A ring starts to roll down the inclined plane of height $h$ without slipping. The velocity with whic

Question

A ring starts to roll down the inclined plane of height $h$ without slipping. The velocity with which it reaches the ground is\

Professor PC Thomas Classes · Answer

D) $v=\sqrt{\frac{2 g h}{1+\frac{K^{2}}{R^{2}}}}$ Where $K$ is the radius of gyration\ For ring, $\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=1$\