University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 333: 10.57


(a) $\alpha = 16.3~rad/s^2$ (b) The angular acceleration decreases as the bar swings down. (c) $\omega = 5.70~rad/s$

Work Step by Step

(a) We can find the moment of inertia of the bar with one ball on the end. $I = \frac{1}{12}ML^2+m(\frac{L}{2})^2$ $I = \frac{1}{12}(3.80~kg)(0.800~m)^2+(2.50~kg)(\frac{0.800~m}{2})^2$ $I = 0.6027~kg~m^2$ We can find the angular acceleration just after the other ball falls off. Note that the torque comes from the weight of the remaining ball on the end of the bar. $\tau = I \alpha$ $\alpha = \frac{\tau}{I}$ $\alpha = \frac{(mg)(\frac{L}{2})}{I}$ $\alpha = \frac{(2.50~kg)(9.80~m/s^2)(0.800~m)}{(2)(0.6027~kg~m^2)}$ $\alpha = 16.3~rad/s^2$ (b) $\tau = (\frac{L}{2})(mg)~sin(\theta)$, where $\theta$ is the angle between the bar and the vertical. Since $\theta$ decreases from $90^{\circ}$ to $0^{\circ}$ as the bar swings down, the torque decreases. Therefore the angular acceleration decreases as the bar swings down. (c) We can use conservation of energy to solve this part of the question. The magnitude of the change in potential energy of the ball will be equal to the magnitude of the rotational kinetic energy. $\frac{1}{2}I\omega^2 = mgh$ $\omega^2 = \frac{2mgh}{I}$ $\omega = \sqrt{\frac{2mgh}{I}}$ $\omega = \sqrt{\frac{(2)(2.50~kg)(9.80~m/s^2)(0.400~m)}{0.6027~kg~m^2}}$ $\omega = 5.70~rad/s$
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