problem09_89

University Physics with Modern Physics with Mastering Physics (11th Edition)

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9.89: a) ) m) 10 00 5 kg)( 60 . 1 ( m) 10 kg)(2.50 80 . 0 (( 2 1 2 1 2 1 2 2 2 2 2 2 2 2 1 1 - - × + × = + . R M R M . m kg 10 25 . 2 2 3 × = - b) See Example 9.9. In this case, becomes for expression the so and , 1 v R v = ϖ s. m 40 . 3 )) m) kg)(0.025 50 . 1 ( ) m kg 10 25 . 2 (( 1 ( m) 00 . 2 )( s m 80 . 9 ( 2 ) ( 1 2 2 2 3 2 2 = × + = + = - mR I gh v c) The same calculation, with 2 R instead of 1 R gives s. m 95 . 4 = v This does make sense, because for a given total energy, the disk combination will have a larger fraction of
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Unformatted text preview: the kinetic energy with the string of the larger radius, and with this larger fraction, the disk combination must be moving faster....
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This document was uploaded on 02/04/2008.

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