Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 5 - Exercises and Problems - Page 92: 54


Please see the work below.

Work Step by Step

We know that $\omega=2\pi(\frac{200}{60})=20.94\frac{rad}{s}$ We also know that the force of friction should be equal to the required centripetal force $\mu mg=mr\omega^2$ This can be rearranged as $r=\frac{\mu g}{\omega^2}$ We plug in the known values to obtain: $r=\frac{(1.2)(9.8)}{(20.94)^2}=0.027m$
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