We have a pulley in the shape of a solid disk of mass M = 2.0 kg and radius R = 0.50 m. If we apply a constant force F = 8 N to a string wrapped around the outside of the pulley what is the pulley's angular acceleration? The pulley is mounted on a horizontal frictionless axle.

As usual, start with a free-body diagram.

The free-body diagram has the force F we apply, mg down and a normal force up from the axle. The pulley can only rotate, and the only force associated with a torque about the axle is the force F, so we often just draw that force on the free-body diagram.

Apply Newton's Second Law for Rotation, taking torques about the axcle through the center.

Στ = I α

Apply the equation torque = r F sinθ to get the torque from the force we apply. Using r = R, the pulley radius, and an angle of 90° we get:

R F = I α

What do we use for the moment of inertia? Looking up the expression for the moment of inertia of a solid disk rotating about its center we find that:

I = ˝MR2

This gives:   R F = ˝MR2 α

This gives:   F = ˝MR α
α =
 2 F M R
=
 2*8 2*0.5
= 16.0 rad/s2

Once you know this, you can plug it into the constant angular acceleration equations to find things like the angular displacement and angular velocity after a certain amount of time has gone by.