# exam3_soln - TAM 212 Spring 2004 solutions 1 d W dW dW m x...

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Unformatted text preview: TAM 212, Spring, 2004, solutions 1. d W dW dW m x m x i i i OC 2 2 3 = + = = ∑ ∑ ( ) ( ) sphere c zz rod c zz c zz I I I + = ; ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 , , 10 / 3 / 1 / 3 / 1 2 / / 12 / / d g W L g W L g W L g W r m I I rod C C rod rod C c zz rod c zz = = + = + = → ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) 2 2 2 2 2 , , / 35 25 / 5 / 7 / / 5 / 2 d g W d g W R g W R g W r m I I sphere C C sphere sphere C c zz sphere c zz = = + = + = → ( ) ( ) ( ) ( ) ( )( ) ( )( ) 2 2 2 / 3 / 205 / 3 / 105 / 3 / 100 d g W d g W d g W I I I sphere c zz rod c zz c zz = + = + = 2. KE Work ∆ = . Since the rod is released from rest and at the instant in question, the velocity of A=0, ( ) 2 / 1 2 − = ∆ ω A zz I KE . ( ) ( ) 2 2 2 2 3 / 1 2 / 12 / mL L m mL r m I I AC rod c zz A zz = + = + = , so ( ) 2 2 3 / 1 ω mL KE = ∆ spring gravity Work Work Work + = ; 4 / , mgL h mg Work down c gravity − = ∆ = ; ( ) ( ) ( ) ( ) ( ) mgL L L mg k Work initial final spring = − −...
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## This note was uploaded on 04/09/2008 for the course TAM 212 taught by Professor Keane during the Spring '08 term at University of Illinois at Urbana–Champaign.

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