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A uniform disk of mass $m$ and radius $R$ rolls without slipping down an incline plane of length $l$ and inclination $\theta$. Initially the disk was at rest at the top of the incline plane. Its angular momentum about the point of contact with the inclined plane when it reaches the bottom will be equal to :-
$\sqrt {4{m^2}{R^2}gl\,\sin \,\theta }$
$\sqrt {3{m^2}{R^2}gl\,\sin \,\theta }$
$\sqrt {\frac{{{m^2}{R^2}gl\,\sin \,\theta }}{4}}$
zero
Solution
$m g l \sin \theta=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$
$=\frac{1}{2} m v^{2}+\frac{1}{2} \frac{m R^{2}}{2} \frac{v^{2}}{R^{2}}=\frac{3}{4} m v^{2}$
$\Rightarrow v=\sqrt{\frac{4 g l \sin \theta}{3}}$
$I=I_{c m} \omega+m v R=\frac{M R^{2}}{2} \frac{v}{R}+M v R=\frac{3 M v R}{2}$
$=\frac{3}{2} m R \sqrt{\frac{4 g l \sin \theta}{3}}=\sqrt{3 m^{2} R^{2} g l \sin \theta}$
