If you were to roll a full pop can, and an empty pop can down the same ramp, which would reach the bottom first, or would it be a tie? If you assume the liquid is perfectly fluid (meaning no viscosity), then it shouldn't effect the rotational inertia. The change in potential energy for the full can increases, but the mass increases, but the moment of inertia increases by the same factor. mgh, and xmr^2. On one hand, they are both rings, and while the mass of the liquid, increase the gravitational pull, it has an increased inertia. Everything says to me that it should be a tie, but that just doesn't seem right.
This is an interesting problem. First, for background, remember that viscosity determines the strength of mechanical coupling between the rotating cylinder and the mass of fluid. A low viscosity substance, like a gas, is affected very little by a moving boundary -- for example if you pull a piece of paper edge-on through the air it drags with it only a very thin layer of air at the large faces of the sheet of paper. Doing the same paper-pulling experiment in water would require doing much more work as you would also be moving a significant mass of water, and pulling the piece of paper edge-on through a bucket of cold pancake syrup (very viscous) would drag with it a very thick layer of syrup. In teh last case, you can see the thickness of the layer of syrup being affected by simply pulling a vertical piece of paper stright up out of a bowl of syrup. The paper will be coated with a thick layer of syrup.
Let m be the mass of the can, let R be the radius of the can(s), let M be the combined mass of a can and the fluid, and let theta be the angle that the incline makes to with the horizontal.
So, let's consider the two natural limits of the problem. In particular, all I'm going to do is calculate the angular accelerations of each can.
First, let's think about the case where the fluid has a very small viscosity so that it doesn't participate in rotational motion, and therefore the fluid does not contribute to the moment of inertia. In this case, the moment of inertia is mR^2 for both the full and the empty can. Then you should be able to show that:
So, in the limit of a fluid with very small viscosity, meaning a fluid that flows very very easily, the full can will have a higher angular acceleration by a factor of M/m, and will reach the bottom of the ramp first.
Second, let's consider the case where the fluid is very viscous,
for example cold pancake syrup. Then the fluid rotates in unison with the
can, and the moments of inertia will be close to the result for
a solid cylinder, I = (1/2)*MR^2. You should be able to
show that:
Putting this all together, the final answer to your question will depend on the viscosity of the fluid and the size of the can. For soda pop, the thickness of the layer of soda which is 'dragged' into motion by the rotating walls of the container is probably very small compared to the radius of the can, because water has fairly low viscosity. Consequently, I would guess that the filled soda pop can gets to the bottom of the ramp first. Of course, you should actually do the experiment! By the way, this business of the thickness of the 'layer' of fluid that is affected by the moving walls has a technical name, the viscous penetration length, and it plays a key role in understanding aerodynamics and many other real-life problems involving gas and fluid flow.