Problem 17.1 In Active Example 17.1, suppose thatat a given
instant the hook H is moving downward at2 m/s. What is the angular
velocity of gear A at thatinstant?AH50 mmB200 mm100mmSolution: The
angular velocity of gear B isB = vHrH= 2 m/s0.1 m = 20 rad/s.The
gears are connected through the common velocity of the
contactpointsrBB = rAA A = rBrAB = 0.2 m0.05 m(20 rad/s) = 80
rad/s.A = 80 rad/s counterclockwise.Problem 17.2 The angle (in
radians) is given as afunction of time by = 0.2t2. At t = 4 s,
determinethe magnitudes of (a) the velocity of point A and (b)
thetangential and normal components of acceleration ofpoint A.A2
mSolution: We have = 0.2t2, = ddt= 0.4t, = ddt= 0.4.Then(a) v = r =
(2)(0.4)(4) = 10.1 m/s. v = 10.1 m/s.(b) an = r2= (2)[(0.4)(4)]2=
50.5 m/s2,at = r = (2)(0.4) = 2.51 m/s2.an = 50.5 m/s2,at = 2.51
m/s2.Problem 17.3 The mass A starts from rest at t = 0 andfalls
with a constant acceleration of 8 m/s2. When themass has fallen one
meter, determine the magnitudes of(a) the angular velocity of the
pulley and (b) the tangen-tial and normal components of
acceleration of a point atthe outer edge of the pulley.100
mmASolution: We havea = 8 m/s2, v =2as =_2(8 m/s2)(1 m) = 4 m/s, =
vr= 4 m/s0.1 m = 40 rad/s, = ar= 8 m/s20.1 m = 80 rad/s2.(a) = 40
rad/s.(b) at = r = (0.1 m)(80 rad/s2) = 8 m/s2,an = r2= (0.1 m)(40
rad/s)2= 160 m/s2.at = 8 m/s2,an = 160 m/s2.333c 2008 Pearson
Education South Asia Pte Ltd. All rights reserved. This publication
is protected by Copyright and 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.Problem
17.4 At the instant shown, the left disk hasan angular velocity of
3 rad/s counterclockwise and anangular acceleration of 1
rad/s2clockwise.(a) What are the angular velocity and angular
accel-eration of the right disk? (Assume that there isno relative
motion between the disks at their pointof contact.)(b) What are the
magnitudes of the velocity and accel-eration of point A?2.5 m1 mA2
m3 rad/s1 rad/s2Solution:(a)rLL = rRR R = rRrLL = 1 m2.5 m(3 rad/s)
= 1.2 rad/srLL = rRR R = rRrLL = 1 m2.5 m(1 rad/s2) = 0.4 rad/s2(b)
vA = (2 m)(1.2 rad/s) = 2.4 m/saAt = (2 m)(0.4 rad/s2) = 0.8
m/s2aAn = (2 m)(1.2 rad/s)2= 2.88 m/s2vA = 2.4 m/saA =_(0.8)2+
(2.88)2m/s2= 2.99 m/s2Problem 17.5 The angular velocity of the left
disk isgiven as a function of time by A = 4 + 0.2t rad/s.(a) What
are the angular velocities B and C att = 5 s?(b) Through what angle
does the right disk turn fromt = 0 to t = 5 s?100 mm100 mm200 mm200
mmA B CSolution:A = (4 + 0.2t ) rad/srAA = rBB B = 0.1 m0.2 mA =
0.5 ArBB = rCC C = 0.1 m0.2 mB = 0.25 A(a) At t = 5 s A = (4 +
0.2[5]) rad/s = 5 rad/sB = 0.5(5 rad/s) = 2.5 rad/sC = 0.25(5
rad/s) = 1.25 rad/s(b) C = 0.25 A = 0.25(4 + 0.2t ) rad/s = (1 +
0.05t ) rad/sC =_ 5 s0Cdt = [t + 0.025t2]5 s0 = 5.625 rad334c 2008
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publication is protected by Copyright and 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.Problem 17.6 (a) If the bicycles 120-mm sprocketwheel
rotates through one revolution, through how manyrevolutions does
the 45-mm gear turn? (b) If the angu-lar velocity of the sprocket
wheel is 1 rad/s, what is theangular velocity of the
gear?45mm120mmSolution: The key is that the tangential
accelerations and tangen-tial velocities along the chain are of
constant magnitude(a) B = 2.67 rev(b) vB = rB vA = rAAvA = vBvB =
(0.045)B vA = (0.120)(1)B =_12045_ rad/s = 2.67 rad/srBdBdt=
rAdAdtIntegrating, we getrBB = rAA rA = 0.120 mrB = 0.045 mB
=_12045_(1) rev A = 1 rev.rAvAatAvBatBrBA BProblem 17.7 The rear
wheel of the bicycle in Prob-lem 17.6 has a 330-mm radius and is
rigidly attached tothe 45-mm gear. It the rider turns the pedals,
which arerigidly attached to the 120-mm sprocket wheel, at
onerevolution per second, what is the bicycles velocity?Solution:
The angular velocity of the 120 mm sprocket wheel is = 1 rev/s = 2
rad/s. Use the solution to Problem 17.6. The angularvelocity of the
45 mm gear is45 = 2_12045_= 16.76 rad/s.This is also the angular
velocity of the rear wheel, from which thevelocity of the bicycle
isv = 45(330) = 5.53 m/s.335c 2008 Pearson Education South Asia Pte
Ltd. All rights reserved. This publication is protected by
Copyright and 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.Problem 17.8 The
disk is rotating about the originwith a constant clockwise angular
velocity of 100 rpm.Determine the x and y components of velocity of
pointsA and B (in cm/s).B16 Axy8 cm8 cm cmSolution: = 100 rpm_2
radrev__1 min60 s_= 10.5 rad/s.Working with point ArA =_(8 cm)2+(8
cm)2=11.3 cm, A = tan1_8 cm8 cm_= 45vA = rA = (11.3 cm)(10.5 rad/s)
= 118 cm/svA = vA(cos Ai + sin Aj) = (118 cm/s)(cos 45i + sin
45j)vA = (83.8i + 83.8j)Working with point BrB = 16 in, vB = rB =
(16 cm)(10.5 rad/s) = 168 cm/svB = (168j)Problem 17.9 The disk is
rotating about the originwith a constant clockwise angular velocity
of 100 rpm.Determine the x and y components of acceleration
ofpoints A and B (in cm/s ).2B16 cmAxy8 cm8 cmSolution: = 100 rpm_2
radrev__1 min60 s_= 10.5 rad/s.Working with point ArA =_(8 cm)2+(8
cm)2= 11.3 cm, A = tan1_8 cm8 cm_= 45aA = rA2= (11.3 cm)(10.5
rad/)2= 1240 cm/s2aA = aA(cos Ai sin Aj) = (1240 cm/s2)(cos 45i sin
45j)aA = (877i 877j) 2.Working with point BrB = 16 in, aB = rB2=
(16 cm)(10.5 rad/s)2= 1750 cm/s2aB = (1750i)
2.336cm/s.cm/s.cm/scm/sc 2008 Pearson Education South Asia Pte Ltd.
All rights reserved. This publication is protected by Copyright and
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.Problem 17.10 The radius of the
Corvettes tires isthe brakes, subjecting the car to a deceleration
of 25 m/s .2Assume that the tires continue to roll, not skid, on
theroad surface. At that instant, what are the magnitudesof the
tangential and normal components of acceleration(in m/s ) of a
point at the outer edge of a tire relative2to a nonrotating
coordinate system with its origin at thecenter of the
tire?yxSolution: We have = vr=_80 1000_ = = ar= 25 m/s2=
rad/s2.Relative to the given coordinate system (which is not an
inertial coor-dinate system)at = r = ( rad/s2) = 25 m/s2an = r2= (
)2= 16 2.at = 25 m/s2,an = 1647 2.Problem 17.11 If the bar has a
counterclockwiseangular velocity of 8 rad/s and a clockwise
angularacceleration of 40 rad/s2, what are the magnitudes ofthe
accelerations of points A and B?0.4 m0.4 m 0.4 m0.2 mABSolution:aA
= rA/O 2rA/O= (40 rad/s2k) (0.4i + 0.4j) m (8 rad/s)2(0.4i + 0.4j)
m= (41.6i 9.6j) m/s2aB = rB/O 2rB/O= (40 rad/s2k) (0.4i 0.2j) m (8
rad/s)2(0.4i 0.2j) m= (33.6i 3.2j) m/s2aA =_(41.6)2+ (9.6)2m/s2=
42.7 m/s2aB =_(33.6)2+ (3.2)2m/s2= 33.8 m/s233730 cm. It is
traveling at 80 km/h when the driver applies3600m) 74.1 rad/s,m)
83.33m) (0.3 83.33m) (0.3 74.1 rad/s 47.2 m/s .2 m/s(0.3(0.3c 2008
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publication is protected by Copyright and 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.Problem 17.12 Consider the bar shown in Problem17.11. If
|vA| = 3 m/s and |aA| = 28 m/s2, what are |vB|and |aB|?Solution:vA
= r 3 m/s = _(0.4)2+ (0.4)2m = 5.30 rad/saAn = 2r = (5.3
rad/s)2_(0.4)2+ (0.4)2m = 15.9 m/s2aAt =_aA2 aAn2=_(28)2
(15.9)2m/s2= 23.0 m/s2aAt = r 23.0 m/s2= _(0.4)2+ (0.4)2m = 40.7
rad/svB = _(0.4)2+ (0.2)2m = 2.37 m/saBt = _(0.4)2+ (0.2)2= 18.2
m/s2aBn = 2_(0.4)2+ (0.2)2= 12.6 m/s2aB =_(18.2)2+ (12.6)2m/s2=
22.1 m/s2Problem 17.13 A disk of radius R = 0.5 m rolls ona
horizontal surface. The relationship between thehorizontal distance
x the center of the disk moves andthe angle through which the disk
rotates is x = R.Suppose that the center of the disk is moving to
the rightwith a constant velocity of 2 m/s.(a) What is the disks
angular velocity?(b) Relative to a nonrotating reference frame
withits origin at the center of the disk, what are themagnitudes of
the velocity and acceleration of apoint on the edge of the
disk?xRRSolution:(a) x = R x = R v = R = vR= 2 m/s0.5 m = 4
rad/s(b)v = R = (0.5 m)(4 rad/s) = 2 m/sa = an = R2= (0.5 m)(4
rad/s)2= 8 m/s2Problem 17.14 The turbine rotates relative to the
coor-dinate system at 30 rad/s about a xed axis coincidentwith the
x axis. What is its angular velocity vector?yzx30 rad/sSolution:
The angular velocity vector is parallel to the x axis,with
magnitude 30 rad/s. By the right hand rule, the positive
directioncoincides with the positive direction of the x axis. = 30i
(rad/s).338c 2008 Pearson Education South Asia Pte Ltd. All rights
reserved. This publication is protected by Copyright and 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.Problem 17.15 The rectangular plate swings in
the xyplane from arms of equal length. What is the angularvelocity
of (a) the rectangular plate and (b) the bar AB?yx AB10
rad/sSolution: Denote the upper corners of the plate by B and B
, anddenote the distance between these points (the length of the
plate) byL. Denote the suspension points by A and A
, the distance separatingthem by L
. By inspection, since the arms are of equal length, andsince L
= L
, the gure AA
B
B is a parallelogram. By denition, theopposite sides of a
parallelogram remain parallel, and since the xedside AA
does not rotate, then BB
cannot rotate, so that the plate doesnot rotate andBB = 0
.Similarly, by inspection the angular velocity of the bar AB isAB =
10k (rad/s),where by the right hand rule the direction is along the
positive z axis(out of the paper).BABLLAProblem 17.16 Bar OQ is
rotating in the clockwisedirection at 4 rad/s. What are the angular
velocity vectorsof the bars OQ and PQ?Strategy: Notice that if you
know the angular velocityof bar OQ, you also know the angular
velocity of barPQ.1.2 mOQP1.2 myx4 rad / sSolution: The magnitudes
of the angular velocities are the same.The directions are
oppositeOQ = (4 rad/s)k, PQ = (4 rad/s)k,339c 2008 Pearson
Education South Asia Pte Ltd. All rights reserved. This publication
is protected by Copyright and 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.Problem
17.17 A disk of radius R = 0.5 m rolls ona horizontal surface. The
relationship between the hor-izontal distance x the center of the
disk moves and theangle through which the disk rotates is x = R.
