Question

A uniform sphere of radius $r$ starts rolling down without slipping from the top of another fixed sphere of radius $R$ shown in the figure. Find the angular velocity of sphere of radius $r$ at the instant when it leaves contact with the surface of the fixed sphere.

Answer 1

Initial energy=reference to break off= $=\text{mg(R+r)(1-cos}\theta )$

Energy at break off point= Kinetic energy(rotation +transnational velocity)

$=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2;\text{where}I=\frac{2}{5}m{r}^2=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\omega }^2=\frac{7m{v}^2}{10}$

The total energy at two stages are equal

$mg(R+r)(1-cos\theta )=\frac{7}{10}m{v}^{2}m{v}^2=\frac{10}{7}mg(R+r)(1-cos\theta )$

Condition for break off point:

$\frac{m{v}^2}{(R+r)}\ge \text{mgcos}\theta$

where (R+r) is the radius of centrifugal force of the ball.

$m{v}^2\ge mgcos\theta (R+r)\frac{10}{7}mg(R+r)(1-cos\theta )\ge mgcos\theta (R+r)cos\theta \le \frac{10}{17}$

$m{v}^2=\frac{10}{7}mg(R+r)(1-\frac{10}{7})v^2=\frac{10}{7}mg(R+r)\left(\frac{7}{17}\right)r\omega =\sqrt{\frac{10}{17}(R+r)}\omega =\sqrt{\frac{10}{17r^2}(R+r)}$