Starting from the highest point, a particle slides down the surface...

Let $m$ be the mass of the particle and $R$ be the radius of the sphere.

Let $\theta$ be the angle made by the radius vector of the particle with the vertical at any instant.

The vertical distance descended by the particle is $h = R - R \cos \theta = R(1 - \cos \theta)$.

Applying conservation of mechanical energy:

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

$v^2 = 2 g R (1 - \cos \theta)$

The equation of motion along the radial direction is:

$m g \cos \theta - N = \dfrac{m v^2}{R}$

where $N is the normal reaction.<br><br>The particle loses contact with the sphere when$N = 0$.<br><br>$m g \cos \theta = \dfrac{m v^2}{R}$<br><br>Substituting the value of$v^2$:<br><br>$m g \cos \theta = \dfrac{m}{R} [2 g R (1 - \cos \theta)]$<br><br>$\cos \theta = 2 (1 - \cos \theta)$<br><br>$\cos \theta = 2 - 2 \cos \theta$<br><br>$3 \cos \theta = 2$<br><br>$\cos \theta = \dfrac{2}{3}$<br><br>$\theta = \cos^{-1} $\left(\dfrac{2}{3}\right)$