Question

A uniform ball of radius $r$ rolls without slipping down from the top of a sphere of radius $R$. The angular velocity of the ball when it breaks from the sphere is:

• A. $\sqrt{\frac{5g(R+r)}{17r^2}}$
• B. $\sqrt{\frac{10g(R+r)}{17r^2}}$
• C. $\sqrt{\frac{5g(R-r)}{10r^2}}$
• D. $\sqrt{\frac{10g(R+r)}{7r^2}}$

Answer 1

The correct option is B $\sqrt{\frac{10g(R+r)}{17r^2}}$

Let the ball breaks with the sphere at point B, i.e, $N=0$

Also as there is no slipping at B $⟹v=rw$

From geometry, $A{O}^{\prime }=(R+r)(1-cos\theta )$

Now, Circular motion equation: $\frac{m{v}^2}{R+r}=mgcos\theta$ .....................(1)

Work-energy theorem: $mg(R+r)(1-cos\theta )=\frac{1}{2}I{w}^2+\frac{1}{2}m{v}^2$

Using (1) , $mg(R+r)-m{r}^2{w}^2=\frac{1}{2}\times \frac{2}{5}m{r}^2{w}^2+\frac{1}{2}m{r}^2{w}^2$

$⟹w=\sqrt{\frac{10g(R+r)}{17r^2}}$