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 15 - Oscillations - Exercises and Problems: 77

Answer

$f = \frac{1}{2\pi}~\sqrt{\frac{5g}{2R}}$

Work Step by Step

The moment of inertia is $I = \frac{2}{5}MR^2$. The distance $d$ from the center of mass to the pivot is $R$. We can find an expression for the frequency of small-angle oscillations as: $f = \frac{1}{2\pi}~\sqrt{\frac{Mgd}{I}}$ $f = \frac{1}{2\pi}~\sqrt{\frac{MgR}{\frac{2}{5}MR^2}}$ $f = \frac{1}{2\pi}~\sqrt{\frac{5g}{2R}}$
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.