Sup-pose that the center of the disk is moving to the rightwith a
constant velocity of 2 m/s.(a) What is the disks angular
velocity?(b) What is the disks angular velocity
vector?xRRxySolution:(a) x = R x = R v = R = vR= 2 m/s0.5 m = 4
rad/s CW(b) = (4 rad/s)kProblem 17.18 The rigid body rotates with
angularvelocity = 12 rad/s. The distance rA/B = 0.4 m.(a) Determine
the x and y components of the velocityof A relative to B by
representing the velocity asshown in Fig. 17.10b.(b) What is the
angular velocity vector of therigid body?(c) Use Eq. (17.5) to
determine the velocity of A rel-ative to B.xyB ArA/BSolution:(a)
vAx = 0vAy = (12 rad/s)(0.4 m) = 4.8 m/s(b) = (12 rad/s)k(c) vA =
rA/B = (12 rad/s)k (0.4 m)ivA = (4.8 m/s)jByVAAxrA/B = 0.4 m12
rad/s340c 2008 Pearson Education South Asia Pte Ltd. All rights
reserved. This publication is protected by Copyright and 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.Problem 17.19 The bar is rotating in the
counterclock-wise direction with angular velocity . The magnitude
ofthe velocity of point A is 6 m/s. Determine the velocityof point
B.0.4 m0.4 m0.4 m0.2 mAByxSolution: = vr= 6 m/s2(0.4 m)= 10.6
rad/s.vB = rB = (10.6 rad/s)k (0.4i 0.2j) mvB = (2.12i + 4.24j)
m/s.Problem 17.20 The bar is rotating in the counterclock-wise
direction with angular velocity . The magnitudeof the velocity of
point A relative to point B is 6 m/s.Determine the velocity of
point B.0.4 m0.4 m0.4 m0.2 mAByxSolution:rA/B =_(0.8 m)2+ (0.6 m)2=
1 m = vrA/B= 6 m/s1 m = 6 rad/s.vB = rB = (6 rad/s)k (0.4i 0.2j)
mvB = (1.2i + 2.4j) m/s.Problem 17.21 The bracket is rotating about
point Owith counterclockwise angular velocity . The magni-tude of
the velocity of point A relative to point B is4 m/s. Determine
.AB180 mm48120 mmyxOSolution:rB/A =_(0.18 + 0.12 cos 48)2+ (0.12
sin 48)2= 0.275 m = vB/ArB/A= 4 m/s0.275 m = 14.5 rad/s. = 14.5
rad/s.341c 2008 Pearson Education South Asia Pte Ltd. All rights
reserved. This publication is protected by Copyright and 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.Problem 17.22 Determine the x and y
components ofthe velocity of point A. yxA5 rad/sO2 m30Solution: The
velocity of point A is given by:vA = v0 + rA/O.Hence, vA = 0 +i j
k0 0 52 cos 30 2 sin 30 0= 10 sin 30i + 10 cos 30j,or vA = 5i +
8.66j (m/s).Problem 17.23 If the angular velocity of the bar
inProblem 17.22 is constant, what are the x and y com-ponents of
the velocity of Point A 0.1 s after the instantshown?Solution: The
angular velocity is given by = ddt= 5 rad/s,_ 0d =_ t05 dt,and = 5t
rad.At t = 0.1 s, = 0.5 rad = 28.6.vA = v0 + rA/O = 0 +i j k0 0 52
cos 28.6 2 sin 28.6 0.Hence, vA = 10 sin 28.6i + 10 cos 28.6j =
4.78i + 8.78j.342c 2008 Pearson Education South Asia Pte Ltd. All
rights reserved. This publication is protected by Copyright and
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.Problem 17.24 The disk is
rotating about the z axis at50 rad/s in the clockwise direction.
Determine the x andy components of the velocities of points A, B,
and C.xy100 mmCAB45 45Solution: The velocity of A is given by vA =
v0 + rA/O, orvA = 0 + (50k) (0.1j) = 5i (m/s).For B, we havevB = v0
+ rB/O = 0 +i j k0 0 500.1 cos 45 0.1 sin 45 0= 3.54i 3.54j
(m/s),For C, we havevc = v0 + rC/O = 0 +i j k0 0 500.1 cos 45 0.1
sin 45 0= 3.54i + 3.54j (m/s).yxB C45 45100 mmAProblem 17.25
Consider the rotating disk shown inProblem 17.24. If the magnitude
of the velocity ofpoint A relative to point B is 4 m/s, what is
themagnitude of the disks angular velocity?Solution:v0 = 0 = k r =
0.1 mvB = v0 + k rOB= k (r cos 45i r sin 45j)= (rcos 45)j + (rsin
45)i.vA = v0 + k rOA = k rj= ri.vA = vB + vA/B,vA/B = vA vB,vA/B =
(r rsin 45)i rcos 45j= r(1 sin 45)i rcos 45j.We know|vA/B| = 4 m/s,
r = 0.1 m|vA/B| =_[r(1 sin 45)]2+ [rcos 45]2Solving for , = 21.6
rad/s (direction undetermined).343c 2008 Pearson Education South
Asia Pte Ltd. All rights reserved. This publication is protected by
Copyright and 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.Problem 17.26 The
radius of the Corvettes tires isroll, not skid, on the road
surface.(a) What is the angular velocity of its wheels?(b) In terms
of the earth-xed coordinate system shown,determine the velocity (in
m/s) of the point of thetire with coordinates (yxSolution:(a) =
vr=_ _ = rad/s. =(b)vP = vO + rvP = _ _i + ( )k __ _ i_vP = ( i
j)Problem 17.27 Point A of the rolling disk is movingtoward the
right. The magnitude of the velocity of pointC is 5 m/s. determine
the velocities of points B and D.xyA45CDB0.6 mSolution: Point B is
the center of rotation (zero velocity).rC/B =2(0.6 m) = 0.849 m, =
vCrC/B= 5 m/s0.849 m = 5.89 rad/sThereforevD = rD/B = (5.89 rad/s)k
[0.6 cos 45i + (0.6 + 0.6 sin 45)j]vD = (6.04i + 2.5j) m/s, vB =
0.Problem 17.28 The helicopter is in planar motion inthe xy plane.
At the instant shown, the positionof its center of mass, G, is x =
2 m, y = 2.5 m,and its velocity is vG = 12i + 4j (m/s). The
positionof point T , where the tail rotor is mounted, is x =3.5 m,
y = 4.5 m. The helicopters angular velocity is0.2 (rad/s)
clockwise. What is the velocity of point T ?yxTGSolution: The
position of T relative to G isrT/G = (3.5 2)i + (4.5 2.5)j = 5.5i +
2j (m).The velocity of T isvT = vG + rT/G = 12i + 4j +i j k0 0
0.25.5 2 0= 12.4i + 5.1j (m/s)34430 cm. It is traveling at 100
km/h. Assume that the tires30 cm, 0,0).100 1000 3600(0.3)92.6 92.6
rad/s.100 1000 3600 m/s rad/s 92.6 0.3 m27. 8 27. 8 m/s.c 2008
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publication is protected by Copyright and 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.Problem 17.29 The bar is moving in the xy planeand is
