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$A$ 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 :
$M\sqrt {\frac{{10}}{7}\,g{\text{l}}} \,\,[{\text{l}} + R]$
$M\sqrt {\frac{{10}}{7}\,g{\text{l}}} \,\,\left[ {{\text{l}} - \,\frac{2}{5}R} \right]$
$M\sqrt {\frac{{10}}{7}\,g{\text{l}}} \,\,\left[ {{\text{l}} + \,\frac{7}{5}R} \right]$
$M\sqrt {\frac{{10}}{7}\,g{\text{l}}} \,\,\left[ {{\text{l}} + \,\frac{2}{5}R} \right]$
Solution
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-
$L=I \omega+M v l=\frac{2}{5} M R^{2} \omega+M v l$
$L=I \omega+M v l=\frac{2}{5} M R v+M v l$
$\Longrightarrow \quad L=M \sqrt{\frac{10 g l}{7}}\left[\frac{2 R}{5}+l\right]$
