ME 2330 By K. L. Ting 5. Tangential and Normal components Let R(s) = x(s) i + y(s) j represent a curve and s is the independent parameter representing the arc length of the curve measured from a reference point. At s and s’, a particle moves from P to P’ and the displacement is dR (Fig. 1). If P and P’ is infinitesimally separated, i.e. ds= s’-s → 0, l dR l = ds, dR/ds = ut.

ut is a unit vector (called unit tangent) tangent to the curve at P (Fig. 2). At P and P’, let ut and ut‘ be the unit vectors in the tangent direction. The lines normal to ut and ut‘ intersect at O (Fig. 2). With ds → 0, point O is the center of curvature and ds/dθ = ρ where ρ, called radius of curvature, is the length OP. In Fig. 3, one may observe that dut is normal to ut and since ut and ut’ are unit vectors, dut/dθ is a unit vector, designated as un. un is the unit vector (called unit normal) in the direction from P to O. That is, dut/ds =(dut/dθ)(dθ/ds) = (1/ρ) un or dut/ds = k un

where k (= 1/ρ = dθ/ds) is the curvature at P.

Fig. 1

Fig. 2

Fig. 3

With a Cartesian coordinate system, the radius of curvature at (x, y) can be calculated as

ρ = [1 + (dy/dx)2]3/2/ ld2y/dx2l or ρ=

where the dots on x and y represent the first and second derivatives with respect to any independent parameter, such as t or s. In a circle, the center of curvature is the center of the circle and the radius of curvature is the radius of the circle.

Let a particle move along a path R(s). Differentiating R with respect to time yields V = dR/dt = (dR/ds)(ds/dt) = ut where it again yields a = dV/dt = where = ut )dt = ut + (dut /ds)(ds/dt) = ut + ( 2/ρ)un. at represents the tangential acceleration, which is caused by the is the tangential velocity or the speed of the particle. Differentiating

change of the speed, and 2/ρ = v2/ρ the normal acceleration of the particle. The normal acceleration, which is caused by the change of traveling direction, is always toward the center of curvature. The above equations can be written as V = v ut a = at ut + (v2/ρ)un. v is the speed of the particle and its direction is always tangent to the path. at is the tangential acceleration and (v2/ρ) is the normal acceleration. v2/ρ = 0 if the particle travels on a straight path, i.e. ρ = 0. In a circular motion, (v2/ρ) is called centrifugal acceleration.

Example: A skier skies on a path y = x2/20. At x = 10 m, he has a speed of 6 m/s, which increasing at 2 m/s2. What is the acceleration? Solution: At x = 10, y = 100/20 = 5. The skier skies on a curve with v = 6 m/s and at = 2 m/s2. The acceleration is a = at ut + (v2/ρ)un. Since y = x2/20, at x = 10, dy/dx = x/10 = 1, d2y/dx2 = 0.1 and ρ = [1 + (dy/dx)2]3/2/(d2y/dx2) = (1+12)3/2/(0.1) = 28.28 m a = at ut + (v2/ρ)un = 2 ut + (62/28.8)un = 2 ut + 1.273un φ = Atan2(1.273, 2) = 57.5°

Example: A race car travels around a horizontal circular track of radius 300 ft. It starts from rest and accelerate at a constant rate of 7 ft/s2 . What is the velocity and acceleration at t = 15 s? Solution: Tangential acceleration 7 ft/s2. Tangential velocity: v = v0 + at = 0 + 7t = 7(15) = 105 a = at ut + (v2/ρ)un = 7 ut + (1052/300)un = 7 ut + 36.75un 6. Complex numbers in Polar coordinates Real numbers: a2 ≥ 0 ft/s

All real number may be expressed on the real axis:

Imaginary numbers:

a2 ≤ 0

Then:

All imaginary number may be expressed ona the imaginary axis Example:

16i – 5i = (16 – 5)i = 11i (3i)(4i) = (3·4)(i·i) = (12)(i2) = (12)(–1) = –12 (i)(2i)(–3i) = (2 · –3)(i 3) = 6i Complex numbers: A complex number is an ordered pair of real numbers, expressed as a + bi where a and b are real numbers and i2 = -1. a and b are the real and imaginary parts of the complex number. A complex number may be used to express a point position or a vector on a plane (Fig. 1). The x and y axes in a xy-system are called real and imaginary axes respectively in complex number system. Thus, the position (a, b) of a point in the xy coordinate system is expressed as (a + bi) in complex number (Fig. 2). In other words, the position of a point as well as any vector on a plane can be expressed in complex number, in which the real part is the x-component and the imaginary part the y-component. The operation rules of vectors and complex numbers are different and complex numbers are not applicable in 3D motion. But complex numbers are more convenient to use in planar motion especially with rotation.

