Two Cylinders on Ramp Suppose I have two cylinders: a light one and a heavy one. Now, I let the cylinders roll down a ramp without slipping. My question is, which one will get to the bottom of the ramp first, and why?
 A: Let's take a look at the net force for a cylinder on an inclined plane:
$$ \Sigma F_{\parallel} = mg\sin{\theta} - f\tag{1}$$ where $f$ is force of friction.
Now the torque about the COM (which is the point about which there is rotation) is:
$$\Sigma \tau = Rf \tag{2}$$
where $R$ is the radius of the cylinder. By Newton's second law, Eq (1) and (2) become:
$$ ma = mg\sin{\theta} - f\tag{3}$$
$$I\alpha = Rf \tag{4}$$
Since there is no slipping $a = R \alpha$. We get,
$$I \dfrac{a}{R} = Rf \tag{5}$$
Now here is the important part. Assume the DENSITY is UNIFORM in both cylinders. That does not imply the same mass, but rather that $\rho$ is the same at every point on the cylinder. In that case, the inertia (about the axis going through the COM and each face of the cylinder) is $$I=\dfrac{1}{2}mR^2$$ where $R$ is the radius and $m$ is the mass.
Let's substitute that in (5) and get,
$$\dfrac{1}{2}mR^2 \dfrac{a}{R} = Rf \quad \implies \quad \dfrac{1}{2}ma  = f \tag{6}$$
Now let us combine (6) and (3) to get
$$ ma = mg\sin{\theta} - \dfrac{1}{2}ma.\tag{7}$$
Observe that the masses all cancel, and we are left with
$$a = \dfrac{2}{3} g\sin\theta.\tag{8}$$
Observe that (8) neither depends on mass nor radius. Therefore, both cylinders will experience the same acceleration. Since the acceleration for each cylinder is the same (and they both start from the same spot from rest), both will arrive at the same time, independent of mass or radius (again, assuming uniform density).
A: There's an intuitive proof that two cylinders differing in only density will roll with the same speed: just superimpose two cylinders with the same mass. The cylinders will roll at the same rate. But together they make a cylinder of twice the mass, that is also rolling with the same rate. As a practical matter, giving a meaning to "superimpose" can be a bit complicated, but that doesn't affect the proof from an intuitive standpoint.
A: Assuming that the cylenders are identical in appearence and just made of material of different densities. Then the torque about the COM will be unequal but the acceleration would be equal.
Instead ofequation just percieve it as
$$\tau(torque) \propto Mass$$ as gravitational torque and other parameters are equal
$$ \tau(torque)=I\alpha$$
$$I \propto Mass$$
So you get $\alpha$ independant of mass.
Now as these are equal for both cylenders.
So any kinematical or rotational calculation you would do should be equal for both.
If the cylenders are not identical then please elaborate your question.
