Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 49

Answer

The rotational kinetic energy of the system is 1.75 J

Work Step by Step

We can convert the angular velocity to units of rad/s $\omega = (100~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega = 10.47~rad/s$ We can find the moment of inertia of the system. Note that the center of mass is 13.3 cm from the 600-gram ball and 26.7 cm from the 300-gram ball. $I = M_1~R_1^2+M_2~R_2^2$ $I = (0.60~kg)(0.133~m)^2+(0.30~kg)(0.267~m)^2$ $I = 0.032~kg~m^2$ We can find the rotational kinetic energy of the system. $KE = \frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}(0.032~kg~m^2)(10.47~rad/s)^2$ $KE = 1.75~J$ The rotational kinetic energy of the system is 1.75 J
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