Question

A small solid cylinder of mass $M$ and radius $R$ slides down a smooth curve from height $h$. It gets onto plank of mass $M$, which is resting on a smooth surface. If $\mu$ is coefficient of friction between cylinder and plank, the time at which that cylinder attains pure rolling on plank is found to be $\frac{v_0}{x\mu g}$ where $v_0=\sqrt{2gh}$.
The value of $4x$ is (Assume plank has sufficient length for cylinder to attain pure rolling)

Answer 1

From conservation of energy principle, we can say that the velocity $v_0$ of cylinder just before getting onto plank is,
$Mgh=\frac{1}{2}M{v}_0^2$
$\Rightarrow v_0=\sqrt{2gh}$
Over the plank, the cylinder has retarding linear motion, with retardation $a=-\mu g$
Velocity of cylinder $v=v_0-\mu gt$

and it has angular acceleration, $\alpha =\frac{{\tau }_f}{I_C}=\frac{\mu MgR}{\frac{M{R}^2}{2}}=\frac{2\mu g}{R}$
So, Angular velocity of cylinder $\omega ={\omega }_0+\alpha t=0+\frac{2\mu gt}{R}=\frac{2\mu gt}{R}$
Plank has accelerated motion, with acceleration $a=\mu g$
So velocity of plank, $v^{\prime }=0+\mu gt$

At pure rolling, $v^{\prime }=v-\omega R\Rightarrow \mu gt=v_0-\mu gt-2\mu gt$
$\Rightarrow t=\frac{v_0}{4\mu g}$
Here, $t$ is time taken for pure rolling to start.
$\Rightarrow x=4,4x=16$