Q:

Which diagram best describes the motion of the cylinder?

The diagram (1) shows an inclined plane with a cylinder placed onto it. The cylinder has a mass attached to it represented by the T shape and the centre of gravity is x.
As the cylinder is not in equilibrium, will it restore it's center of gravity and travel upwards prior to down, or simply roll down the plane directly? Upon experimentation, it was unclear to see the preferred route as it varied with using different masses. The smaller the mass, the less effect it generally had on the center of gravity. Could this be explained mathematically?

A:

You are looking for the net torque on the cylinder. The normal situation (with the COM at the center of the cylinder) gives you this diagram for the torque:

As the center of mass moves to the right, it will eventually align with the vertical force of the ramp (note - it is only vertical when there is friction - otherwise you just have the normal force, which is perpendicular to the ramp).
When the center of mass is above the support point there is no torque and the cylinder will not accelerate. If you start with the cylinder at rest with the COM to the right of the support, it will initially "roll uphill", then stop. If you start with the COM to the left, the cylinder will start to roll downhill. While it rolls, the COM may move to the left and this can cause deceleration.
In general, if the cylinder of mass $m$ and radius $R$ has a center of mass that is offset by $r$, and at an angle to the vertical of $\phi$ while on a slope of $\theta$:

the torque on the cylinder can be written as
$$\Gamma = mg\Delta x = mg\left(r\sin\phi-R\sin\theta\right)$$
This tells us that the cylinder is in an oscillatory potential well - and that if the initial $\phi$ is such that
$$r\sin\phi > R\sin\theta$$
then the cylinder will be in a stable equilibrium - it will oscillate about the angle $$\phi' = \sin^{-1}\left(\frac{R}{r}\sin\theta\right)$$

