A sphere of mass M and radius R is attached b | vedclass

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Question

Let at bottom, $v=$ speed and $\omega=$ angular speed

From conservation of energy-

$M g l=\frac{1}{2} M v^{2}+\frac{1}{2} I \omega^{2}$

$\Lon

Vedclass · Accepted Answer

$M\sqrt {\frac{{10}}{7}\,g{	ext{l}}} \,\,\left[ {{	ext{l}} + \,\frac{2}{5}R} ight]$

Vedclass · Answer

Let at bottom, $v=$ speed and $\omega=$ angular speed

From conservation of energy-

$M g l=\frac{1}{2} M v^{2}+\frac{1}{2} I \omega^{2}$

$\Longrightarrow M g l=\frac{1}{2} M v^{2}+\frac{1}{2} \frac{2}{5} M R^{2} \omega^{2}$

And $v=R \omega$

$\Longrightarrow \frac{7}{10} M v^{2}=M g l$

$\Longrightarrow v=\sqrt{\frac{10 g l}{7}}$

Now, the momentum about $P$ is given by-

$L=I \omega+M v l=\frac{2}{5} M R^{2} \omega+M v l$

$L=I \omega+M v l=\frac{2}{5} M R v+M v l$

$\Longrightarrow \quad L=M \sqrt{\frac{10 g l}{7}}\left[\frac{2 R}{5}+l\right]$

$A$ sphere of mass $M$ and radius $R$ is attached by a light rod of length $l$ to $a$ point $P$. The sphere rolls without slipping on a circular track as shown. It is released from the horizontal position. the angular momentum of the system about $P$ when the rod becomes vertical is :

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