2 + κ 12η1 4 2

(

)

Therefore, 2⋅ dE = Mη1 + (κ + 2κ 12 ) η1 η1 + [ Mη2 + κη2 ] η2 dt

(5)

which exactly vanishes by virtue of Eqs. (12.14). When expressed explicitly in terms of the generalized coordinates, it is evident that there is only 2 one term in the energy that has κ 12 as a coefficient (namely, κ 12η1 ), and through Eq. (12.15) we see that this implies that such a term depends on the C1 ’s and ω 1 , but not on the C2 ’s and ω 2 . To understand why this is so, it is sufficient to recall that η1 is associated with the anitsymmetrical mode of oscillation, which obviously must have κ 12 as a parameter. On the other hand, η2 is associated to the symmetric mode, x1 ( t ) = x2 ( t ) , x1 ( t ) = x2 ( t ) , in which both masses move as if linked together with a rigid, massless rod. For this mode, therefore, if the spring connecting the masses is changed, the motion is not affected.

12-5.

We set κ 1 = κ 2 = κ 12 ≡ κ . Then, the equations of motion are m1 x1 + 2κ x1 − κ x2 = 0 m2 x2 + 2κ x2 − κ x1 = 0

(1)

Assuming solutions of the form x1 ( t ) = B1e iωt iω t x2 ( t ) = B2 e

(2)

we find that the equations in (1) become

402

CHAPTER 12

( 2κ − m ω ) B

2 1

− κ B2 = 0 −κ B1 + 2κ − m2ω 2 B2 = 0

1

(

)

(3)

which lead to the secular equation for ω 2 :

( 2κ − m ω )( 2κ − m ω ) = κ

2 2 1 2

2

(4)

Therefore,

ω2 =

κ µ

3µ 1 ± 1 − m1 + m2

(5)

where µ = m1 m2 ( m1 + m2 ) is the reduced mass of the system. Notice that (5) agrees with Eq. the maximum value of 3µ ( m1 + m2 ) is 3 4 . (Show this.) (12.8) for the case m1 = m2 = M and κ 12 = κ . Notice also that ω 2 is always real and positive since

1

Inserting the values for ω 1 and ω 2 into either of the equations in (3), we find

m1 2 − µ and

3µ 1 + 1 − a11 = a21 m1 + m2

(6)

m 3µ a12 = 2 − 1 1 − 1 − a22 µ m1 + m2 Using the orthonormality condition produces a11 = 1 D1

(7)

(8)

2− a21 = where

m1 + m2 3m1m2 1 + 1 − m2 ( m1 + m2 ) 2 D1

(9)

D1 ≡ 2 ( m2 − m1 ) + 2

2 m1 2 2 2 m1 − m2 + m2 m2

(

)

1−

( m1 + m2 ) 2

3m1m2

(10)

The second eigenvector has the components 2− a12 = m1 + m2 3m1m2 1 − 1 − m2 ( m1 + m2 ) 2 D2

(11)

1

Recall that when we use ω = ω 1 , we call the coefficients β1 (ω = ω 1 ) = a11 and β 2 (ω = ω 1 ) = a21 , etc.

COUPLED OSCILLATIONS

403

a22 = where m2 D2 ≡ 2m1 1 + 1 + m2 2 m2

1 D2

(12)

2 3m1 1 − 2 + 2m1 m2

2 m1 3m1 m2 1− 2 1− ( m1 + m2 )2 m2

(13)

The normal coordinates for the case in which q j ( 0 ) = 0 are

η1 ( t ) = ( m1 a11 x10 + m2 a21 x20 ) cos ω 1t η2 ( t ) = ( m1 a12 x10 + m2 a22 x20 ) cos ω 2t

(14)

12-6. m m k

k

x1

x2

If the frictional force acting on mass 1 due to mass 2 is f = − β ( x1 − x2 ) then the equations of motion are mx1 + β ( x1 − x2 ) + κ x1 = 0 mx2 + β ( x2 − x1 ) + κ x2 = 0 Since the system is not conservative, the eigenfrequencies will not be entirely real as in the previous cases. Therefore, we attempt a solution of the form x1 ( t ) = B1eα t ; x2 ( t ) = B2 eα t (3) (2) (1)

where α = λ + iω is a complex quantity to be determined. Substituting (3) into (1), we obtain the following secular equation by setting the determinant of the coefficients of the B’s equal to zero:

( mα from which we find the two solutions

2

+ βα + κ

)

2

= β 2α 2

(4)

α1 = ± i

m 1 2 −β ± β − mκ α2 = m m ;

κ

ω1 = ±

κ

(5)

(

)

The general solution is therefore

404

CHAPTER 12

+ x1 ( t ) = B11 e

i κ mt

− + B11 e

i κ mt

+ + e − βt m B12 e

(

β 2 − mκ t m

− + B12 e

β 2 − mκ t m

)

(6)

and similarly for x2 ( t ) . The first two terms in the expression for x1 ( t ) are purely oscillatory, whereas the last two terms

+ contain the damping factor e − βt . (Notice that the term B12 exp

(

β 2 − mκ t increases with time if

)

+ β 2 > mκ , but B12 is not required to vanish in order to produce physically realizable motion because the damping term, exp(–βt), decreases with time at a more rapid rate; that is

− β + β 2 − mκ < 0 .) To what modes do α 1 and α 2 apply? In Mode 1 there is purely oscillating motion without friction. This can happen only if the two masses have no relative motion. Thus, Mode 1 is the symmetric mode in which the masses move in phase. Mode 2 is the antisymmetric mode in which the masses move out of phase and produce frictional damping. If β 2 < mκ , the motion is one of damped oscillations, whereas if β 2 > mκ , the motion proceeds monotonically to zero amplitude.

12-7.

k m k m x1

x2

We define the coordinates x1 and x2 as in the diagram. Including the constant downward gravitational force on the masses results only in a displacement of the equilibrium positions and does not affect the eigenfrequencies or the normal modes. Therefore, we write the equations of motion without the gravitational terms: mx1 + 2κ x1 − κ x2 = 0 mx2 + κ x2 − κ x1 = 0 Assuming a harmonic time dependence for x1 ( t ) and x2 ( t ) in the usual way, we obtain (1)

( 2κ − mω ) B

2

− κ B2 = 0 −κ B1 + κ − mω 2 B2 = 0

1

(

)

(2)

Solving the secular equation, we find the eigenfrequencies to be

COUPLED OSCILLATIONS

405

2 ω1 =

3+ 5 κ 2 m 3− 5 κ 2 m

(3)

2 ω2 =

Substituting these frequencies into (2), we obtain for the eigenvector components 1− 5 a11 = a21 2 1+ 5 a12 = a22 2 For the initial conditions x1 ( 0 ) = x2 ( 0 ) = 0 , the normal coordinates are

(4)

η1 ( t ) = ma11 x10 +

1− 5 x20 cos ω 1t 2 (5)

1+ 5 x20 cos ω 2t η2 ( t ) = ma12 x10 + 2 Therefore, when x10 = −1.6180 x20 , η2 ( t ) = 0 and the system oscillates in Mode 1, the

antisymmetrical mode. When x10 = 0.6180 x20 , η1 ( t ) = 0 and the system oscillates in Mode 2, the symmetrical mode. When mass 2 is held fixed, the equation of motion of mass 1 is mx1 + 2κ x1 = 0 and the frequency of oscillation is 2κ m (6)

ω 10 =

(7)

When mass 1 is held fixed, the equation of motion of mass 2 is mx2 + κ x2 = 0 and the frequency of oscillation is (8)

ω 20 =

κ m (9)

Comparing these frequencies with ω 1 and ω 2 we find

ω1 = ω2 =

3+ 5 4 3− 5 4

2κ 2κ = 1.1441 > ω 10 m m κ κ = 0.6180 < ω 20 m m

406

CHAPTER 12

Thus, the coupling of the oscillators produces a shift of the frequencies away from the uncoupled frequencies, in agreement with the discussion at the end of Section 12.2.

12-8. The kinetic and potential energies for the double pendulum are given in Problem 7-7. If we specialize these results to the case of small oscillations, we have

T=

1 m 2 U=

2

( 2φ

2 1

2 + φ2 + 2φ1 φ2

)

(1) (2)

1 2 2 mg 2φ1 + φ 2 2

(

)

where φ1 refers to the angular displacement of the upper pendulum and φ 2 to the lower pendulum, as in Problem 7-7. (We have also discarded the constant term in the expression for the potential energy.) Now, according to Eqs. (12.34), T= U= 1 ∑ mjk qj qk 2 j,k 1 ∑ Ajk qj qk 2 j,k (3)

(4)

Therefore, identifying the elements of {m} and {A} , we find

{m} = m 2 1 1 {A} = mg 0 1 and the secular determinant is 2 g − 2ω 2

2

2 1

(5)

2 0

(6)

−ω 2 g =0 −ω

2

(7)

−ω or,

g 2 g 2 4 2 − 2ω − ω − ω = 0 Expanding, we find

(8)

ω −4

4

g

g ω +2 =0

2

2

(9)

which yields

COUPLED OSCILLATIONS

407

ω2 = 2 ± 2 and the eigenfrequencies are

(

)g g (10)

ω1 = 2 + 2 ω2 = 2 − 2

To get the normal modes, we must solve

g

= 1.848 = 0.765

(11) g g

∑(A j jk

− ω r2 m jk a jr = 0

)

For k = 1, this becomes:

(A

For r = 1:

11

− ω r2 m11 a1r + A21 − ω r2 m21 a2 r = 0

)

(

)

g g 2m 2 a11 − 2 + 2 m 2mg − 2 + 2 Upon simplifying, the result is

(

)

(

)

2

a21 = 0

a21 = − 2 a11 Similarly, for r = 2, the result is a22 = 2 a12 The equations x1 = a11 η1 + a12 η2 x2 = a21 η1 + a22 η2

can thus be written as x1 = a11 η1 +

1 a22 η2 2

x1 = − 2 a11 η1 + a22 η2

Solving for η1 and η2 :

η1 =

2 x1 − x2 ; η2 = 2 2 a11

2 x1 + x2 2 a22

408 x2 2

CHAPTER 12

η1 occurs when η2 = 0; i.e. when x1 = − η2 occurs when η1 = 0; i.e. when x1 =

x2 2

Mode 2 is therefore the symmetrical mode in which both pendula are always deflected in the same direction; and Mode 1 is the antisymmetrical mode in which the pendula are always deflected in opposite directions. Notice that Mode 1 (the antisymmetrical mode), has the higher frequency, in agreement with the discussion in Section 12.2.

12-9. The general solutions for x1 ( t ) and x2 ( t ) are given by Eqs. (12.10). For the initial conditions we choose oscillator 1 to be displaced a distance D from its equilibrium position, while oscillator 2 is held at x2 = 0 , and both are released from rest:

x1 ( 0 ) = D, x2 ( 0 ) = 0, x1 ( 0 ) = 0, x2 ( 0 ) = 0

(1)

Substitution of (1) into Eq. (12.10) determines the constants, and we obtain x1 ( t ) = x2 ( t ) = D ( cos ω1t + cos ω 2t ) 2 D ( cos ω 2t − cos ω1t ) 2

(2) (3)

where

ω1

κ + 2κ 12

M

> ω2 =

κ

M

(4)

As an example, take ω 1 = 1.2 ω 2 ; x1 ( t ) vs. x2 ( t ) is plotted below for this case. It is possible to find a rotation in configuration space such that the projection of the system point onto each of the new axes is simple harmonic. By inspection, from (2) and (3), the new coordinates must be x1 ≡ x1 − x2 = D cos ω 1t ′ x2 ≡ x1 + x2 = D cos ω 2t ′

(5) (6)

These new normal axes correspond to the description by the normal modes. They are represented by dashed lines in the graph of the figure.

COUPLED OSCILLATIONS

409 x2(t)/D 0.8 ω 2t = 5π 2 ω 2t = π 2 0.6 0.4 0.2 0.2 ω2t = 2π π ω 2t = 3 0.4 7π 2 –0.5 3π x1′ ω 2t = 0.6 0.8 ω 2t = 0 1.0 x1(t)/D 4π x2′ ω1 = 1.2 ω2

1.0 ω 2t = π

3π 2

12-10.

The equations of motion are mx1 + bx1 + (κ + κ 12 ) x1 − κ 12 x2 = F0 cos ω t mx2 + bx2 + (κ + κ 12 ) x2 − κ 12 x1 = 0

(1)

The normal coordinates are the same as those for the undamped case [see Eqs. (12.11)]:

η1 = x1 − x2 ;

η2 = x1 + x2

(2)

Expressed in terms of these coordinates, the equations of motion become m (η2 + η1 ) + b (η2 + η1 ) + (κ + κ 12 )(η2 + η1 ) − κ 12 (η2 − η1 ) = 2F0 cos ω t m (η2 − η1 ) + b (η2 − η1 ) + (κ + κ 12 )(η2 − η1 ) − κ 12 (η2 + η1 ) = 0

(3)

By adding and subtracting these equations, we obtain the uncoupled equations:

F b κ + 2κ 12 η1 + η1 = 0 cos ω t m m m F b κ η2 + η2 + η2 = 0 cos ω t m m m

η1 +

(4)

With the following definitions,

410 b m

CHAPTER 12

2β =

2 ω1 =

κ + 2κ 12 m ω =

2 2

κ m F0 m

(5)

A= the equations become

2 η1 + 2βη1 + ω 1 η1 = A cos ω t 2 η2 + 2βη2 + ω 2 η2 = A cos ω t

(6)

Referring to Section 3.6, we see that the solutions for η1 ( t ) and η2 ( t ) are exactly the same as that given for x(t) in Eq. (3.62). As a result η1 ( t ) exhibits a resonance at ω = ω 1 and η2 ( t ) exhibits a resonance at ω = ω 2 . Taking a time derivative of the equations gives ( q = I ) LI1 + LI 2 + I1 + MI 2 = 0 C I2 + MI1 = 0 C

12-11.

Assume I1 = B1e iω t , I 2 = B2 e iωt ; and substitute into the previous equations. The result is −ω 2 LB1e iω t + −ω 2 LB2 e iωt + These reduce to 1 B1 − ω 2 L + B2 − Mω 2 = 0 C 1 B1 e iωt − Mω 2 B2 e iωt = 0 C 1 B2 e iωt − Mω 2 B1e iωt = 0 C

(

)

1 B1 − Mω 2 + B2 − ω 2 L = 0 C

(

)

This implies that the determinant of coefficients of B1 and B2 must vanish (for a non-trivial solution). Thus

COUPLED OSCILLATIONS

411

1 − ω 2L C − Mω 2

2

− Mω 2 1 − ω 2L C =0

1 2 2 C − ω L − Mω

(

)

2

=0

1 − ω 2 L = ± Mω 2 C or

ω2 =

Thus

1 C (L ± M)

ω1 = ω2 =

1 C (L + M) 1 C (L − M)

12-12.

