Assume the angle of the slope is $\theta$, mass of the disk is m, radius is R. The momentum of inertia is $I=\frac{1}{2}mR^2$.
Let analysis this problem from the contact point between the disk and the slope.
The normal force between the disk and the slope is $mg \cos\theta$, and the force along the slope is $mg \sin\theta$,
Assume the friction between the disk and the slope is f.
The net force is $mg \sin\theta + f = m a$, where a is the acceleration along the slope.
(the friction force from the slope to the disk is in the same direction as acceleration a)

The condition for rolling without slipping is $a=R\alpha$
The torque is $\tau= R mg\sin\theta = I \alpha =\frac{1}{2}mR^2 \alpha=\frac{1}{2}m R R\alpha=\frac{1}{2}m Ra$.
So $mg\sin\theta=\frac{1}{2}m a$, i.e. $a =2g \sin\theta$ and $f=ma- mg\sin\theta= mg\sin\theta$
Since the maximum static friction force $f_{max}=mg\cos\theta\mu \ge mg\sin\theta$,
it imply that $\mu\ge \tan\theta$ for the disk to rolling without slipping.

If it is a ball instead of a disk, then $I=\frac{2}{5}mR^2$.
$\tau= R mg\sin\theta = I \alpha =\frac{2}{5}mR^2\alpha=\frac{2}{5}m R a$
So $mg\sin\theta=\frac{2}{5}ma$, or $a=\frac{5}{2}g\sin\theta$.
$f=ma -mg\sin\theta=\frac{3}{5}ma=\frac{3}{2}g\sin\theta$.
So the condition for rolling without slipping becomes $\mu \ge \frac{2}{3}\tan\theta$.