rotating in the counterclockwise direction. Thevelocity of point A
relative to the reference frame isvA = 12i 2j (m/s). The magnitude
of the velocity ofpoint A relative to point B is 8 m/s. what is the
velocityof point B relative to the reference frame?xyBA2
m30Solution: = vr= 8 m/s2 m = 4 rad/s,vB = vA + rB/A = (12i 2j) m/s
+ (4 rad/s)k (2 m)(cos 30i + sin 30j)vB = (8i + 4.93j) m/s.Problem
17.30 Points A and B of the 2-m bar slideon the plane surfaces.
Point B is moving to the right at3 m/s. What is the velocity of the
midpoint G of the bar?Strategy: First apply Eq. (17.6) to points A
and Bto determine the bars angular velocity. Then applyEq. (17.6)
to points B and G.xyABG70Solution: Take advantage of the
constraints (B stays on the oor,A stays on the wall)vA = vB +
rA/BvAj = (3 m/s)i + k (2 m)(cos 70i + sin 70j)= (3 1.88 )i +
(0.684 )jEquating i components we nd 3 1.88 = 0 = 1.60 rad/sNow nd
the velocity of point GvG = vB + rG/B= (3 m/s)i + (1.60 rad/s)k (1
m)(cos 70i + sin 70j)vG = (1.5i 0.546j) m/sProblem 17.31 Bar AB
rotates at 6 rad/s in the clock-wise direction. Determine the
velocity (in cm/s) of theslider C.6 rad/sABC4 cm4 cm 10 cm3
cmSolution:vB = vA + AB rB/AvB = 0 (6 rad/s)k (4i + 4j)vB = (24i
24j)vC = vB + BC rC/BvC = (24i 24j) + BCk (10i 7j)vC = (24 + 7 BC)i
+ (24 + 10 BC)jSlider C cannot move in the j direction, thereforeBC
= 2410 = 2.4 rad/s.vC = (24 + 7(2.4))i = (43.8 cm/s)i vC = 43.8
cm/s to the right.345cmcm/s.c 2008 Pearson Education South Asia Pte
Ltd. All rights reserved. This publication is protected by
Copyright and 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.Problem 17.32 If =
45 and the sleeve P is movingto the right at 2 m/s, what are the
angular velocities ofthe bars OQ and PQ?1.2 mOQP1.2 mSolution: From
the gure, v0 = 0, vP = vPi = 2 i (m/s)vQ = v0 + OQk (Lcos i + Lsin
j)_i: vQx = OQLsin (1)j: vQy = OQLcos (2)vP = vQ + QPk (Lcos i Lsin
j)i: 2 = vQx + QPLsin (3)j: 0 = vQy + QPLcos (4)Eqns (1)(4) are 4
eqns in the 4 unknowns OQ, QP, vQx, vQySolving, we getvQx = 1 m/s,
vQy = 1 m/sOQ = 1.18k (rad/s),QP = 1.18k (rad/s).Problem 17.33 In
Active Example 17.2, consider theinstant when bar AB is vertical
and rotating in the clock-wise direction at 10 rad/s. Draw a sketch
showing thepositions of the two bars at that instant. Determine
theangular velocity of bar BC and the velocity of point C.0.4 m10
rad/sABC0.4 m0.8 mSolution: The sketch is shown. We havevB = vA +
AB rB/A= 0 (10 rad/s)k (0.566 m)j= (5.66 m/s)ivC = vB + BC rBC=
(5.66 m/s)i+ BCk (0.693i 0.566j) m= (5.66 + 0.566 BC)i + (0.693
BC)jPoint C cannot move in the j direction, thereforeBC = 0, vC =
(5.66 m/s)i346c 2008 Pearson Education South Asia Pte Ltd. All
rights reserved. This publication is protected by Copyright and
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.Problem 17.34 Bar AB rotates in
the counterclockwisedirection at 6 rad/s. Determine the angular
velocity ofbar BD and the velocity of point D.y0.32 m0.48 m0.16 m
0.24 m 0.32 m6 rad/sA BxDCSolution:vA = 0 AB = 6 kvB = vA + ABk
rB/A = 6k 0.32i = 1.92j m/s.vC = vCi = vB + BDk rC/B :vci = 1.92j +
BDk (0.24i + 0.48j)._i: vC = 0.48BDj: 0 = 1.92 + 0.24BDSolving, BD
= 8,BD = 8k_rads_,vC = 3.84i (m/s).Now for the velocity of point
DvD = vB + BD rD/B= 1.92j + (8k) (0.4i + 0.8j)vD = 6.40i 1.28j
(m/s).Problem 17.35 At the instant shown, the pistons velo-city is
vC = 14i(m/s). What is the angular velocity ofthe crank AB?BA50
mm50 mm175 mmyxCSolution:vB = vA + AB rB/A= 0 + ABk (0.05i + 0.05j)
m= 0.05ABi + 0.05ABjvC = vB + BC r= (0.05ABi + 0.05ABj)+ BCk
(0.175i 0.05j)= (0.05AB + 0.05BC)i + (0.05AB + 0.175BC)jWe can now
separate components and produce two equations in twounknowns14 =
0.05AB + 0.05BC, 0 = 0.05AB + 0.175BCSolving we ndBC = 62.2 rad/s,
AB = 218 rad/s.Thus AB = 218 rad/s counterclockwise.347c 2008
Pearson Education South Asia Pte Ltd. All rights reserved. This
publication is protected by Copyright and 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.Problem 17.36 In Example 17.3, determine the angu-lar
velocity fo the bar AB that would be necessary sothat the downward
velocity of the rack VR = 120 cm/sat the instant shown.DvR6 cmBCA12
cm10 rad/s16 cm 6 cm 6 cm10 cmSolution: We haveCD = vRr= 120 cm/scm
= 20 rad/s.ThenVB = VA + AB rB/A= 0 + ABk (6i + 12j) = 12ABi +
6ABjVC = VB + BC rC/B= (12ABi + 6ABj) + BCk (16i 2j)= (12AB + 2BC)i
+ (6AB + 16BC)jVD = VC + CD rD/C= [(12AB + 2BC)i + (6AB + 16BC)j]
(20)k (6i 10j)= (12AB + 2BC 200)i + (6AB + 16BC 120)jPoint D cannot
move, therefore12AB + 2BC 200 = 0, 6AB + 16BC 120 = 0.Solving, we
ndAB = 14.5 rad/s, BC = 12.9 rad/s.AB = 14.5 rad/s
counterclockwise.3486c 2008 Pearson Education South Asia Pte Ltd.
All rights reserved. This publication is protected by Copyright and
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.Problem 17.37 Bar AB rotates at
12 rad/s in theclockwise direction. Determine the angular
velocities ofbars BC and CD.ACBD350mm200mm300 mm 350 mm12
rad/sSolution: The strategy is analogous to that used in Problem
17.36.The radius vector AB is rB/A = 200j (mm). The angular
velocity ofAB is = 12k (rad/s). The velocity of point B isvB = vA +
rB/A = 0 +__i j k0 0 120 200 0__= 2400i (mm/s).The radius vector BC
is rC/B = 300i + (350 200)j = 300i + 150j(mm). The velocity of
point C isvC = vB + BC rC/B = vB +__i j k0 0 BC300 150 0__= (2400
150BC)i + BC300j (mm/s).The radius vector DC is rC/D = 350i + 350j
(mm). The velocity ofpoint C isvC = vD + CD rC/D = 0 +__i j k0 0
CD350 350 0__= 350CD(i + j).Equate the two expressions for vC, and
separate components:(2400 150BC + 350CD)i = 0,and (300BC + 350CD)j
= 0.Solve: BC = 5.33 rad/s,BC = 5.33k (rad/s) .CD = 4.57 rad/s,CD =
4.57k (rad/s) .349c 2008 Pearson Education South Asia Pte Ltd. All
rights reserved. This publication is protected by Copyright and
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.Problem 17.38 Bar AB is
rotating at 10 rad/s in thecounterclockwise direction. The disk
rolls on the circularsurface. Determine the angular velocities of
bar BC andthe disk at the instant shown.2 mABC3 m3 m1 mSolution:
The point D at the bottom of the wheel has zero veloc-ity.vB = vA +
AB rB/A= 0 + (10)k (1i 2j) = (20i + 10j) m/s.vC = vB + BC rC/B=
(20i + 10j) + BCk (3i) = (20)i + (10 + 3BC)jvD = vC + CD rD/C=
(20)i + (10 + 3BC)j + CDk (1j) = (20 + CD)i + (10 + 3BC)jSince the
velocity of D is zero, we can set the components of velocityequal
to zero and solve to nddisk = CD = 20 rad/s, BC = 3.33 rad/s.disk =
20 rad/s clockwise, BC = 3.33 rad/s clockwise.Problem 17.39 Bar AB
rotates at 2 rad/s in the coun-terclockwise direction. Determine
the velocity of themidpoint G of bar BC.AC BD2 rad/s4530Gyx305
mm254 mmSolution: We rst need to nd the angular velocities of BCand
CDvB = vA + AB rB/A = (2 rad/s)k ( )(cos 45i + sin 45j)= ( i + j)vC
= vB +BC rC/B =( i + j) + BCk ( )i= [( )i + ( + { }BC)j]vD = vC +
CD rD/C+ CDk ( )[sin 45 cot 30i sin 45j]= [( + { }CD)i + (+ { m} BC
+ { }CD)j]Since D is xed, we set both components to zero and solve
for theangular velocities + { }CD =0+ { } BC + { }CD=0 BC =3.22
rad/sCD=2.00 rad/sNow we can nd the velocity of point G.vG = vB +
BC rG/B= ( 4i + j) + (3.22 rad/s)k ( )ivG = ( i .132j)3500.254
m0.354 0.354 m/s0.354 0.354 m/s 0.305 m0.354 m/s 0.354 m/s 0.305 m=
[( )i + ( + { }BC)j] 0.354 m/s 0.354 m/s 0.305 m0.254 m0.354 m/s
0.18 m 0.354 m/s0.305 0.31 m0.354 m/s 0.18 m0.354 m/s 0.305 m 0.31
m0.35 0.354 m/s 0.152 mm0.354 0 m/sc 2008 Pearson Education South
Asia Pte Ltd. All rights reserved. This publication is protected by
Copyright and 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.Problem 17.40 Bar
AB rotates at 10 rad/s in thecounterclockwise direction. Determine
the velocity ofpoint E.xyABC10 rad/s E D400 mm700 mm 700 mm
400mmSolution: The radius vector AB is rB/A = 400j (mm). The
angu-lar velocity of bar AB is AB = 10k (rad/s). The velocity of
point BisvB = AB rAB =__i j k0 0 100 400 0__= 4000i (mm/s).The
radius vector BC is rC/B = 700i 400j (mm). The velocity ofpoint C
isvC = vB + BC rC/B = 4000i +__i j k0 0 BC700 400 0__= (4000 +
400BC)i + 700BCj.The radius vector CD is rC/D = 400i (mm). The
point D is xed(cannot translate). The velocity at point C isvC = CD
rC/D = (CD(k) (400i)) =__i j k0 0 CD400 0 0__= 400CDj.Equate the
two expressions for the velocity at point C, and
separatecomponents: 0 = (4000 + 400BC)i, 0 = (700BC +
400CD)j.Solve: BC = 10 rad/s, CD = 17.5 rad/s. The radius vector
DEis rD/E = 700i (mm). The velocity of point E isvE = CD rD/E =__i
j k0 0 17.5700 0 0__vE = 12250j (mm/s) .351c 2008 Pearson Education
South Asia Pte Ltd. All rights reserved. This publication is
protected by Copyright and 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.Problem
17.41 Bar AB rotates at 4 rad/s in thecounterclockwise direction.
Determine the velocity ofpoint C.ABCDExy600 mm600 mm400 mm500
mm300mm300mm200mmSolution: The angular velocity of bar AB is = 4k
(rad/s). Theradius vector AB is rB/A = 300i + 600j (mm). The
velocity of point BisvB = AB rB/A =__i j k0 0 4300 600 0__,from
which vB = 2400i + 1200j (mm/s). The vector radius from Bto C is
rC/B = 600i + (900 600)j = 600i + 300j (mm). The veloc-ity of point
C isvC = vB +__i j k0 0 BC600 300 0__= (2400 300BC)i + (1200 +
600BC)j (mm/s).The radius vector from C to D is rD/C = 200i 400j
(mm). Thevelocity of point D isvD = vC +__i j k0 0 BC200 400 0__=
vC + 400BCi + 200BCj (mm/s).The radius vector from E to D is rD/E =
300i + 500j (mm). Thevelocity of point D isvD = DE rD/E =__i j k0 0
DE300 500 0__= 500DEi 300DEj (mm/s).Equate the expressions for the
velocity of point D; solve for vC, toobtain one of two expressions
for the velocity of point C. Equate thetwo expressions for vC, and
separate components: 0 = (500DE 100BC + 2400)i, 0 = (1200 + 300DE +
800BC)j. Solve DE =5.51 rad/s, BC = 3.57 rad/s. Substitute into the
expression for thevelocity of point C to obtainvC = 1330i 941j
(mm/s).352c 2008 Pearson Education South Asia Pte Ltd. All rights
reserved. This publication is protected by Copyright and 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.Problem 17.42 The upper grip and jaw of the
pli-ers ABC is stationary. The lower grip DEF is rotatingat 0.2
rad/s in the clockwise direction. At the instantshown, what is the
angular velocity of the lower jawCFG?G70 mmB ADC30 mm30 mmE F30
mmStationarySolution:vE = vB + BE rE/B = 0 + BEk (0.07i 0.03j) m=
BE(0.03i + 0.07j) mvF = vE + DEF rF/E = BE(0.03i + 0.07j) m+ (0.2
rad/s)k (0.03 m)i= [(0.03 m)BEi + (0.006 m/s + {0.07 m}BE)j]vC = vF
+ CFG rC/G= [(0.03 m)BEi + (0.006 m/s + {0.07 m}BE)j]+ CFGk (0.03
m)j= [(0.03 m)(BE CFG)i + (0.006 m/s + {0.07 m}BE)j]Since C is xed
we have(0.03 m)(BE CFG) = 00.006 m/s + {0.07 m}BE = 0 BE = 0.0857
rad/sCFG = 0.0857 rad/sSo we have CFG = 0.0857 rad/s CCWProblem
17.43 The horizontal member ADE support-ing the scoop is
stationary. If the link BD is rotating inthe clockwise direction at
1 rad/s, what is the angularvelocity of the scoop?CBD EAScoop0.61
m0.46 m1.52 mm0.31 mm0.76 mSolution: The velocity of B is vB = vD +
BD rB/D. Expand-ing, we getvB = 0 +i j k0 0 0= i j ( ).The velocity
of C isvC = vB + BC rC/B = i j +i j k0 0 +BC. 0. 0(1).We can also
express the velocity of C as vC = vE + CE rC/E orvC = 0 +i j k0 0
+CE0 0.(2).Equating i and j components in Equations (1) and (2) and
solving, weobtain BC = 0.4 rad/s and CE = 1.47 rad/s.1
rad/sBDCEyxCEBC3530.31 0.610. 3.1 0.61 0.31 m/s0.61 0.310 76 1546
0c 2008 Pearson Education South Asia Pte Ltd. All rights reserved.
