# Chapter 7 - Exercises and Problems - Page 127: 60

#### 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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