A uniform sphere of radius r rolls without slipping down the top of a sphere of radius R. The initial velocity of the sphere is negligible. The angular velocity of the moving sphere at the moment when it breaks off from the other sphere is

When the line joining the centers of the two spheres makes an angle $\mathrm{\theta }$ with the vertical, then
$\mathrm{mgcos}\mathrm{\theta }-\mathrm{N}=\frac{{\mathrm{mv}}^2}{\mathrm{R}+\mathrm{r}}$
When the contact breaks off N= 0
$\mathrm{mgcos}\mathrm{\theta }=\frac{{\mathrm{mv}}^2}{\mathrm{R}+\mathrm{r}}=\frac{{\mathrm{m\omega }}^2{\mathrm{r}}^2}{\mathrm{R}+\mathrm{r}}$  $\cdots \left(1\right)[\because \mathrm{v}=\mathrm{\omega r}]$
From COE principle
$\begin{array}{l}\mathrm{mg}(\mathrm{R}+\mathrm{r})=\mathrm{mg}(\mathrm{R}+\mathrm{r})\cos \mathrm{\theta }+\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2\\ \mathrm{mg}(\mathrm{R}+\mathrm{r})=(\mathrm{R}+\mathrm{r})\frac{{\mathrm{m\omega }}^2{\mathrm{r}}^2}{\mathrm{R}+\mathrm{r}}+\frac{{\mathrm{m\omega }}^2{\mathrm{r}}^2}{2}+\frac{1}{2}\cdot \frac{2}{5}{\mathrm{mr}}^2{\mathrm{\omega }}^2\\ \text{ or }\mathrm{g}(\mathrm{R}+\mathrm{r})=\frac{17}{10}{\mathrm{\omega }}^2{\mathrm{r}}^2\\ \therefore \mathrm{\omega }=\sqrt{\frac{10\mathrm{g}(\mathrm{R}+\mathrm{r})}{17{\mathrm{r}}^2}}\end{array}$