Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 9 - Rotational Dynamics - Problems - Page 246: 37


a) The magnitude of $\alpha$ is $5.94rad/s^2$ b) The magnitude of the frictional force $f$ is $44N$

Work Step by Step

a) Find the angular deceleration of the cylinder We have $\omega_0=76rad/s$, $\omega=38rad/s$ and $t=6.4s$. Therefore, $$\alpha=\frac{\omega-\omega_0}{t}=-5.94rad/s^2$$ The magnitude of $\alpha$ is $5.94rad/s^2$ b) The torque acting on the cylinder is $$\tau=I\alpha=0.615\times5.94=3.65N.m$$ This torque is created by frictional force $f$, and it has lever arm equal to the cylinder's radius $r=0.083m$. Therefore, $$f=\frac{\tau}{r}=44N$$
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.