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 $\omega =\frac{v}{r}=\sqrt{\frac{10g(R+r)}{x{r}^2}}$. Find $x$. Assume initial velocity negligible.

Answer 1

$\frac{m{v}^2}{(R+r)}=mgcos\theta ;mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
$mg(R+r)(1-cos\theta )=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=\frac{7}{10}m{v}^2$
$\frac{10}{7}mg(1-cos\theta )=mgcos\theta$
$m{v}^2=\frac{10}{7}mg(R+r)(1-cos\theta )\frac{10}{7}=\frac{17}{7}cos\theta$
Or, $cos\theta =\frac{10}{17}$
$v=\sqrt{g(R+r)cos\theta }=\sqrt{\frac{10}{17}g(R+r)}$
and
$\omega =\frac{v}{r}=\sqrt{\frac{10g(R+r)}{17r^2}}$