Question

A block of mass $M$ with a semicircular track of radius $R$ rests on a horizontal frictionless surface. A uniform cylinder of radius $r$ and mass $m$ is released from rest from the top point A. The cylinder slips on these semicircular frictionless tracks. The distance traveled by the block when the cylinder reaches the point B is :

• A. $\left(\frac{M(R-r)}{M+m}\right)$
• B. $\left(\frac{m(R-r)}{M+m}\right)$
• C. $\left(\frac{(M+m)R}{M}\right)$
• D. None of the above

Answer 1

Answer: (B)

$\left(\frac{m(R-r)}{M+m}\right)$

Given that initially the system is at rest so $v_{cm}$ = 0. Now in slipping as no horizontal force is applied, the horizontal component of velocity of centre of mass of the system must remain constant. So, if $v$ and $V$ are the velocities of cylinder and block at B relative to the ground.

$\frac{mv+MV}{m+M}=const.=0$

i.e. $mv+MV=0$ ............(1)

$\frac{m\Delta r_1}{\Delta t}+\frac{\Delta r_2}{\Delta t}M=0$

as , $v_1=\frac{\Delta r_1}{\Delta t}$

or, $m\Delta r_1+M\Delta r_2=0$

as $\Delta r_1=d$

or, $m{d}_1-M{d}_2=0$ .................(2)

as $d_2$ is opposite to $d_1$

Now, when the cylinder reaches from A and B, it is displaced by a distance $(R-r)$ horizontally to the right relative to the block. So, if the displacement of the block relative to ground is $d_2$ to the left, the displacement of cylinder relative to ground to the right will be $d_1=(R-r)-d_2$......(3)

Substituting $d_1$ from Equation (3) in (2),

we get $M\left[\right(R-r)-d_2]-m{d}_2=0$

$d_2=\frac{m(R-r)}{M+m}$