Physics (10th Edition)

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

Chapter 9 - Rotational Dynamics - Problems - Page 246: 41


It takes $2s$ for the 2nd sheet to reach the same angular velocity.

Work Step by Step

Taking a look at Table 9.1, the moment of inertia for a thin rectangular sheet with the axis along one edge is $I=1/3ML^2$ According to Newton's 2nd law for rotational motion, $$\tau=I\alpha$$ $$\alpha=\frac{\tau}{I}=\frac{3\tau}{ML^2} (1)$$ Angular acceleration $\alpha$ can also be calculated by $$\alpha=\frac{\omega-\omega_0}{t}(2)$$ Equating (1) and (2): $$\frac{3\tau}{ML^2}=\frac{\omega-\omega_0}{t}$$ $$t=\frac{ML^2(\omega-\omega_0)}{3\tau}$$ In the 1st sheet, the axis lies along the 0.2-m side, so $L_1=0.4m$. In the 2nd sheet, the axis lies along the 0.4-m side, so $L_2=0.2m$ Both sheets have the same mass $M$, final angular velocity $\omega$ and torque $\tau$. They both start from rest, so $\omega_0=0$ $$\frac{t_2}{t_1}=\frac{\frac{ML_2^2(\omega-0)}{3\tau}}{\frac{ML_1^2(\omega-0)}{3\tau}}=\Big(\frac{L_2}{L_1}\Big)^2=\frac{1}{4}$$ $$t_2=\frac{1}{4}\times(8s)=2s$$
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