A uniform ball of radius$r$ rolls without slipping down from the top of a sphere of radius $R$. Find the angular velocity of the ball at the moment it breaks off the sphere. The initial velocity of the ball is negligible.

Let $v$ is the velocity and $O$ is the angle made by the radius vector with the vertical at the instant when the ball break-off the sphere.

Let it happens at a vertical height $h$ below the top.Therefore we have,

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

where $h=(R+r)(1-\cos \theta )$

$$\begin{gathered} \therefore mg\left(R+r\right)(1+\cos \theta )=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2 \\ mg\left(R+r\right)(1+\cos \theta )=\frac{1}{2}m{\left(\omega v\right)}^2+\frac{1}{2}I{\omega }^2 \dots \left(1\right) \end{gathered}$$

At the break-off

$mg\cos \theta =N+\frac{m{v}^2}{(R+r)}$

Substituting$N=0$ and $v=\omega r$, we have

$mg \cos \theta =\frac{m{\left(\omega r\right)}^2}{(R+r)}$

Solving equation $\left(1\right)$ and $\left(2\right)$, we get

$\omega =\sqrt{\frac{10g\left(R+r\right)}{17r^2}}$