Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 7 - Exercises and Problems: 60

Answer

Please see the work below.

Work Step by Step

We know that the speed of the bug at a distance of y from the top is given as $mgy=(\frac{1}{2})mv^2$ This simplifies to: $v=\sqrt{2gy}$ We also know that the centripetal force is given as: $mgcos(\theta)=mv^2R$ $\implies mg(\frac{y}{R})=mv^2R$ $\implies mg(\frac{y}{R})=m(\sqrt{2gy})^2{R}$ $mg(\frac{R-y}{R})=m(\frac{2gy}{R})$ $R-y=2y$ $y=\frac{R}{3}$
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