A solid sphere of mass M and radius r slips on a rough horizontal plane. At some instant it has translational velocity $v_{0}$ and rotational velocity about the centre $\frac{v _{0}}{2 r}$ . The translational velocity after the sphere starts pure rolling is

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TestHub · Accepted Answer

Friction acts at the lowest point, so angular momentum about it is conserved

$m v_{0} r + \frac{2}{5} m r^{2} \cdot \frac{v _{0}}{2 r} = m v^{'} r + \frac{2}{5} m r^{2} \cdot \frac{v ^{'}}{r}$

$\frac{6}{5} m v_{0} r = \frac{7}{5} m v^{'} r$

$v^{'} = \frac{6 v _{0}}{7}$ forward

Answer: A

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