**Unformatted text preview: **10.101: a) The friction force is f = k n = k Mg , so a = k g . The magnitude of the angular acceleration is gives t= and
2 R0 1 1 R 20 = d = at 2 = ( k g ) 3 g 18 g . 2 2 k k 2 fR I = ( 1 2 ) MR 2 k MgR = 2 k g R . b) Setting v = at = R = ( 0 - t ) R and solving for t R0 R0 R0 = = , a + R k g + 2 k g 3 k g c) The final kinetic energy is ( 3 4 ) Mv 2 = ( 3 4) M ( at ) 2 , so the change in kinetic energy is 3 R0 1 1 2 2 - MR 20 = MR 2 0 . M k g 4 4 3 k g 6
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