I will try and give you an answer without providing a solution to the problem!
The no slipping condition is that $v_{\rm com} = r \omega$ which when differentiated with respect to time gives $a_{\rm com} = r \alpha$ where $a_{\rm com}$ is the linear acceleration of the centre of mass and $\alpha$ is the angular acceleration.
So it is a "balance" between two different accelerations.
What determines the linear acceleration of the centre of mass, $a$?
$ma = W \sin \theta-F$
$mg\sin \theta$ is the component of the weight down the slope, $F$ the frictional force up the slope, $m$, the mass of the cylinder, and $\theta$ is the angle of the incline.
What determines the angular acceleration, $\alpha$?
$\tau = F\,r=I_{\rm com} \alpha$
$\tau$ is the torque on the cylinder, $r$ the radius of the cylinder and $I_{\rm com}$ the moment of inertia of the cylinder about its centre of mass.
There is a limitation on the magnitude of the frictional force $F\le\mu N$ where $N$ is the normal force, $N=m\,g\cos \theta$, ie $F\le\mu m\,g\cos \theta$.
This means that the maximum value of $\alpha$ is $\dfrac{\mu\,N\,r}{I_{\rm com}} = \dfrac{\mu\,m\,g\cos \theta\,r}{I_{\rm com}}$
Consider the cylinder rolling down the slope without slipping.
Now consider what happens if the incline angle $\theta$ is increased.
The normal force decreases and so the maximum value of the angular acceleration,$\alpha$, decreases.
There will come a time when $a> r\,\alpha$ because of the limiting value of $\alpha$ and the cylinder will start to slip, ie its increase in angular velocity is not sufficient to keep up with the increase in linear velocity.
Without giving too much away, you can show that for a cylinder to roll down an incline without slipping the coefficient of static friction $\mu \ge \frac 13 \tan \theta$.
