Kinematics and DynamicsME 230

Homework #2Week 3

2.1 Using Figure 1 (2.2 in text), verify that er = cos(θ)Ex + sin(θ)Ey and eθ = − sin(θ)Ex +cos(θ)Ey. Then, by considering cases where er lies in the second, third, and fourth quadrants,verify that these definitions are valid for all values of θ.

Figure 1: The unit vectors er and eθ

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1

2.4 The position vector of a particle of mass m that is placed at the end of a rotating telescopingrod is r = 6ter, where θ = 10t + 5 (radians). Calculate the velocity and acceleration vectors ofthe particle, and determine the force F needed to sustain the motion of the particle. What is theforce that the particle exerts on the telescoping rod?

Given, the position vector of the particle,

r = rer = 6ter

Differentiating with respect to time,

v =drdt

= 6er + 6ter = 6er + 6tθeθ

Now,θ = 10

Thus, the velocity vector is given by,

v = 6er + 60teθ

And, the acceleration vector is given by,

a =dvdt

= (r− rθ2)er + (rθ + 2rθ)eθ

a = −600ter + 120eθ

The force needed to sustain this motion is given by,

F = ma = −600mter + 120meθ

The force exerted by the particle on the telescoping rod is equal in magnitude and opposite indirection. Thus,

Fp = −F = −ma = 600mter − 120meθ

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2

2.5 In solving a problem, one person uses cylindrical polar coordinates whereas another usesCartesian coordinates. To check that their answers are identical, they need to examine therelationship between the Cartesian and cylindrical polar components of a certain vector, sayb = brer + bθeθ. To this end, show that

bx = b · Ex = br cos(θ)− bθ sin(θ), by = b · Ey = br sin(θ) + bθ cos(θ).

Given the vector b. The vector is the same when expressed in any coordinate system. Thisimplies,

b = bxEx + byEy = brEr + bθEθ (1)

Now, taking dot product with Ex throughout,

b · Ex = bxEx · Ex + byEy · Ex = brEr · Ex + bθEθ · Ex

⇒ bx = b · Ex = br cos θ − bθ sin θ

The dot products Er · Ex and Eθ · Ex is obtained from Figure 1 (Figure 2.2 in OReilly). Similarly,taking dot product with Ey throughout in equation (1), we have,

b · Ey = bxEx · Ey + byEy · Ey = brEr · Ey + bθEθ · Ey

⇒ by = b · Ey = br sin θ + bθ cos θ

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3

3.1 For the space curve r = xEx + axEy, show that

et =1√

1 + a2

(Ex + aEy

), κ = 0.

Given the position vector of a point on the space curve,

r = xEx + axEy

Now,drdt

=dxdt

Ex + adxdt

Ey

Thus, (dsdt

)2

=

(dxdt

)2

+ a2(

dxdt

)2

and, assuming that s increases in the direction of increasing x,

dsdt

=dxdt

√1 + a2

⇒ dxds

=1√

1 + a2

Now, from the definition of the tangent vector,

et = et(x) =drds

=drdx

dxds

=1

sqrt1 + a2 Ex +a√

1 + a2Ey

⇒ et =1√

1 + a2

(Ex + aEy

)Now, using the definition for en,

κen =det

ds=

det

dxdxds

= 0

⇒ κ = 0

since, en is a unit vector of magnitude 1. �

4

12–163.

A car is traveling along the circular curve of radius At the instant shown, its angular rate of rotation is

which is increasing at the rate ofDetermine the magnitudes of the car’s velocity andacceleration at this instant.

u$

= 0.2 rad>s2.u#

= 0.4 rad>s,

r = 300 ft.

SOLUTIONVelocity: Applying Eq. 12–25, we have

Thus, the magnitude of the velocity of the car is

Ans.

Acceleration: Applying Eq. 12–29, we have

Thus, the magnitude of the acceleration of the car is

Ans.a = a2r + a2

u = (-48.0)2 + 60.02 = 76.8 ft s2

au = ru$

+ 2r#u#

= 300(0.2) + 2(0)(0.4) = 60.0 ft>s2

ar = r$ - ru

#2 = 0 - 300 A0.42 B = -48.0 ft>s2

v = 2v2r + v2

u = 202 + 1202 = 120 ft>s

vr = r# = 0 vu = ru

#= 300(0.4) = 120 ft>s

r = 300 ft

A

= 0.2 rad/s2..

