Question

A ring of mass $M$ and radius $R$ is rotating with angular speed about a fixed vertical axis passing through its centre $O$ with two point masses each of mass $\frac{M}{8}$ at rest at $O$. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of system is $\frac{8}{9}\omega$ and one of the masses is at a distance of $\frac{3}{5}R$ from $O$. At this instant the distance of the other mass from $O$ is

• A.
  $\frac{2}{3}R$
• B.
  $\frac{1}{3}R$
• C.
  $\frac{3}{5}R$
• D.
  $\frac{4}{5}R$

Answer 1

The correct option is D

$\frac{4}{5}R$
at intially both masses are at the center of ring. So initial angular momentum $L_i=M{r}^2\omega$
at a later moment when mass free from the origen they ran in a opposite direction. Let in this condition whole system rotates by $\frac{8}{9}\omega$ angular speed.
Because there is no external torque so the complet angular momentum will conserved.

On applying the conservation of angular momentum $L_i=L_f$
$L_i=M{R}^2\omega$
$L_f=[M{R}^2+\frac{M}{8}(\frac{3}{5}R{)}^2+\frac{M}{8}{x}^2]\frac{8}{9}\omega$
$L_f=M{R}^2\frac{8\omega }{9}+\frac{M}{8}(\frac{3}{5}R{)}^2\frac{8\omega }{9}+\frac{M}{8}{x}^2\frac{8\omega }{9}$ where $x$ is the distance of the $2^{nd}$ mass from center.
So, $L_i=L_{f}M{R}^2\omega =M{R}^2\frac{8}{9}\omega +\frac{M}{8}(\frac{3}{5}R{)}^2\frac{8}{9}\omega +\frac{M}{8}{x}^2\frac{8}{9}\omega$
Solving we get $x=\frac{4R}{5}$