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 $\frac{8\omega}{9}$ and one of the masses is at a distance of $\frac{3R}{5}$ from $O$. At this instant the distance of the other mass from $O$ is

Step 1: Determine the initial moment of inertia.

Initially, both point masses of mass $\frac{M}{8}$ are at the center $O$, so only the ring contributes to the moment of inertia.

$$I_0 = MR^2$$

Step 2: Determine the moment of inertia at the given instant.

At this instant, the two masses are at distances $r_1$ and $r_2$ from the center.

$$I(t) = MR^2 + \frac{M}{8}\left(r_1^2 + r_2^2\right)$$

Step 3: Apply conservation of angular momentum.

Since there is no external torque acting on the system about the vertical axis, the angular momentum remains constant.

$$I_0 \omega = I(t) \omega'$$

Step 4: Substitute the given values into the conservation equation.

Substitute $\omega' = \frac{8}{9}\omega$ and the expressions for the moments of inertia.

$$MR^2 \omega = \left[MR^2 + \frac{M}{8}\left(r_1^2 + r_2^2\right)\right] \frac{8}{9}\omega$$

Step 5: Simplify the relation between the distances.

Cancel the common terms $M$ and $\omega$ from both sides.

$$R^2 = \frac{8}{9}R^2 + \frac{1}{9}\left(r_1^2 + r_2^2\right)$$ $$\frac{1}{9}R^2 = \frac{1}{9}\left(r_1^2 + r_2^2\right)$$ $$r_1^2 + r_2^2 = R^2$$

Step 6: Calculate the distance of the second mass.

Substitute $r_1 = \frac{3}{5}R$ into the simplified relation.

$$\left(\frac{3}{5}R\right)^2 + r_2^2 = R^2$$ $$r_2^2 = R^2 - \frac{9}{25}R^2$$ $$r_2^2 = \frac{16}{25}R^2$$ $$r_2 = \frac{4}{5}R$$

Answer: $\dfrac{4}{5}R$