Physics - Rotation Question with Solution | TestHub

When solid sphere rolls down on an inclined plane, then it has both rotational and translational kinetic energy or $\begin{array}{ll}& K=K_{\text{rot }}+K_{\text{trans }}\\ & K=\frac{1}{2}l{\omega }^2+\frac{1}{2}M{v}^2\end{array}$ where, $I=$ moment of inertia of solid sphere $=\frac{2}{5}M{R}^2$ $\begin{array}{rlr}\therefore K& =\frac{1}{2}\left(\frac{2}{5}M(2R{)}^2\right){\omega }^2+\frac{1}{2}M{v}^2& [\because R=2R]\\ & =\frac{4}{5}M{R}^2{\left(\frac{v}{2R}\right)}^2+\frac{7}{2}M{v}^2& [\because v=r\omega ]\end{array}$ $=\frac{1}{5}M{v}^2+\frac{1}{2}M{v}^2=\frac{1}{10}M{v}^2$ Now, gain in $\mathrm{KE}=$ loss in PE $\begin{array}{rl}\frac{7}{10}M{v}^2& =Mgh\\ v& =\sqrt{\frac{10}{7}gh}\end{array}$