A solid cylinder of mass $\mathrm{m} \&$ radius $\mathrm{R}$ rolls down inclined plane without slipping. The…

A solid cylinder of mass $\mathrm{m} \&$ radius $\mathrm{R}$ rolls down inclined plane without slipping. The speed of its C.M. when it reaches the bottom is
  1. $\sqrt{2 \mathrm{gh}}$
  2. $\sqrt{4 \mathrm{gh} / 3}$
  3. $\sqrt{3 / 4 \mathrm{gh}}$
  4. $\sqrt{4 \mathrm{gh}}$

Solution

By energy conservation $(\mathrm{K} . \mathrm{E})_{\mathrm{i}}+(\mathrm{P.E})_{\mathrm{i}}=(\mathrm{K} \cdot \mathrm{E})_{\mathrm{f}}+(\mathrm{P} \cdot \mathrm{E})_{\mathrm{f}}$
$(\mathrm{K} . \mathrm{E})_{\mathrm{i}}=0,(\mathrm{P} . \mathrm{E})_{\mathrm{i}}=\mathrm{mgh},(\mathrm{P} . \mathrm{E})_{\mathrm{f}}=0$
$(\mathrm{K} . \mathrm{E})_{\mathrm{f}}=1 / 2 \mathrm{I} \omega^{2}+1 / 2 \mathrm{mv}_{\mathrm{cm}}^{2}$
Where I (moment of inertia) $=1 / 2 \mathrm{mR}^{2}$ (for solid cylinder) so $m g h=1 / 2\left(1 / 2 m R^{2}\right)\left(\frac{v_{\mathrm{cm}}^{2}}{R^{2}}\right)+1 / 2 m v_{\mathrm{cm}}^{2}$
$\Rightarrow v_{\mathrm{cm}}=\sqrt{4 \mathrm{gh} / 3}$

Asked in: JEE Mains - Rotational Motion - Test 3

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