Physics - Rotation Question with Solution | TestHub

Moment of inertia of a circular disc about a diametric axis is $I_d=\frac{1}{4}M{R}^2$.

Using the parallel axis theorem, the moment of inertia of the disc about a tangential axis in its plane is $I_1=I_d+M{R}^2=\frac{1}{4}M{R}^2+M{R}^2=\frac{5}{4}M{R}^2$.

Radius of gyration of the disc is $K_1=\sqrt{\frac{I_1}{M}}=\sqrt{\frac{5}{4}}R=\frac{\sqrt{5}}{2}R$.

Moment of inertia of a circular ring about a diametric axis is $I_r=\frac{1}{2}M{R}^2$.

Using the parallel axis theorem, the moment of inertia of the ring about a tangential axis in its plane is $I_2=I_r+M{R}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$.

Radius of gyration of the ring is $K_2=\sqrt{\frac{I_2}{M}}=\sqrt{\frac{3}{2}}R=\frac{\sqrt{3}}{\sqrt{2}}R$.

The required ratio is $\frac{K_1}{K_2}=\frac{\frac{\sqrt{5}}{2}R}{\frac{\sqrt{3}}{\sqrt{2}}R}=\frac{\sqrt{5}}{2}\times \frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{5}}{\sqrt{6}}$.

Answer: $\sqrt{5}:\sqrt{6}$