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 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 199: 13


$v = \sqrt{\frac{m_2~g~r}{m_1}}$

Work Step by Step

The weight of $m_2$ should be equal to the centripetal force required to keep $m_1$ moving in a circle. We can find the required speed of $m_1$ as; $F_c = \frac{m_1~v^2}{r} = m_2~g$ $v^2 = \frac{m_2~g~r}{m_1}$ $v = \sqrt{\frac{m_2~g~r}{m_1}}$
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.