Fig. 1

Fig. 2

Fig. 3

A complex number may also be expressed in polar form as (Fig. 3) R = r eiθ, where i2 = -1 and eiθ = cosθ + sinθ i. All angles are measured from the real axis and positive if measured counterclockwise and negative if clockwise. If then and R = r eiθ = x + y i, x = r cosθ, r2 = x2 + y2, y = r sinθ θ = tan-1 (y/x) or in Excel, θ = atan2(x, y).

Atan2(x, y) is a function used in EXCEL and MATLAB. In C/C++, it is written as atan2(y, x). Atan2(x, y) gives the angle in the proper quadrant. Example: R1 = 1 + i, R2 = 1 - i, R3 = -1 + i, R4 = -1 - i. Express them in polar form. Solution: R1 = 1 + i = R2 = 1 - i = R3 = -1 + i = θ3 = R4 = -1 - i = θ4 = , where = atan2(-1, -1) , where θ1 = , where θ2 = , where or θ1 = atan2(1, 1) = . .

or θ2 = atan2(1, -1) =-

Fig. 4 Complex number operation: Generally speaking, real number operation rules have their counterparts in complex numbers except that i2= -1. Complex number operations are available in C/C++, Fortran, Matlab, and spread sheet, such as EXCEL. For any two complex numbers, say A1 = x1 + y1 i = r1 eiθ1 and A2 = x2 + y2 i = r2 eiθ2 Addition and subtraction: A1 ± A2 = (x1 ± x2) + (y2 ± y2) i Or A1 ± A2 =(r1 cosθ1 + r1 sinθ1 i) ± (r2 cosθ2 + r2 sinθ2 i)

= (r1 cosθ1 ± r2 cosθ2 ) + (r1 sinθ1 ± r2 sinθ2 ) i Product: A1 A2 = (x1+y1i) (x2+ y2i) =(x1x2 +y1y2 i2)+(x1y2 i + x2y1 i)=(x1x2 - y1y2) +(x1y2 + x2y1) i Or A1 A2 = (r1 eiθ1)(r2 eiθ2) = (r1r2) ei(θ1+θ2)

Since A1 eiθ2 = (r1 eiθ1)(eiθ2) = (r1) ei(θ1+θ2), A1 eiθ2 represents the rotation operation of A1 through θ2 counterclockwise (Figure 5).

Fig. 5

Fig. 6

If θ2 = 90°, then ei90° = cos 90° + i sin 90° = i. Therefore, eiθ(ei90°) = i eiθ. That is, if eiθ represents a unit vector ur, then ieiθ represents the unit vector uθ, which is in the direction obtained by rotating ur through 90° counterclockwise (Figure 6).

Division:

A1 A2 = (r1 eiθ1)/(r2 eiθ2) = (r1/r2) ei(θ1-θ2) There is no inner or outer product as in vectors.

Differentiation: Let R = x + yi, i2 = -1. Since i is a constant, dR/dt = (dx/dt) + (dy/dt) i d(eθ)/dθ = eθ d(eiθ)/dθ = [d(eu)/du](du/dθ) = eu (i) = i eiθ , where u = iθ and du/dθ = i d(eiθ)/dt = [d(eiθ)/dθ](dθ/dt) = (i eiθ ) = (ieiθ)

Example: Let A = x + yi = aeiα , B = u + v i = beiβ Example: Let R = reθ describe the position of a particle, where r and θ are functions of time. The velocity and acceleration of the particle can be described as V = dR/dt = d(reiθ)/dt = (dr/dt)eiθ + r d(eiθ)/dt = eiθ + r (i eiθ ). Let ur and uθ be the unit vectors along the radial and transverse directions. Since and are the complex numbers representing the unit vectors in the radial and transverse directions,

is the radial component of the velocity. r is the transverse component of the velocity.

Differentiating the above equation yields a = dR2/dt2 = d2(reiθ)/dt2 = d[ eiθ + r (i eiθ )]/dt = d( eiθ)/dt + d[r (i eiθ )]/dt

= = = =[

eiθ + d(eiθ)/dt] + [( eiθ + -r eiθ - r -r (2 ] + [( + (2 (2 eiθ ] eiθ

eiθ+ r eiθ+ r ]

]

is the radial component of the acceleration. is the transverse component of the acceleration.

In vectors, the position, velocity, and acceleration of the particle can be expressed as R = r ur V = ur + r uθ a= -r ur+ (2 uθ

If the particle travels on a circular path, = = 0, then V = dR/dt = d(reiθ)/dt = r (i eiθ ). a = dR2/dt2 = - r Or, in the vector form, V = r uθ a=-r r un ur + uθ = ut + r un + eiθ)

As shown in Fig. 7, in a circular motion, (-eiθ) and eiθ) are the complex number representation of the unit normal (radial) and unit tangent (transverse).

