Physics (10th Edition)

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

Chapter 8 - Rotational Kinematics - Problems - Page 216: 72


The angular displacement of the wheel is $+270.8rad$

Work Step by Step

We are given final angular velocity $\omega=+1.88rad/s$, angular acceleration $\alpha=-5.04rad/s^2$ and duration $t=10s$ The angular displacement of the wheel is $$\theta=\omega_0t+\frac{1}{2}\alpha t^2$$ We also know that $\omega_0=\omega-\alpha t$ $$\theta=(\omega-\alpha t)t+\frac{1}{2}\alpha t^2$$ $$\theta=\omega t-\frac{1}{2}\alpha t^2=+270.8rad$$
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