## Physics (10th Edition)

Published by Wiley

# Chapter 8 - Rotational Kinematics - Problems - Page 217: 78

#### Answer

i) In case a), angular acceleration is positive. The magnitude of $\vec{\alpha}$ is $1.94 rad/s^2$ and it is directed counterclockwise. ii) In case b), angular acceleration is negative. The magnitude of $\vec{\alpha}$ is $1.23 rad/s^2$ and it is directed clockwise.

#### Work Step by Step

We have $$\alpha=\frac{\omega-\omega_0}{\Delta t}$$ i) In the case when the angular speed increases, the final angular speed is greater than the initial one, making $(\omega-\omega_0)\gt0$. Therefore, $\alpha\gt0$, or the angular acceleration is positive. $$\alpha=\frac{28.5-21.7}{3.5}=+1.94rad/s^2$$ The magnitude of $\vec{\alpha}$ is $1.94 rad/s^2$. The positive sign indicates $\vec{\alpha}$ is directed counterclockwise. ii) In the case when the angular speed decreases, the final angular speed is less than the initial one, making $(\omega-\omega_0)\lt0$. Therefore, $\alpha\lt0$, or the angular acceleration is negative. $$\alpha=\frac{15.3-28.5}{10.7}=-1.23rad/s^2$$ The magnitude of $\vec{\alpha}$ is $1.23 rad/s^2$. The negative sign indicates $\vec{\alpha}$ is directed clockwise.

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