Physics - Rotation Question with Solution | TestHub

When a body rolls i.e. have rotational motion, the total kinetic energy of the system will be $$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2\left(1+\frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}\right)$$ where, $\mathrm{m}=$ mass of body, $\mathrm{v}=$ velocity and $\mathrm{k}=$ radius of gyration
Given, $\mathrm{m}=2\mathrm{kg},\theta =30^{\circ },\mathrm{v}=4{\mathrm{ms}}^{-1}$ Let $h$ be the height of the inclined plane, then from law of conservation of energy, $$\begin{array}{rl}\mathrm{KE}& =\mathrm{PE}\\ \frac{1}{2}{\mathrm{mv}}^2\left(1+\frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}\right)& =\mathrm{mgh}\end{array}$$ Substituting the given values in the above equation, we get $$\begin{array}{rl}& \frac{1}{2}\times 2\times 16\left(1+\frac{1}{2}\right)=2\times 10\times \mathrm{h}\left[\because \text{ For cylinder }\frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}=\frac{1}{2}\right]\\ & \Rightarrow 8\times \frac{3}{2}=10\mathrm{h}\Rightarrow \mathrm{h}=1.2\mathrm{m}\end{array}$$ From the above diagram $$\begin{array}{c}\sin \theta =\frac{\mathrm{h}}{\mathrm{x}}\\ \Rightarrow \mathrm{x}=\frac{\mathrm{h}}{\sin \theta }=\frac{1.2}{\sin 30^{\circ }}=1.2\times 2=2.4\mathrm{m}\left[\because \sin 30^{\circ }=\frac{1}{2}\right]\end{array}$$