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If the angular momentum of a particle of mass $m$ rotating along a circular path of radius $r$ with uniform speed is $L$, the centripetal force acting on the particle is
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The correct answer is:
$\frac{L^2}{m r^2}$
From the relation, angular momentum, $L=m v r$
$\mid \left.v=\frac{L}{m r} \right\rvert\,$ Centripetal force acting on the particle $F=\frac{m v^2}{r}=\frac{m\left(\frac{L}{m r}\right)^2}{r}=\frac{L^2}{m r^3}$
$\mid \left.v=\frac{L}{m r} \right\rvert\,$ Centripetal force acting on the particle $F=\frac{m v^2}{r}=\frac{m\left(\frac{L}{m r}\right)^2}{r}=\frac{L^2}{m r^3}$
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