Question

A ring of mass $M$ and radius $R$ is rotating with angular speed $\omega$ 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 the 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. $\frac{3}{5}R$
• B. $\frac{2}{3}R$
• C. $\frac{1}{3}R$
• D. $\frac{4}{5}R$

Answer 1

The correct option is D $\frac{4}{5}R$

$I_1{\omega }_1=I_2{\omega }_2$

$\Rightarrow M{R}^2\times \omega =(M{R}^2+\frac{M}{8}\times \frac{9}{25}{R}^2+\frac{M}{8}{x}^2)\times \frac{8}{9}\omega$

$\Rightarrow \frac{9}{8}{R}^2=R^2+\frac{9}{200}{R}^2+\frac{x^2}{8}$

$\Rightarrow \frac{x^2}{8}=\frac{9}{8}{R}^2-R^2-\frac{9}{200}{R}^2$

$\Rightarrow x^2=\frac{16}{25}{R}^2$

$\Rightarrow x=\frac{4}{5}R$