Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems - Page 114: 68


(a) t = 6.32 s (b) The motor turns through an angle of 84.3 radians.

Work Step by Step

(a) At the moment when the motor changes direction, the angular velocity will be zero. We can find the time when the angular velocity is zero. $\omega(t) = (20-\frac{1}{2}t^2)~rad/s = 0$ $t^2 = 40$ $t = \sqrt{40}$ $t = 6.32~s$ (b) The angle that the motor turns is equal to the area under the omega versus time graph. We can integrate to find this area. $\Delta \theta = \int_{0}^{\sqrt{40}}\omega(t)~dt$ $\Delta \theta = \int_{0}^{\sqrt{40}}(20-\frac{1}{2}t^2)~dt$ $\Delta \theta = (20t-\frac{1}{6}t^3)\vert_{0}^{\sqrt{40}}$ $\Delta \theta = (20)(\sqrt{40})-\frac{1}{6}(\sqrt{40})^3$ $\Delta \theta = 84.3~rad$ The motor turns through an angle of 84.3 radians.
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.