A body is rolling down an inclined plane. If | vedclass

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Question

Rotational kinetic energy is

$\mathrm{K}_{\mathrm{R}}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{Mk}^{2}\left(\frac{\mathrm{v}}{\mathrm{

Vedclass · Accepted Answer

Solid ball

Vedclass · Answer

Rotational kinetic energy is

$\mathrm{K}_{\mathrm{R}}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{Mk}^{2}\left(\frac{\mathrm{v}}{\mathrm{R}}ight)^{2}\left(\because \mathrm{I}=\mathrm{Mk}^{2} 	ext { and } \mathrm{v}=\mathrm{R} \omegaight)$

$=\frac{1}{2} \mathrm{Mv}^{2}\left(\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}ight)$

Translational kinetic energy is

$\mathrm{K}_{\mathrm{T}}=\frac{1}{2} \mathrm{Mv}^{2}$

As per questions, $\mathrm{K}_{\mathrm{R}}=40 \% \mathrm{K}_{\mathrm{T}}$

$	herefore \frac{1}{2} \mathrm{Mv}^{2}\left(\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}ight)=40 \% \frac{1}{2} \mathrm{Mv}^{2}$ or $\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}=\frac{40}{100}=\frac{2}{5}$

For solid sphere, $\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}=\frac{2}{5}$

Hence, the body is solid ball.

A body is rolling down an inclined plane. If kinetic energy of rotation is $40\%$ of translational kinetic energy, then the body is a

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