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: 17

Answer

(a) $\alpha = -96~rad/s^2$ (b) The engine makes 98 revolutions.

Work Step by Step

(a) $\Delta \omega = 1200~rpm - 3500~rpm = -2300~rpm$ $\Delta \omega = (-2300~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s})$ $\Delta \omega = \frac{-230~\pi}{3}~rad/s$ We can find the angular acceleration $\alpha$ as: $\alpha = \frac{\Delta \omega}{t} = \frac{\frac{-230~\pi}{3}~rad/s}{2.5~s}$ $\alpha = -96~rad/s^2$ (b) $\omega_0 = (3500~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = \frac{350\pi}{3}~rad/s$ $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ $\theta = (\frac{350\pi}{3}~rad/s)(2.5~s) + \frac{1}{2}(-96~rad/s^2)(2.5~s)^2$ $\theta = 616~radians$ We can then find the number of revolutions: $rev = \frac{\theta}{2\pi} = \frac{616~rad}{2\pi} = 98~rev$
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