From problem 12-11: LI1 + LI 2 + 1 I1 + MI 2 = 0 C 1 I 2 + MI1 = 0 C (1) (2)

Solving for I1 in (1) and substituting into (2) and similarly for I 2 , we have M2 M 1 L− I1 + C I1 − CL I 2 = 0 L M2 M 1 L − L I 2 + C I 2 − CL I1 = 0 If we identify M κ 12 = LC M 1 κ = 1 − C L m= L− M2 L

(3)

(4)

412 then the equations in (3) become mI1 + (κ + κ 12 ) I1 − κ 12 I 2 = 0 mI 2 + (κ + κ 12 ) I 2 − κ 12 I1 = 0 which are identical in form to Eqs. (12.1). Then, using Eqs. (12.8) for the characteristic frequencies, we can write M 1 L = ω1 = 2 M C (L − M) CL − L 1+

CHAPTER 12

(5)

(6) M 1− 1 L ω2 = = 2 M C (L + M) CL − L which agree with the results of the previous problem.

12-13.

q1 C1

L1 q2 C2 L12 I2 L2

I1

The Kirchhoff circuit equations are q1 + L12 I1 − I 2 = 0 C1 q L2 I 2 + 2 + L12 I 2 − I1 = 0 C2 L1 I1 +

(

)

(

)

(1)

Differentiating these equations using q = I , we can write 1 I1 − L12 I 2 = 0 C1 1 ( L2 + L12 ) I 2 + I 2 − L12 I1 = 0 C2

( L1 + L12 ) I1 +

(2)

As usual, we try solutions of the form I1 ( t ) = B1 e iωt ; which lead to I 2 ( t ) = B2 e iωt (3)

COUPLED OSCILLATIONS

413

2 1 2 ω ( L1 + L12 ) − B1 − L12 ω B2 = 0 C1 2 1 2 − L12 ω B1 + ω ( L2 + L12 ) − B2 = 0 C1

(4)

Setting the determinant of the coefficients of the B’s equal to zero, we obtain

2 1 2 1 4 2 ω ( L1 + L12 ) − ω ( L2 + L12 ) − = ω L12 C1 C2 with the solution (5)

ω

2

( L1 + L12 ) C1 + ( L2 + L12 ) C2 ± ( L1 + L12 ) C1 − ( L2 + L12 ) C2 = 2C1C2 ( L1 + L12 )( L2 + L12 ) − L2 12

2

+ 4 L2 C1C2 12

(6)

We observe that in the limit of weak coupling ( L12 → 0 ) and L1 = L2 = L , C1 = C2 = C , the frequency reduces to

ω=

1 LC

(7)

which is just the frequency of uncoupled oscillations [Eq. (3.78)].

12-14.

L1

C1 C12

L2 C2

I1

I2

The Kirchhoff circuit equations are (after differentiating and using q = I ) 1 1 1 L1I1 + + I2 = 0 I1 − C12 C1 C12 1 1 1 L2 I 2 + + I1 = 0 I2 − C12 C2 C12 Using a harmonic time dependence for I1 ( t ) and I 2 ( t ) , the secular equation is found to be C1 + C12 C2 + C12 1 2 2 L1ω − L2ω − = 2 C1 C12 C2 C12 C12 Solving for the frequency,

2 2 C1L1 ( C2 + C12 ) + C2 L2 ( C1 + C12 ) ± C1L1 ( C2 + C12 ) − C2 L2 ( C1 + C12 ) + 4C1 C2 L1L2 ω = 2L1L2C1C2C12 2 2

(1)

(2)

(3)

414

CHAPTER 12

Because the characteristic frequencies are given by this complicated expression, we examine the normal modes for the special case in which L1 = L2 = L and C1 = C2 = C . Then,

2 ω1 =

2C + C12 LCC12

(4)

1 2 ω2 = LC Observe that ω 2 corresponds to the case of uncoupled oscillations. The equations for this simplified circuit can be set in the same form as Eq. (12.1), and consequently the normal modes can be found in the same way as in Section 12.2. There will be two possible modes of oscillation: (1) out of phase, with frequency ω 1 , and (2) in phase, with frequency ω 2 . Mode 1 corresponds to the currents I1 and I 2 oscillating always out of phase:

;

I1

I2

I1

I2

Mode 2 corresponds to the currents I1 and I 2 oscillating always in phase:

;

I1

I2

I1

I2

(The analogy with two oscillators coupled by a spring can be seen by associating case 1 with Fig. 12-2 for ω = ω 1 and case 2 with Fig. 12-2 for ω = ω 2 .) If we now let L1 ≠ L2 and C1 ≠ C2 , we do not have pure symmetrical and antisymmetrical symmetrical modes, but we can associate ω 2 with the mode of highest degree of symmetry and ω 1 with that of lowest degree of symmetry.

12-15.

C1

L1 R C2 L2

I1

I2

Setting up the Kirchhoff circuit equations, differentiating, and using q = I , we find 1 I1 = 0 C1 1 L2 I 2 + R I 2 − I1 + I2 = 0 C2 L1I1 + R I1 − I 2 +

(

)

(

)

(1)

Using a harmonic time dependence for I1 ( t ) and I 2 ( t ) , the secular equation is 2 2 1 1 2 2 ω L1 − C − iω R ω L2 − C − iω R + ω R = 0

1 2

(2)

COUPLED OSCILLATIONS

415

From this expression it is clear that the oscillations will be damped because ω will have an imaginary part. (The resistor in the circuit dissipates energy.) In order to simplify the analysis, we choose the special case in which L1 = L2 = L and C1 = C2 = C . Then, (2) reduces to 1 2 2 2 ω L − − iω R + ω R = 0 C which can be solved as in Problem 12-6. We find

2

(3)

ω1 = ± i ω2 = L The general solution for I1 ( t ) is

+ I1 ( t ) = B11e

1 LC L 2 R ± R − C (4)

i 1 LC t

− + B11 e

− i 1 LC t

+ + e − Rt L B12 e

R2 − L C t L

− + B12 e

− R2 − L C t L

(5)

and similarly for I 2 ( t ) . The implications of these results follow closely the arguments presented in Problem 12-6. Mode 1 is purely oscillatory with no damping. Since there is a resistor in the circuit, this means that I1 and I 2 flow in opposite senses in the two parts of the circuit and cancel in R. Mode 2 is the mode in which both currents flow in the same direction through R and energy is dissipated. If R2 < L C , there will be damped oscillations of I1 and I 2 , whereas if R2 > L C , the currents will decrease monotonically without oscillation.

12-16.

y O R θ P φ Mg R Q(x,y) Mg x

Let O be the fixed point on the hoop and the origin of the coordinate system. P is the center of mass of the hoop and Q(x,y) is the position of the mass M. The coordinates of Q are x = R ( sin θ + sin φ ) y = − R ( cos θ + cos φ ) (1)

The rotational inertia of the hoop through O is

416

I O = I CM + MR 2 = 2 MR 2 The potential energy of the system is

CHAPTER 12

(2)

U = U hoop + U mass = − MgR ( 2 cos θ + cos φ ) (3)

Since θ and φ are small angles, we can use cos x ≅ 1 − x 2 2 . Then, discarding the constant term in U, we have U= The kinetic energy of the system is T = Thoop + Tmass = 1 1 I Oθ 2 + M x 2 + y 2 2 2 1 MgR 2θ 2 + φ 2 2

(

)

(4)

(

)

(5)

= MR 2θ 2 +

1 MR 2 θ 2 + φ 2 + 2θφ 2

where we have again used the small-angle approximations for θ and φ. Thus, T= Using Eqs. (12.34), T= U= we identify the elements of {m} and {A} : 1 ∑ mjk qj qk 2 j,k 1 ∑ Ajk qj qk 2 j,k (7) 1 MR 2 3θ 2 + φ 2 + 2θφ 2 (6)

(8)

{m} = MR2 {A} = MgR

The secular determinant is 2 g − 3ω 2 R −ω from which

2

3 1 1 1

(9)

2 0 0 1

(10)

−ω 2 g − ω2 R =0 (11)

COUPLED OSCILLATIONS

417

g 2 g 2 4 2 − 3ω − ω − ω = 0 R R

(12)

Solving for the eigenfrequencies, we find

ω1 = 2 ω2 =

To get the normal modes, we must solve: 2 2

g R g R

(13)

∑(A j jk

− ω r2 m jk a jr = 0

)

For k = r = 1, this becomes: g g 2 2 2mgR − 2 3mR a11 − 2 mR a21 = 0 R R or a21 = −2a11 For k = 1, r = 2, the result is a12 = a22 Thus the equations x1 = a11 η1 + a12 η2 x2 = a21 η1 + a22 η2

can be written as x1 = a11 η1 + a22 η2 x2 = −2 a11 η1 + a22 η2

Solving for η1 , η2

η1 =

x1 − x2 ; 3a11

η1 =

2x1 + x2 3a22 1 2

η1 occurs when the initial conditions are such that η2 = 0 ; i.e., x10 = − x20

This is the antisymmetrical mode in which the CM of the hoop and the mass are on opposite sides of the vertical through the pivot point.

η2 occurs when the initial conditions are such that η1 = 0 ; i.e., x10 = x20

418

CHAPTER 12

This is the symmetrical mode in which the pivot point, the CM of the hoop, and the mass always lie on a straight line.

12-17.

k m x1 k

m x2 k

m

k x3 Following the procedure outlined in section 12.6:

T=

1 1 1 2 2 2 mx1 + mx2 + mx3 2 2 2

U=

1 2 1 1 1 2 2 2 kx1 + k ( x2 − x1 ) + k ( x3 − x2 ) + kx3 2 2 2 2

2 2 2 = k x1 + x2 + x3 − x1 x2 − x2 x3

Thus m 0 0 m= 0 m 0 0 0 m 2k − k 0 A = − k 2k − k 0 − k 2k Thus we must solve 2k − ω 2 m −k 0 This reduces to −k 2k − ω m

2

0 −k 2k − ω m

2

=0

−k

( 2 k − ω m)

2

3

− 2k 2 2k − ω 2 m = 0

(

)

or

( 2k − ω m) ( 2k − ω m)

2 2

2

− 2k 2 = 0

If the first term is zero, then we have

ω1 =

If the second term is zero, then

2k m

2k − ω 2 m = ± 2 k

COUPLED OSCILLATIONS

419

which leads to

ω2 =

(2 + 2 ) k ; m ω3 =

(2 − 2) k m To get the normal modes, we must solve

∑(A j jk

− ω r2 m jk a jr = 0

)

For k = 1 this gives:

( 2 k − ω m) a

2 r

1r

+ ( − k ) a2 r = 0

Substituting for each value of r gives

r = 1: r = 2: r = 3:

( 2k − 2k ) a11 − ka21 = 0 → a21 = 0

(− (

2 k a12 − ka22 = 0 → a22 = − 2 a12 2 k a13 − ka23 = 0 → a23 = 2 a13

)

)

Doing the same for k = 2 and 3 yields a11 = − a31 a12 = a32 a13 = a33 The equations a21 = 0 a22 = − 2 a32 a23 = 2 a33

x1 = a11 η1 + a12 η2 + a13 η3 x2 = a21 η1 + a22 η2 + a23 η3 x3 = a31 η1 + a32 η2 + a33 η3 can thus be written as x1 = a11 η1 − 1 a22 η2 + a33 η3 2

x2 = a22 η2 + 2 a33 η3 x3 = − a11 η1 − 1 a22 η2 + a33 η3 2

We get the normal modes by solving these three equations for η1 , η2 , η3 :

420 x1 − x3 2a11 − x1 + 2 x2 − x3 2 2 a22

CHAPTER 12

η1 = η2 = and η3 =

The normal mode motion is as follows

x1 + 2 x2 + x3 4 a33

η1 :

• →

•

← •

x1 = − x3 x2 = − 2 x1 = − 2 x3

η2 : ← • η3 :

• →

• → ← • • →

• → x2 = 2 x1 = 2 x3

12-18. x M

θ

b m

y1

x1

x1 = x + b sin θ ; x1 = x + bθ cos θ y1 = b − b cos θ ; Thus

T= = 1 1 2 2 Mx 2 + m x1 + y1 2 2

y1 = bθ sin θ

(

)

1 1 Mx 2 + m x 2 + b 2 θ 2 + 2b xθ cos θ 2 2

(

)

U = mgy1 = mgb (1 − cos θ ) For small θ, cos θ 1−

θ2

2

. Substituting and neglecting the term of order θ 2θ gives

COUPLED OSCILLATIONS

421

1 1 ( M + m) x 2 + m b 2θ 2 + 2b xθ 2 2 mgb 2 θ 2 M + m mb m= mb 2 mb 0 0 A= 0 mgb

T= U= Thus

(

)

We must solve −ω 2 ( M + m) −ω 2 mb which gives −ω 2 mb mgb − ω 2 mb 2 =0

ω 2 ( M + m) (ω 2 mb 2 − mgb ) − ω 4 m2 b 2 = 0 ω 2 ω 2 Mb 2 − mgb ( m + M ) = 0

Thus

ω1 = 0 ω2 = g ( M + m) mb − ω r2 m jk a jr = 0

∑(A j jk

)

Substituting into this equation gives a21 = 0 a12 = − Thus the equations x = a11 η1 + a12 η2 bm a ( m + M ) 22

( k = 2, r = 1) ( k = 2, r = 2)

θ = a21 η1 + a22 η2 become x = a11 η1 − mb a η ( m + M ) 22 2

422

CHAPTER 12

θ = a22 η2

Solving for η1 , η2 :

η2 =

θ a22 x+ bm θ (m + M) a11

n1 =

n1 occurs when n2 = 0; or θ = 0 n2 occurs when n1 = 0; or x = −

(m + M)

bm

θ

12-19.

With the given expression for U, we see that { A} has the form 1 {A} = −ε12 − ε 13 − ε 12 1 − ε 23 − ε 13 − ε 23 1

(1)

The kinetic energy is T= so that {m} is 1 0 0 {m} = 0 1 0 0 0 1 The secular determinant is 1− ω2 − ε 12 − ε 13 Thus, − ε 12 1−ω − ε 23

2

1 2 2 2 θ1 + θ 2 + θ 3 2

(

)

(2)

(3)

− ε 13 − ε 23 = 0 1−ω

2

(4)

(1 − ω ) − (1 − ω )( ε

2 3 2

2 12

2 2 + ε 13 + ε 23 − 2ε 12 ε 13 ε 23 = 0

)

(5)

This equation is of the form (with 1 − ω 2 ≡ x ) x 3 − 3α 2 x − 2β 2 = 0 which has a double root if and only if (6)

(α )

2 32

= β2

(7)

COUPLED OSCILLATIONS

423

Therefore, (5) will have a double root if and only if

2 2 2 ε 12 + ε 13 + ε 23 3 32

= ε 12 ε 13 ε 23

(8)

This equation is satisfied only if

ε 12 = ε13 = ε 23

Consequently, there will be no degeneracy unless the three coupling coefficients are identical.

12-20.

