Physics (10th Edition)

Published by Wiley

Chapter 9 - Rotational Dynamics - Problems - Page 249: 73

Answer

$\alpha=0.5rad/s^2$

Work Step by Step

According to Newton's 2nd law of rotational motion, $$\tau=I\alpha$$ Here, force $F=68N$, when applied to one section, produces a torque with lever arm $l=1.2m$. Therefore, $\tau=68N\times1.2m=81.6N.m$ Each section is a thin rectangle rotating around the axis at one end, so the total moment of inertia of 4 sections is $$\sum I=4\times\frac{1}{3}MR^2=\frac{4}{3}(85kg)(1.2m)^2=163.2kg.m^2$$ Therefore, $$\alpha=\frac{\tau}{\sum I}=0.5rad/s^2$$

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