Physics - Rotation Question with Solution | TestHub

Radius of gyration is given by, $K=\sqrt{\frac{I}{m}}$ We know moment of inertia for the ring $I_{\text{disc }}=\frac{1}{2}m{R}^2$ and the disc $I_{\text{ring }}=m{R}^2$ about its central axis perpendicular to its plane. Using parallel axis theorem, the moment of inertia about a tangential axis perpendicular to the plane can be easily obtained using, $I^{\prime }=I+m{R}^2$ The radius of gyration is defined as: $\begin{array}{rl}& K=\sqrt{\frac{I}{m}}\\ & \therefore K_{\text{disc }}=\sqrt{\frac{\frac{1}{2}m{R}^2+m{R}^2}{m}}=\sqrt{\frac{3}{2}}R\\ & \therefore K_{\text{ring }}=\sqrt{\frac{m{R}^2+m{R}^2}{m}}=\sqrt{2}R\end{array}$ Thus, the ratio of the radius of gyration of a ring to that of a disc of same radius and mass, about a tangential axis perpendicular to the plane is: $\frac{K_{\text{disc }}}{K_{\text{ring }}}=\frac{\sqrt{3/2}}{\sqrt{2}}=\frac{\sqrt{3}}{2}$