College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 8 - Problems - Page 311: 7


The motor needs to do $0.0512~J$ of work.

Work Step by Step

We can find the angular velocity $\omega$: $\omega = \frac{v}{r} = \frac{1.20~m/s}{0.020~m} = 60~rad/s$ We can find the rotational kinetic energy: $KE_{rot} = \frac{1}{2}I~\omega^2$ $KE_{rot} = \frac{1}{2}(\frac{1}{2})MR^2~\omega^2$ $KE_{rot} = \frac{1}{4}(0.0158~kg)(0.060~m)^2(60~rad/s)^2$ $KE_{rot} = 0.0512~J$ Since the change in energy of the disk is $0.0512~J$, the motor needs to do $0.0512~J$ of work.
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