Question

A block of mass $M$ with a semicircular track of radius $R$ rest on a horizontal frictionless surface. A uniform cylinder of radius $r$ and mass $m$ is released from rest at the top point of the semicircular track. The cylinder slips in the semicircular frictionless track. How far the block moved with the cylinder reaches the bottom of the track?

• A. $\frac{m}{M+m}\left(Rr\right)$
• B. $\frac{4m}{M+m}(R-r)$
• C. $\frac{m}{M+m}(R-r)$
• D. $\frac{m^2}{M+m}(R-r)$

Answer 1

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

Considering external force.

In y direction In x - direction

$\downarrow$ $\downarrow$ No forces (external)

mg Mg $\overrightarrow{P}$ is conserved

$\to$ since there no velocity or external force in x-driection

[Centre of mass] $V_x=0$

$COM{V}_x=0$

which gives, $xcom\to constant\Rightarrow \Delta xcom=0$

$xcom=\frac{m{x}_1+M{x}_2}{m+M}$ $x_1\to$ position of com pf $m$

$\Delta xCOM=\frac{m\Delta x_1+M\Delta x_2}{m+M}$ $x_2\to$ potion of COM of $M$

$0=m\Delta x_1+m\Delta x_2$

${\overrightarrow{x}}_{1/g}={\overrightarrow{x}}_{1/2}+{\overrightarrow{x}}_{2/g}\Rightarrow \Delta x_1=\Delta$

$x_{1/2}+\Delta x_{2/g}$ $g\to$ ground

$0=m(\Delta x_{1/2}+\Delta x_{2/g})+M\Delta x_{2/g}$

$0=m\left[\right(R-r)+\Delta x_{2/g}]+m\Delta x_{2/g}$

$\Delta x_{2/g}=\frac{-m(R-r)}{m+M}$

Block will move by $\Delta x_{2/g}=\frac{m(R-r)}{m+M}$ m left.