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 87: 33



Work Step by Step

We know that the centripetal force is given by $\frac{mv^2}{r}$. We know that the force of friction is the centripetal force, so, using a reasonable coefficient of friction of .25, we find: $\mu =.25 \times \frac{60^2}{40^2}=.56$
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