ch10-p115 - 115. We employ energy methods in this solution;...

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(a) The speed of the box is related to the angular speed of the wheel by v = R ω , so that Km v v K m box box box box m/s = ¡ == 1 2 2 141 2 . implies that the angular speed is = 1.41/0.20 = 0.71 rad/s. Thus, the kinetic energy of rotation is 1 2 2 10 0 I = .J . (b) Since it was released from rest at what we will consider to be the reference position for gravitational potential, then (with SI units understood) energy conservation requires
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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