This publication is protected by Copyright and 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.Problem 17.44 The diameter of the disk is 1 m, andthe
length of the bar AB is 1 m. The disk is rolling, andpoint B slides
on the plane surface. Determine the angu-lar velocity of the bar AB
and the velocity of point B.4 rad/sABSolution: Choose a coordinate
system with the origin at O, thecenter of the disk, with x axis
parallel to the horizontal surface. Thepoint P of contact with the
surface is stationary, from whichvP = 0 = v0 + 0 R = v0 +__i j k0 0
00 0.5 0__= v0 + 2i,from which v0 = 2i (m/s). The velocity at A is
vA = v0 + 0 rA/O.vA = 2i +__i j k0 0 00.5 0 0__= 2i + 2j (m/s).The
vector from B to A is rA/B = i cos + j sin (m), where =sin10.5 =
30. The motion at point B is parallel to the x axis. Thevelocity at
A isvA = vBi + rA/B =__i j k0 0 AB0.866 0.5 0__= (vB 0.5AB)i
0.866ABj (m/s).Equate and solve: (2 0.866AB)j = 0, (vB 0.5AB + 2)i
= 0,from which AB = 2.31k (rad/s), vB = 3.15i (m/s).yxBPAO354c 2008
Pearson Education South Asia Pte Ltd. All rights reserved. This
publication is protected by Copyright and 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.Problem 17.45 A motor rotates the circular diskmounted at
A, moving the saw back and forth. (The sawis supported by a
horizontal slot so that point C moveshorizontally). The radiusAB is
101.6 mm, and the link BC is =45 and thelink BC is horizo. ntal. If
the angular velocity of the diskis one revolution per second
counterclockwise, what isthe velocity of the saw?xyBACSolution: The
radius vector from A to B isrB/A = (i cos 45 + j sin 45) = 2(i + j)
(m).The angular velocity of B isvB = vA + AB rB/A,vB = 0 + 2(2)__i
j k0 0 11 1 0__= 2(i + j) ( /s).The radius vector from B to C is
rC/B = ( cos 45 )i.The velocity of point C isvC = vB + BC rC/B =
vB__i j k0 0 BC14 0 0__= The saw is constrained to move parallel to
the x axis, henceBC = 0, and the saw velocity isvS = i (m/s)
.Problem 17.46 In Problem 17.45, if the angularvelocity of the disk
is one revolution per second counter-clockwise and = 270, what is
the velocity of the saw?Solution: The radius vector from A to B is
rB/A = 4j (m).The velocity of B isvB = rB/A = 2( )__i j k0 0 10 1
0__= i (m/s).The coordinates of point C are( , + sin 45) = ( , )
,where = sin1_ (1 + sin 45)_= 29.19.The coordinates of point B are
(0, ) in. The vector from C to B isrC/B = ( . 0) i + ( ( ))j = i +
. j (m)The velocity at point C isvC = vB + BC rC/B = vB +__i j k0 0
BC . .__= ( . BC)i BCj.Since the saw is constrained to move
parallel to the x axis, BCj = 0, from which BC = 0, and the
velocity of the saw isvC = i = i (m/s)[Note: Since the vertical
velocity at B reverses direction at = 270,the angular velocity BC =
0 can be determined on physical groundsby inspection, simplifying
the solution.]CRRsin45 AB355355. 6 mm long. In the position
shown,0.1016 0.0510.051 0.102 m0.1016 0.356 = ( cos 45 )i. 0.1016
0.356 = i 0.284+0.284 0.453 i + 0.453 j + j . ( 0 284 ) BC= 0.453 i
+ 0.453 j . ( 0 284 ) BC0.453 . 0 2840.4530.1016 0.2030.356 cos
0.1016 0.31 0.072 m0.10160.3560.10160 31 0.072 0.1016 0.31 0 1730
31 0 173 00.203 0 173 0.310.310.203 0.638c 2008 Pearson Education
South Asia Pte Ltd. All rights reserved. This publication is
protected by Copyright and 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.Problem
17.47 The disks roll on a plane surface.The angular velocity of the
left disk is 2 rad/s in theclockwise direction. What is the angular
velocity of theright disk?2 rad/s0.91 m0.31 m 0.31 mSolution: The
velocity at the point of contact P of the left diskis zero. The
vector from this point of contact to the center of the leftdisk is
rO/P = j (m). The velocity of the center of the left disk isvO =
rO/P =__i j k0 0 20 0__= i (m/s).The vector from the center of the
left disk to the point of attachmentof the rod is rL/O= i (m). The
velocity of the point of attachmentof the rod to the left disk isvL
= vO + rL/O = i +__i j k0 0 20 0__= i j (m/s),The vector from the
point of attachment of the left disk to the pointof attachment of
the right disk isrR/L = (i cos + j sin ) (m),where = sin1_ _=
19.47.The velocity of the point on attachment on the right disk
isvR = vL + rod rR/L = vL +__i j k0 0 rod.863 1 0__= ( rod)i + ( +
rod)j (m/s).The velocity of point R is also expressed in terms of
the contactpoint Q,vR = RO rR/O = RO(2)__i j k0 0 10 0__= ROi
(m/s).Equate the two expressions for the velocity vR and
separatecomponents:( rod + 2RO)i = 0,( + .863rod)j = 0,from which
RO = 0.65k (rad/s)and rod = 0.707 rad/s.2 rad/syxO
LQRrodRP3560.310.610.310.310.610.310.61 0.610.910.310.9100.61 0.61
.863 00.310.610.610.61 0c 2008 Pearson Education South Asia Pte
Ltd. All rights reserved. This publication is protected by
Copyright and 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.Problem 17.48 The
disk rolls on the curved surface.The bar rotates at 10 rad/s in the
counterclockwisedirection. Determine the velocity of point A.10
rad/s40 mmAxy120 mmSolution: The radius vector from the left point
of attachment ofthe bar to the center of the disk is rbar = 120i
(mm). The velocity ofthe center of the disk isvO = bar rbar =
10(120)__i j k0 0 11 0 0__= 1200j (mm/s).The radius vector from the
point of contact with the disk and the curvedsurface to the center
of the disk is rO/P = 40i (m). The velocity ofthe point of contact
of the disk with the curved surface is zero, fromwhichvO = O rO/P
=__i j k0 0 O40 0 0__= 40Oj.Equate the two expressions for the
velocity of the center of the diskand solve: O = 30 rad/s. The
radius vector from the center of thedisk to point A is rA/O = 40j
(mm). The velocity of point A isvA = vO + O rA/O = 1200j
(30)(40)__i j k0 0 10 1 0__= 1200i + 1200j (mm/s)yxPO10
rad/sProblem 17.49 If AB = 2 rad/s and BC = 4 rad/s,what is the
velocity of point C, where the excavatorsbucket is attached?xyBC5
m5.5 m1.6 mA4 m 3 m 2.3 mBCABvvSolution: The radius vector AB
isrB/A = 3i + (5.5 1.6)j = 3i + 3.9j (m).The velocity of point B
isvB = AB rB/A =__i j k0 0 23 3.9 0__= 7.8i + 6j (m/s).The radius
vector BC is rC/B = 2.3i + (5 5.5)j = 2.3i 0.5j (m).The velocity at
point C isvC = vB + BC rC/B = 7.8i + 6j +__i j k0 0 42.3 0.5 0__=
9.8i 3.2j (m/s)357c 2008 Pearson Education South Asia Pte Ltd. All
rights reserved. This publication is protected by Copyright and
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.Problem 17.50 In Problem 17.49,
if AB = 2 rad/s,what clockwise angular velocity BC will cause
thevertical component of the velocity of point C to be zero?What is
the resulting velocity of point C?xyBC5 m5.5 m1.6 mA4 m 3 m 2.3
mBCABvvSolution: Use the solution to Problem 17.49. The velocity
ofpoint B isvB = 7.8i + 6j (m/s).The velocity of point C isvC = vB
+ BC rC/B= 7.8i + 6j +__i j k0 0 BC2.3 0.5 0__,vC = (7.8 0.5BC)i +
(6 2.3BC)j (m/s).For the vertical component to be zero,BC = 62.3 =
2.61 rad/s clockwise.The velocity of point C isvC = 9.1i
(m/s)Problem 17.51 The steering linkage of a car is shown.Member DE
rotates about xed pin E. The right brakedisk is rigidly attached to
member DE. The tie rod CD ispinned at C and D. At the instant
shown, the Pitman armAB has a counterclockwise angular velocity of
1 rad/s.What is the angular velocity of the right brake disk?180
mm220 mm100 mm460mm340mm70mm200mmSteering linkBrake
disksBACDESolution: Note that the steering link moves in
translation only.Thus vB = vC.vC = vB = vA + AB rB/A = 0 + (1k)
(0.18j) = 0.18ivD = vC + CD rD/C = 0.18i + CDk (0.34i 0.08j)= (0.18
+ 0.08CD)i + (0.34CD)jvE = vD + DE rE/D= (0.18 + 0.08CD)i +
(0.34CD)j + DEk (0.07i + 0.2j)= (0.18 + 0.08CD 0.2DE)i + (0.34CD +
0.07DE)jPoint E is not moving. Equating the components of the
velocity ofpoint E to zero, we have0.18 + 0.08CD 0.2DE = 0, 0.34CD
+ 0.07DE = 0.Solving these equations simultaneously, we nd thatCD =
0.171 rad/s, DE = 0.832 rad/s.The angular velocity of the right
brake disk is thendisk = DE = 0.832 rad/s counterclockwise.358c
2008 Pearson Education South Asia Pte Ltd. All rights reserved.
This publication is protected by Copyright and 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.Problem 17.52 An athlete exercises his arm by raisingthe
mass m. The shoulder joint A is stationary. Thedistance AB is 300
mm, and the distance BC is 400 mm.At the instant shown, AB = 1
rad/s and BC = 2 rad/s.How fast is the mass m
rising?ABmCABBC3060Solution: The magnitude of the velocity of the
point C parallel tothe cable at C is also the magnitude of the
velocity of the mass m. Theradius vector AB is rB/A = 300i (mm).
