A colleague and I were discussing the following fluid mechanics experiment, which most people have probably tried in their kitchen: take a bottle full of water (a 2-litre transparent soda pop bottle works well), invert it, and swirl it around to impart some angular momentum to the fluid. It will quickly form a vortex, which allows the bottle to drain rapidly.
The question that neither of us knows the answer to is this: is there some mechanism by which additional angular momentum is transferred into the system from outside as the water drains, or is it just that the initial angular momentum (which you add by swirling) becomes more concentrated?
Probably equivalently, if I were to cut the bottom off the bottle and then do the same experiment while continually adding water from the tap, would the vortex continue to spin indefinitely, or would the angular momentum in the system eventually be depleted, killing the vortex?*
If angular momentum is transferred to the system from outside the bottle, what is the mechanism for this? I assume that the Coriolis force is not relevant on such a small scale (the fact that a bottle vortex can be formed with either chirality would seem to back this up) so I would imagine that any transfer of angular momentum would be due to the fluid exerting a torque on the bottle it's in. Obviously frictional forces can only exert a torque in the direction that would slow the rotation down, but perhaps there are pressure gradient effects that can exert one in the opposite direction.
* this is obviously a rather easy experiment, and my only excuse for not doing it is that I don't have a pair of scissors handy.
Best Answer
I think the answer to the question is that there is no mechanism by which angular momentum is transferred to the water from outside, after you stop the initial swirling. Rather, the vortex takes angular momentum from the draining water and transfers it to that remaining in the bottle. I give a fuller explanation below - it talks about water draining from a bath but the principles are the same.
Three basic physical principles apply:
First, angular momentum is conserved: at any time during the draining of the bath the total angular momentum of the water (inside and outside the bath) stays the same.
Second, the only energy input to the system is that available from gravity as the water drains.
Third, the water is viscous, so that any rotating volume of water within the bath will tend to transfer its angular momentum to the body of water as a whole.
The presence of the vortex tube with a water/air surface shows that the force experienced by the water at that interface points radially outward and must be balanced by the normal pressure of the water in the bath. As water exits the bottom of the vortex, pressure from the body of bath water outside pushes more water inward towards the vortex centre.
Conservation of angular momentum means that as this water moves in, its angular velocity and consequent rotational energy increase. At the same time viscosity acts to decrease the shear in water velocities, transferring angular momentum from the fast rotating water to the main body of water away from the vortex. The two effects combine to build rotational energy near the vortex but transfer angular momentum away from it.
What appears to the watcher to be the adding of angular momentum to the water is, if this model is right, rather its transfer from one part of the water volume to another. The liquid that falls out of the bottom end of the vortex has high angular energy but low angular momentum. Most of its original angular momentum remains in the liquid still in the vessel; as the amount of liquid in the vessel decreases so its angular speed increases.
The increase in angular energy is fuelled by the decrease in gravitational potential energy as the water falls down the vortex.
Some numbers. If the air tube at the centre of the vortex in a draining sink has a radius of 3mm, and the gradient of the vortex tube surface is 9, then the angular velocity of the water must be:
$\omega^2 \cdot 3\times 10^{-3} = 9g$
that is, $\omega=171$ rad/s; quite rapid. Conversely, a speck of water rotating with the same angular momentum at a distance of 20cm from the vortex centre would have $\omega=3.9\times 10^{-2}$ rad/s (2.2 deg/s); almost too slow to notice. Though the two specks of water have the same angular momentum, their rotational energies differ by a factor of 4325.