A body rolls down an inclined plane. It its kinetic energy of rotation is $40 \%$ of its kinetic ene

Question

A body rolls down an inclined plane. It its kinetic energy of rotation is $40 \%$ of its kinetic energy of translation, then the body is\

Professor PC Thomas Classes · Answer

Kinetic energy of rotation $=40 \%$ of KE of translation Kinetic energy of rotation\ $=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{mK}^{2} 	imes \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}$\ Kinetic energy of translation $=\frac{1}{2} m v^{2}$\ or, $\frac{1}{2} \mathrm{mK}^{2} \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}=\frac{40}{100} 	imes \frac{1}{2} \mathrm{mv}^{2}$\ or $\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{2}{5} \Rightarrow \mathrm{~K}^{2}=\frac{2}{5} \mathrm{R}^{2}$ Moment of inertia of a solid sphere about its diameter is $\frac{2}{5} \mathrm{MR}^{2} . 	herefore$ The body is a solid sphere.\