Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 11 - Rolling, Torque, and Angular Momentum - Problems - Page 320: 2a

Answer

The angular speed of the tires is $~~59.2~rad/s$

Work Step by Step

We can convert the speed of the car into units of $m/s$: $v = (80.0~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 22.2~m/s$ The tangential speed of the tire's circumference is equal to the speed of the car. We can find the angular speed of the tires: $\omega ~R = v$ $\omega = \frac{v}{R}$ $\omega = \frac{22.2~m/s}{0.375~m}$ $\omega = 59.2~rad/s$ The angular speed of the tires is $~~59.2~rad/s$
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