Question

A particle slides on the surface of a fixed smooth sphere starting from the top most point. Find the angle rotated by the radius through the particle, when it leaves contact with the sphere.

Answer 1

As shown in the figure, the particle falls through a height of $h=R(1-\cos \theta )$, where $R$ is the radius of the smooth sphere, and $\theta$ is the angle through which it falls before losing contact with the sphere.

Seen from the inertial reference frame of the fixed sphere, two forces act on the particle during its descent: gravitational force $mg$ downwards, and normal reaction $N$ due to the sphere along a radial direction away from the center of the sphere.

The resultant of radial components of forces provide the necessary centripetal force to the particle to maintain circular motion. Thus:

$m\frac{v^2}{R}=mg\cos \theta -N$

When the particle is about to lose contact with the sphere, $N$ becomes $0$.

$\therefore \frac{v^2}{R}=g\cos \theta$

Applying work-energy theorem for the descent of the particle down the sphere:

$mgh=\frac{1}{2}m{v}^2$

$\Rightarrow gR(1-\cos \theta )=v^2$

$\Rightarrow \frac{v^2}{R}=2g(1-\cos \theta )$

Substituting with $\frac{v^2}{R}=g\cos \theta$:

$\Rightarrow g\cos \theta =2g(1-\cos \theta )$

$\Rightarrow \cos \theta =\frac{2}{3}$

$\therefore \theta ={\cos }^{-1}\left(\frac{2}{3}\right)$