Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 348: 11


(a) $I = 0.032~kg~m^2$ (b) The rotational kinetic energy is 15.8 J

Work Step by Step

(a) We can find $R$, the distance from each ball to the axis of rotation. $\frac{0.20~m}{R} = cos(30^{\circ})$ $R = \frac{0.20~m}{cos(30^{\circ})}$ $R = 0.23~m$ We can find the moment of inertia of the system. $I = (3)(M~R^2)$ $I = (3)(0.20~kg)(0.23~m)^2$ $I = 0.032~kg~m^2$ (b) We can find the angular velocity. $\omega = (5.0~rev/s)(2\pi~rad/rev)$ $\omega = 10\pi~rad/s$ We can find the rotational kinetic energy. $KE_{rot} = \frac{1}{2}I~\omega^2$ $KE_{rot} = \frac{1}{2}(0.032~kg~m^2)(10\pi~rad/s)^2$ $KE_{rot} = 15.8~J$ The rotational kinetic energy is 15.8 J.
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