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A solid cylinder of mass $\mathrm{m}$ and radius $\mathrm{R}$ rolls down inclined plane without slipping. The speed of its C.M. when it reaches the bottom is
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Verified Answer
The correct answer is:
$\sqrt{4 \operatorname{gh} / 3}$
By energy conservation
$$
\begin{array}{l}
(\mathrm{K} . \mathrm{E})_{\mathrm{i}}+(\mathrm{P} . \mathrm{E})_{\mathrm{i}}=(\mathrm{K} . \mathrm{E})_{\mathrm{f}}+(\mathrm{P} . \mathrm{E})_{\mathrm{f}} \\
(\mathrm{K} . \mathrm{E})_{\mathrm{i}}=0,(\mathrm{P} . \mathrm{E})_{\mathrm{i}}=\mathrm{mgh},(\mathrm{P} . \mathrm{E})_{\mathrm{f}}=0 \\
(\mathrm{~K} . \mathrm{E})_{\mathrm{f}}=1 / 2 \mathrm{I} \omega^{2}+1 / 2 \mathrm{mv}_{\mathrm{cm}}^{2}
\end{array}
$$
For solid cylinder, moment of inertia, $\mathrm{I}=1 / 2 \mathrm{mR}^{2}$
$(\mathrm{K} . \mathrm{E})_{\mathrm{i}}+(\mathrm{P} . \mathrm{E})_{\mathrm{i}}=(\mathrm{K} . \mathrm{E})_{\mathrm{f}}+(\mathrm{P} . \mathrm{E})_{\mathrm{f}}$
$$
\begin{array}{l}
\text { so, } \mathrm{mgh}=1 / 2\left(1 / 2 \mathrm{mR}^{2}\right)\left(\frac{\mathrm{v}_{\mathrm{cm}}^{2}}{\mathrm{R}^{2}}\right)+1 / 2 \mathrm{mv}_{\mathrm{cm}}^{2} \\
\Rightarrow \mathrm{v}_{\mathrm{cm}}=\sqrt{4 \mathrm{gh} / 3}
\end{array}
$$
$$
\begin{array}{l}
(\mathrm{K} . \mathrm{E})_{\mathrm{i}}+(\mathrm{P} . \mathrm{E})_{\mathrm{i}}=(\mathrm{K} . \mathrm{E})_{\mathrm{f}}+(\mathrm{P} . \mathrm{E})_{\mathrm{f}} \\
(\mathrm{K} . \mathrm{E})_{\mathrm{i}}=0,(\mathrm{P} . \mathrm{E})_{\mathrm{i}}=\mathrm{mgh},(\mathrm{P} . \mathrm{E})_{\mathrm{f}}=0 \\
(\mathrm{~K} . \mathrm{E})_{\mathrm{f}}=1 / 2 \mathrm{I} \omega^{2}+1 / 2 \mathrm{mv}_{\mathrm{cm}}^{2}
\end{array}
$$
For solid cylinder, moment of inertia, $\mathrm{I}=1 / 2 \mathrm{mR}^{2}$
$(\mathrm{K} . \mathrm{E})_{\mathrm{i}}+(\mathrm{P} . \mathrm{E})_{\mathrm{i}}=(\mathrm{K} . \mathrm{E})_{\mathrm{f}}+(\mathrm{P} . \mathrm{E})_{\mathrm{f}}$
$$
\begin{array}{l}
\text { so, } \mathrm{mgh}=1 / 2\left(1 / 2 \mathrm{mR}^{2}\right)\left(\frac{\mathrm{v}_{\mathrm{cm}}^{2}}{\mathrm{R}^{2}}\right)+1 / 2 \mathrm{mv}_{\mathrm{cm}}^{2} \\
\Rightarrow \mathrm{v}_{\mathrm{cm}}=\sqrt{4 \mathrm{gh} / 3}
\end{array}
$$
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