Physics - Rotation Question with Solution | TestHub

We know that total kinetic energy of a body rolling without slipping $K_{\text{total }}=K_{\text{rot }}+K_{\text{trans }}$ For solid spherical ball, $I=\frac{2}{5}m{R}^2$ (along to diameter) and $v=R\omega$ where $R$ is radius of spherical ball So. $$K_{\text{total }}=\frac{1}{2}\left(\frac{2}{5}m{R}^2\right){\omega }^2+\frac{1}{2}m{R}^2{\omega }^2$$ $$\begin{array}{l}=\frac{7}{10}m{R}^2{\omega }^2\\ K=\frac{7}{10}m{v}^2\end{array}$$ Potential energy = Kinetic energy $$\begin{array}{rl}mgh& =\frac{7}{10}m{v}^2\\ v^2& =\frac{10}{7}gh\end{array}$$ For vertical projection, $$\begin{array}{c}{v}^2=u^2+2g{h}^{\prime }\\ \frac{10}{7}gh=0+2g{h}^{\prime }\Rightarrow h^{\prime }=\frac{5}{7}h\end{array}$$