## Physics (10th Edition)

Published by Wiley

# Chapter 9 - Rotational Dynamics - Problems - Page 246: 44

#### Answer

$t_B=2.12s$

#### Work Step by Step

From Newton's 2nd law for rotational motion, the angular acceleration of each door can be calculated by $$\alpha=\frac{\tau}{I}=\frac{Fl}{I} (1)$$ We also have $$\theta=\omega_0t+\frac{1}{2}\alpha t^2$$ Here, both doors start from rest, so $\omega_0=0$ $$\theta=\frac{1}{2}\alpha t^2$$ $$\alpha=\frac{2\theta}{t^2} (2)$$ Equating (1) and (2): $$\frac{Fl}{I}=\frac{2\theta}{t^2}$$ $$t=\sqrt{\frac{2I\theta}{Fl}}$$ Both doors rotate through the same angle $\theta$ and are under the same force $F$. Therefore, $$\frac{t_B}{t_A}=\sqrt{\frac{\frac{2I_B\theta}{Fl_B}}{\frac{2I_A\theta}{F(l_A)}}}=\sqrt{\frac{I_B}{I_A}\times\frac{l_A}{l_B}}$$ From table 9.1, we also have $I_A=1/3ML_A^2$ and $I_B=1/12ML_B^2$. Therefore, $\frac{I_B}{I_A}=\frac{1}{4}$ The lever arm in door A is twice that in door B, so $\frac{l_A}{l_B}=2$ Therefore, $$\frac{t_B}{t_A}=\sqrt{\frac{1}{2}}$$ $$t_B=(3s)\sqrt{\frac{1}{2}}=2.12s$$

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