Consider the stepped cylinder in the figure with the radii R1 and R2 given as 0.5 m and 1 m, respectively. Its mass m is 50 kg and its radius of gyration k is 0.5 m. A rope wound on the hub is pulled by a force T as shown. The coefficients for static and dynamic friction are given as 0.15 and 0.1, respectively. For the value of T given below, what would be the acceleration of the centre of mass?

Hint: The static friction is the friction that resists the motion. Once the motion starts, the dynamic friction, i.e. the friction during sliding, is usually smaller than the static friction. This is the reason why two coefficients are given for friction in this question.

These are the hints
1. Assume no slipping and determine the angular acceleration on this basis
2. Determine the friction, F, required to balance the forces at this acceleration
3. If F is less than the normal force N, the wheel will not slip. Just convert the angular acceleration to the linear acceleration of the centre of mass.
4. If F exceeds static friction, then the wheel will slip. The friction force is then F=0.1 N.
5. Knowing T and knowing F, recalculate the acceleration that will apply in this case.

T[N] = 160;

http://micko.dyndns.org/stepped_cylinder.jpg