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 math_failure (math_image_error): 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.