Rolling Spheres and Cylinders Consider the following three diagrams. The first one shows the velocity vectors for an object experiencing only pure translation. The center diagram is for an object that is experiencing only rotation. The final diagram is a combination of the two  both translation and rotation.
Notice how the vectors add together. At the bottom of the object that is both rotating and translating, the contact point is instantaneously at rest. This is why static friction is used to calculate the torque that produces rotational motion. Also notice that the very top point on the wheel is moving with speed 2v_{CM}  faster than any other point on the wheel.
Rotational Dynamics Now consider an object rolling down an incline plane. The first thing that we are going to do is draw a freebody diagram of the forces acting on this mass and then resolve those forces into their components which act parallel and perpendicular to the plane.
Let's consider the axis of rotation passing through the disk's center of mass, cm. Notice in this case that only the instantaneous static friction force will supply a torque since the lines of action of the other two forces (normal and weight) act through the center of mass and cannot produce a torque. As long as the mass rolls without slipping, we can use the relationships: v = rω and a = rα.
Rotationally,
net τ = I_{CM}α
τ = f_{s}r
f_{s} = I_{CM}(α/r)
Remember linearly,
net F = ma mg sinθ  f_{s} = mrα where a = rα
Simultaneously eliminating f_{s} and solving for α yields:
mg sinθ  I_{CM}(α/r) = mrα α = g sinθ /(r + I_{CM}/mr)
The moment of inertia for a disk, or solid cylinder (see chart provided below), equals ½mr^{2}. Substituting in this value and simplifying gives us
α = 2g sinθ/(3r)
Since this angular acceleration is uniform, you would be free to use any of the rotational kinematics equations to solve for final angular velocity, time to travel down the incline, or the number of rotations it completes as it rolls along the incline's surface.