The velocity of point B isvB = AB rB/A =__i j k0 0 AB300 0 0__=
300j (mm/s).The radius vector BC is rC/B = 400(i cos 60 + j sin 60)
= 200i +346.4j (mm). The velocity of point C isvC = vB + BC rC/B =
300j +__i j k0 0 2200 346.4 0__= 692.8i + 700j (mm/s).The unit
vector parallel to the cable at C is eC = i cos 30 +j sin 30 =
0.866i + 0.5j. The component of the velocity parallel tothe cable
at C isvC eC = 950 mm/s ,which is the velocity of the mass
m.Problem 17.53 The distance AB is 305mm, the distanceBC is 406. 4
mm, AB = 0.6 rad/s, and the mass m isrising at 610 mm/s. What is
the angular velocity BC?Solution: The radius vector AB is rB/A = i
(m). The velocityat point B isvB = AB rB/A =__i j k0 0 0.60 0__= .
j ( /s).The radius vector BC isrC/B = (i cos 60 + j sin 60) = i + .
j (m)The velocity at C isvC = vB + BC rC/B = 7.2j +__i j k0 0
BC0__= BCi + ( + BC)j.The unit vector parallel to the cable at C
iseC = i cos 30 + j sin 60 = 0.866i + 0.5j.The component of the
velocity at C parallel to the cable is|vCP| = vC eC = + . BC + BC +
(m/s).This is also the velocity of the rising mass, from which. BC
+ = ,BC = 1.28 rad/s3590.3050.3052 19 m0.4064 0.2032 0 35310.2032
0.35310.3531 2.19 0.20323 67 1.22 1.14 89 1.1 0.61c 2008 Pearson
Education South Asia Pte Ltd. All rights reserved. This publication
is protected by Copyright and 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.Problem
17.54 Points B and C are in the x y plane.The angular velocity
vectors of the arms AB and BC areAB = 0.2k (rad/s), and BC = 0.4k
(rad/s). What isthe velocity of point C.Ayxz3040BC920 mm760
mmSolution: Locations of Points:A: (0, 0, 0) mB: (0.76 cos 40, 0.76
sin 40, 0) mC: (xB + 0.92 cos 30, yB 0.92 sin 30, 0) mor B: (0.582,
0.489, 0),C: (1.379, 0.0285, 0) mrB/A = 0.582i + 0.489j (m)rC/B =
0.797i 0.460j (m)vA = 0, AB = 0.2k_rads_, BC = 0.4k_rads_vB = vA +
AB rB/AvC = vB + BC rC/BvB = (0.2k) (0.582i + 0.489j)vB = 0.0977i
0.116j (m/s).vC = vB + BC rC/BvC = vB + 0.184i + 0.319j (m/s)vC =
0.282i + 0.202j (m/s).360c 2008 Pearson Education South Asia Pte
Ltd. All rights reserved. This publication is protected by
Copyright and 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.Problem 17.55 If
the velocity at point C of therobotic arm shown in Problem 17.54 is
vC = 0.15i +0.42j (m/s), what are the angular velocities of the
armsAB and BC?Solution: From the solution to Problem 17.54,rB/A =
0.582i + 0.489j (m)rC/B = 0.797i 0.460j (m)vB = ABk rB/A (vA = 0)vC
= vB + BCk rC/BWe are givenvC = 0.15i + 0.42j + 0k (m/s).Thus, we
know everything in the vC equation except AB and BC.vC = ABk rB/A +
BCk rC/BThis yields two scalar equations in two unknowns i and j
components.Solving, we getAB = 0.476k (rad/s),BC = 0.179k
(rad/s).Problem 17.56 The link AB of the robots arm is rotat-ing at
2 rad/s in the counterclockwise direction, the linkBC is rotating
at 3 rad/s in the clockwise direction, andthe link CD is rotating
at 4 rad/s in the counterclockwisedirection. What is the velocity
of point D?xDCyB3020A250 mm300 mm250 mmSolution: The velocity of B
isvB = vA + AB rB/A,or vB = 0 +i j k0 0 20.3 cos 30 0.3 sin 30 0=
0.3i + 0.520j (m/s).The velocity of C isvC = vB + BC rC/Bor vC =
0.3i + 0.520j +i j k0 0 30.25 cos 20 0.25 sin 20 0= 0.557i 0.185j
(m/s).The velocity of D isvD = vC + CD rD/C = 0.557i 0.185j +i j k0
0 40.25 0 0,or vD = 0.557i + 0.815j (m/s).361c 2008 Pearson
Education South Asia Pte Ltd. All rights reserved. This publication
is protected by Copyright and 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.Problem
17.57 The person squeezes the grips of theshears, causing the
angular velocities shown. What isthe resulting angular velocity of
the jaw BD?0.12 rad/s0.12 rad/sB CDE25 mm18 mm25 mmSolution:vD = vC
+ CD rD/C = 0 (0.12 rad/s)k (0.025i + 0.018j) m= (0.00216i 0.003j)
m/svB = vD + BD rB/D= (0.00216i 0.003j) m/s + BDk (0.05i 0.018j) m=
(0.00216 m/s+{0.018 m}BD)i+(0.003 m/s{0.05 m}BD)jFrom symmetry we
know that C and B do not move vertically0.003 m/s {0.05 m}BD = 0 BD
= 0.06 rad/sBD = 0.06 rad/s CWProblem 17.58 Determine the velocity
vW and theangular velocity of the small pulley.50 mm0.6 m/svW100
mmSolution: Since the radius of the bottom pulley is not given,
wecannot use Eq (17.6) (or the equivalent). The strategy is to use
the fact(derived from elementary principles) that the velocity of
the center ofa pulley is the mean of the velocities of the extreme
edges, wherethe edges lie on a line normal to the motion, taking
into account thedirections of the velocities at the extreme edges.
The center rope fromthe bottom pulley to the upper pulley moves
upward at a velocityof vW. Since the small pulley is xed, the
velocity of the center iszero, and the rope to the left moves
downward at a velocity vW,from which the left edge of the bottom
pulley is moving at a velocityvW downward. The right edge of the
bottom pulley moves upwardat a velocity of 0.6 m/s. The velocity of
the center of the bottompulley is the mean of the velocities at the
extreme edges, from whichvW = 0.6 vW2 .Solve: vW = 0.63 = 0.2 m/s
.The angular velocity of the small pulley is = vWr= 0.20.05 = 4
rad/s362c 2008 Pearson Education South Asia Pte Ltd. All rights
reserved. This publication is protected by Copyright and 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.Problem 17.59 Determine the velocity of the
blockand the angular velocity of the small pulley. 51 mm228.6
mm/s76.2 mmSolution: Denote the velocity of the block by vB. The
strategyis to determine the velocities of the extreme edges of a
pulley bydetermining the velocity of the element of rope in contact
with thepulley. The upper rope is xed to the block, so that it
moves to the rightat the velocity of the block, from which the
upper edge of the smallpulley moves to the right at the velocity of
the block. The xed endof the rope at the bottom is stationary, so
that the bottom edge ofthe large pulley is stationary. The center
of the large pulley moves atthe velocity of the block, from which
the upper edge of the bottompulley moves at twice the velocity of
the block (since the velocity ofthe center is equal to the mean of
the velocities of the extreme edges,one of which is stationary)
from which the bottom edge of the smallpulley moves at twice the
velocity of the block. The center of the smallpulley moves to the
right at 9 in/s. The velocity of the center of thesmall pulley is
the mean of the velocities at the extreme edges, fromwhich= vB +
vB= vB,from whichvB = = .The angular velocity of small pulley is
given byi = vBi + j =__i j k0 0 0 0__= vBi i,from which = = 1.5
rad/s3630.2286 0.0510.0510.025 0.0760.0510.0510.076 0.2286 0.152
m/s0.2286 0.051 0.0510.0510.051 0.0510.305 0.22860.051c 2008
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publication is protected by Copyright and 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.Problem 17.60 The device shown is used in the
semi-conductor industry to polish silicon wafers. The wafersare
placed on the faces of the carriers. The outer andinner rings are
then rotated, causing the wafers to moveand rotate against an
abrasive surface. If the outer ringrotates in the clockwise
direction at 7 rpm and the innerring rotates in the
counterclockwise direction at 12 rpm,what is the angular velocity
of the carriers?0.6 mInner ring1.0 mCarriers (3)Outer ring12 rpm7
rpmSolution: The velocity of pt. B is vB = (1 m)0i = 0i.
Thevelocity of pt. A is vA = (0.6 m)ii. ThenvB = vA + C rB/A : 0i =
0.6ii +i j k0 0 C0 0.4 0.The i component of this equation is 0 =
0.6i 0.4C,so C = 0.6i 00.4= 0.6(12 rpm) 7 rpm0.4= 35.5
rpm.xCyBA0ciProblem 17.61 In Problem 17.60, suppose that theouter
ring rotates in the clockwise direction at 5 rpmand you want the
centerpoints of the carriers to remainstationary during the
polishing process. What is thenecessary angular velocity of the
inner ring?0.6 mInner ring1.0 mCarriers (3)Outer ring12 rpm7
rpmSolution: See the solution of Problem 17.60. The velocity of pt.
Bis vB = 0i and the angular velocity of the carrier isC = 0.6i 00.4
.We want the velocity of pt. C to be zero:vC = 0 = vB + C rC/B = 0i
+i j k0 0 C0 0.2 0.From this equation we see that C = 50. Therefore
the velocity ofpt. A isvA = vC + C rA/C= 0 + (50k) (0.2j)= 0iWe
also know that vA = (0.6 m)ii,so i = 00.6 = 5 rpm0.6 = 8.33
rpm.364c 2008 Pearson Education South Asia Pte Ltd. All rights
reserved. This publication is protected by Copyright and 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.Problem 17.62 The ring gear is xed and the
huband planet gears are bonded together. The connectingrod rotates
in the counterclockwise direction at 60 rpm.Determine the angular
velocity of the sun gear and themagnitude of the velocity of point
A.A240 mm720 mm340mmPlanet gearConnectingrodHub gear140 mmSun
gearRing gearSolution: Denote the centers of the sun, hub and
planet gears bythe subscripts Sun, Hub, and Planet, respectively.
