Question:
A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle rotated by the radius through the particle, when it leaves contact with the sphere.
Solution:
From figure,
$\left(m v^{2}\right) / R=m g \cos \theta$
$v=(\sqrt{g} R \cos \theta) \ldots \ldots 1$
and
Change in K.E. = Work done
$\frac{1}{2} m v^{2}-0=m g(R-R \cos \theta)$
$v=(\sqrt{2} g R[1-\cos \theta]) \ldots . .2$
from 1 and 2
$g R \cos \theta=2 g R(1-\cos \theta)$
$\theta=\cos ^{-1}\left(\frac{2}{3}\right)$
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