Question

A ring of mass $m$ and radius $R$ is rotating with angular speed $\omega$ about a fixed vertical axis passing through it's 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. $\frac{2}{3}R$
• B. $\frac{1}{3}R$
• C. $\frac{3}{5}R$
• D. $\frac{4}{5}R$

Answer 1

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

From principle of conservation of angular momentum-
$L_i=L_f$
$m{R}^2\omega =\frac{8}{9}\omega (I_1+I_2+I_{\text{ring}})$
$m{R}^2\omega =\frac{8}{9}\omega (\frac{m}{8}{\left(\frac{3}{5}R\right)}^2+\frac{m}{8}{x}^2+m{R}^2)$
$9R^2=8(\frac{1}{8}{\left(\frac{3}{5}R\right)}^2+\frac{1}{8}{x}^2+R^2)$
$9R^2=({\left(\frac{3}{5}R\right)}^2+x^2+8R^2)$
$R^2={\left(\frac{3}{5}R\right)}^2+x^2$
$R^2-\frac{9R^2}{25}=x^2$
$\frac{16R^2}{25}=x^2$
On solving, we get
$x=\frac{4R}{5}$