A uniform solid sphere of mass $M$ and radius $R$ is projected along a rough horizontal surface with an initial linear velocity $v_0$ and zero angular velocity. If the coefficient of kinetic friction is $\mu$, the time taken for the sphere to start pure rolling is

Solution

1. The kinetic friction force $f=\mu Mg$ causes a linear deceleration $a=\mu g$ and a torque $$\begin{array}{rl}\tau & =fR\\ & =\mu MgR\end{array}$$.
2. Using the moment of inertia for a solid sphere $I=\frac{2}{5}M{R}^2$, the angular acceleration is $$\begin{array}{rl}\alpha & =\frac{\tau }{I}\\ & =\frac{\mu MgR}{\frac{2}{5}M{R}^2}\\ & =\frac{5\mu g}{2R}\end{array}$$.
3. At the time $t$ when pure rolling begins, the linear velocity $v=v_0-\mu gt$ must equal the tangential velocity $$\begin{array}{rl}v& =\omega R\\ & =(\alpha t)R\end{array}$$.
4. Substituting the expressions, we get $v_0-\mu gt=(\frac{5\mu g}{2R}t)R$, which simplifies to $v_0=\frac{7}{2}\mu gt$.
5. Solving for time gives $t=\frac{2v_0}{7\mu g}$.
6. Hence the answer is (A).