Physics - Rotation Question with Solution | TestHub

Let$\mathrm{I}$bt the moment of inertia of the body. Then total$\mathrm{K}\mathrm{E}=\frac{1}{2}\mathrm{ }{\mathrm{m}\mathrm{v}}^2+\frac{1}{2}\mathbf{ }\mathrm{I}{\mathrm{\omega }}^2$

or$\mathrm{K}\mathrm{E}=\frac{1}{2}\mathrm{ }{\mathrm{m}\mathrm{v}}^2+\frac{1}{2}\mathrm{ }\mathrm{I}\frac{v^2}{R^2}\left(\mathrm{\omega }=\frac{v}{R}\right)$

According to energy conservation loss in$\mathrm{K}\mathrm{E}=\mathrm{g}\mathrm{a}\mathrm{i}\mathrm{n}$in$\mathrm{P}\mathrm{E}$.

or$\frac{1}{2}\left(m+\frac{I}{R^2}\right){\mathrm{v}}^2=\mathrm{m}\mathrm{g}\mathrm{h}=\mathrm{m}\mathrm{g}\left(\frac{3v^2}{4g}\right)$

Solving this, we get$\mathrm{I}=\frac{1}{2}{\mathrm{m}\mathrm{R}}^2$

i.e., the solid body is a disc