Question

A solid sphere of mass $M$ and radius $R$ is lying on a rough horizontal plane. A constant force $F=4Mg$ acts vertically at point $P$ such that $OP$ makes $600$ with horizontal. Find the minimum value of the coefficient of friction $\mu$ so that sphere starts pure rolling.

Answer 1

$F_y=4Mg$

$\theta ={60}^{\circ }$

$F_y=F\sin \theta =F\sqrt{\frac{3}{2}}=4mg$

$\therefore F=\frac{4\times 2}{\sqrt{3}}=\frac{8}{\sqrt{3}}Mg$

$\therefore F_x=F\cos \theta =\frac{8}{\sqrt{3}}\cos {60}^{\circ }=\frac{8}{\sqrt{3}}\times \frac{1}{2}=\frac{4}{\sqrt{3}}Mg$

Now, for rolling,

$F_x=\frac{7}{2}M\mu g$

$\therefore \mu =\frac{2F_x}{7mg}=\frac{2\times 4}{7\sqrt{3}Mg}=\frac{8\times 1}{7\sqrt{3}Mg}$