Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 26

Answer

A torque of $0.28~N~m$ will bring the balls to a halt in 5.0 seconds.

Work Step by Step

We first express the initial angular velocity in units of rad/s. Note that since the rotation is clockwise, the angular velocity is negative. $\omega_0 = -(20~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega_0 = -2.094~rad/s$ We then find the angular acceleration if the object comes to rest in 5.0 seconds. $\alpha = \frac{\omega_f-\omega_0}{t}$ $\alpha = \frac{0-(-2.094~rad/s)}{5.0~s}$ $\alpha = 0.419~rad/s^2$ Next, we find the moment of inertia of the object. Note that the center of mass is 33 cm from the 2.0-kg ball. $I = m_1~R_1^2+m_2~R_2^2$ $I = (2.0~kg)(0.33~m)^2+(1.0~kg)(0.67~m)^2$ $I = 0.67~kg~m^2$ We then find the required torque: $\tau = I~\alpha$ $\tau = (0.67~kg~m^2)(0.419~rad/s^2)$ $\tau = 0.28~N~m$ A torque of $0.28~N~m$ will bring the balls to a halt in 5.0 seconds.
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.