(9)

If we require a11 = 2a21 , then Eq. (12.122) gives a31 = −3a21 , and from Eq. (12.126) we 14 . Therefore, 1 3 2 , ,− a1 = 14 14 14 (1)

obtain a21 = 1

The components of a2 can be readily found by substituting the components of a1 above into Eq. (12.125) and using Eqs. (12.123) and (12.127): −5 1 4 , , a2 = 42 42 42 These eigenvectors correspond to the following cases: (2)

a1

a2

12-21.

The tensors {A} and {m} are: κ1 1 {A} = κ 3 2 0

1 κ3 2

κ2

1 κ3 2

0 1 κ3 2 κ1

(1)

m 0 0 {m} = 0 m 0 0 0 m

(2)

thus, the secular determinant is

424 1 κ3 2

CHAPTER 12

κ 1 − mω 2

1 κ3 2 0 from which

0 1 κ3 2

=0

κ 2 − mω 2

1 κ3 2

(3)

κ 1 − mω 2

(κ

1

− mω 2

) (κ

2

2

1 2 − mω 2 − κ 3 κ 1 − mω 2 = 0 2

)

(

)

(4)

2 In order to find the roots of this equation, we first set (1 2)κ 3 = κ 1κ 2 and then factor:

(κ (κ (κ

Therefore, the roots are

1

− mω 2 κ 1 − mω 2 κ 2 − mω 2 − κ 1κ 2 = 0 − mω 2 m2ω 4 − (κ 1 + κ 2 ) mω 2 = 0 − mω 2 mω 2 mω 2 − (κ 1 + κ 2 ) = 0

)( ) )

)(

)

1

(5)

1

ω1 = ω2 = ω3 = 0

κ1 m κ1 +κ 2 m (6)

Consider the case ω 3 = 0 . The equation of motion is

2 η3 + ω 3 η 3 = 0

(7)

so that

η3 = 0 with the solution

(8)

η3 ( t ) = at + b

That is, the zero-frequency mode corresponds to a translation of the system with oscillation.

(9)

COUPLED OSCILLATIONS

425

12-22. The equilibrium configuration is shown in diagram (a) below, and the nonequilibrium configurations are shown in diagrams (b) and (c).

(a)

x3

4

3

O

1

2A x1

2B

2

x2

(b)

1

x3

2

O′ A O

A

θ

}Aθ

x3

x2

(c)

4

x3

1

Bφ {

φ

B x

3

O′ B O

x2

The kinetic energy of the system is T= 1 1 1 2 Mx3 + I1θ 2 + I 2φ 2 2 2 2 (1)

where I1 = (1 3) MA2 and I 2 = (1 3) MB2 . The potential energy is

1 2 2 2 2 U = κ ( x3 − Aθ − Bφ ) + ( x3 + Aθ − Bφ ) + ( x3 + Aθ + Bφ ) + ( x3 − Aθ + Bφ ) 2 1 2 = κ 4 x3 + 4 A2θ 2 + 4B2φ 2 2 Therefore, the tensors {m} and {A} are M {m} = 0 0 4κ {A} = 0 0 The secular equation is 0 1 MA2 3 0 0 4κ A 0

2

(

)

0 1 MB2 3 0 0 0 4κ B2

(2)

(3)

(4)

426

CHAPTER 12

( 4κ − Mω ) 4κ A

2

2

−

1 1 MA2ω 2 4κ B2 − MB2ω 2 = 0 3 3

(5)

Hence, the characteristic frequencies are

ω1 = 2 ω2 = 2 ω3 = 2

We see that ω 2 = ω 3 , so the system is degenerate.

κ

M 3κ M 3κ = ω2 M (6)

The eigenvector components are found from the equation

∑(A j jk

− ω r2 m jk a jr = 0

)

(7)

Setting a32 = 0 to remove the indeterminacy, we find 0 1 M 0 a1 = 0 ; a 2 = 3 MA2 ; a 3 = 0 0 0 3 MB2 The normal coordinates are (for x3 ( 0 ) = θ ( 0 ) = φ ( 0 ) = 0 )

(8)

η1 ( t ) = x30 M cos ω 1t η2 ( t ) = η3 ( t ) = θ0 A M

3 cos ω 2t cos ω 3t (9)

φ0 B M

3

Mode 1 corresponds to the simple vertical oscillations of the plate (without tipping). Mode 2 corresponds to rotational oscillations around the x1 axis, and Mode 3 corresponds to rotational oscillations around the x2 -axis. The degeneracy of the system can be removed if the symmetry is broken. For example, if we place a bar of mass m and length 2A along the x2 -axis of the plate, then the moment of inertia around the x1 -axis is changed: I1 = ′ The new eigenfrequencies are 1 ( M + m) A2 3 (10)

COUPLED OSCILLATIONS

427

ω1 = 2 ω2 = 2 ω3 = 2 and there is no longer any degeneracy.

12-23.

κ

M 3κ M+m 3κ M (11)

The total energy of the r-th normal mode is Er = Tr + U r = 1 2 1 2 2 ηr + ω r ηr 2 2 (1)

where

ηr = β r e iωr t

Thus,

(2)

ηr = iω r β r e iωr t

In order to calculate Tr and U r , we must take the squares of the real parts of ηr and ηr :

(3)

ηr2 = ( Re ηr ) = Re i ω r ( µ r + iν r ) ( cos ω r t + i sin ω r t )

2

2

= −ω r ν r cos ω r t − ω r µ r sin ω r t so that Tr = Also

2

(4)

1 2 2 ω r ν r cos ω r t + µr sin ω r t 2

2

(5)

ηr2 = ( Re ηr ) = Re ( µ r + iν r ) ( cos ω r t + i sin ω r t )

2

= µr cos ω r t − ν r sin ω r t so that Ur =

2

(6)

1 2 2 ω r µ r cos ω r t − ν r xin ω r t 2

(7)

Expanding the squares in Tr and U r , and then adding, we find

428

Er = Tr + U r = Thus, Er = 1 2 ω r βr 2

2

CHAPTER 12

1 2 2 ω r µ r + ν r2 2

(

)

(8)

So that the total energy associated with each normal mode is separately conserved. For the case of Example 12.3, we have for Mode 1

η1 =

Thus,

M ( x10 − x20 ) cos ω1t 2

(9)

η1 = −ω 1

Therefore, E1 = But

M ( x10 − x20 ) sin ω1t 2

(10)

1 2 1 2 2 η1 + ω 1 η1 2 2

(11)

2 ω1 =

κ + 2κ 12

M

(12)

so that E1 = = 1 κ + 2κ 12 M 1 κ + 2κ M ( x10 − x20 ) 2 sin 2 ω1t + 2 M 12 2 ( x10 − x20 ) 2 cos2 ω1t M 2 2 1 (κ + 2κ 12 ) ( x10 − x20 ) 2 4 (13)

which is recognized as the value of the potential energy at t = 0. [At t = 0, x1 = x2 = 0 , so that the total energy is U1 ( t = 0 ) .]

12-24. Refer to Fig. 12-9. If the particles move along the line of the string, the equation of motion of the j-th particle is

mx j = −κ x j − x j −1 − κ x j − x j+ 1 Rearranging, we find xj =

(

) (

)

(1)

κ m (x

j −1

− 2x j + x j +1

)

(2)

which is just Eq. (12.131) if we identify τ md with κ m .

COUPLED OSCILLATIONS

429

12-25.

The initial conditions are q1 ( 0 ) = q2 ( 0 ) = q3 ( 0 ) = a q1 ( 0 ) = q2 ( 0 ) = q3 ( 0 ) = 0 (1)

Since the initial velocities are zero, all of the ν r [see Eq. (12.161b)] vanish, and the µr are given by [see Eq. (12.161a)]

µr = so that

a 2

πr πr 3π r sin 4 + sin 2 + sin 4

2 +1 a 2 µ2 = 0 2 −1 µ3 = a 2

(2)

µ1 =

(3)

The quantities sin jrπ ( n + 1) are the same as in Example 12.7 and are given in Eq. (12.165). The displacements of the particles are q1 ( t ) = q2 ( t ) = q3 ( t ) = 1 2 a ( cos ω 1t + cos ω 3t ) + a ( cos ω 1t − cos ω 3t ) 2 4 2 1 a ( cos ω 1t − cos ω 3t ) + a ( cos ω 1t + cos ω 3t ) 2 2 1 2 a ( cos ω 1t − cos ω 3t ) + a ( cos ω 1t + cos ω 3t ) 2 4 (4)

where the characteristic frequencies are [see Eq. (12.152)]

ωr = 2

rπ sin , r = 1, 2, 3 md 8

τ

(5)

Because all three particles were initially displaced, there can exist no normal modes in which any one of the particles is located at a node. For three particles on a string, there is only one normal mode in which a particle is located at a node. This is the mode ω = ω 2 (see Figure 12-11) and so this mode is absent. mb 2 ⇒ [ m] = + mb 2

12-26.

mb 2 2 2 2 Kinetic energy T = θ1 + θ 2 + θ 3 2

(

)

mb

2

Potential energy

430

CHAPTER 12

k 2 2 U = mgb (1 − cos θ 1 ) + (1 − cos θ 2 ) + (1 − cos θ 3 ) + b 2 ( sin θ 2 − sin θ 1 ) + ( sin θ 3 − sin θ 2 ) 2 ≈ mgb 2 kb 2 2 2 2 2 2 θ1 + θ 2 + θ 3 + θ 1 + 2θ 2 + θ 3 − 2θ 1θ 2 − 2θ 2θ 3 2 2

(

)

(

)

mgb + kb 2 ⇒ [ A] = − kb 2 0

− kb 2 mgb + kb − kb 2

2

− kb mgb + kb 2 0

2

The proper frequencies are solutions of the equation mgb + kb 2 − mb 2ω 2 − kb 2 0 = Det [ A] − ω 2 [ m] = D et 0

(

)

(

)

( mgb + 2kb

− kb 2

2

− mb 2ω 2

)

− kb 2

(

− kb 2 mgb + kb 2 − mb 2ω 2 0

)

We obtain 3 different proper frequencies

2 ω1 =

mg + kb mb

⇒ ω1 = ⇒ ω2 = ⇒ ω3 =

mg + kb = 4.64 rad/s mb mg + 3kb = 4.81 rad/s mb g = 4.57 rad/s f

2 ω2 =

mg + 3kb mb

2 ω3 =

mg mb

Actually those values are very close to one another, because k is very small.

12-27.

The coordinates of the system are given in the figure:

θ1 L1

m1 θ2 L2 m2

Kinetic energy: T= 1 2 2 2 1 2 2 m1 θ 1 L1 + m2 L2 θ 1 + L2 θ 2 − 2L1L2θ 1θ 2 cos (θ 1 − θ 2 ) 1 2 2 2

(

)

1 2 1 1 2 ≈ θ 1 m1L2 + m2 L2 + m2 L2 θ 2 − m2 L1L2θ 1θ 2 = ∑ m jk θ j θ k 1 1 2 2 2 2 jk

(

)

COUPLED OSCILLATIONS

431

( m + m2 ) L2 1 ⇒ m jk = 1 − m2 L1 L2 − m2 L1L2 m2 L2 2

Potential energy: U = m1 gL1 (1 − cos θ 1 ) + m2 g L1 (1 − cos θ 1 ) + L2 (1 − cos θ 2 ) ≈ ( m1 + m2 ) gL1

2 θ1

2

+ m2 gL2

2 θ2

2

=

1 ∑ Ajkθ jθ k 2 jk m2 gL2 0

( m + m2 ) gL1 ⇒ Ajk = 1 0

Proper oscillation frequencies are solutions of the equation Det [ A] − ω 2 [ m] = 0 ⇒ ω 1,2 =

(

)

( m1 + m2 ) g ( L1 + L2 ) + ( m1 + m2 ) g 2 m1 ( L1 − L2 )2 + m2 ( L1 + L2 ) 2

2m1 L1 L2

a11 The eigenstate corresponding to ω 1 is where a21

a21

( m + m2 ) L1 1 − = 1 m1L2

×a 11 2 2 ( m1 + m2 ) g ( L1 + L2 ) + ( m1 + m2 ) g 2 m1 ( L1 − L2 ) + m2 ( L1 + L2 ) 2 gm1 L2

a12 The eigenstate corresponding to ω 2 is where a22

a22

( m + m2 ) L1 1 − = 1 m1L2

×a 12 2 2 ( m1 + m2 ) g ( L1 + L2 ) − ( m1 + m2 ) g 2 m1 ( L1 − L2 ) + m2 ( L1 + L2 ) 2 gm1 L2

These expressions are rather complicated; we just need to note that a11 and a21 have the same a a sign 11 > 0 while a12 and a22 have opposite sign 11 < 0 . a21 a21 The relationship between coordinates (θ 1 , θ 2 ) and normal coordinates η1 , η2 are

θ1 = a11 η1 + a12 η2 θ 2 = a21 η1 + a22 η2

⇔

a12 η1 ~ θ 1 − a θ 2 22 a11 η2 ~ θ 1 − θ2 a21

432

CHAPTER 12

To visualize the normal coordinate η1 , let η2 = 0 . Then to visualize the normal coordinate η2 , a a we let η1 = 0 . Because 11 > 0 and 12 < 0 , we see that these normal coordinates describe two a21 a22 oscillation modes. In the first one, the two bobs move in opposite directions and in the second, the two bobs move in the same direction.

m2 b 2 0

12-28.

Kinetic energy:

T=

1 1 2 2 m1b 2θ 1 + m2 b 2θ 2 2 2

⇒

[ m] =

0 m2 b 2

Potential energy:

U = m1 gb (1 − cos θ 1 ) + m2 gb (1 − cos θ 2 ) + ⇒

k ( b sin θ1 − b sin θ 2 ) 2

[ A] ≈

m1 gb + kb 2 − kb

2

m2 gb + kb − kb 2

2

Solving the equation, Det [ A] − ω 2 [ m] = 0 , gives us the proper frequencies of oscillation,

2 ω1 =

(

)

g = 25 (rad/s)2 b

2 ω2 =

g k k + + = 25.11 (rad/s)2 b m1 m2

a11 The eigenstate corresponding to ω 1 is with a21 = 7.44 a11 a22 a12 The eigenstate corresponding to ω 2 is with a22 = 8.55 a12 a22

From the solution of problem 12-27 above, we see that the normal coordinates are

η1 ~ θ1 − η2 ~ θ 1 −

a12 θ 2 = θ 1 + 0.12 θ 2 a22 a11 θ 2 = θ 1 + 0.13 θ 2 a21

Evidently η1 then characterizes the in-phase oscillation of two bobs, and η2 characterizes the out-of-phase oscillation of two bobs. Now to incorporate the initial conditions, let us write the most general oscillation form:

θ1 = Re (α a11 e iω1t − iδ1 + α a12 e iω2t − iδ 2 ) θ 2 = Re (α a21 e iω1t − iδ1 + α a22 e iω 2t − iδ 2 )

= Re 7.44α a11 e iω1t − iδ1 − 8.35α a12 e iω 2t − iδ 2

(

)

where α is a real normalization constant. The initial conditions helps to determine parameters α’s, a’s, δ’s.

