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: 21

Answer

A point on the edge of the wheel travels 33 meters.

Work Step by Step

$\Delta \omega = 360~rpm-240~rpm = 120~rpm$ $\Delta \omega = (120~rpm)(\frac{1~min}{60~s})$ $\Delta \omega = 2.0~rev/s$ We can find the angular acceleration: $\alpha = \frac{\Delta \omega}{t}$ $\alpha = \frac{2.0~rev/s}{6.8~s}$ $\alpha = 0.294~rev/s^2$ We can convert $\omega_0 = 240~rpm$ to $rev/s$: $\omega_0 = (240~rpm)(\frac{1~min}{60~s})$ $\omega_0 = 4.0~rev/s$ We can find the angular displacement: $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ $\theta = (4.0~rev/s)(6.8~s) + \frac{1}{2}(0.294~rev/s^2)(6.8~s)^2$ $\theta = 34.0~rev$ We can find the distance $d$ traveled by a point on the edge of the wheel: $d = \theta ~(2\pi~r)$ $d = (34.0~rev)(2)(\pi)(0.155~m)$ $d = 33~m$ A point on the edge of the wheel travels 33 meters.
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