Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems: 23

Answer

(a) $\alpha = 0.53~rad/s^2$ (b) The pottery wheel reaches a speed of 65 rpm in 13 seconds.

Work Step by Step

(a) The angular accelerations of the two wheels are different, but the tangential acceleration is equal since their edges touch without slipping. We can find the tangential acceleration $a_{tan}$ of the small wheel. $a_{tan} = \alpha ~r = (7.2~rad/s^2)(0.020~m)$ We can use the same tangential acceleration to find the angular acceleration $\alpha$ of the pottery wheel. $\alpha = a_{tan} / r = (7.2~rad/s^2)(0.020~m)/(0.270~m)$ $\alpha = 0.53~rad/s^2$ (b) $\omega = (65~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = \frac{13\pi}{6}~rad/s$ We can then find the time to reach this angular speed: $t = \frac{\omega}{\alpha} = \frac{\frac{13\pi}{6}~rad/s}{0.53~rad/s^2} = 13~s$ The pottery wheel reaches a speed of 65 rpm in 13 seconds.
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