Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 5 - Exercises and Problems - Page 88: 63


$17 \ min^{-1}$

Work Step by Step

We know that the gravitational and frictional forces must be at least equal. Thus: $mg = \mu F_n$ We know that the normal force is the centripetal force, so it follows: $mg = \frac{\mu mv^2}{r} \\ v =\sqrt{\frac{gr}{\mu}}=\sqrt{\frac{(9.81)(5.1)}{.62}}=8.98 \ m/s$ We convert this to rotations per minute: $\omega=\frac{v}{r}=\frac{8.98 \ m/s}{5.1}=1.76\ rads/s = 17 \ min^{-1}$
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