COUPLED OSCILLATIONS

433

Re θ 1 ( t = 0 ) = −7 ° ⇒ α a11 cos δ 1 + α a12 cos δ 2 = −0.122 rad Re (θ 2 ( t = 0 ) = 0° ⇒ 7.44α a11 cos δ 1 + 8.35α a12 cos δ 2 = 0 ) ⇒ sin δ 1 = sin δ 1 = 0 . Then

(

)

θ1 = α a11 cos ω 1t + α a12 cos ω 2t = −0.065 cos ω 1t − 0.057 cos ω 2 θ 2 = 7.44α a11 cos ω 1t + 8.35α a12 cos ω 2t = 0.48 ( cos ω 2t − cos ω 1t ) where ω 1 = 5.03 rad/s , ω 2 = 4.98 rad/s (found earlier) Approximately, the maximum angle θ 2 is 0.096 rad and it happens when

cos ω 2t = 1 cos ω 1t = −1

which gives

ω 1t = ( 2k + 1) π because ω 2 t = 2n π

⇒

ω 1 2k + 1 = 2n ω2

ω 1 101 100π = we finally find k = n = 50 and t = = 63 s . ω 2 100 ω2

Note: θ 2 max = 0.96 rad and at this value the small-angle approximation breaks down, and the value θ 2max we found is just a rough estimate.

434

CHAPTER 12

CHAPTER

13

Continuous Systems; Waves

13-1. The initial velocities are zero and so all of the ν r vanish [see Eq. (13.8b)]. The µr are given by [see Eq. (13.8a)]

µr =

2A 3π x rπ x sin sin dx L ∫ L L 0

L

(1)

= Aδ 3 r so that

µ3 = A

µr = 0, r ≠ 3

(2)

The characteristic frequency ω 3 is [see Eq. (13.11)]

ω3 = and therefore,

3π L

τ ρ

(3)

3π q ( x , t ) = A cos L

τ ρ

3π x t sin L

(4)

For the particular set of initial conditions used, only one normal mode is excited. Why?

13-2.

L 3 h

L

435

436

CHAPTER 13

The initial conditions are

L 3h L x, 0 ≤ x ≤ 3 q ( x , 0) = 3h L ( L − x) , ≤x≤L 3 2L q ( x , 0) = 0 Because q ( x , 0 ) = 0 , all of the ν r vanish. The µr are given by

(1)

(2)

6h µr = 2 L = rπ

2

L3

∫

0

rπ x 3h x sin dx + 2 L L rπ 3

L3

∫ ( L − x ) sin

L

rπ x dx L (3)

9h

2

sin

We see that µr = 0 for r = 3, 6, 9, etc. The displacement function is q ( x, t) = where 9 3h 2π 2 1 2π x 1 4π x πx sin L cos ω 1t + 4 sin L × cos ω 2t − 16 sin L cos ω 4 t − … (4)

ωr =

rπ L

τ ρ

(5)

The frequencies ω 3 , ω 6 , ω 9 , etc. are absent because the initial displacement at L 3 prevents that point from being a node. Thus, none of the harmonics with a node at L 3 are excited.

13-3.

The displacement function is q ( x, t) πx 1 3π x 1 5π x = sin cos ω 1t − sin cos ω 3t + sin cos ω 5t + … 8h π 2 L 9 L 25 L (1)

where

ω1 =

π

L

τ ρ

(2)

ω r = rω 1

For t = 0,

q ( x, 0) πx 1 3π x 1 5π x = sin − sin + +… sin 2 8h π L 9 L 25 L The figure below shows the first term, the first two terms, and the first three terms of this function. It is evident that the triangular shape is well represented by the first three terms.

(3)

CONTINUOUS SYSTEMS; WAVES

437

1 term 1 0 1 0 1 0

L 2 terms

L 3 terms

L

The time development of q(x,t) is shown below at intervals of 1 8 of the fundamental period. t = 0, T

t=

1 7 T, T 8 8

t=

1 3 T, T 4 4

t=

3 5 T, T 8 8

t=

1 T 2

13-4.

The coefficients ν r are all zero and the µr are given by Eq. (13.8a): 8 rπ x µr = 2 ∫ x ( L − x ) sin dx L 0 L

L

=

1 − ( −1) r rπ

3 3

16

(1)

so that 0, µ r = 32 , 3 rπ 3 Since r even (2) r odd

438

CHAPTER 13

q ( x , t ) = ∑ µ r sin r rπ x cos ω r t L

(3)

the amplitude of the n-th mode is just µn . The characteristic frequencies are given by Eq. (13.11): nπ L

ωn =

τ ρ

(4)

13-5.

The initial conditions are q ( x , 0) = 0 v0 , q ( x, t) = 0,

( L 2) + s

1 ≤s 2 otherwise x−

(1)

The µr are all zero and the ν r are given by [see Eq. (13.8b)]

νr = −

= from which

2 rπ x v sin dx ω r L ( L ∫) − s 0 L 2 (2)

4 v0 rπ rπ s sin sin 2 rπω r L

0, ν r = 4v rπ s 0 , ( −1)( r −1) 2 sin − rπω r L

r even r odd (3)

(Notice that the even modes are all missing, as expected from the symmetrical nature of the initial conditions.) Now, from Eq. (13.11),

ω1 = and ω r = rω 1 . Therefore,

π

L

τ ρ

(4)

νr =

According to Eq. (13.5),

−4v0 rπ s , r odd ( −1)( r −1) 2 sin r 2πω 1 L

(5)

CONTINUOUS SYSTEMS; WAVES

439

q ( x , t ) = ∑ β r e iω r t sin r rπ x L rπ x L (6)

= − ∑ ν r sin ω r t sin r Therefore,

q ( x, t) =

πω 1

4 v0

sin ω 1t sin

πs

L

sin

πx

L

1 3π s 3π x − sin ω 3t sin sin + … 9 L L

(7)

Notice that some of the odd modes—those for which sin ( 3π s L) = 0 —are absent.

13-6.

The initial conditions are q ( x , 0) = 0 4 v0 L 4v q ( x , 0) = 0 L 0 x L 2 − x 0≤x≤ L 4 L L ≤x≤ 4 2 L ≤x≤L 2

(1)

The velocity at t = 0 along the string, q ( x , 0 ) , is shown in the diagram. v v0

L 4

L 2

3L 4

L

The µr are identically zero and the ν r are given by:

νr = −

=

2 rπ x ∫ q ( x , 0) sin L dx rLω 1 0

L

rπ rπ sin 2 − 2 sin 4 r π ω1 8 v0

2 3

(2)

Observe that for r = 4n, ν r is zero. This happens because at t = 0 the string was struck at L 4 , and none of the harmonics with modes at that point can be excited. Evaluation of the first few ν r gives

440

CHAPTER 13

ν1 = −0.414 ⋅ ν2 = − ⋅ ν3 = − and so,

π ω1

2

8 v0

ν4 = 0

1 8 v0 4 π 2ω 1

2.414 8v0 ⋅ 27 π 2ω 1

2.414 8v0 ⋅ ν5 = 125 π 2ω 1 2 8 v0 ⋅ 2 ν6 = 216 π ω 1

(3)

q ( x, t) =

8 v0 2π x πx 1 + sin ω 2t sin 0.414 sin ω 1t sin 2 L 4 L π ω1 2.414 3π x 2.414 5π x + sin ω 3t sin − sin ω 5t sin − … L L 27 135

(4)

From these amplitudes we can find how many db down the fundamental are the various harmonics: Second harmonic: 0.250 = −4.4 db 10 log 0.414 Third harmonic: 2.414 27 10 log = −13.3 db 0.414

2 2

(5)

(6)

These values are much smaller than those found for the case of example (13.1). Why is this so? (Compare the degree of symmetry of the initial conditions in each problem.)

13-7.

3 L 7 h O h L

3 L 7

Since q ( x , 0 ) = 0 , we know that all of the ν r are zero and the µr are given by Eq. (13.8a):

µr =

The initial condition on q ( x , t ) is

2 rπ x ∫ q ( x , 0) sin L dx L0

L

(1)

CONTINUOUS SYSTEMS; WAVES

441

7h − 3L x , 7h q ( x , 0 ) = ( 2 x − L) , L 7h 3L ( L − x ) , Evaluating the µr we find

0≤x≤

3 4 L ≤ x ≤ L 7 7 4 L≤ x≤L 7 3 L 7

(2)

µr =

98 h 3 r 2π 2

4 rπ 3rπ sin 7 − sin 7

(3)

Obviously, µr = 0 when 4 r 7 and 3r 7 simultaneously are integers. This will occur when r is any multiple of 7 and so we conclude that the modes with frequencies that are multiples of 7ω 1 will be absent.

13-8.

For the loaded string, we have [see Eq. (12.152)]

ωr = 2

Using ρ = m d and L = ( n + 1) d , we have

τ md sin

rπ 2 ( n + 1)

(1)

ωr =

=

The function

2 d

τ rπ sin 2 ( n + 1) ρ rπ τ sin 2 ( n + 1) ρ (2)

2 ( n + 1) L

ωr

2 L

τ ρ

= ( n + 1) sin

rπ 2 ( n + 1)

(3)

is plotted in the figure for n = 3, 5, and 10. For comparison, the characteristic frequency for a continuous string is also plotted:

ωr

2 L

τ ρ

=

rπ 2

(4)

442

CHAPTER 13

12 n = 10 10 ss tri ng Co nti nu ou

ωr 2 L τ ρ

8

6

n=5

4

n=3

2

0

0

2

4

6

8

10

Of course, the curves have meaning only at the points for which r is an integer.

13-9.

From Eq. (13.49), we have:

β=

D s 2π 2τ 2 ; ω0 = 2ρ ρb

(1)

From section 3.5, we know that underdamped motion requires:

2 β 2 < ω0

Using (1) this becomes D 2 s 2π 2τ < 4ρ 2 ρb or D2 < 4 ρ s 2 π 2τ b 4 ρ s 2π 2 τ b 4 ρ s 2 π 2τ b underdamped critically damped overdamped

Likewise D 2 = D2 >

The complementary solution to Eq. (13.48) for underdamped motion can be written down using Eq. (3.40). The result is:

ηs ( t ) = Cs e − βt cos (ω 1t − φs )

CONTINUOUS SYSTEMS; WAVES

443

2 2 where ω 1 = ω 0 − β 2 , ω 0 and β are as defined in (1), and Cs and φs are arbitrary constants depending on the initial conditions. The complete solution to Eq. (13.48) is the sum of the particular and complementary solutions (analogous to Eq. (13.50)):

ηs ( t ) = C s e − β t

sπ 2 F0 sin cos (ω t − δ s ) 2 cos (ω 1t − φ s ) + 2 2 s π τ D − ω2 + ω2 ρb ρb ρ

where

δ s = tan −1

Dω 2 s π τ 2 ρ −ω ρb

2

From Eq. (13.40): q ( x , t ) = ∑ ηr ( t ) sin r rπ x b

Thus

rπ s 2π 2τ D 2 2 F0 sin 2 cos (ω t − δ r ) Dt rπ x q ( x , t ) = ∑ Cr exp − cos − 2 t − φr + sin b 2 2 ρb 2ρ 4ρ r ρb r π τ − ω 2 + D ω 2 ρb ρ (underdamped)

13-10.

From Eq. (13.44) the equation for the driving Fourier coefficient is: f s ( t ) = ∫ F ( x , t ) sin

0 b

sπ x dx b

If the point x is a node for normal coordinate s, then x n = where n is an integer ≤ s b s (This comes from the fact that normal mode s has s-half wavelengths in length b.) For x n = , b s sin sπ x = sin nπ = 0; hence f s ( t ) = 0 b

444

CHAPTER 13

Thus, if the string is driven at an arbitrary point, none of the normal modes with nodes at the driving point will be excited.

13-11.

From Eq. (13.44) f s ( t ) = ∫ F ( x , t ) sin

0

b

sπ x dx b

(1)

where F ( x , t ) is the driving force, and f s ( t ) is the Fourier coefficient of the Fourier expansion of coordinate s. Thus, we desire F ( x , t ) such that

F ( x , t ) . Eq. (13.45) shows that f s ( t ) is the component of F ( x , t ) effective in driving normal f s ( t ) = 0 for s ≠ n ≠ 0 for s = n

From the form of (1), we are led to try a solution of the form

F ( x , t ) = g ( t ) sin nπ x b

where g(t) is a function of t only. Thus f s ( t ) = ∫ g ( t ) sin

0

b

nπ x sπ x sin dx b b

; hence f s ( t ) = 0 for s ≠ n. x=0 b For n ≠ s, the integral is proportional to sin For n = s, we have

( n ± s) π x b f s ( t ) = g ( t ) ∫ sin 2

0

b

nπ x b dx = g ( t ) ≠ 0 2 b

Only the nth normal coordinate will be driven. Thus, to drive the nth harmonic only, nπ x F ( x , t ) = g ( t ) sin b

CONTINUOUS SYSTEMS; WAVES

445

13-12.

The equation to be solved is

ηs +

Compare this equation to Eq. (3.35):

D

ρ

ηs +

s 2π 2τ η =0 ρb s

(1)

2 x + 2β x + ω 0 x = 0

The solution to Eq. (3.35) is Eq. (3.37): x ( t ) = e − βt A1 exp

(

2 2 β 2 − ω 0 t + A2 exp − β 2 − ω 0 t

)

(

)

Thus, by analogy, the solution to (1) is

D 2 s 2π 2τ D 2 s 2 π 2τ − − ns ( t ) = e − Dt 2 ρ A1 exp t + A2 exp − t 2 2 ρb ρb 4ρ 4ρ Assuming k is real, while ω and v are complex, the wave function becomes

13-13.

ψ ( x , t ) = Ae i(αt + iβt − kx)

= Ae (α t − kx ) e − βt whose real part is (1)

ψ ( x , t ) = Ae − βt cos (α t − kx ) and the wave is damped in time, with damping coefficient β. From the relation k2 = we obtain

(2)

ω2 v2 (3)

(α + iβ ) 2 = k 2 ( u + iw ) 2

(4)

By equating the real and imaginary part of this equation we can solve for α and β in terms of u and w:

α= and k 2 uw

β

(5)

kw β= iku

(6)

Since we have assumed β to be real, we choose the solution

446

CHAPTER 13

β = kw

Substituting this into (5), we have

(7)

α = ku as expected. Then, the phase velocity is obtained from the oscillatory factor in (2) by its definition:

V=

(8)

Re ω α = k k

V=u

(9)

That is,

13-14.

