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Question Number 32236 by momo last updated on 22/Mar/18

Answered by ajfour last updated on 22/Mar/18

$$a=\frac{Mg\sin {\theta }_S}{\frac{7}{5}{MR}^2}=\frac{mg\sin {\theta }_C}{\frac{3}{2}{mR}^2}$$$$\Rightarrow \frac{\sin {\theta }_C}{\sin {\theta }_S}=\frac{15}{14}.$$

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