problem10_94

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

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10.94: a), g) b) Using the vector product form for the angular momentum, so , and 2 1 2 1 r r v v - = - = , 1 1 2 2 v r v r × = × m m so the angular momenta are the same. c) Let . ˆ j ω ϖ = Then, ( 29 and , ˆ ˆ 1 1 k i v x z r - ϖ = × ϖ = ( 29 ( 29 ( 29 ( 29 . ˆ ˆ ˆ 2 2 1 1 1 k j i v L xR y x xR r m + + + - = × = With , 2 2 2 R y x = + the magnitude of , and , 2 is 2 2 1 2 R m R m ϖ = ω L L and so . and , cos 6 2 1 ) )( 2 ( 2 2 2 π θ = = = R m R m This is true for 2 L as well, so the total angular momentum makes an angle of 6 with the + y -axis. d) From the intermediate calculation of part (c), , 2 1 mvR R m L y = = so the total y -component of angular momentum is y y L mvR L e) . 2 = is constant, so the net y- component of torque is zero. f)
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This document was uploaded on 02/04/2008.

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