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: 50

Answer

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: $r=\frac{(1.2)(9.8)}{(20.94)^2}=0.027m$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.