A 0.108kg Frisbee is 0.12 meter in radius and has about half its mass spread uniformly in a disk, while the other half is concentrated in the rim of the Frisbee. With a quarter-tum flish of the wrist, a student starts the frisbee rotating at 550 revolutions per minute.

What is the moment of inertia of the Frisbee? I wonder what the coefficient would be

What is the magnitude of the torque, assumed constant, that the student applies? How can I get the angular acceleration from the rpm?

You can find the moment of inertia, I, as the sum of two separate moments of inertia, I_1 + I_2. I_1 is the moment of inertia of a disk with half the total mass of the frisbee (so 0.054 kg of mass with a moment of inertia of I_1 = mr^2). Meanwhile, I_2 is the moment of inertia of a ring with the other half of the total mass (so 0.054 kg of mass with a moment of inertia of I_2 = (1/2)mr^2. Again, the total resultant moment of inertia will be the sum of the two individual ones. By the way, this phenomenon, where something can be computed as the sum of two independent things, is something you'll see repeatedly throughout physics and engineering (and life in general). It's referred to as "superposition". In this case you'd say that the total moment of inertia is the superposition of the two individual moments of inertia (that of the disk and that of the ring).

Once you have the total moment of inertia, you can compute the torque by finding the rotational energy of the spinning frisbee (1/2*I*w^2, where w is the angular velocity in radians per second = 2*pi*rpm/60). If that's the rotational energy that the frisbee ended up with, it must be equal to the amount of work done by the thrower. Work (in a rotational sense) can be calculated as torque * angular distance. In this case, the angular distance is pi/2 radians.

Just set work = energy, and solve for the torque.

Does that all make sense?