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. $\sqrt{\frac{10g(R+r)}{17r^2}}$

Answer 1

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

let the radis od sphere=$R$
and radis od ball=$r$
Force calculation:
$F=mg$ (downwards)
centrifugal force =$\frac{m{v}^2}{r+R}$

So, from the force equation,

$\frac{m{v}^2}{r+R}+N=mgcos\theta$
at a point of no contact, $N=0$
$\frac{m{v}^2}{r+R}=mgcos\theta v^2=(R+r)gcos\theta$ ..(1)

Using the energy conservation theorem:

let the hight of ball at intial=$h$
then the total hight og the ball from point $O$=$(R+rcos\theta )$
total hight $h=(R+r)-(R+r)cos\theta$

potntial energy= $mg(R+r)-(R+r)cos\theta$

now, by the energy conservation potrential energy converts into linear kinetic energy + rotational kinetic energy
$mg(R+r)-(R+r)cos\theta =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

where, $I=\frac{2}{5}M{r}^2$ for ball
$mg(R+r)(1-cos\theta )=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{r}^2\times \frac{v^2}{r^2}$
$g(R+r)(1-cos\theta )=\frac{7}{10}{v}^2$
$g(R+r)-g(R+r)cos\theta )=\frac{7}{10}{v}^2$

using eq. (1)
$\frac{7}{10}{v}^2+v^2=g(R+r)v=\sqrt{\frac{10g(R+r)}{17}}$

Now, $v=r\omega \omega =\frac{v}{r}\Rightarrow \sqrt{\frac{10g(R+r)}{17r^2}}$