Vn–2

In–2 L′

Vn–1

In L′

Vn

In+1 L′

Qn

Vn+1

In+2 L′

Vn+2

Qn–2 C′

Qn–1

Qn+1 C′

C′

Qn+2

C′

C′

Consider the above circuit. The circuit in the nth inductor is I n , and the voltage above ground at the point between the nth elements is Vn . Thus we have

Vn = Qn C′

and

L′ dI n = Vn −1 − Vn dt = Qn −1 Qn − C′ C′

(1)

We may also write dQn = I n − I n+1 dt

(2)

Differentiating (1) with respect to time and using (2) gives

L′ d2 I n 1 = [ I n −1 − 2I n + I n + 1 ] 2 dt C′

(3)

or d2 In 1 = [ I n − 1 − 2I n + I n + 1 ] 2 dt L′ C ′

(4)

CONTINUOUS SYSTEMS; WAVES

447

Let us define a parameter x which increases by ∆x in going from one loop to the next (this will become the coordinate x in the continuous case), and let us also define

L≡

L′ ; ∆x

C≡

C′ ∆x

(5)

which will become the inductance and the capacitance, respectively, per unit length in the limit ∆x → 0 . From the above definitions and

∆I r = I r + 1 − I r (4) becomes d2 In 1 + ( ∆I n−1 − ∆I n ) = 0 2 dt L′C ′ or, d 2 I n ∆ ( ∆I n ) − =0 dt 2 L′C ′ Dividing by ( ∆x ) , and multiplying by (–L′C′), we find

2

(6)

(7)

(8)

∆ ( ∆I n )

( ∆x )

2

− LC

d2 In =0 dt 2

(9)

But by virtue of the above definitions, we can now pass to the continuous limit expressed by In (t) → I ( x, t) Then, ∆ ( ∆I ( x , t ) ) ∆x 2 and for ∆x → 0 , we obtain ∂2I 1 ∂2I − =0 ∂x 2 v 2 ∂t 2 where v= 1 LC (13) (12) − LC ∂2 I ( x, t) =0 ∂t 2 (11) (10)

13-15.

Consider the wave functions

ψ 1 = A exp i (ω t − kx )

ψ 2 = B exp i ( (ω + ∆ω ) t − ( k + ∆k ) x )

(1)

where ∆ω

ω ; ∆k

k . A and B are complex constants:

448

CHAPTER 13

A = A exp ( iφ a ) B = B exp ( iφb ) The superposition of ψ 1 and ψ 2 is given by

(2)

ψ = ψ1 + ψ2

∆ω t −∆kx ∆ω t −∆kx −i i ∆ω ∆k = exp i ω + t − k + x × A exp ( iφ a ) e 2 + B exp ( iφb ) e 2 2 2

(3)

which can be rewritten as

ψ = exp i w +

Define

∆ω t −∆kx − φ a + φb ∆ω t −∆kx + φb − φ a i −i φ + φb ∆ω ∆k 2 2 ×A e + B e t −k + x + a 2 2 2

(4)

t∆ω − x∆k ≡ δ φb − φ a ≡ α and A e − i(δ +α ) 2 + B e i(δ +α ) 2 = Γ e iθ Therefore,

(5)

(6)

Γ =2 A + B

2 2

(

2

)

(δ + a )

2

(7) (8)

cos θ =

2 A + B

2 2

(

A+B

)

12

cos

sin θ =

2 A 2 + B 2

(

B− A

)

12

sin

(δ + a )

2

(9)

That is, θ is a function of ( ∆ω ) t − ( ∆k ) x . Using (6) and (7) – (9), we can rewrite (4) as φ a + φb iθ ∆ω ∆k ψ = Γ exp i ω + e t −k + x + 2 2 2 and then, φ a + φb ∆ω ∆k Re ψ = Γ cos ω + t − k + x cos θ + 2 2 2 ∆ω ∆k − sin ω + t − k + x sin θ + 2 2 (10)

φ a + φb

2

(11)

CONTINUOUS SYSTEMS; WAVES

449

From this expression we see that the wave function is modulated and that the phenomenon of beats occurs, but for A ≠ B, the waves never beat to zero amplitude; the minimum amplitude is, from Eq. (11), A − B , and the maximum amplitude is A + B . The wave function has the form shown in the figure.

A+B

Reψ A−B

wt – kx

13-16. As explained at the end of section 13.6, the wave will be reflected at x = x0 and will then propagate in the –x direction. 13-17.

We let m′ , mj = m′′ , j = 2n j = 2n + 1 (1)

where n is an integer. Following the procedure in Section 12.9, we write F2 n = m′ q2 n =

τ d ( q2 n−1 − 2q2n + q2n+1 ) τ d

(2a) (2b)

F2 n +1 = m′′ q2 n +1 = Assume solutions of the form

( q2n − 2q2n+1 + q2n+ 2 )

q2 n = Ae i(ωt − 2 nkd) q2 n + 1 = Be [ Substituting (3a,b) into (2a,b), we obtain −ω 2 A = −ω B =

2

i ω t − ( 2 n + 1) kd ]

(3a) (3b)

τ m′ d

( Be

ikd

− 2 A + Be − ikd − 2B + Ae

) )

τ m′′ d

( Ae

ikd

− ikd

(4)

from which we can write

450

CHAPTER 13

2τ 2τ − ω2 − B A cos kd = 0 m′d m′ d 2τ 2τ cos kd + B −A − ω2 = 0 m′′ d m′′ d The solution to this set of coupled equations is obtained by setting the determinant of the coefficients equal to zero. We then obtain the secular equation 1 2τ 2 2τ 2 m′ d − ω m′′ d − ω − m′ m′′ Solving for ω, we find

2 1 1 1 4 ± + ω = + sin 2 kd − d m′ m′′ m′ m′′ m′ m′′ 2

(5)

2τ d cos kd = 0

2

(6)

τ 1

12

(7)

from which we find the two solutions

2 1 1 1 4 + + sin 2 kd ω = + − d m′ m′′ m′ m′′ m′ m′′ 2 1

τ 1

12

12 2 τ 1 1 1 1 4 2 2 + ω 2 = + sin kd − − d m′ m′′ m′ m′′ m′ m′′

(8)

If m′ < m″, and if we define

ωa ≡

2τ , ωb ≡ m′′ d

2τ 2 2 , ωc = ω a + ωb m′ d

(9)

Then the ω vs. k curve has the form shown below in which two branches appear, the lower branch being similar to that for m′ = m″ (see Fig. 13-5). ωc ω ωb ωa π/2d

0

k

Using (9) we can write (6) as sin 2 kd =

ω2 2 ω 2 + ω b − ω 2 ) ≡ W (ω ) 2 2 ( a ωaωb

(10)

From this expression and the figure above we see that for ω > ω c and for ω a < ω < ω b , the wave number k is complex. If we let k = κ + iβ , we then obtain from (10) sin 2 (κ + iβ ) d = sin 2 κ d cosh 2 β d − cos 2 κ d sinh 2 β d + 2i sin κ d cos κ d sinh β d cosh β d = W (ω ) Equating the real and imaginary parts, we find (11)

CONTINUOUS SYSTEMS; WAVES

451

2 2 2 2 sin κ d cosh β d − cos κ d sinh β d = W (ω ) We have the following possibilities that will satisfy the first of these equations:

a) sin κd = 0, which gives κ = 0. This condition also means that cos κd = 1; then β is determined from the second equation in (12):

sin κ d cos κ d sinh β d cosh β d = 0

(12)

− sinh 2 β d = W (ω ) Thus, ω > ω c , and κ is purely imaginary in this region.

(13)

b) cos κd = 0, which gives κ = π/2d. Then, sin κd = 1, and cosh 2 β d = W (ω ) . Thus, ω a < ω < ω b ,

and κ is constant at the value π/2d in this region.

c) sinh βd = 0, which gives β = 0. Then, sin 2 κ d = W (ω ) . Thus, ω < ω a or ω b < ω < ω c , and κ is real in this region.

Altogether we have the situation illustrated in the diagram. k π 2d

κ β β

κ ωa ωb

κ κ ωc ω

13-18.

The phase and group velocities for the propagation of waves along a loaded string are V (k) = U (k) =

ω k =

ω c d sin ( kd 2)

2 kd 2

(1) (2)

dω ω c d = cos ( kd 2) dk 2

where

ω = ω c sin ( kd 2)

The phase and group velocities have the form shown below. ωc d 2

(3)

V(k) U(k) π/d

V,U 0

k

When k = π d , U = 0 but V = ω c d π . In this situation, the group (i.e., the wave envelope) is stationary, but the wavelets (i.e., the wave structure inside the envelope) move forward with the velocity V.

452 13-19.

CHAPTER 13

The linear mass density of the string is described by

ρ1 ρ= ρ2 > ρ1

I ρ1 0 II ρ2

if x < 0; x > L if 0 < x < L

III ρ1 x

Consider the string to be divided in three different parts: I for x < 0, II for 0 < x < L, and III for x > L. Let φ = A e ( i ω t − k1 x )

be a wave train, oscillating with frequency ω, incident from the left on II. We can write

for the different zones the corresponding wave functions as follows:

ψ I = Ae i(ωt − k1x ) + Be i(ωt + k1x) = e iωt Ae − ik1x + Be ik1x ψ II = Ce i(ωt − k2 x) + De i(ωt + k2 x ) = e iωt Ce − ik2 x ψ III = Ee i(ωt − k1x )

Where

+ De ik2 x

(1)

k1 =

ω

V1

, k2 =

ω

V2

, V1 =

τ τ , V2 = ρ1 ρ2

(2)

and where τ is the tension in the string (constant throughout). To solve the problem we need to state first the boundary conditions; these will be given by the continuity of the wave function and its derivative at the boundaries x = 0 and x = L. For x = 0, we have

ψ I ( x = 0 ) = ψ II ( x = 0 )

∂ψ I ∂x and for x = L, the conditions are

= x=0 ∂ψ II ∂x

x=0

(3)

ψ II ( x = 0 ) = ψ III ( x = L)

∂ψ II ∂x = x=L ∂ψ III ∂x

x=L

(4)

Substituting ψ as given by (1) into (3) and (4), we have

k2 − A + B = ( −C + D) k1 and A+B=C+D

(5)

CONTINUOUS SYSTEMS; WAVES

453

C e − ik2 L + D e ik2 L = E e − ik1L −C e

From (6) we obtain

− ik2 L

+De

ik2 L

k = − 1 E e − ik1L k2

(6)

C=

k 1 1 + 1 E e i( k2 − k1 ) L 2 k2 1 k D = 1 − 1 E e − i( k2 + k1 ) L 2 k2 k 2 + k1 i 2 k 2 L e D k 2 − k1

(7)

Hence,

C=

From (5) we have

(8)

A=

1 k2 1 k2 1 + k C + 2 1 − k D 2 1 1

(9)

Using (7) and rearranging the above equation

2 1 ( k1 + k 2 ) i 2 k2 L A= e − ( k 2 − k1 ) D 2k1 ( k 2 − k1 )

(10)

In the same way

B=

From (10) and (11) we obtain

1 − ( k 2 + k1 ) e i 2 k2 L + ( k1 + k 2 ) D 2 k1

(11)

2 2 k1 − k 2 e i 2 k2 L − 1 B = A ( k1 + k 2 ) 2 e i 2 k 2 L − ( k1 − k 2 ) 2

(

)

(12)

On the other hand, from (6) and (8) we have

E= which, together with (10) gives

2k 2 D i( k2 + k1 ) L e k 2 − k1

(13)

4 k1 k 2 e i( k1 + k2 ) L E = A ( k1 + k 2 ) 2 e i 2 k 2 L − ( k1 − k 2 ) 2

Since the incident intensity I 0 is proportional to A , the reflected intensity is I r = B , and the total

2 2

(14)

transmitted intensity is I t = E , we can write

2

454

2 2 2 2

CHAPTER 13

Ir = I0

B A

, It = I 0

E A

(15)

Substituting (12) and (14) into (15), we have, for the reflected intensity,

2 − k 2 e i 2 k2 L − 1 Ir = I0 ( k1 + k 2 ) 2 e i 2 k 2 L − ( k1 − k 2 ) 2 2 1

(k

)

2

(16)

From which

2 2 2 k1 − k 2 (1 − cos 2k 2 L) Ir = I0 4 4 2 2 2 2 k1 + k 2 + 6 k1 k 2 − k1 − k 2 2 cos 2k2 L

(

)

(

)

(17)

and for the transmitted intensity, we have

It = I0 so that

( k1 + k 2 ) 2 e i 2 k L − ( k1 − k 2 ) 2

2

4 k1k 2 e i( k1 + k2 ) L

2

(18)

It = I0

2 2 8 k1 k 2 4 4 2 2 2 2 k1 + k 2 + 6 k1 k 2 − 2 k1 − k 2

(

)

2

(19)

cos 2k 2 L

We observe that I r + I t = I 0 , as it must. For maximum transmission we need minimum reflection; that is, the case of best possible transmission is that in which

It = I0 Ir = 0

In order that I r = 0 , (17) shows that L must satisfy the requirement

(20)

1 − cos 2k2 L = 0 so that we have

(21)

L=

mπ mπ = k2 ω

τ , m = 0,1, 2,… ρ2

(22)

The optical analog to the reflection and transmission of waves on a string is the behavior of light waves which are incident on a medium that consists of two parts of different optical densities (i.e., different indices of refraction). If a lens is given a coating of precisely the correct thickness of a material with the proper index of refraction, there will be almost no reflected wave.

CONTINUOUS SYSTEMS; WAVES

455

13-20. y I M 0 II

We divide the string into two zones: I: x < 0 II: x > 0 Then,

ψ I = A1e i(ωt − kx) + B1e i(ωt + kx) ψ II = A2 e

The boundary condition is i (ω t − kx )

(1)

ψ I ( x = 0 ) = ψ II ( x = 0 )

That is, the string is continuous at x = 0. But because the mass M is attached at x = 0, the derivative of the wave function will not be continuous at this point. The condition on the derivative is obtained by integrating the wave equation from x = –ε to x = +ε and then taking the limit ε → 0. Thus, M ∂ 2ψ ∂t 2 ∂ψ I ∂ψ = τ II − ∂x ∂x

(2)

(3) x=0 x=0

Substituting the wave functions from (1), we find A1 + B1 = A2 ikτ ( − A2 + A1 − B1 ) = −ω 2 MA2 which can be rewritten as A1 − B1 = A2 From (4) and (6) we obtain A1 + B1 ikτ = A1 − B1 ikτ − ω 2 M from which we write (7) (4) (5)

( ikτ − ω M)

2

ikτ

(6)

ω 2 M 2ikτ B1 ω2M = = A1 2ikτ − ω 2 M 1 − ω 2 M 2ikτ

Define

(8)

456

CHAPTER 13

ω2M = P = tan θ 2kτ

Then, we can rewrite (8) as − iP B1 = A1 1 + iP And if we substitute this result in (4), we obtain a relation between A1 and A2 : A2 1 = A1 1 + iP B The reflection coefficient, R = 1 , will be, from (10), A1 tan 2 θ B P2 = R= 1 = 1 + P 2 1 + tan 2 θ A1 or,

2 2

(9)

(10)

(11)

(12)

R = sin 2 θ A and the transmission coefficient, T = 2 , will be from (10) A1 T= or, T = cos 2 θ A2 A1

2 2

(13)

=

1 1 = 2 1+ P 1 + tan 2 θ

(14)

(15)

The phase changes for the reflected and transmitted waves can be calculated directly from (10) and (11) if we substitute B1 = B1 e iφ B1 iφ A1 = A1 e A1 A2 = A2 e iφ A2

(16)

Then, B i(φ −φ ) B1 P i tan −1 ( 1 P ) = 1 e B1 A1 = e A1 A1 1 + P2 and

−1 A2 A2 i(φA2 −φA1 ) 1 = = e e i tan ( − P) A1 A1 1 + P2

(17)

(18)

CONTINUOUS SYSTEMS; WAVES

457

Hence, the phase changes are

φB1 − φ A1 = tan −1 = tan −1 ( cot θ ) P φ A2 − φ A1 = − tan ( P ) = − tan

−1 −1

1

(19)

( tan θ ) = −θ

13-21.