Denote the contactpoints between the sun gear and the planet gear
by the subscript SPand the point of contact between the hub gear
and the ring gear by thesubscript HR. The angular velocity of the
connecting rod is CR =6.28 rad/s. The vector distance from the
center of the sun gear to thecenter of the hub gear is rHub/Sun =
(720 140)j = 580j (mm). Thevelocity of the center of the hub gear
isvHub = CR rHub/Sun =__i j k0 0 20 580 0__= 3644i (mm/s)The
angular velocity of the hub gear is found fromvHR = 0 = vHub + Hub
140j =__i j k0 0 Hub0 140 0__+ vHub= 3644i 140Hubi,from whichHub =
3644140 = 26.03 rad/s.This is also the angular velocity of the
planet gear. The linear velocityof point A isvA = Hub (340 140)j
=__i j k0 0 26.030 200 0__= 5206i (mm/s)The velocity of the point
of contact with the sun gear isvPS = Hub (480j) =__i j k0 0 26.030
480 0__= 12494.6i (mm/s).The angular velocity of the sun gear is
found fromvPS = 12494.6i = Sun (240j) =__i j k0 0 Sun0 240 0__=
240Suni,from which Sun = 12494.6240 = 52.06 rad/sAHRHubSunCRHub365c
2008 Pearson Education South Asia Pte Ltd. All rights reserved.
This publication is protected by Copyright and 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.Problem 17.63 The large gear is xed. Bar AB has
acounterclockwise angular velocity of 2 rad/s. What arethe angular
velocities of bars CD and DE?4 cm2 rad/s16 cmAB C DE10 cm4 cm10
cmSolution: The strategy is to express vector velocity of point D
interms of the unknown angular velocities of CD and DE, and thento
solve the resulting vector equations for the unknowns. The
vectordistance AB is rB/A = 14j (cm) The linear velocity of point B
isvB = rB/A =__i j k0 0 20 14 0__= 28i (cm/s).The lower edge of
gear B is stationary. The radius vector from thelower edge to B is
rB = 4j (cm), The angular velocity of B isvB = B rB =__i j k0 0 B0
4 0__= 4Bi (cm/s),from which B = vB4 = 7 rad/s. The vector distance
from B to Cis rC/B = 4i (cm). The velocity of point C isvC = vB + B
rC/B = 28i +__i j k0 0 74 0 0__= 28i + 28j (cm/s).The vector
distance from C to D is rD/C = 16i (cm), and from E toD is rD/E =
10i + 14j (cm). The linear velocity of point D isvD = vC + CD rD/C
= 28i + 28j +__i j k0 0 CD16 0 0__= 28i + (16CD + 28)j (cm/s).The
velocity of point D is also given byvD = DE rD/E =__i j k0 0 DE10
14 0__= 14DEi 10DEj (cm/s).Equate components:(28 + 14DE)i = 0,(16CD
+ 28 + 10DE)j = 0.Solve: DE = 2 rad/s , CD = 3 rad/s .The negative
sign means a clockwise rotation.366c 2008 Pearson Education South
Asia Pte Ltd. All rights reserved. This publication is protected by
Copyright and 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.Problem 17.64 If
the bar has a clockwise angularvelocity of 10 rad/s and vA = 20
m/s, what are thecoordinates of its instantaneous center of the
bar, andwhat is the value of vB?1 m 1 mA ByxA Bv vSolution: Assume
that the coordinates of the instantaneous centerare (xC, yC), = k =
10k. The distance to point A is rA/C =(1 xC)i + yCj. The velocity
at A isvA = 20j = rA/C =__i j k0 0 1 xC yC 0__= yCi (1 xC)j,from
which yCi = 0, and (20 + (1 xC))j = 0.Substitute = 10 rad/s to
obtain yC = 0 and xC = 3 m. The coordi-nates of the instantaneous
center are (3, 0) (m) . The vector distancefrom C to B is rB/C = (2
3)i = i (m). The velocity of point B isvB = rB/C =__i j k0 0 101 0
0__= 10(j) = 10j (m/s)Problem 17.65 In Problem 17.64, if vA = 24
m/sand vB = 36 m/s, what are the coordinates of theinstantaneous
center of the bar, and what is its angularvelocity?1 m 1 mA ByxA Bv
vSolution: Let (xC, yC) be the coordinates of the instantaneous
cen-ter. The vectors from the instantaneous center and the points A
andB are rA/C = (1 xC)i + yCj (m) and rB/C = (2 xC)i + yCj.
Thevelocity of A is given byvA = 24j = AB rA/C =__i j k0 0 AB1 xC
yC 0__= AByCi + AB(1 xC)j (m/s)The velocity of B isvB = 36j = AB
rB/C =__i j k0 0 AB2 xC yC 0__= yCABi + AB(2 xC)j (m/s).Separate
components:24 AB(1 xC) = 0,36 AB(2 xC) = 0,AByC = 0.Solve: xC = 1,
yC = 0,and AB = 12 rad/s counter clockwise.367c 2008 Pearson
Education South Asia Pte Ltd. All rights reserved. This publication
is protected by Copyright and 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.Problem
17.66 The velocity of point O of the bat isvO = i j (m/s), and the
bat rotates about the zaxis with a counterclockwise angular
velocity of 4 rad/s.What are the x and y coordinates of the bats
instanta-neous center?yxOSolution: Let (xC, yC) be the coordinates
of the instantaneous cen-ter. The vector from the instantaneous
center to point O is rO/C =xCi yCj (m). The velocity of point O
isv0 = i . j = rO/C =__i j k0 0 xC yC 0__= yCi xCj (m/s).Equate
terms and solve:yC = = 4 = ,xC = .= .4 = . ,from which the
coordinates are ( . , . )3681.83 4.271.83 4 271.83 1.830.46 m4 27 4
271 07 m1 07 0 46 mc 2008 Pearson Education South Asia Pte Ltd. All
rights reserved. This publication is protected by Copyright and
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.Problem 17.67 Points A and B of
the 1-m bar slide onthe plane surfaces. The velocity of B is vB =
2i (m/s).(a) What are the coordinates of the instantaneous cen-ter
of the bar?(b) Use the instantaneous center to determine
thevelocity at A.xyABG70Solution:(a) A is constrained to move
parallel to the y axis, and B is con-strained to move parallel to
the x axis. Draw perpendiculars to thevelocity vectors at A and B.
From geometry, the perpendicularsintersect at(cos 70, sin 70) =
(0.3420, 0.9397) m .(b) The vector from the instantaneous center to
point B isrB/C = rB rC = 0.3420i (0.3420i + 0.9397j) = 0.9397jThe
angular velocity of bar AB is obtained fromvB = 2i = AB rB/C =__i j
k0 0 AB0 0.9397 0__= AB(0.9397)i,from which AB = 20.9397 = 2.13
rad/s.The vector from the instantaneous center to point A is rA/C
=rA rC = 0.3420i (m). The velocity at A isvA = AB rA/C =__i j k0 0
2.130.3420 0 0__= 0.7279j (m/s).70yACxB369c 2008 Pearson Education
South Asia Pte Ltd. All rights reserved. This publication is
protected by Copyright and 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.Problem
17.68 The bar is in two-dimensional motionin the x y plane. The
velocity of point A is vA =8i (m/s), and B is moving in the
direction parallel to thebar. Determine the velocity of B (a) by
using Eq. (17.6)and (b) by using the instantaneous center of the
bar.xyBA4 m30Solution:(a) The unit vector parallel to the bar iseAB
= (i cos 30 + j sin 30) = 0.866i + 0.5j.The vector from A to B is
rB/A = 4eAB = 3.46i + 2j ( ). Thevelocity of point B isvB = vA + AB
rB/A = 8i +__i j k0 0 AB3.46 2 0__vB = (8 2AB)i + 3.46ABj.But vB is
also moving parallel to the bar,vB = vBeAB = vB(0.866i +
0.5j).Equate, and separate components:(8 2AB 0.866vB)i = 0,(3.46AB
0.5vB)j = 0.Solve: AB = 1 rad/s, vB = 6.93 m/s, from whichvB =
vBeAB = 6i + 3.46j (m/s)(b) Let (xC, yC) be the coordinates of the
instantaneous center. Thevector from the center to A isrA/C = rA rC
= rC = xCi yCj (m).The vector from the instantaneous center to B
isrB/C = rB rC = (3.46 xC)i + (2 yC)j.The velocity of point A isvA
= 8i = AB rA/C =__i j k0 0 ABxC yC 0__= AByCi ABxCj (m/s).From
which xC = 0 , and AByC = 8. The velocity ofpoint B isvB = vBeAB =
AB rB/C =__i j k0 0 AB3.46 xC 2 yC 0__= AB(2 yC)i + AB(3.46
xC)j.Equate terms and substituteAByC = 8, and xC = 0, to obtain:
(0.866vB + 2AB 8)i = 0,and (0.5vC 3.46AB)j = 0. These equations are
algebraicallyidentical with those obtained in Part (a) above (as
can be shownby multiplying all terms by 1). Thus AB = 1 rad/s, vB
=6.93 (m/s), and the velocity of B is that obtained in Part (a)vB =
vBeAB = 6i + 3.46j ( /s) .370mmc 2008 Pearson Education South Asia
Pte Ltd. All rights reserved. This publication is protected by
Copyright and 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.Problem 17.69 Point
A of the bar is moving at 8 m/sin the direction of the unit vector
0.966i 0.259j, andpoint B is moving in the direction of the unit
vector0.766i + 0.643j.(a) What are the coordinates of the bars
instanta-neous center?(b) What is the bars angular velocity?xyBA2
m30Solution: Assume the instantaneous center Q is located at (x,
y).ThenvA = rA/Q, vB = rB/Q(8 m/s)(0.966i 0.259j) = k (xi
yj)vB(0.766i + 0.643j) = k ([{2 m} cos 30 x]i+ [{2 m/s} sin 30
y]j)Expanding we have the four equations7.73 m/s = y2.07 m/s =
x0.766vB = (y 1 m)0.643vB = (1.73 m x)___ x = 0.623 my = 2.32 m371c
2008 Pearson Education South Asia Pte Ltd. All rights reserved.
This publication is protected by Copyright and 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.Problem 17.70 Bar AB rotates with a counterclock-wise
angular velocity of 10 rad/s. At the instant shown,what are the
angular velocities of bars BC and CD? (SeeActive Example 17.4.)A10
rad/sDB C2 m1 m2 mSolution: The location of the instantaneous
center for BC is shown,along with the relevant distances. Using the
concept of the instanta-neous centers we havevB = (10)(2) = BC(4)BC
= 5 rad/svC = (4.472)(5) = 2.236(CD)CD = 10 rad/sWe determine the
directions by inspectionBC = 5 rad/s clockwise.CD = 10 rad/s
counterclockwise.Problem 17.71 Use instantaneous centers to
determinethe horizontal velocity of B.6 cm12 cmAB1 rad/sOSolution:
The instantaneous center of OA lies at O, by denition,since O is
the point of zero velocity, and the velocity at point A isparallel
to the x-axis:vA = OA rA/O =__i j k0 0 OA0 6 0__= 6i ( ).A line
perpendicular to this motion is parallel to the y axis. The point
Bis constrained to move on the x axis, and a line perpendicular to
thismotion is also parallel to the y axis. These two lines will not
intersectat any nite distance from the origin, hence at the instant
shown theinstantaneous center of bar AB is at innity and the
angular velocity ofbar AB is zero. At the instant shown, the bar AB
translates only, fromwhich the horizontal velocity of B is the
horizontal velocity at A:vB = vA = 6i ( ) .372cm/scm/sc 2008
Pearson Education South Asia Pte Ltd. All rights reserved. This
publication is protected by Copyright and 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.Problem 17.72 When the mechanism in Problem17.71 is in the
position shown here, use instantaneouscenters to determine the
horizontal velocity of B.AB1 rad/sOSolution: The strategy is to
determine the intersection of linesperpendicular to the motions at
A and B. The velocity of A is parallelto the bar AB. A line
perpendicular to the motion at A will be parallelto the bar OA.
