rigid body dynamics (ch17-21)

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Problem 17.1 In Active Example 17.1, suppose that at a given instant the hook H is moving downward at 2 m/s. What is the angular velocity of gear A at that instant? A H 50 mm B 200 mm 100 mm Solution: The angular velocity of gear B is ω B = v H r H = 2 m/s 0.1m = 20 rad/s. The gears are connected through the common velocity of the contact points r B ω B = r A ω A ω A = r B r A ω B = 0.2m 0.05 m (20 rad/s) = 80 rad/s. ω A = 80 rad/s counterclockwise. Problem 17.2 The angle θ (in radians) is given as a function of time by θ = 0.2πt 2 . At t = 4 s, determine the magnitudes of (a) the velocity of point A and (b) the tangential and normal components of acceleration of point A. A 2 m θ Solution: We have θ = 0.2πt 2 , ω = dt = 0.4πt, α = dt = 0.4π. Then (a) v = = (2)(0.4π)(4) = 10.1 m/s. v = 10.1 m/s. (b) a n = 2 = (2)[(0.4π)(4)] 2 = 50.5 m/s 2 , a t = = (2)(0.4π) = 2.51 m/s 2 . a n = 50.5 m/s 2 , a t = 2.51 m/s 2 . Problem 17.3 The mass A starts from rest at t = 0 and falls with a constant acceleration of 8 m/s 2 . When the mass 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 at the outer edge of the pulley. 100 mm A Solution: We have a = 8 m/s 2 ,v = 2as = 2(8 m/s 2 )(1m) = 4 m/s, ω = v r = 4 m/s 0.1 m = 40 rad/s, α = a r = 8 m/s 2 0.1 m = 80 rad/s 2 . (a) ω = 40 rad/s. (b) a t = = (0.1 m)(80 rad/s 2 ) = 8 m/s 2 , a n = 2 = (0.1 m)(40 rad/s) 2 = 160 m/s 2 . a t = 8 m/s 2 , a n = 160 m/s 2 . 333 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.

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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 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 17.6 (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 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 17.12 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 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 17.29 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 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 17.60 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 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 17.76 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 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 17.77 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 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 17.81 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 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 17.86 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 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 17.88 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