Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 10 - Rotation - Problems - Page 292: 80b

Answer

$\alpha = 1.67~rad/s^2$

Work Step by Step

We can find the average angular speed: $\omega_{ave} = \frac{\theta_2-\theta_1}{t}$ $\omega_{ave} = \frac{70.0~rad-10.0~rad}{6.00~s}$ $\omega_{ave} = 10.0~rad/s$ We can find $\omega_1$ the angular velocity at $\theta_1$: $\omega_{ave} = \frac{\omega_1+\omega_2}{2} = 10.0~rad/s$ $\omega_1 = (2)(10.0~rad/s)-\omega_2$ $\omega_1 = (2)(10.0~rad/s)-(15.0~rad/s)$ $\omega_1 = 5.00~rad/s$ The angular velocity at $\theta_1$ is $~~5.00~rad/s$ We can find the angular acceleration $\alpha$: $\omega_2 = \omega_1+\alpha~t$ $\alpha = \frac{\omega_2 - \omega_1}{t}$ $\alpha = \frac{15.0~rad/s - 5.00~rad/s}{6.00~s}$ $\alpha = 1.67~rad/s^2$
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