From the dimensions given in Problem 17.71, the lengthof bar AB is
rAB = 62+ 122= 13.42 cm. Consider the triangle OAB.The interior
angle at B is = tan1_ 6rAB_= 24.1,and the interior angle at O is =
90 = 65.9. The unit vectorparallel to the handle OA is eOA = i cos
+ j sin , and a point on theline is LOA = LOAeOA, where LOA is the
magnitude of the distanceof the point from the origin. A line
perpendicular to the motion at Bis parallel to the y axis. At the
intersection of the two linesLOA cos = rABcos ,from which LOA = 36
cm. The coordinates of the instantaneous centerare (14.7, 32.9)
(in.).Check: From geometry, the triangle OAB and the triangle
formed bythe intersecting lines and the base are similar, and thus
the interiorangles are known for the larger triangle. From the law
of sinesLOAsin 90 = rOBsin = rABsin cos = 36 cm,and the coordinates
follow immediately from LOA = LOAeOA. check.The vector distance
from O to A is rA/O = 6(i cos + j sin ) =2.450i + 5.478j ( ). The
angular velocity of the bar AB is determinedfrom the known linear
velocity at A.vA = OA rA/O =__i j k0 0 12.450 5.477 0__= 5.48i
2.45j (cm/s).The vector from the instantaneous center to point A
isrA/C = rOA rC = 6eOA (14.7i + 32.86j)= 12.25i 27.39j ( )The
velocity at point A isvA = AB rA/C =__i j k0 0 AB12.25 27.39 0__=
AB(27.39i 12.25j) ( ).CBAEquate the two expressions for the
velocity at point A and separatecomponents, 5.48i = 27.39AB, 2.45j
= 12.25ABj (one of theseconditions is superuous) and solve to
obtain AB = 0.2 rad/s, coun-terclockwise.[Check: The distance OA is
6 cm. The magnitude of the velocity at Ais OA(6) = (1)(6) = 6 cm/s.
The distance to the instantaneous cen-ter from O is 14.72+ 32.92=
36 cm, and from C to A is (36 6) = 30 cm from which 30 AB = 6 cm/s,
from which AB = 0.2 rad/s.check.]. The vector from the
instantaneous center to point B isrB/C = rB rC = 14.7i (14.7i +
32.86j = 32.86j) ( )The velocity at point B isvB = AB rB/C =__i j
k0 0 0.20 32.86 0__ = 6.57i ( )373cmcm/scmcm/scmc 2008 Pearson
Education South Asia Pte Ltd. All rights reserved. This publication
is protected by Copyright and 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.Problem
17.73 The angle = 45, and the bar OQ isrotating in the
counterclockwise direction at 0.2 rad/s.Use instantaneous centers
to determine the velocity ofthe sleeve P.2 mOQP2 mSolution: The
velocity of Q isvQ = 20Q = 2(0.2) = 0.4 m/s.Therefore|PQ| = vQ2 m =
0.42 = 0.2 rad/s(clockwise) and |vP| = 22PQ = 0.566 m/s (vP is to
the left).P22 PQOQOQCInstantaneouscenter ofbar PQpQmProblem 17.74
Bar AB is rotating in the counterclock-wise direction at 5 rad/s.
The disk rolls on the horizontalsurface. Determine the angular
velocity of bar BC.C0.2 m0.4 m 0.2 m 0.2 mDAB5 rad/sSolution: First
locate the instantaneous centerFrom the geometry we haveBCAE=
QBQA0.6 m0.4 m = QBQB 0.22+ 0.42mSolving we nd BQ = 1.342 mNowvB =
AB(AB) = BC(QB)BC = ABQB(5 rad/s) =0.22+0.42m1.342 m (5 rad/s)
=1.67 rad/s CCWABCEQ374c 2008 Pearson Education South Asia Pte Ltd.
All rights reserved. This publication is protected by Copyright and
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.Problem 17.75 Bar AB rotates at
6 rad/s in the clock-wise direction. Use instantaneous centers to
determinethe angular velocity of bar BC.6 rad/sABC4 cm4 cm 10 cm3
cmSolution: Choose a coordinate system with origin at A and y
axisvertical. Let C
denote the instantaneous center. The instantaneouscenter for bar
AB is the point A, by denition, since A is the point ofzero
velocity. The vector AB is rB/A = 4i + 4j (cm). The velocity atB
isvB = AB rB/A =__i j k0 0 64 4 0__= 24i 24j (cm/s).The unit vector
parallel to AB is also the unit vector perpendicular tothe velocity
at B,eAB = 12(i + j).The vector location of a point on a line
perpendicular to the velocityat B is LAB = LABeAB, where LAB is the
magnitude of the distancefrom point A to the point on the line. The
vector location of a pointon a perpendicular to the velocity at C
is LC = (14i + yj) where y isthe y-coordinate of the point
referenced to an origin at A. When thetwo lines intersect,LAB2 i =
14i,and y = LAB2 = 14from which LAB = 19.8 , and the coordinates of
the instantaneouscenter are (14, 14) (cm).[Check: The line AC
is the hypotenuse of a right triangle with a baseof 14 cm and
interior angles of 45, from which the coordinates of C
are (14, 14) cm check.]. The angular velocity of bar BC is
determinedfrom the known velocity at B. The vector from the
instantaneous centerto point B isrB/C = rB rC = 4i + 4j 14i 14j =
10i 10j.The velocity of point B isvB = BC rB/C =__i j k0 0 BC10 10
0__= BC(10i 10j) (cm/s).Equate the two expressions for the
velocity: 24 = 10BC, from whichBC = 2.4 rad/s ,ABCC375cmc 2008
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publication is protected by Copyright and 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.Problem 17.76 The crank AB is rotating in the clock-wise
direction at 2000 rpm (revolutions per minute).(a) At the instant
shown, what are the coordinates ofthe instantaneous center of the
connecting rod BC?(b) Use instantaneous centers to determine the
angularvelocity of the connecting rod BC at theinstant shown. BA50
mm50 mm175 mmyxCSolution:AB = 2000 rpm_2 radrev__min60 s_= 209
rad/s(a) The instantaneous center of BC is located at point QQ =
(225, 225) mm(b) vB = AB(AB) = BC(QB)BC = ABQBAB = 502 mm(225 50)2
mm(209 rad/s) = 59.8 rad/svC = BC(QC) = (59.8 rad/s)(0.225 m) =
13.5 m/svC = (13.5 m/s)iABCQ225 mm225 mm376c 2008 Pearson Education
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protected by Copyright and 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.Problem
17.77 The disks roll on the plane surface.The left disk rotates at
2 rad/s in the clockwise direction.Use the instantaneous centers to
determine the angularvelocities of the bar and the right disk.2
rad/s1 1 m3 mmSolution: Choose a coordinate system with the origin
at the pointof contact of the left disk with the surface, and the x
axis parallel tothe plane surface. Denote the point of attachment
of the bar to the leftdisk by A, and the point of attachment to the
right disk by B. Theinstantaneous center of the left disk is the
point of contact with thesurface. The vector distance from the
point of contact to the point A isrA/P = i + j (m). The velocity of
point A isvA = LD rA/P =__i j k0 0 21 1 0__= 2i 2j (m/s).The point
on a line perpendicular to the velocity at A is LA = LA(i +j),
where LA is the distance of the point from the origin. The point
Bis at the top of the right disk, and the velocity is constrained
to beparallel to the x axis. A point on a line perpendicular to the
velocityat B is LB = (1 + 3 cos )i + yj ( ), where = sin1_13_=
19.5.At the intersection of these two lines LA = 1 + 3 cos = 3.83
ft,and the coordinates of the instantaneous center of the bar are
(3.83,3.83) (m). The angular velocity of the bar is determined from
theknown velocity of point A. The vector from the instantaneous
centerto point A isrA/C = rA rC = i + j 3.83i 3.83j = 2.83i 2.83j
(m).The velocity of point A isvA = AB rA/C =__i j k0 0 AB2.83 2.83
0__= AB(2.83i 2.83j) (m/s).Equate the two expressions and solve:AB
= 22.83 = 0.7071 (rad/s) counterclockwise.The vector from the
instantaneous center to point B isrB/C = rB rC = (1 + 3 cos )i + 2j
3.83i 3.83j = 1.83j.The velocity of point B isvB = AB rB/A =__i j
k0 0 0.70710 1.83 0__= 1.294i (m/s).Using the xed center at point
of contact:vB = RD rB/P =__i j k0 0 RD0 2 0__= 2RDi (m/s).Equate
the two expressions for vB and solve:RD = 0.647 rad/s,
clockwise.ABP PC377mc 2008 Pearson Education South Asia Pte Ltd.
All rights reserved. This publication is protected by Copyright and
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.Problem 17.78 Bar AB rotates at
12 rad/s in the clock-wise direction. Use instantaneous centers to
determinethe angular velocities of bars BC and CD.ACBD350mm200mm300
mm 350 mm12 rad/sSolution: Choose a coordinate system with the
origin at A and thex axis parallel to AD. The instantaneous center
of bar AB is point A,by denition. The velocity of point B is normal
to the bar AB. Usingthe instantaneous center A and the known
angular velocity of bar ABthe velocity of B isvB = rB/A =__i j k0 0
120 200 0__= 2400i (mm/s).The unit vector perpendicular to the
velocity of B is eAB = j, anda point on a line perpendicular to the
velocity at B is LAB = LABj(mm). The instantaneous center of bar CD
is point D, by denition.The velocity of point C is constrained to
be normal to bar CD. Theinterior angle at D is 45, by inspection.
The unit vector parallel toDC (and perpendicular to the velocity at
C) iseDC = i cos 45 + j sin 45 =_ 12_(i + j).The point on a line
parallel to DC isLDC =_650 LDC2_i + LDC2 j (mm).At the intersection
of these lines LAB = LDC, from which_650 LDC2_= 0and LAB = LDC2
,from which LDC = 919.2 mm, and LAB = 650 mm. The coordinatesof the
instantaneous center of bar BC are (0, 650) (mm). Denote thiscenter
by C
. The vector from C
to point B isrB/C = rB rC = 200j 650j = 450j.The vector from
C
to point C isrC/C = 300i + 350j 650j = 300i 300j (mm).The
velocity of point B isvB = BC rB/C =__i j k0 0 BC0 450 0__= 450BCi
(mm/s).Equate and solve: 2400 = 450BC, from whichBC = 2400450 =
5.33 (rad/s) .The angular velocity of bar CD is determined from the
known velocityat point C. The velocity at C isvC = BC rC/C =__i j
k0 0 5.33300 300 0__= 1600i + 1600j (mm/s).The vector from D to
point C is rC/D = 350i + 350j (mm). Thevelocity at C isvC = CD rC/D
=__i j k0 0 CD350 350 0__= 350CDi 350CDj (mm/s).Equate and solve:
CD = 4.57 rad/s clockwise.CBD AC378c 2008 Pearson Education South
Asia Pte Ltd. All rights reserved. This publication is protected by
Copyright and 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.Problem 17.79 The
horizontal member ADE support-ing the scoop is stationary. The link
BD is rotating in theclockwise direction at 1 rad/s. Use
instantaneous centersto determine the angular velocity of the
scoop.CBD EAScoop0.61 m0.45 m1.52 m0.31 m0.76 mSolution: The
distance from D to B is rBD = 2+ 2= .The distance from B to H isrBH
= cos 63.4 rBD = . ,and the distance from C to H is rCH = . tan
63.4 rCE = .The velocity of B is vB = rBDBD = ( )(1) = .BC = vBrBH=
..= 0.4 rad/s.The velocity of C is vc = rCHBC = ( )(0.4) = .angular
velocity of the scoop isCE = vCrCE= .= 1.47
rad/sInstantaneouscenter ofbar BCBCCEBD = 1rad/sCErCHvCrBHHvBDB
63.41.07 m3790.31 0.61 0 68 m.1.071 7 m1 07 1 68 m.0.68 0 68 m/s.
