The normal force for circular motion is from normal component of gravitation force, and it should be larger than the centripetal force
i.e. $N= mg cos\theta\ge m \frac{v^2}{R}$, where R is the radius.
Since it starts to roll down from the top,
the kinetic energy is coming from changes of potential energy
$\frac{1}{2}mv^2=mgR(1-cos\theta)$
So $\frac{v^2}{R}=2mg(1-cos\theta)\le mg cos\theta$
$2mg\le3mg cos\theta$
The result is $\theta\le cos^{-1}\frac{2}{3}$