Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems: 74

Answer

$.4347$

Work Step by Step

We first find the angular velocity: $\omega = \frac{2\pi}{1.8}=3.491$ We now find the coefficient of friction: $\omega^2 A m_2 =\mu m_2g$ $\mu = \frac{\omega^2 A}{g}$ $\mu = \frac{3.491^2 (.35)}{9.81}=.4347$
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