Question

A block of mass $M$ with a semicircular groove 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 the semicircular frictionless track. Then find the distance travelled by the block when the cylinder reaches the bottom-most point $B$.

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

Answer 1

The correct option is B $\frac{m(R-r)}{M+m}$

From the FBD of block, it is clear that the block will move towards $-vex-$ axis due to force $N_x$. In vertical direction, $N_y,N^{\prime }\&Mg$ are balanced.


$\because F_{\text{ext}}=0$ in horizontal direction on the system of (Cylinder + Block), hence centre of mass of the system will not get displaced in horizontal direction.
$\Rightarrow \Delta x_{\text{CM}}=0$


When cylinder reaches $B$, then displacement of $\text{COM}$ of block$\left(M\right)$ along $-vex-$ axis is $x^{\prime }$ w.r.t ground.
i.e $\Delta x_{B,G}=\Delta x_1=-x^{\prime }...\left(i\right)$

When cylinder reaches $B$, then displacement of its $\text{COM}$ along $+vex-$axis relative to the block is,
$\Delta x_{C,B}=(R-r)$
Hence, with respect to ground, displacement of its $\text{COM}$ is,
$\Delta x_2=\Delta x_{C,B}+\Delta x_{B,G}$
$\therefore \Delta x_2=(R-r)-x^{\prime }...\left(ii\right)$

Applying the equation for displacement of centre of mass:
$\Delta x_{\text{CM}}=\frac{m_1\Delta x_1+m_2\Delta x_2}{m_1+m_2}$
Here, $m_1=M,m_2=m$
$\Rightarrow 0=\frac{M(-x^{\prime })+m\left(\right(R-r)-x^{\prime })}{M+m}$
$\Rightarrow m(R-r)=x^{\prime }(M+m)$
$\therefore x^{\prime }=\frac{m(R-r)}{M+m}$
Here $x^{\prime }$ is the required displacement or distance covered by the block in leftward direction.