= 0.4 rad/s.

θθ

θ

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12–175.

SOLUTION

Since

Thus,

Ans.

Ans.

When

Ans.

Ans.vu =20

2= 14.1 ft s

vr = ¢ -20

22≤ = -14.1 ft>s

u = 1 rad

vu = ru#

= a10ub £ 2u2

21 + u2≥ =

20u

21 + u2

vr = r# = - a10

u2 b £ 2u2

21 + u2≥ = -

20

21 + u2

u =2u2

21 + u2

(20)2 = a102

u4 b(1 + u2)u#2

(20)2 = a102

u4 bu#2 + a102

u2 bu#2

v2 = r# 2 + aru

# b2

r = - a10u2 bu

#

r =10u

A particle P moves along the spiral path where is in radians. If it maintains a constant speed of

determine the magnitudes and as functionsof and evaluate each at u = 1 rad.u

vuvrv = 20 ft>s,u

r = 110>u2 ft,

r

r = θ

P

v

10––θ

#

#

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12–179.

SOLUTION

Ans.

Ans.a = r$ - ru

#2 + ru

$+ 2 r

#u#

2 = [4 - 2(6)2 + [0 + 2(4)(6)]2

= 83.2 m s2

v = 3Ar# B + Aru# B2 = 2 (4)2 + [2(6)]2 = 12.6 m>sr = 2t2 D10 = 2 m

L1

0dr = L

1

04t dt

u = 6 u$

= 0

r = 4t|t = 1 = 4 r# = 4

A block moves outward along the slot in the platform witha speed of where t is in seconds. The platformrotates at a constant rate of 6 rad/s. If the block starts fromrest at the center, determine the magnitudes of its velocityand acceleration when t = 1 s.

r# = 14t2 m>s,

θ

θ· = 6 rad/sr

$

#

2

2 ]2

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13–86.

Determine the magnitude of the resultant force acting on a5-kg particle at the instant , if the particle is movingalong a horizontal path defined by the equations

and rad, where t is inseconds.

u = (1.5t2 - 6t)r = (2t + 10) m

t = 2 s

SOLUTION

Hence,

Ans.F = 2(Fr)2 + (Fu)

2 = 210 N

©Fu = mau; Fu = 5(42) = 210 N

©Fr = mar; Fr = 5(0) = 0

au = ru$

+ 2r#u#= 14(3) + 0 = 42

ar = r$ - ru#2 = 0 - 0 = 0

u$

= 3

u#= 3t - 6�t=2 s = 0

u = 1.5t2 - 6t

r$ = 0

r# = 2

r = 2t + 10|t=2 s = 14

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13–95.

SOLUTION

Thus,

Ans.

Since

Ans. T = 8 N

a Fr = mar; -T = 2(-4)

ar = r$ - r# (u)2 = 0 - 0.25(4.00)2 = -4 m>s2

r# = -0.2 m>s, r$ = 0

u#= 4.00 rad>s

(0.5)2(1) = C = (0.25)2u#

r2u#= C

d(r2u#) = 0

a Fu = mau; 0 = m[ru$

+ 2r#u#] = m c1

r d

dt (r2u

#) d = 0

The ball has a mass of 2 kg and a negligible size. It is originallytraveling around the horizontal circular path of radius

such that the angular rate of rotation isIf the attached cord ABC is drawn down

through the hole at a constant speed of determine thetension the cord exerts on the ball at the instantAlso, compute the angular velocity of the ball at this instant.Neglect the effects of friction between the ball and horizontalplane. Hint: First show that the equation of motion in thedirection yieldsWhen integrated, where the constant c is determinedfrom the problem data.

r2u#= c,au = ru

$+ 2r

#u#= 11>r21d1r2u# 2>dt2 = 0.

u

r = 0.25 m.0.2 m>s,

u#0 = 1 rad>s.r0 = 0.5 m

C

F

r

r0

B

A

0.2 m/s

·uu0

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