Asked by: Lei Bao

Assume the 2 cans have the same mass (actually, the mass doesn't matter, it will cancel out of the equations anyway, but the argument will be more clear this way). As the cans sit at the top of the ramp, they are at the same height (h) with the same mass (m), and so have the same gravitational potential energy: PE =mgh. As they move down the ramp, this energy will convert into 2 things: translational kinetic energy heading down the ramp with velocity=v and KEt =(1/2) mv^{2}, and rotational kinetic energy as the material of the cans spin around their central axes KEr = (1/2)I^{2} where I is the moment of inertia of the cans and omega is the angular velocity, or rate of spin, of the cans. For the empty can, 100% of the mass of the can spins as the can rolls, but for the full can, much of the liquid in the can is effectively sliding down the ramp without spinning. Thus 100% of the mass of the sliding liquid goes into translational kinetic energy, and the full can will have a greater speed.

Now for in reality there will be some internal friction between the walls of the can and the liquid, and a viscous liquid will have considerable internal friction as well, but these should not suck away enough of the kinetic energy to slow the full can down to the speed of the empty can.

Answered by:
Rob Landolfi, Science Teacher, Washington, DC

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