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 - Page 222: 14

Answer

(a) $\alpha = 4.2~rad/s^2$ (b) $a_R = 130~m/s^2$ $a_{tan} = 1.3~m/s^2$

Work Step by Step

(a) $\omega_0 = (120~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = 4\pi~rad/s$ $\omega_f = (280~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = \frac{28\pi}{3}~rad/s$ $\alpha = \frac{\Delta \omega}{t} = \frac{\frac{28\pi}{3}~rad/s ~- ~4\pi~rad/s}{4.0~s}$ $\alpha = \frac{4\pi}{3}~rad/s^2 = 4.2~rad/s^2$ (b) 2.0 seconds after the wheel started accelerating: $\omega = \omega_0 + \alpha ~t$ $\omega = (4\pi~rad/s) + (\frac{4\pi}{3}~rad/s^2)(2.0~s)$ $\omega =\frac{20\pi}{3} ~rad/s$ We can find the radial acceleration $a_R$. $a_R = \omega^2 ~r = (\frac{20\pi}{3} ~rad/s)^2(0.61~m/2)$ $a_R = 130~m/s^2$ We can find the tangential acceleration $a_{tan}$. $a_{tan} = \alpha ~r = (\frac{4\pi}{3}~rad/s^2)(0.61~m/2)$ $a_{tan} = 1.3~m/s^2$
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