A solid cylinder of mass M and radius R& | vedclass

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Question

Potential energy of the solid cylinder at height $\mathrm{h}=\mathrm{Mgh}$

$K.E.$ of centre of mass when reached at bottom

$=\frac{1}{2} M v^{2}

Vedclass · Accepted Answer

$\sqrt {\frac{4}{3}gh} $

Vedclass · Answer

Potential energy of the solid cylinder at height $\mathrm{h}=\mathrm{Mgh}$

$K.E.$ of centre of mass when reached at bottom

$=\frac{1}{2} M v^{2}+\frac{1}{2} I \omega^{2}=\frac{1}{2} M v^{2}+\frac{1}{2} M k^{2} v^{2} / R^{2}$

$=\frac{1}{2} M v^{2}\left(1+\frac{k^{2}}{R^{2}}ight)$

For a solid cylinder $\frac{k^{2}}{R^{2}}=\frac{1}{2}$

$	herefore \quad \mathrm{K.E.}=\frac{3}{4} M v^{2}$

$	herefore \quad M g h=\frac{3}{4} M v^{2}$

$v=\sqrt{\frac{4}{3} g h}$

A solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane without slipping. The speed of its centre of mass when it reaches the bottom is ...

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