## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems - Page 112: 34

#### Answer

(a) $\omega = 98~rpm$ (b) The wheel makes 12 revolutions during this time.

#### Work Step by Step

We can find the change in angular velocity. $\Delta \omega = \alpha ~t$ $\Delta \omega = (0.50~rad/s^2)(10~s)$ $\Delta \omega = 5.0~rad/s$ We can convert this to units of rpm. $\Delta \omega = (5.0~rad/s)(\frac{1~rev}{2\pi~rad})(\frac{60~s}{1~min})$ $\Delta \omega = 48~rpm$ We can find the angular velocity after 10 seconds. $\omega = \omega_0+\Delta \omega$ $\omega = 50~rpm+48~rpm$ $\omega = 98~rpm$ (b) We can find the number of revolutions that the wheel rotates through in this time. $\theta = (\frac{\omega_0+\omega_f}{2})(10~s)$ $\theta = (\frac{50~rpm+98~rpm}{2})(\frac{1}{6}~min)$ $\theta = 12~rev$ The wheel makes 12 revolutions during this time.

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