Question: If I had a Cylinder, a Cone and a Hemisphere all of the same volume, filled with water; Which would maintain a vortex longer?
That is, if you applied the same force/torque to the water, which shape keeps the vortex going longest?
(assume that the actual container is larger than the volume of water to allow for rise in the liquid)

OK, by 'vortex' I assume you mean a whirlpool-like rotation of water around the axis of symmetry.

Let's also assume that this water gets its velocity either by some means imparted to it - e.g. a pipe pushing water in tangential to the axis of rotation at the furthest distance out - i.e. at the surface of the water - or by spinning the container until the entire body of water is rotating the same way and then stopping the container.

I'm also assuming that we're not emptying water out of the container, we're just wanting to see how long it keeps its rotational momentum.

Using the formula at http://www.mathleague.com/help/geometry/3space.htm

The main thing slowing the water down is friction with the surface of the container. Therefore my first idea is to minimise the amount of surface area. A sphere has the minimum surface area for a given volume, so therefore I'd rank them in order cone, cylinder and hemisphere being able to sustain the vortex for longest.

Friction is usually A + BX, where X is the lateral component of the relative velocity of the two objects. In other words, there's a constant drag but the friction will increase the faster the two objects are rubbing against each other. The mechanics are a bit more complex than that, though - you also have to take into account how much pressure is forcing the two surfaces into contact with each other. You can reduce friction by making the two slide by each other rather than grind into each other.

As you know from centripetal acceleration, the tighter the circle the more the moving body wants to go out of the turn and the more force you need to apply to keep turning that circle. You can do a ninety-degree turn in ten
metres or so at 30KPH, but if you tried the same corner at 100 you would probably roll the car. So a very long thin cylinder, where the angle of turn is tight and therefore the outward force is high, would suffer more friction than a big flat disc of the same volume and surface area but where the water is turning a wide circle at the outside.

The total surface area of the cylinder that the water is nominally rotating against is 2 Pi R H + 2 Pi R (because we're assuming that the water doesn't touch the top of the cylinder), whereas the area of the hemisphere is going to be 4 Pi R^2. As you can see in the attached graph, as the radius of the hemisphere rises the hemisphere's surface area goes above (and stays above) the equivalent volume cylinders. The Cyl Rad calculations always stay lower than the other two

The single variable in the graph is the radius of the hemisphere. The two cylinders are then made to be the same volume as the hemisphere. I've used two different sizes of cylinder - one that keeps the radius of the cylinder the same size as the hemisphere's radius (Cyl Rad), and the other which makes the cylinder's height match the hemisphere's radius (Cyl Hei). The other dimension of the cylinder is then calculated backward from the volume in each case.

It is possible to produce formulae where the cylinder's height is the same as its diameter - in other words, as wide as it is tall - and calculate the surface area out from there. This is left as an exercise for the reader. It is not known, however, if this minimises the surface area of the cylinder in relation to its volume.

So most of the time we can construct a cylinder that contains the same amount of water as a given hemisphere but a smaller surface area. This to me is a compelling argument that the cylinder is in fact the best shape to keep the vortex going for longest. The cylinder is also going to have more water moving at the fastest speed possible in the cylinder (thus having the most momentum) compared with a hemisphere of the same volume.

Also consider that there will be less turbulence caused by water being forced to move out of a plane - at the base of the hemisphere the water will slide upward by the surface rather than just being turned around the axis. This will make it push the other water upward - i.e. the water is colliding with itself rather than just spinning around. A cylinder does not suffer this problem as much, since at the bottom a turbulent layer will form across the bottom but this will not force water up the sides of the container in the same way as a hemisphere's 'base' will.

None of this takes into account the vertical and lateral motion of the water. Now, as you know when stirring a cup of tea, the fluid is pushed up higher at the side and lower in the middle. Considered from side on, the water moves up the outside, across the top to the middle, down the middle, and back across the bottom to the outside. The vortex of tea leaves you see in the bottom of the cup is caused by the centre being at lower pressure than the outside, and brownian motion and other particulate factors cause the particles to be deposited in the area of lowest pressure.

>The Cylinder may have problems as it has a corner which may create
>turbulence. The Cone may have a better shape for the vortex but once it
>collpses, there is turbulence.

I'm not sure what you mean by turbulence here. I'm assuming you mean 'friction against the surface'.

I think the only way to really sort out all the physics in one easy go is to test it out.

Thanks for the distracting hour or so,

Paul