A sphere of mass M and radius r slips on a ro | vedclass

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(D) The angular momentum of the sphere about the point of contact with the ground remains conserved because the frictional force acts through the poin

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$\frac{6V_0}{7}$

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(D) The angular momentum of the sphere about the point of contact with the ground remains conserved because the frictional force acts through the point of contact,exerting no torque about it.Initial angular momentum about the point of contact:$L_i = M V_0 r + I_{cm} \omega = M V_0 r + (\frac{2}{5} M r^2) (\frac{V_0}{2r}) = M V_0 r + \frac{1}{5} M V_0 r = \frac{6}{5} M V_0 r$Final angular momentum when pure rolling starts (let final velocity be $v'$ and angular velocity be $\omega' = \frac{v'}{r}$):$L_f = M v' r + I_{cm} \omega' = M v' r + (\frac{2}{5} M r^2) (\frac{v'}{r}) = M v' r + \frac{2}{5} M v' r = \frac{7}{5} M v' r$Equating $L_i = L_f$:$\frac{6}{5} M V_0 r = \frac{7}{5} M v' r$$v' = \frac{6}{7} V_0$

$A$ 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 $\frac{V_0}{2r}$. The translational velocity when the sphere starts pure rolling motion is

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