Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 33

Answer

(a) $\omega = 640~rpm$ (b) v = 40 m/s (c) v = 0

Work Step by Step

(a) We can find the angular velocity in rad/s $\omega = \frac{v}{r}$ $\omega = \frac{20~m/s}{0.30~m}$ $\omega = 66.7~rad/s$ We can convert the angular velocity to units of rpm. $\omega = (66.7~rad/s)(\frac{60~s}{1~min})(\frac{1~rev}{2\pi~rad})$ $\omega = 640~rpm$ (b) At the top edge of the tire, the speed from the rotation is in the same direction as the speed $v_{cm}$ of the center of mass of the tire. We can find the speed of a point at the top of the tire. $v = \omega~r+v_{cm}$ $v = (66.7~rad/s)(0.30~m)+20~m/s$ $v = 40~m/s$ (c) At the bottom edge of the tire, the speed from the rotation is in the opposite direction from the speed $v_{cm}$ of the center of mass of the tire. We can find the speed of a point at the top of the tire. $v = v_{cm}-\omega~r$ $v = (20~m/s) - (66.7~rad/s)(0.30~m)$ $v = 0$
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