Physics - Rotation Question with Solution | TestHub

Translational kinetic energy of a ring is: $${\mathrm{KE}}_{\text{Trans }}=\frac{1}{2}{\mathrm{mv}}^2$$ Rotational kinetic energy of the ring is: $$\begin{array}{rl}{\mathrm{KE}}_{\text{Rolling }}& =\frac{1}{2}\mathrm{I}{\omega }^2\\ & =\frac{1}{2}{\mathrm{mR}}^2{\omega }^2\\ {\mathrm{KE}}_{\text{Rolling }}& =\frac{1}{2}{\mathrm{mv}}^2(\because \mathrm{v}=\omega \mathrm{R})\end{array}$$