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 8 - Dynamics II: Motion in a Plane - Exercises and Problems: 48

Answer

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

Work Step by Step

For the hanging mass $m_2$ to remain at rest, the tension $T$ in the string must be equal to the weight of $m_2$ $T = m_2~g$ The tension $T$ provides the centripetal force to keep $m_1$ moving around in a circle. We can find an expression for the speed of $m_1$. $T = \frac{m_1~v^2}{r}$ $m_2~g = \frac{m_1~v^2}{r}$ $v^2 = \frac{m_2~g~r}{m_1}$ $v = \sqrt{\frac{m_2~g~r}{m_1}}$
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