The wave function can be written as {see Eq. (13.111a)]

ψ ( x, t) =

+∞

−∞

∫ A(k) e k0 +∆k k0 −∆k

i (ω t − kx )

dk

(1)

Since A(k) has a non-vanishing value only in the vicinity of k = k0 , (1) becomes

ψ ( x, t) =

According to Eq. (13.113),

∫

e i(ωt − kx ) dk

(2)

ω = ω 0 + ω 0 ( k − k0 ) ′

Therefore, (2) can now be expressed as k0 +∆k k0 −∆k

(3)

ψ ( x , t ) = e i(ω0 − w0′ k0 )t

=e(

i ω 0 − ω 0 k0 ) t ′

∫

e(

i ω0t − x) k ′

dk

e i( k0 +∆k )(ω0′ t − x ) − e i( k0 −∆k )(ω 0′ t − x) i (ω 0 t − x ) ′ e i(ω 0′ t − x) ∆k − e i(ω 0′ t − x) ∆k 2i (4)

=

2e (

i ω 0 −ω 0 k0 ) t ′

ω 0t − x ′

and writing the term in the brackets as a sine, we have

ψ ( x, t) =

′ 2 sin (ω 0 t − x ) ∆k i(ω 0t − k0 x) e ω 0t − x ′

(5)

The real part of the wave function at t = 0 is Re ψ ( x , 0 ) = 2 sin ( x∆k ) cos k0 x x (6)

If ∆k k0 , the cosine term will undergo many oscillations in one period of the sine term. That is, the sine term plays the role of a slowly varying amplitude and we have the situation in the figure below.

458

ReΨ(x,0)

CHAPTER 13

x

13-22. a)

Using Eq. (13.111a), we can write (for t = 0)

ψ ( x, 0) =

+∞

−∞

∫ A(k) e

+∞ −∞

− ikx

dk

=B

∫e

− σ ( k − k0 )

2

e − ikx dk

2

= Be

− ik0 x

+∞

−∞ +∞

∫e ∫e

− σ ( k − k0 )

−i k−k x e ( 0 ) dk

= Be − ik0 x

− σ u2

e − iux du

(1)

−∞

This integral can be evaluated by completing the square in the exponent:

+∞

−∞

∫e

− ax 2

+∞

e dx = bx −∞

∫e ∫ e b − a x 2 − x a

dx b2 =

2 +∞ − a x 2 − b x + b a 4 a2

e 4 a dx

−∞ b b 2 +∞ − a x − 2a 4a −∞

2

=e and letting y = x − b 2a , we have

+∞

∫e

dx

(2)

−∞

∫e

− ax 2

e dx = e bx b 2 +∞ − ay 2 4a −∞

∫e π a

dy

(3)

Using Eq. (E.18c) in Appendix E, we have

+∞

−∞

∫e

− ax 2

e dx = bx e

b2 4a

(4)

Therefore,

CONTINUOUS SYSTEMS; WAVES

459

ψ ( x , 0) = B

π − ik0 x − x2 4σ e e σ

(5)

The form of ψ ( x , 0 ) (the wave packet) is Gaussian with a 1 e width of 4 σ , as indicated in the diagram below.

Ψ ( x, 0 )

B

π σ

−2 σ

2 σ

1 π ⋅B e σ x

b) The frequency can be expressed as in Eq. (13.113a):

ω ( k ) = ω 0 + ω 0 ( k − k0 ) + … ′ and so,

+∞

(6)

ψ ( x, t) =

=

−∞ +∞

∫ A(k) e ∫ A(k) e

i (ω t − kx )

dk

i ω 0 t + ω 0 ( k − k0 ) t − kx ′

dk

−∞

= Be ( = Be (

i ω 0 t − k0 x )

+∞

−∞ i ω 0 t − k0 x ) +∞

∫e ∫e

− σ ( k − k0 )

2

e

i ω 0 ( k − k0 ) t − ( k − k0 ) x ′

dk

− σ u2

e(

i ω0t − x) u ′

du

(7)

−∞

Using the same integral as before, we find

ψ ( x, t) = B

π i(ω0t − k0 x) − (ω0′ t − x )2 e e σ

4σ

(8)

c) Retaining the second-order term in the Taylor expansion of ω (k), we have

ω ( k ) = ω 0 + ω 0 ( k − k0 ) + ω 0 ( k − k0 ) + … ′ ′′

2

1 2

(9)

Then,

ψ ( x , t ) = e i (ω 0 t − k 0 x )

= Be (

+∞

−∞ i ω 0 t − k0 x ) +∞

∫

A(k) e

1 2 i ω 0 ( k − k0 ) t + ω 0 ( k − k0 ) t − ( k − k0 ) x ′ ′′ 2

dk

−∞

∫

e

ω ′′t − σ − i 0 u2 2

e(

i w0 t − x ) u ′

du

(10)

We notice that if we make the change σ − iω 0 t 2 → σ , then (10) becomes identical to (7). ′′ Therefore,

460

CHAPTER 13

ψ ( x, t) = B where 2π i ω t−k x i ( 0 0 ) e −α ( x ,t) 2σ − iw0 t ′′

(11)

α ( x, t) =

(ω 0′t − x ) 2 σ + 2 iω 0′′t 1 4σ 2 + (ω 0 ) t 2 ′′

2

(12)

The 1 e width of the wave packet will now be w1 e ( t ) = 2 or, W1 e ( t ) = 4 σ ω ′′t 1+ 0 2σ

2

4σ 2 + (ω 0 ) t 2 ′′

2

σ

(13)

(14)

In first order, W1 e , shown in the figure above, does not depend upon the time, but in second order, W1 e depends upon t through the expression (14). But, as can be seen from (8) and (11), the group velocity is ω 0 , and is the same in both cases. Thus, the wave packet propagates with ′ velocity ω 0 but it spreads out as a function of time, as illustrated below. ′

Ψ (x, t) t=0 t = t1 O x

CHAPTER

14

(1)

The Special Theory of Relativity

14-1.

Substitute Eq. (14.12) into Eqs. (14.9) and (14.10):

v x1 = γ x1 − x1 ′ c v x1 = γ x1 + x1 ′ ′ c

From (1)

(γ = γ ′ )

(2)

x1 ′ v = γ 1 − x1 c From (2)

x1 ′ = x1 So

1 v γ 1 + c

v 1 γ 1 − = c v γ 1+ c or

γ=

1 1 − v2 c2

461

462

CHAPTER 14

14-2. We introduce cosh α ≅ y , sinh α ≅ y v c and substitute these expressions into Eqs. (14.14); then

x1 = x1 cosh α − ct sinh α ′ x1 t ′ = t cosh a − sinh α c x2 = x2 ; x3 = x3 ′ ′ Now, if we use cosh α = cos (iα) and i sinh α = sin (iα), we can rewrite (1) as

(1)

ict ′ = − x1 sin ( iα ) + ict cos ( iα ) Comparing these equations with the relation between the rotated system and the original system in ordinary three-dimensional space, x1 = x1 cos θ + x2 sin θ ′ x2 = − x1 sin θ + x2 cos θ ′ x3 = x3 ′ x2′ x2

x1 = x1 cos ( iα ) + ict sin ( iα ) ′

(2)

(3)

x1′ θ x1

We can see that (2) corresponds to a rotation of the x1 − ict plane through the angle iα. If the equation ∇ 2ψ ( x , ict ) −

14-3.

1 ∂ 2 ψ ( x , ict ) =0 c2 ∂t 2 1 ∂ 2 ψ ( x ′ , ict ′ ) =0 c2 ∂t ′ 2

(1)

is Lorentz invariant, then in the transformed system we must have

∇ ′ 2ψ ( x ′ , ict ′ ) −

(2)

where

∇′2 = ∂2 ∂2 ∂2 + + 2 2 ∂x ′ ∂y ′ ∂z ′ 2

(3)

We can rewrite (2) as

∂ 2ψ ( x ′ , ict ′ ) ∑ ∂x ′ 2 = 0 µ =1 µ

4

(4)

THE SPECIAL THEORY OF RELATIVITY

463

Now, we first determine how the operator We know the following relations:

∂2 ∑ ∂x ′ 2 is related to the original operator µ µ

∂2 ∑ ∂x 2 . µ µ (5) (6) (7)

xµ = ∑ λ µν xν ′ ν xν = ∑ λ µν xµ ′ µ ∑ λµν λµλ = δ νλ µ Then,

∂ ∂ ∂xν ∂ =∑ = ∑ λ µν ∂x µ ∂xν ′ ′ ν ∂xν ∂xµ ν ∂2 ∂ = ∑ λ µν 2 ∂x µ ∂xν ′ ν

Therefore,

∂2 ∂ ∂ ∑ ∂x ′ 2 = ∑∑∑ λµν λµλ ∂x ∂x µ ν λ µ µ ν λ = ∑∑ δ νλ ν λ

(8)

∑ λ µλ λ ∂ ∂ ∂ = ∑∑ λ µν λ µλ ∂x λ ∂xν ∂xλ ν λ

(9)

∂ ∂ ∂xν ∂xλ

=∑

∂2 2 ∂x λ

(10)

Since µ and λ are dummy indices, we see that the operator Lorentz transformation. So we have

∂ 2ψ ( x ′ , ict ′ ) ∑ ∂x ′ 2 = 0 µ µ

∑∂

2

2 ∂xµ is invariant under a

(11)

This equation means that the function ψ taken at the transformed point (x′,ict′) satisfies the same equation as the original function ψ (x,ict) and therefore the equation is invariant. In a Galilean transformation, the coordinates become x ′ = x − vx t y ′ = y − vy t z ′ = z − vz t t′ = t

(12)

Using these relations, we have

464

CHAPTER 14

∂ ∂ ∂x ∂ ∂ t ∂ 1 ∂ = + = − ∂ x ′ ∂x ∂ x ′ ∂t ∂ x ′ ∂x v x ∂ t 1 ∂ ∂ ∂ = − ∂ y ′ ∂y v y ∂t 1 ∂ ∂ ∂ = − ∂z ′ ∂y v z ∂t ∂ ∂ = ∂t ′ ∂t Therefore,

∂2 1 ∂2 1 ∂2 1 1 1 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 + + − 2 = 2 + 2 + 2 − 2 2+ 2 + 2 + 2 2 ∂ x ′ 2 ∂y ′ 2 ∂ z ′ 2 c ∂ t ′ 2 ∂x ∂y ∂z c ∂t v x v y v z ∂t 1 ∂ 1 ∂ 1 ∂ − 2 + + v x ∂x ∂t v y ∂y ∂t v z ∂ z ∂ t

2 2 2

(13)

(14)

This means that the function ψ (x′,ict′) does not satisfy the same form of equation as does ψ ( x , ict) , and the equation is not invariant under a Galilean transformation.

14-4. In the K system the rod is at rest with its ends at x1 and x2 . The K′ system moves with a velocity v (along the x axis) relative to K.

K′

K

x1

x2

If the observer measures the time for the ends of the rod to pass over a fixed point in the K′ system, we have

v t1 − 2 x1 c v 1− 2 c v 1 t2 = ′ t − 2 x2 2 2 c v 1− 2 c t1 = ′ 1

2

(1)

where t1 and t2 are measured in the K′ system. From (1), we have ′ ′

THE SPECIAL THEORY OF RELATIVITY

465

t1 − t2 = ′ ′

v ( t1 − t2 ) − c 2 ( x1 − x2 ) v 1− 2 c 1

2

(2)

We also have x1 − x2 = v ( t1 − t2 ) = v ( t1 − t2 ) = ′ ′ ′ Multiplying (2) by v and using (3), (4), and (5), we obtain the FitzGerald-Lorentz contraction:

′= 1− v2 c2 (6)

(3) (4) (5)

14-5. The “apparent shape” of the cube is that shape which would be recorded at a certain instant by the eye or by a camera (with an infinitesimally short shutter speed!). That is, we must find the positions that the various points of the cube occupy such that light emitted from these points arrives simultaneously at the eye of the observer. Those parts of the cube that are farther from the observer must then emit light earlier than those parts that are closer to the observer. An observer, looking directly at a cube at rest, would see just the front face, i.e., a square.

When in motion, the edges of the cube are distorted, as indicated in the figures below, where the observer is assumed to be on the line passing through the center of the cube. We also note that the face of the cube in (a) is actually bowed toward the observer (i.e., the face appears convex), and conversely in (b).

(a)

Cube moving toward the observer.

(a)

Cube moving away from the observer.

466 14-6.

K

CHAPTER 14

K′ v

x1

x2

We transform the time t at the points x1 and x2 in the K system into the K′ system. Then,

vx t1 = γ t − 21 ′ c vx t2 = γ t − 22 ′ c

From these equations, we have

∆t ′ = t1 − t2 = −γ v ′ ′

(1)

( x1 − x2 ) = −γ v∆x c 2

1 c2

(2)

14-7.

K

K′ v x

Suppose the origin of the K′ system is at a distance x from the origin of the K system after a time t measured in the K system. When the observer sees the clock in the K′ system at that time, he actually sees the clock as it was located at an earlier time because it takes a certain time for a light signal to travel to 0. Suppose we see the clock when it is a distance from the origin of the K system and the time is t1 in K and t′ in K′. Then we have 1

v t1 = γ t1 − 2 ′ c c ( t − t1 ) = tv = x t1v =

We eliminate , t1 , and x from these equations and we find

(1)

THE SPECIAL THEORY OF RELATIVITY

467

v t1 = γ 1 − t ′ c

This is the time the observer reads by means of a telescope.

14-8.

