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PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 445 PROBLEM 12.92 Two 2.6-lb collars A and B can slide without friction on a frame, consisting of the horizontal rod OE and the vertical rod CD, which is free to rotate about CD. The two collars are connected by a cord running over a pulley that is attached to the frame at O and a stop prevents collar B from moving. The frame is rotating at the rate 12 rad/sθ = and 0.6 ftr = when the stop is removed allowing collar A to move out along rod OE. Neglecting friction and the mass of the frame, determine, for the position 1.2 ft,r = (a) the transverse component of the velocity of collar A, (b) the tension in the cord and the acceleration of collar A relative to the rod OE. SOLUTION Masses: 22.6 0.08075 lb s /ft 32.2A Bm m= = = ⋅ (a) Conservation of angular momentum of collar A: 0 2 0 1( ) ( )H H= 1 1 2 2( ) ( )A Am r v m r vθ θ= 2 2 1 1 1 1 2 2 2 ( ) (0.6) (12) ( ) 3.6 1.2 r v r v r r θ θ θ= = = = 2( ) 3.60 ft/svθ = 2 2 ( ) 3.6 3.00 rad/s 1.2A v r θθ = = = (b) Let y be the position coordinate of B, positive upward with origin at O. Constraint of the cord: constant or r y y r− = = Kinematics: 2( ) and ( )B y A ra y r a r rθ= = = − Collar B: :y B B B B BF m a T W m y m rΣ = − = = (1) Collar A: 2( ) : ( )r A A r AF m a T m r rθΣ = − = − (2) Adding (1) and (2) to eliminate T, 2( )B A B AW m m r m rθ− = + + 2 2 2 /rod (0.08075)(1.2)(3.00) (2.6) 10.70 ft/s 0.08075 0.08075 A B A A B m r W a r m m θ − −= = = = − + + ( ) (0.08075)( 10.70 32.2)BT m r g= + = − + 1.736 lbT = 2 / rod 10.70 ft/sAa = radially inward.