Therefore0 681 71.68 0 67, so the0 670.46c 2008 Pearson Education
South Asia Pte Ltd. All rights reserved. This publication is
protected by Copyright and 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.Problem
17.80 The disk is in planar motion. Thedirections of the velocities
of points A and B are shown.The velocity of point A is vA = 2
m/s.(a) What are the coordinates of the disks instanta-neous
center?(b) Determine the velocity vB and the disks angularvelocity.
BAxy(0.5, 0.4)m30 70BASolution: = krc/A = xci + ycjrc/B = (xc xB)i
+ (yc yB)jThe velocity of C, the instantaneous center, is zero.vc =
0 = vA + k rc/A_0 = vAx yc (1)0 = vAy + xc (2)where vAx = vA cos 30
= 2 cos 30 m/svAy = vA sin 30 = 1 m/svBx = vB cos 70vBy = vB sin
70Also vc = 0 = vB + k rc/B0 = vB cos 70 (yc yB) (3)0 = vB sin 70 +
(xc xB) (4)Eqns (1) (4) are 4 eqns in the four unknowns , vB, xc,
and yc.Solving, = 2.351 rad/s, = 2.351k rad/s,vB = 2.31 m/s,xc =
0.425 m,yc = 0.737 m.ABCvAvBrC/ArC/Byx7030(0.5, 0.4) m380c 2008
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publication is protected by Copyright and 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.Problem 17.81 The rigid body rotates about the z axiswith
counterclockwise angular velocity = 4 rad/s andcounterclockwise
angular acceleration = 2 rad/s2. Thedistance rA/B = 0.6 m.(a) What
are the rigid bodys angular velocity andangular acceleration
vectors?(b) Determine the acceleration of point A relative topoint
B, rst by using Eq. (17.9) and then by usingEq. (17.10).xyB
ArA/BSolution:(a) By denition, = 4k, = 6k.(b) aA/B = ( rA/B) +
rA/BaA/B = 4k (4k 0.6i) + 2k 0.6iaA/B = 9.6i + 1.2j (m/s2).Using
Eq. (17.10),aA/B = rA/B 2rA/B= 2k 0.6i 16(0.6)iaA/B = 9.6i + 1.2j
(m/s2).yx+381c 2008 Pearson Education South Asia Pte Ltd. All
rights reserved. This publication is protected by Copyright and
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.Problem 17.82 The bar rotates
with a counterclock-wise angular velocity of 5 rad/s and a
counterclockwiseangular acceleration of 30 rad/s2. Determine the
accel-eration of A (a) by using Eq. (17.9) and (b) by usingEq.
(17.10).xyA2 m30 rad/s25 rad/s30Solution:(a) Eq. (17.9): aA = aB +
rA/B + ( rA/B).Substitute values:aB = 0. = 30k (rad/s2),rA/B = 2(i
cos 30 + j sin 30) = 1.732i + j (m). = 5k (rad/s).Expand the cross
products: rA/B =__i j k0 0 301.732 1 0__= 30i + 52j (m/s2). rA/B
=__i j k0 0 51.732 1 0__= 5i + 8.66j (m/s). ( rA/B) =__i j k0 0 55
8.66 0__= 43.3i 25j (m/s2).Collect terms: aA = 73.3i + 27j (m/s2)
.(b) Eq. (17.10): aA = aB + rA/B 2rA/B.Substitute values, and
expand the cross product as in Part (b) toobtainaA = 30i + 52j
(52)(1.732i + j) = 73.3i + 27j (m/s2)382c 2008 Pearson Education
South Asia Pte Ltd. All rights reserved. This publication is
protected by Copyright and 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.Problem
17.83 The bar rotates with a counterclock-wise angular velocity of
20 rad/s and a counterclockwiseangular acceleration of 6 rad/s2.(a)
By applying Eq. (17.10) to point A and the xedpoint O, determine
the acceleration of A.(b) By using the result of part (a) and Eq.
(17.10),to points A and B, determine the accelerationpoint B. xyA
BO1 m 1 m20 rad/s 6 rad/s2Solution:(a) aA = a0 + rA/O 2rA/Owhere =
20k rad/s = 6k rad/s2rA/O = 1i, and aA = 0aA = O + 6k 1i 400(1i)aA
= 400i + 6j (m/s2).(b) aB = aA + rB/A 2rB/AwhererB/A = 1iaB = 400i
+ 6j + 6k 1i 400(1i)aB = 800i + 12j (m/s2).Problem 17.84 The
helicopter is in planar motion inthe x y plane. At the instant
shown, the position of itscenter of mass G is x = 2 m, y = 2.5 m,
its velocity isvG = 12i + 4j (m/s), and its acceleration is aG = 2i
+3j (m/s2). The position of point T where the tail rotoris mounted
is x = 3.5 m, y = 4.5 m. The helicoptersangular velocity is 0.2
rad/s clockwise, and its angularacceleration is 0.1
rad/s2counterclockwise. What is theacceleration of point T
?yxTGSolution: The acceleration of T isaT = aG + rT/G 2rT/G;aT = 2i
+ 3j +i j k0 0 0.15.5 2 0 (0.2)2(5.5i + 2j)= 2.02i + 2.37j
(m/s2).383c 2008 Pearson Education South Asia Pte Ltd. All rights
reserved. This publication is protected by Copyright and 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.Problem 17.85 Point A of the rolling disk is
movingtoward the right and accelerating toward the right.
Themagnitude of the velocity of point C is 2 m/s, and themagnitude
of the acceleration of point C is 14 m/s2.Determine the
acceleration of points B and D. (SeeActive Example
17.5.)xyA45CDB300 mmSolution: First the velocity analysisvC = vA +
rC/A= ri k (ri) = ri + rjvc =_(r)2+ (r)2=2r = v2r= 2 m/s2(0.3 m) =
4.71 rad/s.Now the acceleration analysisaC = aA + rC/A 2rC/A= ri k
(ri) 2(ri) = (r + 2r)i + (r)jaC =_(r + 2r)2+ (r)2= r_( + 2)2+ 2(14
m/s2)2= (0.3 m)2[( + [4.71]2)2+ 2]Solving this quadratic equation
for we nd = 20.0 rad/s2.NowaB = r2j = (0.3 m)(4.71 rad/s)2j aB =
6.67j (m/s2)aD = aA + rD/A 2rD/A= ri k (r cos 45i + 4 sin 45j) 2(r
cos 45i + 4 sin 45j)= (r[1 + sin 45] + r2cos 45)i + (r cos 45 r2sin
45)jPutting in the numbers we nd aD = 14.9i 0.480j (m/s2).384c 2008
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publication is protected by Copyright and 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.Problem 17.86 The disk rolls on the circular surfacewith a
constant clockwise angular velocity of 1 rad/s.What are the
accelerations of points A and B?Strategy: Begin by determining the
acceleration of thecenter of the disk. Notice that the center moves
in a cir-cular path and the magnitude of its velocity is
constant.0.4 mAB 1.2 myxSolution:vB = 0v0 = vB + k rO/B = (1k)
(0.4j)v0 = 0.4 i m/sPoint O moves in a circle at constant speed.
The acceleration of O isa0 = v20/(R + r)j = (0.16)/(1.2 + 0.4)ja0 =
0.1j (m/s2).aB = a0 2rB/O = 0.1j (1)2(0.4)jaB = 0.3j (m/s2).aA = a0
2rA/O = 0.1j (1)2(0.4)jaA = 0.5j (m/s2).Problem 17.87 The length of
the bar is L = 4 m andthe angle = 30. The bars angular velocity is
=1.8 rad/s and its angular acceleration is = 6 rad/s2.The endpoints
of the bar slide on the plane surfaces.Determine the acceleration
of the midpoint G.Strategy: Begin by applying Eq. (17.10) to the
end-points of the bar to determine their
accelerations.LGvauyxSolution: Call the top point D and the bottom
point B.aD = aB + rD/B 2rD/BPut in the known constraintsaDj = aBi +
k L(sin i + cos j) 2L(sin i + cos j)aDj = (aB Lcos + 2Lsin )i +
(Lsin 2Lcos )jEquating components we haveaB = Lcos 2Lsin = (6)(4)
cos 30 (1.8)2(4) sin 30= 14.3 m/s2aD = Lsin 2Lcos = (6)(4) sin 30
(1.8)2(4) cos 30= 23.2 m/s2Now we can use either point as a base
point to nd the accelerationof point G. We will use B as the base
point.aG = aB + rG/B 2rGB= 14.3i + 6k (2)(sin 30i + cos 30j)
(1.8)2(2)(sin 30i + cos 30j)aG = 7.15i 11.6j( 2.)385m/sc 2008
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publication is protected by Copyright and 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.Problem 17.88 The angular velocity and angularacceleration
of bar AB are AB = 2 rad/s and AB =10 rad/s2. The dimensions of the
rectangular plate are1 m 2 m. What are the angular velocity and
angularacceleration of the rectangular plate?1 mABCD4545 ABAB 1.67
mSolution: The instantaneous center for bar AB is point B, by
def-inition. The instantaneous center for bar CD is point D, by
denition.The velocities at points A and C are normal to the bars AB
andCD, respectively. However, by inspection these bars are parallel
atthe instant shown, so that lines perpendicular to the velocities
at Aand C will never intersect the instantaneous center of the
plate ACis at innity, hence the plate only translates at the
instant shown, andAC = 0 . If the plate is not rotating, the
velocity at every point onthe plate must be the same, and in
particular, the vector velocity at Aand C must be identical. The
vector A/B isrA/B = i cos 45 j sin 45 =_12_(i + j) (m).The velocity
at point A isvA = AB rA/B = AB2__i j k0 0 11 1 0__= 2(i j)
(m/s).The vector C/D isrC/D = (i cos 45 j sin 45) = 1.179(i + j)
(m).The velocity at point C isvC = 1.179CD__i j k0 0 11 1 0__=
1.179CD(i j) (m/s).Equate the velocities vC = vA, separate
components and solve: CD =1.2 rad/s. Use Eq. (17.10) to determine
the accelerations. The accel-eration of point A isaA = AB rA/B
2ABrA/B = 102__i j k0 0 11 1 0__+_ 222_(i + j)= 9.9i 4.24j ( 2).The
acceleration of point C relative to point A isaC = aA + AC rC/A =
aA +__i j k0 0 AC2 0 0__= 9.9i + (2AC 4.24)j ( 2).The acceleration
of point C relative to point D is aC = aD + CD rC/D 2CDrC/D.
Not