homework10-solutions

homework10-solutions - ME 104 Homework 10 Solutions Prof...

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ME 104: Homework 10 Solutions Prof. Karl Hedrick: University of California, Berkeley

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Chapter 17, Solution 3 FOR ANY DISK: 2 2 4 ( ) 1 2 1 2 m r t I mr tr    Moment of inertia. Disk A : 4 1 2 A I br   Disk B : 4 4 4 4 total 4 1 (3 )( ) 2 1 3 2 3 (1 3 ) B A A B A I b nr n br n I I I I n I    Work-energy. 1 1 2 2 2 total 2 0 (4 rad) 1 2 T U M M T I 4 2 1 1 2 2 2 1 : 0 (4 ) (1 3 ) 2 A T U T M n I 2 2 4 8 (1 3 ) A M n I For Point D on rim of disk B 2 ( ) D v nr or 2 2 2 2 2 2 2 4 8 1 3 D A Mr n v n r I n Value of n for maximum final speed. For maximum 2 4 : 0 1 3 D d n v dn n 2 3 4 4 2 5 5 4 1 [ (12 ) (1 3 )(2 )] 0 (1 3 ) 12 2 6 0 2 (3 1) 0 n n n n n n n n n n 0 n and 0.25 1 0.7598 3 n 0.760 n
Chapter 17, Solution 7 Work of external friction force on disk A . Only force doing work is F . Since its moment about A is , M rF we have 1 2 ( ) k U M rF r mg Kinetic energy of disk A . Angular velocity becomes constant when 2 1 2 2 2 2 2 2 0 1 2 1 1 2 2 4 v r T T I v mr r mv    Principle of work and energy for disk A . 2 1 1 2 2 : 0 4 k mv T U T r mg Angle change. 2 rad 4 k v r g 2 rev 8 k v r g  

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Chapter 17, Solution 28 The point of contact with ground is the instantaneous center. Position 1 . Point B is at the top. 1 1 1 2 2 2 1 2 2 2 2 1 1 1 2 2 1 2 1 1 1 2 2 2 1 1 1 (2 ) ( ) ( ) 2 2 2 3 B A B A v r v r T mv mv I m r m r mr mr Position 2 . Point B is at the bottom. 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 1 0 1 1 1 2 2 2 1 1 0 ( ) ( ) 2 2 (3 ) 9 B A B A v v r T mv mv I m r mr mr mr mr Work.
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