Question

A semi-circular track of radius $R=62.5cm$ is cut in a block. Mass of block, having track, is $M=1kg$ and rests over a smooth horizontal floor. A cylinder of radius $r=10cm$ and mass $m=0.5kg$ is hanging by a thread such that axes of cylinder and track are in same level and surface of cylinder is in contact with the track as shown in figure.
When the thread is burnt, cylinder starts to move down the track. Sufficient friction exists between surface of cylinder and track, so that cylinder does not slip.
Calculate velocity of axis of cylinder (in m/s) when it reaches bottom of the track. ($g=10{ms}^{-2}$).

Answer 1

Momentum of the cylinder-block system is conserved.

$m{v}_1+M{v}_2=0$

$v_1$ is the velocity of cylinder, $v_2$ is the velocity of block.

Velocity of cylinder w.r.t. block is

$v=v_1-v_2$

As it is pure rolling,

$w=\frac{v_1-v_2}{R}$

Since there is no sliding, energy is conserved,

$\frac{1}{2}{I}_c{w}^2+\frac{1}{2}m{v}_1^2+\frac{1}{2}M{v}_2^2-mgR=0$

$I_c$ is the moment of inertia of the cylinder about it's axis,

$I_c=m{r}^2/2$

Therefore, substituting $w$ and $v_2$ in energy equation, we get,

$\frac{1}{2}\frac{m{r}^2}{2}(\frac{v_1-v_2}{r}{)}^2+\frac{1}{2}m{v}_1^2+\frac{1}{2}\frac{m}{M}{v}_1^2=mgR$

$v_1=\sqrt{\frac{gR}{\frac{1}{4}(1+\frac{m}{M}{)}^2+\frac{1}{2}(1+\frac{m}{M})}}$

$v_1=\sqrt{\frac{10\times 0.625}{\frac{1}{4}(1+0.5{)}^2+\frac{1}{2}(1+0.5)}}\approx 2$

Answer: 2