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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
�
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
�
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
�
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
#
#
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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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
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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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
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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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
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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