Physics (10th Edition)

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

Chapter 8 - Rotational Kinematics - Problems - Page 214: 43


a) $\omega=3.61rad/s$ b) $\alpha=6.53rad/s^2$

Work Step by Step

a) The object pulls the rod downward due to gravitational force. We assume gravitational force is the only force at play here, so the principle of mechanical energy conservation can be applied: $$\frac{1}{2}m(v^2-v_0^2)+mg(h-h_0)=0$$ $$\frac{1}{2}(v^2-v_0^2)+g(h-h_0)=0$$ We know $v_0=0, g=9.8m/s^2$. The change in height equals the rod's length, so $h-h_0=-1.5m$. $$\frac{1}{2}v^2+9.8(-1.5)=0$$ $$v=5.42m/s$$ This is the tangential speed of the rod just before it strikes the floor. The radius of rotation is $r=1.5m$. The angular speed just before the rod strikes the floor is $$\omega=\frac{v}{r}=3.61rad/s$$ b) The rod's tangential speed changes from $0$ to $5.42m/s$ with $g=9.8m/s^2$. The time it takes the rod to achieve such speed is $$\Delta t=\frac{5.42}{9.8}=0.553s$$ The rod's angular acceleration is $$\alpha=\frac{\omega-\omega_0}{\Delta t}=\frac{3.61}{0.553}=6.53rad/s^2$$
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