Fig. 7 Some complex number functions used in FORTRAN, C/C++, MATLAB (or Ch), and EXCEL are listed below.
FORTRAN, C/C++ COMPLEX T,P P=CMPLX(x,y) PRINT P x=REAL(P) r=CABS(P) q=ATAN2(y,x) P=CONJ(P) Q=CEXP(P)
P= r * CEXP(CMPLX(0,q))

EXCEL =Complex(x, y)

MATLAB, Ch P = x + i*y

Remarks To declare T and P as complex numbers To form the complex number x + iy To print a complex number(two ordered real numbers, x and y)

=IMREAL(P) =IMABS(P) =IMARGUMENT(P) or = atan2(x,y) =IMCONJUGATE(P) =IMEXP(P)
IMPRODUCT(r, IMEXP(COMPLEX (0,q)) )

x = real(P) y = imag(P) r = abs(P) q = atan2(x,y) P = conj(P) Q = exp(P) P = r * exp(i * q)

To obtain the real part of P To obtain the imaginary part of P To obtain the magnitude(modulus) of P To obtain the argument angle q of P = x + iy To obtain the complex conjugate P of P. To calculate Q = e
T (i*q)

y=AIMAG(P) = IMAGINARY(P)

To calculate P = r*e

*For EXCEL:

IMSUM(_,_), which calculates the complex sum of two complex numbers IMSUB(_,_), which calculates the complex difference of two complex numbers IMPRODUCT(_,_), which calculates the complex product of two complex numbers IMDIV(_,_), which calculates the complex quotient of two complex numbers Exercise: Let A1 = x1 + y1i = r1 eiθ1 and A2 = x2 + y2 i = r2 eiθ2. 1. A1 = 2 – 4i, A2 = -3 -2i. Find r1, α1 and r2, α2. 2. r1 = 2, α1 = 45°, r2 = 4, α2 = 2 rad. Find x1, y1, x2, y2. 3. In the above problem, find A1A2 and A1/A2. Express them in polar form with α in degrees.

Example: A child is on a merry-go-round, which rotates at a constant speed of 60 rpm (cw) while the child moves on the platform at a uniform speed of 1 ft/s toward the center of rotation. What are the velocity and acceleration of the child? r = 4 ft, = =0 = -1 ft/s, = - 60 rpm = -1 rev/s = -2π rad/s

= (4)(-2π) = -8π r 2 r = 4(-2π)2 16π2 = 2(-1)(-2π) = 4π =0 or V = -1 ur - 8 π uθ + 2(4π) eiθ = -16π2 + 8π eiθ

V = eiθ + r (i eiθ ) = -1 eiθ - 8π (i eiθ ) a= or -r + [(2

eiθ = (0 -16π2)

a = -16π2 ur + 8π uθ.

Example 12.18 A rod rotates in the horizontal plane such that θ = t 3 rad and the collar B is sliding outward along OA so that r = 100 t2 mm. t is in seconds. Determine the velocity and acceleration of the collar when t = 1 s.

At t = 1, r = 100 t2 = 100, θ = t3 = 1 rad, R = r eiθ

= 200 t = 200,

= 200

= 3 t2 = 3 rad/s,

= 6 t = 6 rad2/s2

V = eiθ + r (i eiθ ) = 200 eiθ + 100(3) (i eiθ ).

a=

-r = -800

+ (2 + 1800 ( eiθ)

eiθ = [200 – (100)(

+ [(2)(200)(3)

eiθ

Example 12.20: A ball slides in the rotating forked rod and travels around the slotted path, described as r = 0.5(1-cosθ) ft.

1. At θ = 180°, the ball velocity is v = 4 ft/s, and acceleration a = 30 ft/s2,. Find of the fork. 2. If of the ball at θ = 180°. Solution: R = r eiθ V = eiθ + r (i eiθ ) a= -r + (2 eiθ eiθ = -1 = -0.5 /s2 ccw. Find the velocity and acceleration

At θ = 180°, r = 0.5(1-cosθ) = 1, = 0.5 sin θ = 0, = 0.5 sin θ + 0.5 1. V = eiθ + r (i eiθ ) = 0 + (1) (-i) = -4 i, a= -r = 24 (24)2 + 2. If V = eiθ + r (i eiθ ) = 0 + (1)(-5)(-i) = 5 i a= -r +(2 eiθ
2

= 4 rad/s – (1)(16)](-1) + [0 + (1) ] (-i)

+ (2

eiθ = [-0.5

= (30)2,

= 18 rad/s2

ft/s

=[-0.5 4. Cylindrical coordinate

2

– (1)(16)](-1)+[0 + (1)

] (-i)= 28.5 – 20 i

ft/s2

If z-axis is normal to the plane of ur and uθ, a cylindrical coordinate system is formed. The position of a particle in a cylindrical coordinate system is expressed as R = r ur + z uz. Since uz does not change the direction, its derivative is zero. Therefore, V = ur + r uθ + uz a= -r ur+ (2 uθ + uz