( : + ) -4 = 8-0.6(v) Ans. v A = v B + v A>B. ( : + ) v A = 8-20(0.3) Ans. v A = 2 ft>s : Also, -4i = 8i + (vk) * (0.6j) -4 = 8-0.6v. Ans.

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2 Pinion gear A rolls on the gear racks B and C. If B is moving to the right at 8 ft>s and C is moving to the left at 4 ft>s, determine the angular velocity of the pinion gear and the velocity of its center A. v C A 0.3 ft v C = v B + v C>B ( : + ) -4 = 8-0.6(v) B v = 20 rad>s v A = v B + v A>B ( : + ) v A = 8-20(0.3) v A = 2 ft>s : Also, v C = v B + v * r C>B -4i = 8i + (vk) * (0.6j) -4 = 8-0.6v v = 20 rad>s v A = v B + v * r A>B v A i = 8i + 20k * (0.3j) v A = 2 ft>s :

3 Determine the velocity of point A on the outer rim of the spool at the instant shown when the cable is pulled to the right with a velocity of v. The spool rolls without slipping. R A O r v Kinematic Diagram: Since the spool rolls without slipping, the velocity of the contact point P is zero. The kinematic diagram of the spool is shown in Fig. a. General Plane Motion: Applying the relative velocity equation and referring to Fig. a, v B = v P + v * r B>D vi = 0 + (-vk) * C(R - r)jd vi = v(r - r)i Equating the i components, yields v = v(r - r) v = v R - r Using this result, v A = v P + v * r A>P v = k * 2Rj R - r = B 2R R - r vri Thus, v A = 2R R - r v :

4 A bowling ball is cast on the alley with a backspin of v = 10 rad>s while its center O has a forward velocity of v O = 8 m>s. Determine the velocity of the contact point A in contact with the alley. v A = v O + v A>O O 120 mm v 10 rad/s v O 8 m/s a : + b v A = (0.12) Also, v A = 9.20 m>s : A v A = v O + v * r A>O v A i = 8i + (10k) * (-0.12j) a : + b v A = 9.20 m>s :

5 * The planetary gear system is used in an automatic transmission for an automobile. By locking or releasing certain gears, it has the advantage of operating the car at different speeds. Consider the case where the ring gear R is held fixed, v R = 0, and the sun gear S is rotating at v S = 5 rad>s. Determine the angular velocity of each of the planet gears P and shaft A. v R R P v S 40 mm S A v A = 5(80) = 400 mm>s ; 80 mm v B = 0 v B = v A + v * r B>A 0 = -400i + (v p k) * (80j) 40 mm 0 = -400i - 80v p i v P = -5 rad>s = 5 rad>s v C = v B + v * r C>B v C = 0 + (-5k) * (-40j) = -200i v A = 200 = 1.67 rad>s 120

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10 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher At the instant shown, ball B is rolling along the slot in the disk with a velocity of 600 mm>s and an acceleration of 150 mm>s 2, both measured relative to the disk and directed away from O. If at the same instant the disk has the angular velocity and angular acceleration shown, determine the velocity and acceleration of the ball at this instant. 0.8 m B z v 6 rad/s a 3 rad/s 2 O 0.4 m y x Kinematic Equations: v B = v O +Æ*r B>O + (v B>O ) xyz a B = a O +Æ # * r B>O + Æ * (Æ * r B>O ) + 2Æ * (v B>O ) xyz + (a B>O ) xyz (1) (2) v O = 0 a O = 0 Æ={6k} rad>s Æ={3k} rad>s 2 r B>O = {0.4 i } m (v B>O ) xyz = {0.6i} m>s (a B>O ) xyz = {0.15i} m>s 2 Substitute the data into Eqs.(1) and (2) yields: v B = 0 + (6k) * (0.4i) + (0.6i) = {0.6i + 2.4j} m>s a B = 0 + (3k) * (0.4i) + (6k) * [(6k) * (0.4i) ] + 2 (6k) * (0.6i) + (0.15i) = {-14.2i j} m>s 2 616

11 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The disk rolls without slipping and at a given instant has the angular motion shown. Determine the angular velocity and angular acceleration of the slotted link BC at this instant. The peg at A is fixed to the disk. 2 ft A C v 2 rad/s 0.5 ft a 4 rad/s 2 O 0.7 ft B v A = -(1.2)(2)i = -2.4i ft>s a A = a O + a * r A>O - v 2 r A>O a A = -4(0.7)i + (4k) * (0.5j) - (2) 2 (0.5j) a A = -4.8 i - 2j v A = v B +Æ*r A>B + (v A>B ) xyz -2.4i = 0 + (v BC k) * (1.6i + 1.2j) + v A>B a 4 5 bi + v A>B a 3 5 bj -2.4i = 1.6 v BC j v BC i + 0.8v A>B i + 0.6v A>B j -2.4 = -1.2 v BC v A>B 0 = 1.6v BC + 0.6v A>B Solving, v BC = rad>s v A>B = ft>s d a A = a B +Æ*r A>B +Æ*(Æ *r A>B ) + 2Æ *(v A>B ) xyz + (a A>B ) xyz -4.8i - 2j = 0 + (a BC k) * (1.6i + 1.2j) + (0.72k) * (0.72k * (1.6i + 1.2j)) +2(0.72k) * C -(0.8)(1.92)i - 0.6(1.92)jD a B>A i + 0.6a B>A j -4.8i - 2j = 1.6a BC j - 1.2a BC i i j j i + 0.8a B>A i + 0.6a B>A j -4.8 = -1.2a BC a B>A -2 = 1.6a BC a B>A Solving, a BC = 2.02 rad>s 2 d a B>A = ft>s 2 626