Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 9 - Rotational Dynamics - Problems - Page 248: 71



Work Step by Step

According to Newton's 2nd law of rotational motion, $$\tau=I\alpha$$ Here, the force $F=45N$, when applied to the disk, produces a torque with lever arm $r=0.15m$. Therefore, $\tau=45N\times0.15m=6.75N.m$ A solid cylindrical disk has $I=\frac{1}{2}mr^2$. The disk's angular acceleration $\alpha=120rad/s^2$ $$\frac{1}{2}mr^2\alpha=\tau$$ $$m=\frac{2\tau}{r^2\alpha}=5kg$$
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