Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems - Page 238: 35


a) 1.66 rads b) 2.42 radians per second

Work Step by Step

a) We simplify the equation for energy to find the maximum angular displacement: $\theta_{max} = \sqrt{\frac{2E}{k}}= \sqrt{\frac{2(4.7)}{3.4}}=\fbox {1.66 rads}$ b) We find the maximum angular speed: $\omega= \sqrt{\frac{k}{I}} = \sqrt{\frac{3.4}{1.6}}=1.46$ This means that the maximum speed is: $\omega\times \theta=1.46 \times 1.66 = \fbox{2.42 radians per second}.$
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.