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A solid sphere of mass m and radius R is lying on a rough horizontal plane. A constant force $4 m g$ acts vertically at point P such that OP makes an angle $60^{0}$ with horizontal. Minimum coefficient of friction is $\frac{n}{7}$ so that sphere starts pure rolling. Then the value of $n$ is____

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answer is 2.

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Detailed Solution

$π_{0} = Ι_{0} α \Rightarrow \frac{2}{5} m R α + f = 2 m g & f = m a = m a R \Rightarrow f = \frac{10}{7} m g \Rightarrow \frac{10}{7} m g \leq μ N \Rightarrow \frac{10}{7} m g \leq μ (5 m g), μ \geq (\frac{2}{7})$

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