Question

A block of mass $M$ with semicircular track of radius $R$ rests on a horizontal smooth surface. A cylinder of radius $r$ slips on the track. If the cylinder is released $(u=0)$ from top, the distance moved by block when cylinder reaches the bottom of the track is :

• A. $R-r$
• B. $\frac{M(R-r)}{M+m}$
• C. $\frac{M}{M+m}(R+r)$
• D. $\frac{m(R-r)}{M+m}$

Answer 1

Answer: (D)

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

Let the block and cylinder are system cylinder more (R - r) distance right.

Let the system have center of mass at x.

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

finally when cylinder is at bottom then the x-coordinate of center of mass of block and cylinder system is same x.

let block move distance d.

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

$\Rightarrow m(R-r)-md+M(R-r)-Md=M(R-r)$

$m(R-r)-md+M(R-r)-Md=M(R-r)$

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