University Physics with Modern Physics (14th Edition)

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

Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 295: 9.12

Answer

(a) $\omega^2 = \omega_0^2 + 2\alpha~(\theta - \theta_0)$ (b) $\alpha = 8.00~rad/s^2$

Work Step by Step

(a) Equation 9.7: $\omega = \omega_0+\alpha ~t$ $t = \frac{\omega - \omega_0}{\alpha}$ We can replace $t$ in equation 9.11: $\theta = \theta_0+ \omega_0 ~t + \frac{1}{2}\alpha ~t^2$ $\theta = \theta_0+ \omega_0 (\frac{\omega - \omega_0}{\alpha}) + \frac{1}{2}\alpha (\frac{\omega - \omega_0}{\alpha})^2$ $\alpha~(\theta - \theta_0) = \omega_0 ~\omega - \omega_0^2 + \frac{1}{2}(\omega^2 -2~\omega~\omega_0+ \omega_0^2)$ $2\alpha~(\theta - \theta_0) = 2\omega_0 ~\omega - 2\omega_0^2 + \omega^2 -2~\omega~\omega_0+ \omega_0^2$ $2\alpha~(\theta - \theta_0) = \omega^2 - \omega_0^2$ $\omega^2 = \omega_0^2 + 2\alpha~(\theta - \theta_0)$ (b) $\omega^2 = \omega_0^2 + 2\alpha~(\theta - \theta_0)$ $\alpha = \frac{\omega^2 - \omega_0^2}{2~(\theta - \theta_0)}$ $\alpha = \frac{(16.0~rad/s)^2 - (12.0~rad/s)^2}{2~(7.00~rad)}$ $\alpha = 8.00~rad/s^2$
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