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.

Explanation

Let us write the equation of motion for the centre of sphere at the moment of breaking-off:
`mv^2//(R+r)=mgcostheta`,
where v is the velocity of the centre of the sphere at that moment, and `theta` is the corresponding angle (figure). The velocity v can be found from the energy conservation law:
`mgh=1/2mv^2+1/2Iomega^2`,
where I is the moment of inertia of the sphere relative to the axis passing through the sphere's centre. i.e. `I=2/5mr^2`. In addition,
`v=omegar`, `h=(R+r)(1-costheta)`.
From these four equations we obtain
`omega=sqrt(10g(R+r)17r^2)`.