A solid cylinder of radius 'R' and mass 'M' rolls down an inclined plane of height 'h'. When it reaches the bottom of the plane, its rotational kinetic energy is $(\mathrm{g}=$ acceleration due to gravity)

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Aicharya · Accepted Answer

$\begin{aligned} 	ext { Total } \mathrm{KE} &=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2} \ &=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2}\left(\frac{1}{2} \mathrm{MR}^{2}ight) \mathrm{v}^{2} \ &=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{4} \mathrm{Mv}^{2}=\frac{3}{4} \mathrm{Mv}^{2}=\mathrm{Mgh} \end{aligned}$

A solid cylinder of radius 'R' and mass 'M' rolls down an inclined plane of height 'h'. When it reaches the…

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