A cylinder of radius a and mass m contains a point mass, also of mass m, located a distance ##a/2## from the symmetry axis. The cylinder is placed on an incline, which is initially horizontal, but is very slowly raised. Assuming the cylinder cannot slide on the incline, at what inclination angle ##\alpha## does the cylinder begin to roll down?
The Attempt at a Solution
I tried to write the torque around the contact point, with ##\theta## being the angle where the point mass is located, from the line connecting the contact point with the center of the cylinder. I have this: $$\tau = mg(a-a/2 cos(\theta))sin(\alpha)+mga sin(\alpha)=mga(2- cos(\theta)/2)sin(\alpha)$$ But i don't know what condition to put. I know I have to use the no slipping, which would be $$x=\theta a$$ where x is the distance along the plane, but I am not sure what equation involving x should I use.