(2)

The velocity of a point on the surface of the Earth at the equator is

v=

2π Re

τ

=

2π × 6.38 × 108 cm 8.64 × 10 sec

4

(

)

(1)

= 4.65 × 10 4 cm/sec which gives v 4.65 × 10 4 cm/sec = = 1.55 × 10 −6 c 3 × 1010 cm/sec 1 ≅ ∆t 1 + β 2 2 1− β

2

β=

(2)

According to Eq. (14.20), the relationship between the polar and equatorial time intervals is ∆t ′ = ∆t (3)

so that the accumulated time difference is

∆ = ∆t ′ − ∆t = Supplying the values, we find ∆= Thus, ∆ = 0.0038 sec (6) 1 × 1.55 × 10 −6 × 3.156 × 107 sec/yr × 10 2 yr 2 1 2 β ∆t 2 (4)

(

) (

) (

)

(5)

14-9. w dm′ m + dm v + dv

The unsurprising part of the solution to the problem of the relativistic rocket requires that we apply conservation of momentum, as was done for the nonrelativistic case. The surprising, and key, part of the solution is that we not assume the mass of the ejected fuel is the same as the mass lost from the rocket. Hence p = γ mv = (γ + dγ ) ( m + dm)( v + dv ) + γ w dm′ w where –dm is the mass lost from the rocket, dm′ is the mass of the ejected fuel, w ≡ ( v − V ) 1 − vV c 2 is the velocity of the exhaust with respect to the inertial frame, and (1)

(

)

γw ≡1

1 − w 2 c 2 . One can easily calculate dγ = γ 3 β dβ , ad after some algebra one obtains

468

CHAPTER 14

γ 2 m dv + v dm +

γ w dm′ w γ

(2)

where we of course keep infinitesimals only to first order. The additional unknown dm′ is unalarming because of another conservation law

E = γ mc 2 = (γ + dγ ) ( m + dm) c 2 + γ w dm′ c 2

Subsequent substitution of dm′ into (2) gives, in one of its many intermediate forms

(3)

γ 2 m dv 1 −

βw

+ dm ( v − w ) = 0 c

(4)

and will finally come to its desired form after dividing by dt

m

dv dm +V 1− β2 = 0 dt dt

(

)

(5)

The quantity dt can be measured in any inertial frame, but would presumably only make sense for the particular one in which we measure v. Interestingly, it is not important for the ejected fuel to have an especially large kinetic energy but rather that it be near light speed, a nontrivial distinction. For such a case, a rocket can reach 0.6c by ejecting half its mass.

14-10.

From Eq. (14.14)

x1 = γ ( x1 − vt ) ′ v t ′ = γ t − 2 x1 c

Solving (1) for x1 and substituting into (2) gives

(1) (2)

v x′ t ′ = γ t − 2 1 + vt c γ t′ + v v2 t x1 = γ t − γ 2 t = ′ 2 γ c c v t = γ t ′ + 2 x1 ′ c

Solving (2) for t and substituting into (1) gives

t′ v x1 = γ x1 − v + 2 x1 ′ γ c or x1 = γ ( x1 + vt ′ ) ′

THE SPECIAL THEORY OF RELATIVITY

469

14-11.

θ

x1

From example 14.1 we know that, to an observer in motion relative to an object, the dimensions of objects are contracted by a factor of component of the stick will be 1 − v 2 c 2 in the direction of motion. Thus, the x′ 1 cos θ 1 − v 2 c 2 while the perpendicular component will be unchanged: sin θ So, to the observer in K′, the length and orientation of the stick are

′ = sin 2 θ + 1 − v 2 c 2 cos 2 θ

(

)

12

θ ′ = tan −1 or

cos θ 1 − v 2 c 2 sin θ

cos 2 θ ′ = sin 2 θ + γ2 tan θ ′ = γ tan θ

12

14-12.

The ground observer measures the speed to be

v=

100 m = 2.5 × 108 m/s .4 µsec

The length between the markers as measured by the racer is

′=

1 − v2 c2

2

2.5 = 100 m 1 − = 55.3 meters 3 The time measured in the racer’s frame is given by

470

CHAPTER 14

v t ′ = γ t − 2 x1 c 2.5 × 10 8 m/s (100 m ) .4 µsec − c2 1 − ( 2.5 3)

2

(

)

=

= .22 µsec The speed observed by the racer is v= ′ = = 2.5 × 108 m/s t′ t

14-13.

∆t ′ = γ ∆t ∆t = 1.5 µs

γ = ( 1 − 0.9992 )

Therefore ∆t ′ 34 µs .

−1 2

22.4

14-14.

K K′

v source receiver

In K, the energy and momentum of each photon emitted are

E = hν 0

and

p=

hν 0 c

Using Eq. (14.92) to transform to K′: hν E′ = hν = γ ( E − vp1 ) ; p1 = − 0 c v = γ hν 0 + hν 0 c

THE SPECIAL THEORY OF RELATIVITY

471

So v ν = ν0 γ 1 + c = ν0 = ν0

1+ β 1− β

2

1+ β 1− β

which agrees with Eq. (14.31).

14-15.

From Eq. (14.33)

ν=

Since λ = c ν

1− β 1+ β

ν0

λ0 = or 1− β 1+ β

λ

λ=

With λ 0 = 656.3 nm and β =

1− β 1+ β

λ0

4 × 10 4 , λ = 656.4 nm. 3 × 10 8

So the shift is 0.1 nm toward the red (longer wavelength).

14-16.

K′ v

K

θ′

star

θ Earth

Consider a photon sent from the star to the Earth. From Eq. (14.92)

E′ = γ ( E − vp1 ) also E = γ ( E′ + vp1 ) ′

472

CHAPTER 14

Now

E = hν , E′ = hν 0 ,

Substituting yields

p1 = −

hν cos θ , c

p1 = − ′

hν 0 cos θ ′ c

ν 0 = ν γ (1 + β cos θ ) and ν = γ ν 0 (1 − β cos θ ′ )

Thus

(1 + β cos θ )(1 − β cos θ ′ ) = γ −2

1 + β cos θ − β cos θ ′ − β 2 cos θ cos θ ′ = 1 − β 2 cos θ − cos θ ′ − β cos θ cos θ ′ = − β Solving for cos θ yields cos θ = where cos θ ′ − β 1 − β cos θ ′

β =v c θ = angle in earth’s frame θ ′ = angle in star’s frame

14-17.

From Eq. (14.33)

ν=

Since

1− β ν0 1+ β

ν=c λ, λ=

We have λ = 1.5 λ 0 . This gives β = or 5 13 1+ β λ0 1− β

v = 1.2 × 10 8 m/sec

THE SPECIAL THEORY OF RELATIVITY

473

14-18.

K′ v source

K

θ

light

observer

Proceeding as in example 14.11, we treat the light as a photon of energy hν. In K ′ : E′ = hν 0 , p ′ = In K :

hν 0 c

E = hν = γ ( E′ + ν p1 ) hν 0 c

1+ β 1− β

For the source approaching the observer at an early time we have

p1 =

Thus

ν = γ ν0 + ν0 = ν0

v c

For the source receding from the observer (at a much later time) we have

p1 = − and hν 0 c

1− β 1+ β

ν = ν0

So

ν = ν0 ν = ν0

1+ β source approaching observer 1− β 1− β source receding from observer 1+ β

474 14-19.

K

CHAPTER 14

source

K′

θ v observer

Proceeding as in the previous problem, we have In K ′ : E′ = hν

p1 = − ′

In K : So

hν hν cos θ = − c c

βr β + βt2

2 r

E = γ ( E′ + ν p1 ) = hν 0 ′

hν 0 = or

hν β r hν − c β r2 + βt2 2 2 c β2 + β2 1 − β r − βt r t 1

ν0 =

ν (1 − β r )

1 − β r2 − βt2

1 − β r2 − β t2 ν λ0 = = ν0 λ 1 − βr For λ > λ 0 , we have

(1 − β r )2 > 1 − βr2 − βt2 βt2 > 2β r − 2β r2 βt2 > 2β r (1 − β r )

14-20.

As measured by observers on Earth, the entire trip takes 4 lightyears 80 2 = 3 years 0.3 c

The people on earth age the astronaut ages

80 years. The astronaut’s clock is ticking slower by a factor of γ. Thus, 3

THE SPECIAL THEORY OF RELATIVITY

475

80 80 1 − 0.32 = 0.95 years 3 3 So Those on Earth age 26.7 years. The astronaut ages 25.4 years.

14-21.

1 − ( −2β ) β mv v 2 d 0 +v F= = m0 32 dt 1 − β 2 1− β2 1 − β2

(

)

(1)

v vββ = m0 + 32 1 − β2 1 − β2

(

)

If we take v = v1e1 (this does not mean v2 = v3 = 0 ), we have v v v1 1 1 v1 m0 c c = F1 = m0 + 2 2 32 2 1− β 1− β 1− β

(

)

(

)

32

v1 = m v1

(2)

F2 =

m0 1 − β2 m0 1 − β2

v2 = mt v2

(3)

F3 =

v3 = mt v3

(4)

14-22.

The total energy output of the sun is dE = 1.4 × 10 3 W ⋅ m −2 × 4π R 2 dt

(

)

(1)

where R = 1.50 × 1011 m is the mean radius of the Earth’s orbit around the sun. Therefore, dE dt The corresponding rate of mass decrease is dm 1 dE = dt c 2 dt 4.4 × 109 kg ⋅ s −1 (3) 3.96 × 10 26 W (2)

The mass of the sun is approximately 1.99 × 10 30 kg , so this rate of mass decrease can continue for a time

476

1.99 × 10 30 yr 4.4 × 109 kg ⋅ s −1

CHAPTER 14

T=

1.4 × 1013 yr

(4)

Actually, the lifetime of the sun is limited by other factors and the sun is expected to expire about 4.5 × 10 9 years from now.

14-23.

From Eq. (14.67)

2 p 2 c 2 = E2 − E0 2 = ( E0 + T ) − E0 2

= 2E0 T + T 2 p 2 c 2 = 2T mc 2 + T 2

14-24. The minimum energy will occur when the four particles are all at rest in the center of the mass system after the collision.

Conservation of energy gives (in the CM system) 2Ep = 4 mp c 2 or Ep ,CM = 2mp c 2 = 2E0 which implies γ = 2 or β = 3 2 To find the energy required in the lab system (one proton at rest initially), we transform back to the lab E = γ ( E′ + vp1 ) ′ (1)

The velocity of K′(CM) with respect to K(lab) is just the velocity of the proton in the K′ system. So u = v. Then vp1 = v ( pCM ) = v (γ mu) = γ mv 2 = γ mc 2 β 2 ′ Since γ = 2, β = 3 2 , vp1 = ′ Substituting into (1) 3 E0 2

THE SPECIAL THEORY OF RELATIVITY

477

3 7 Elab = γ 2E0 + E0 = 2 E0 = 7E0 2 2

The minimum proton energy in the lab system is 7 mp c 2 , of which 6 mp c 2 is kinetic energy.

14-25.

Let B = B0 z v = vx i + v y j

Then i qv × B = q v x 0 j vy 0 k

0 B0

= q vy B0 i − vx B0 j

F = qv × B =

d d ( p) = γ m ( v ) gives dt dt qB d v = 0 v y i − vx j dt γm

(

)

Define ω ≡ q B0 γ m Thus vx = ω vy and vy = −ω vx

or vx = ω vy = −ω 2 vx

and vy = −ω vx = −ω 2 vy

So vx = A cos ω t + B sin ω t vy = C cos ω t + D sin ω t

Take vx ( 0 ) = v , vy ( 0 ) = 0 . Then A = v, C = 0. Then vx ( 0 ) = ω vy ( 0 ) = 0 vy ( 0 ) = −ω vx ( 0 ) = −ω v

→ B = 0, D = − v

478

Thus

CHAPTER 14

v = i v cos ω t − j v sin ω t

Then

r=i

The path is a circle of radius v v

ω

sin ω t + j

v

ω

cos ω t

ω r= v q B0 γ m

=

γ mv q B0

=

p q B0

From problem 14-22

T2 p = 2Tm + 2 c

12

So

T2 2Tm + 2 c r= q B0

12

14-26. below

Suppose a photon traveling in the x-direction is converted into an e − and e + as shown e+ θ θ

e– before

after

Cons. of energy gives pp c = 2Ee where pp = momentum of the photon Ee = energy of e + = energy of e − Cons. of px gives pp = 2 pe cos θ Dividing gives

(p

e

= momentum of e + , e −

)

THE SPECIAL THEORY OF RELATIVITY

479 pp c pp Ee pe cos θ

=c=

or

2 pe c 2 cos 2 θ = Ee2 2 But Ee2 > pe c 2 , so (1) cannot be satisfied for cos 2 θ ≤ 1 .

(1)

An isolated photon cannot be converted into an electron-positron pair. This result can also be seen by transforming to a frame where px = 0 after the collision. But, before the collision, px = pp c ≠ 0 in any frame moving along the x-axis. So, without another object nearby, momentum cannot be conserved; thus, the process cannot take place.

14-27. The minimum energy required occurs when the p and p are at rest after the collision. By conservation of energy

2Ee = 2 ( 938 MeV ) Ee = 938 MeV = T + E0 Since Ee = 0.5 MeV , Te+ = Te− = 937.5 MeV

14-28.

Tclassical =

1 mv 2 2

Trel = (γ − 1) mc 2 ≥ Tclassical

We desire

Trel − Tclassical ≤ 0.01 Trel

1 mv 2 2 1− ≤ 0.01 (γ − 1) mc2 1 2 v 2 ≥ 0.99 (γ − 1) c 2

β2 γ −1

Putting γ = 1 − β 2

≥ 1.98

(

)

−1 2

and solving gives

480 v ≤ 0.115 c

CHAPTER 14

The classical kinetic energy will be within 1% of the correct value for 0 ≤ v ≤ 3.5 × 107 m/sec, independent of mass.

14-29.

E = γ E0 E = 30 × 109 eV E0

For

0.51 × 106 eV, 5.88 × 10 4 1 1 − β2 1 2γ

2

γ γ=

or β = 1 − γ −2

= 1 − 1.4 × 10 −10

(

)

12

β 1−

v = 1 − 1.4 × 10 −10 c = 0.99999999986 c

(

)

14-30. A neutron at rest has an energy of 939.6 MeV. Subtracting the rest energies of the proton (938.3 MeV) and the electron (0.5 MeV) leaves 0.8 MeV.

Other than rest energies 0.8 MeV is available.

14-31.

0.98c θ

θ

Conservation of energy gives Eπ = 2Ep where Ep = energy of each photon (Cons. of py implies that the photons have the same energy).

THE SPECIAL THEORY OF RELATIVITY

481

Thus

γ E0 = 2Ep

Ep =

γ E0

2

=

135 MeV 2 1 − 0.98 2

= 339 MeV

The energy of each photon is 339 MeV. Conservation of px gives

γ mv = 2 pp cos θ where pp = momentum of each photon cos θ =

(135 Mev/c ) ( 0.98 c)

2

2 1 − 0.98 2 ( 339 MeV/c )

= 0.98

θ = cos −1 0.98 = 11.3°

14-32.

