If L is the length of the ramp, then the bottom point of any symmetric object would be
at height [itex]L\sin \theta[/itex]. Using conservation of energy, the potential energy
at the top should be equal to the sum of translational and rotational kinetic energies
at the bottom. For an object with moment of inertia about its center of mass I,
[tex]mg(L \sin \theta)= \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2[/tex]
for pure rolling motion, [itex]\omega = v/R[/itex] so
[tex]mg(L \sin \theta)= \frac{1}{2}mv^2 + \frac{1}{2}I \frac{v^2}{R^2}[/tex]
solving for v, we have
[tex]v = \left[\frac{2gL \sin \theta}{1+(I/mR^2)}\right]^{1/2}[/tex]
from here, you can derive special cases for different bodies, in your case, a solid
cylinder (or is it hollow) and a hoop.