Q:

If you have 2 plastic balls with the same outer surface but one is filled with water and the other one is empty and you released them, from rest, down the same slope, the ball with the water would accelerate more due to viscous flow right? If that is the case then would a ball that is fairly(but not completely) filled with sand accelerate faster than the empty ball and why? would there be a similar effect?Thanks

- Jack (age 19)

UK

- Jack (age 19)

UK

A:

First, let's think about the ball with water in it. It's easy to solve two cases- zero viscosity and infinite viscosity. Infinite viscosity would just be a solid ball rather than a hollow shell. It's well known that a solid rolling sphere accelerates down a bit faster than a hollow shell, by a factor of 25/21, because for a given size and weight the shell has a higher moment of inertia, and thus more of the energy goes into rotation rather than into downward motion. For zero viscosity, the weight of the internal fluid helps drive things down, but since it doesn't rotate you get faster downward acceleration. If almost all the mass is in the fluid, the downward acceleration is the same as for a mass sliding down a frictionless surface, higher than for a rolling mass.

What about with real water, with some finite viscosity? I haven't been able to solve that problem. Initially, the behavior resembles the zero-viscosity case, so the acceleration is faster than for a hollow shell. The behavior gradually crosses over to another limit on a time scale of roughly radius^{2}_{*}density/viscosity. After long times, the filled ball ends up accelerating just like a solid ball, with a very minor correction for the different densities of the plastic and the water. It's not that the water all rotate together with the shell but rather that the pattern of how much the inside lags behind becomes fixed. So that again is more rapid acceleration than for the hollow ball. I haven't proven that for intermediate times it also accelerates faster than for the hollow ball.

Meanwhile, the question about the ball partly filled with sand is easy, because we have a very similar demonstration- a cylinder partly filled with sand. The ball will accelerate and toward a surprisingly low terminal velocity. The sand, lifted up and then dropped back down by the rotations, keeps dissipating energy, heating up. At the terminal speed that energy dissipation just matches the power input from rolling down in a gravitational field.

The part-filling with sand isn't like full-filling with water, because for rolling on a flat surface the full-filled case leads gradually to uniform rolling and no more dissipation. The part-filled case keeps dissipating inside so that even on a flat surface the ball comes to rest.

For the filled ball rolling downhill, the internal rotation isn't rigid although after a while the acceleration nearly is. There's a lag between the rotation rate of the middle and the outside. That means that internal parts of the fluid continue rubbing against neighboring parts, dissipating energy. The ball will keep heating up. How can that happen even though the downward acceleration ends up looking just like that of a rigid filled sphere? The rotational energy goes as the square of the rotation rate. After a while the average rotation rate grows linearly in time. The lag of the middle becomes fixed. So the missing rotational energy due to that lag is the product of a fixed lag pattern and an average rate that grow linearly in time. The missing rotational energy, compared to a solid sphere, grows linearly in time. That means that the rate of heating becomes constant. That agrees with our assumed fixed pattern of internal rotation differences.

Mike W.

*(published on 02/02/2015)*