Like a rolling stone

Note that the axis through the point of contact of the hollow sphere is the instantaneous axis of rotation.

So, about the instantaneous axis of rotation,

$\frac{mg\sin \theta R}{\frac{5}{3}mR^2}=\frac{fr}{\frac{2}{3}mR^2} \\ f=\frac{2}{5}mg \sin \theta ≤ \mu m g\cos \theta \\ \boxed{\mu ≥ \frac{2}{5} \tan \theta}$

Here $f$ denotes friction force and $m$ denotes mass of the hollow sphere.