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 mass-less rods fixed on the ring as shown in 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 \(\frac{3}{5}R\) from point \(O\). At this instant, the distance of the other mass from point \(O\) is
Show Hint
No external torque acts, so angular momentum is conserved; write the moment of inertia with the two radial masses.
Step 1: Understanding the Concept
There is no external torque about the vertical axis, so angular momentum \(I\omega\) is conserved.
Step 2: Initial state
The two masses are at the centre, so \(I_{1}=MR^{2}\) and the angular speed is \(\omega\).
Step 3: Final state
With one mass at \(\frac{3R}{5}\) and the other at \(x\), \(I_{2}=MR^{2}+\frac{M}{8}\left(\frac{9R^{2}}{25}+x^{2}\right)\) and the angular speed is \(\frac{8\omega}{9}\).
Step 4: Calculation
\(MR^{2}\omega=\frac{8\omega}{9}\left[MR^{2}+\frac{M}{8}\left(\frac{9R^{2}}{25}+x^{2}\right)\right]\), so \(9R^{2}=8R^{2}+\frac{9R^{2}}{25}+x^{2}\). Then \(x^{2}=\frac{16R^{2}}{25}\) and \(x=\frac{4R}{5}\), which is option (D).
Final Answer:
The other mass is at a distance \(\frac{4R}{5}\) from \(O\).
\[ \boxed{\text{(D) }\dfrac{4R}{5}} \]