From Eq. (14.67) we have

2 E2 − E0 = p 2 c 2

With E = E0 + T , this reduces to 2E0 T + T 2 = p 2 c 2 Using the quadratic formula (taking the + root since T ≥ 0) gives

2 T = E0 + p 2 c 2 − E0

Substituting pc = 1000 MeV E0 ( electron ) = 0.5 MeV E0 ( proton ) = 938 MeV gives

Telectron = 999.5 MeV Tproton = 433 MeV

482

14-33.

e 120˚ n ν before after p 120˚

CHAPTER 14

Conservation of py gives pe sin 60° = pν sin 60° or pe = pν Conservation of px gives pp = pe cos 60° + pν cos 60° = pe So pe = pp = pν ≡ p Conservation of energy gives E0 n = Ee + Ep + Eν

2 2 E0 n = E0 e + p 2 c 2 + E0 p + p 2 c 2 + pc

(1)

Substituting

E0 n = 939.6 MeV E0 p = 938.3 MeV E0 e = 0.5 MeV

and solving for pc gives p = 0.554 MeV/c pp = pe = pν = 0.554 MeV/c

Substituting into

T = E − E0

2 = E0 + p 2 c 2 − E0

gives ( E0ν = 0 )

Tν = 0.554 MeV Tp = 2 × 10 −4 MeV, or 200 eV Te = 0.25 MeV

THE SPECIAL THEORY OF RELATIVITY

483

∆s′ 2 = − c 2t ′ 2 + x1 2 + x2 2 + x3 2 ′ ′ ′

14-34.

Using the Lorentz transformation this becomes

∆s′ 2 = −c 2t 2 −

2 v 2 x1 + 2 x1vt x 2 + v 2t 2 − 2x1vt 2 2 c2 + 1 + x2 + x3 1 − v2 c2 1 − v2 c2

2 2 v 2 x1 2 2 v 2 2 x1 − c 2 − c t − c 2 t + x2 + x2 = 2 3 2 2 1− v c 2 2 2 = − c 2t 2 + x1 + x2 + x3

So ∆s′ 2 = ∆s2

14-35. Let the frame of Saturn be the unprimed frame, and let the frame of the first spacecraft be the primed frame. From Eq. (14.17a) (switch primed and unprimed variables and change the sign of v)

u1 = Substituting v = 0.9 c

u1 + v ′ u ′v 1 + 12 c

u1 = 0.2 c ′ gives u1 = 0.93 c

14-36.

Since Fµ = d dτ dX µ m dτ and X µ = ( x1 , x2 , x3 , ict )

we have F1 = d dτ d2 x dx1 = m 21 m dτ dτ d 2 x2 dτ 2 F3 = m d 2 x3 dτ 2

F2 = m F4 = d dτ

d 2t d ( ict ) = icm 2 m dτ dτ

484

Thus

F1′ = m d 2 x1 d2 − m 2 γ ( x1 − vt ) dτ 2 dτ d 2 x1 d 2t ′ = γ mv 2 = γ ( F1 + iβ F4 ) dτ 2 dτ

CHAPTER 14

=γm F2′ = m

d 2 x2 d 2 x2 ′ =m = F2 ; F3′ = F3 dτ 2 dτ 2 d dτ 2 vx1 γ t − c 2

F4′ = icm

= γ icm

d 2t d2 x − γ iβ m 21 dτ 2 dτ

= γ ( F4 − iβ F1 ) Thus the required transformation equations are shown.

14-37.

From the Lagrangian L = mc 2 1 − 1 − β 2 −

(

)

1 2 kx 2

(1)

we compute ∂L = − kx ∂x ∂ L ∂β ∂ L β = = mc ∂v ∂ v ∂β 1 − β2 Then, from (2) and (3), the Lagrange equation of motion is d dt from which mcβ + kx = 0 1 − β2 mcβ (4) (2) (3)

(1 − β )

Using the relation cβ = we can rewrite (4) as

2 32

+ kx = 0

(5)

dv dv dx dv = =v dt dx dt dx

(6)

THE SPECIAL THEORY OF RELATIVITY

485 mc 2 β

2 32

(1 − β )

This is easily integrated to give mc 2

dβ + kx = 0 dx

(7)

1− β where E is the constant of integration.

2

+

1 2 kx = E 2

(8)

β = 0:

The value of E is evaluated for some particular point in phase space, the easiest being x = a; 1 2 ka 2

E = mc 2 + From (8) and (9), mc 2 1− β Eliminating β 2 from (10), we have

2

(9)

+

1 2 1 kx = mc 2 + ka 2 2 2

(10)

β2 = 1 −

m2 c 4 2 2 1 mc + k a 2 − x 2 2

(

)

= k a2 − x 2

(

)

2 k 2 2 mc + 4 a − x 2 2 k 2 mc + a − x 2 2

(

)

(

)

(11)

and, therefore, 1 dx β= = c dt k a2 − x 2

(

)

mc 2 + k a 2 − x 2 4

mc + k a − x

2 2

(

(

2

)

)

2

(12)

The period will then be four times the integral of dt = dt(x) from x = 0 to x = a: m τ=4 k

∫

0

a

k 2 2 1 + 2mc 2 a − x dx k 2 2 2 2 a − x 1+ a −x 4 mc 2

(

)

(

)

(13)

Since x varies between 0 and a, the variable x a takes on values in the interval 0 to 1, and therefore, we can define sin φ = from which x a

(14)

486 a2 − x 2 a

CHAPTER 14

cos φ = and

(15)

dx = a 2 − x 2 dφ

(16)

We also define the dimensionless parameter,

κ≡

Using (14) – (17), (13) transforms into

a 2

k mc 2

(17)

τ=

Since ka 2 mc 2

2a κc

π 2

∫

0

(1 + 2κ

2

cos 2 φ

1 + κ 2 cos 2 φ

) dφ

(18)

1 for the weakly relativistic case, we can expand the integrand of (18) in a

series of powers of κ :

(1 + 2κ cos φ ) ≅ 1 + 2κ ( (1 + κ cos φ )

2 2 2 2 12

2

κ2 cos 2 φ 1 − cos 2 φ 2

)

1 ≅ 1 + 2 − κ 2 cos 2 φ 2 3 = 1 + κ 2 cos 2 φ 2 Substitution of (19) into (18) yields

2a τ≅ κc = π 2

(19)

∫ 1 + 2 κ

0

3

2

cos 2 φ dφ π 2

aπ 3κ a + κ c 2c

1 φ + 2 sin 2φ 0

(20)

Evaluating (20) and substituting the expression for κ from (17), we obtain

τ = 2π or, m 3π a 2 + k 8c 2

k m

(21)

τ = τ 0 1 +

3 ka 2 16 mc 2

(22)

THE SPECIAL THEORY OF RELATIVITY

487

14-38.

F=

dp d = (γ mu) dt dt d (γ u) dt d u dt 1 − u 2 c 2 (for m = constant)

=m =m

2 2 1− u c = m

(

)

12

u − u − 2 1 − u2 c 2 c 1 − u2 c2

(

( )

)

−1 2

du dt

= m 1 − u2 c 2 Thus F=m

(

)

−3 2

du dt

du 1 − u2 c 2 dt

(

)

−3 2

14-39.

The kinetic energy is

2 T = p 2 c 2 + m0 c 4 − m0 c 2

(1)

For a momentum of 100 MeV/c, Tproton = 10 4 + ( 931) − 931 ≅ 936 − 931 = 5 MeV

2

(2) (3)

Telectron = 10 4 + ( 0.51) − 0.51 ≅ 100 − 0.5 = 99.5 MeV

2

In order to obtain γ and β, we use the relation E = mc 2 = γ m0 c 2 = so that m0 c 2 1 − β2 (4)

γ= and E m0 c 2

(5)

β = 1− γ electron =

1

γ2

(6) (7)

100 ≅ 200 0.51

488

CHAPTER 14

βelectron = 1 − 200 ≅ 0.999988 1 This is a relativistic velocity.

2

(8)

γ proton =

936 ≅ 1.0054 931 1

2

(9)

β proton = 1 − 1.0053 ≅ 0.1 This is a nonrelativistic velocity.

14-40.

(10)

If we write the velocity components of the center-of-mass system as v j , the v j Eα pα , j = γ pα , j − 2 ′ c

transformation of pα , j into the center-of-mass system becomes (1) = 0 must be satisfied, we have

where γ =

1 1− v2 j c2

. Since in the center-of-mass system,

′ ∑ pα α ,j

′ ∑ pα α ,j

v j Eα = ∑ γ pα , j − 2 = 0 c α

(2)

or,

vj c

∑ pα c = α ∑ Eα

,j

(3)

α

14-41.

We want to compute

T1 E1 − m0 c 2 = T0 E0 − m0 c 2

(1)

where T and E represent the kinetic and total energy in the laboratory system, respectively, the subscripts 0 and 1 indicate the initial and final states, and m0 is the rest mass of the incident particle. The expression for E0 in terms of γ 1 is

E0 − m0 c 2γ 1

(2)

E1 can be related to E′ (total energy of particle 1 in the center of momentum reference frame 1 after the collision) through the Lorentz transformation [cf. Eq. (14.92)] (remembering that for the inverse transformation we switch the primed and unprimed variables and change the sign of v):

THE SPECIAL THEORY OF RELATIVITY

489

E1 = γ 1 ( E1 + cβ1 p1 cos θ ) ′ ′ ′ ′

(3)

where p1 = m0 cβ1γ 1 and E1 = m0 c 2γ 1 : ′ ′ ′ ′

E1 = m0 c 2γ 1 2 1 + β1 2 cos θ ′ ′

Then, from (1), (2), and (4),

(

)

(4)

′ ′ ′ T1 γ 1 2 + γ 1 2 β1 2 cos θ − 1 = T0 γ1 −1

For the case of collision between two particles of equal mass, we have, from Eq. (14.127), 1+ γ1 2

(5)

γ 12 = ′ and, consequently,

(6)

γ 1 2 β1 2 = γ 1 2 − 1 = ′ ′ ′

Thus, with the help of (6) and (7), (5) becomes

γ1 −1

2

(7)

T1 γ 1 − 1 + (γ 1 − 1) cos θ = T0 2 (γ 1 − 1) = 1 + cos θ 2 (8)

We must now relate the scattering angle θ in the center of momentum system to the angle ψ in the lab system. Squaring Eq. (14.128), which is valid only for m1 = m2 , we obtain an equation quadratic in cos θ. Solving for cos θ in terms of tan 2 ψ , we obtain cos θ = −

γ1 +1

2 2

tan 2 ψ ± 1 tan ψ

2

1+

γ1 +1

(9)

One of the roots given in (9) corresponds to θ = π, i.e., the incident particle reverses its path and is projected back along the incident direction. Substitution of the other root into (8) gives 2 cos 2 ψ T1 1 = = T0 1 + γ 1 + 1 tan 2 ψ 2 cos 2 ψ + (γ 1 + 1) sin 2 ψ 2 An elementary manipulation with the denominator of (10), namely, (10)

490

2 cos 2 ψ + (γ 1 + 1) sin 2 ψ = 2 cos 2 ψ + γ 1 1 − cos 2 ψ + sin 2 ψ

= γ 1 + sin 2 ψ + cos 2 ψ − γ 1 cos 2 ψ + cos 2 ψ = γ 1 + 1 − γ 1 cos 2 ψ + cos 2 ψ = (γ 1 + 1) − (γ 1 − 1) cos 2 ψ

CHAPTER 14

(

)

(11)

provides us with the desired result: 2 cos 2 ψ T1 = T0 (γ 1 + 1) − (γ 1 − 1) cos 2 ψ Notice that the shape of the curve changes when T1 > m0 c 2 , i.e., when γ 1 > 2 .

T1 T0

(12)

1.0

T1 = 0.1 GeV

0.8

T1 = 1 GeV

0.6

T1 = 10 GeV

0.4

0.2 0

0˚

30˚

60˚

90˚

ψ

14-42. y γmec2 x

hν

φ θ

hν′

From conservation of energy, we have hν + me c 2 = γ me c 2 + hν ′

(1)

Momentum conservation along the x axis gives hν hν ′ cos θ + γ me v cos φ = c c

(2)

Momentum conservation along the y axis gives

γ me v sin φ =

hν ′ sin θ c

(3)

THE SPECIAL THEORY OF RELATIVITY

491

In order to eliminate φ, we use (2) and (3) to obtain cos φ = 1 hν hν ′ c − c cos θ γ me v hν ′ sin φ = sin θ γ me v

(4)

Then, 1 cos φ + sin φ = 1 = 2 2 2 γ me v

2 2

hν 2 hν ′ 2 hν hν ′ cos θ − 2 + + c c c c

(5)

Since γ =

1 1− v c2

2

and v =

c

γ

γ 2 − 1 we have γ 2 v 2 = c 2 ( γ 2 − 1)

(6)

Substituting γ from (1) into (6), we have

γ 2v2 =

2h h2 ( ν − ν ′ ) + 2 2 (ν − ν ′ ) 2 me me c

(7)

From (5) and (7), we can find the equation for ν′: h2 hν hν ′ hν hν ′ 2 cos θ = 2 hme (ν − ν ′ ) + 2 (ν − ν ′ ) + −2 c c c c c

2 2

(8)

or,

2me c 2 2me c 2 ν + 2ν (1 − cos θ ) ν ′ = h h

(9)

Then,

1 ν ν′ = 1 + hν (1 − cos θ ) me c 2

(10)

or,

E 1 − cos θ ) E′ = E 1 + 2 ( me c

−1

(11)

The kinetic energy of the electron is

492

1 T = γ me c 2 − me c 2 = hν − hν ′ = E 1 − E 1 − cos θ ) 1+ 2 ( me c

CHAPTER 14

T=

1 − cos θ E2 2 me c 1 + E 1 − cos θ ( ) me c 2

(12)

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...1 KHYBER PAKHTUNKHWA PUBLIC SERVICE COMMISSION SYLLABUS FOR PROVINCIAL MANAGEMENT SERVICE COMPETITIVE EXAMINATION The Syllabus and standard for the Competitive Examination for the Provincial Management Service shall be as under : 1. The Examination shall include compulsory and optional subjects, and every candidate shall take all the compulsory subjects and opt for three of the optional subjects carrying 600 marks in all but not more than 200 marks from a single group. 2. A candidate shall answer the language papers in the language concerned. The question paper in Islamiat is to be answered in Urdu or English. All other papers must be answered in English. Violation of this instruction shall incur cancellation of the concerned paper(s) and consequently award of Zero. 3. The compulsory and optional subjects and maximum marks fixed for each subject shall be as below: Sr. No. 1 2 3 COMPULSORY SUBJECTS Subjects English (Précis & Composition) English Essay General Knowledge (a) Current Affairs 100 (b) Every Day Science 100 (c) Pakistan Affairs 100 Islamiat Viva Voce Total Maximum Marks 100 100 300 100 300 900 600 120 4 5 Qualifying marks in the aggregate of written papers: Qualifying marks in the Viva Voce: The non-Muslim candidates will have the option to take Islamiat as a compulsory subject or otherwise Pakistan Affairs (General Knowledge PaperIII) will be treated of 200 marks and counted in lieu of Islamiat. A candidate who fails to appear in any of the......

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