A uniform disc of mass $M$ and radius $R$ is rotating with angular velocity $\omega$ about its central axis. It is placed gently on a rough horizontal surface. The time taken for the disc to start pure rolling is:

Solution

1. The friction force $f=\mu Mg$ acts at the contact point.
2. Linear acceleration $$\begin{array}{rl}a& =f/M\\ & =\mu g\end{array}$$.
3. Angular deceleration $$\begin{array}{rl}\alpha & =\tau /I\\ & =(fR)/(1/2M{R}^2)\\ & =2\mu g/R\end{array}$$.
4. At time $t$, $$\begin{array}{rl}v& =at\\ & =\mu gt\end{array}$$ and $$\begin{array}{rl}{\omega }_t& =\omega -\alpha t\\ & =\omega -(2\mu g/R)t\end{array}$$.
5. Pure rolling condition $v={\omega }_{t}R$ implies $$\begin{array}{rl}\mu gt& =R(\omega -(2\mu g/R)t)\\ & =R\omega -2\mu gt\end{array}$$.
6. Solving for $t$ gives $3\mu gt=R\omega$, so $t=R\omega /(3\mu g)$.
7. Hence the answer is (A).