What will be the shape states of water in a centrifuge in 0G

Consider the concept of these videos http://www.youtube.com/watch?v=Zip9ft1PgV0&sns=em and http://www.youtube.com/watch?v=cUhgKFV5Ri4&sns=em but set in a zero gravity environment and in a fully encapsulated and high pressure container. My questions are as follows: What shapes will the water conform to at increasing speeds? Will the water conform to a parabola and then ultimately a hollow cylinder as it does on earth?

Is it possible that without an opening to spit and by continuously adding more water to the centrifuge, that the water could ultimately fold in on itself and conform to various states of torus, by creating vortices within itself?

 Also, taking into account 0G, will the water even conform to its bounding container in a manner in which it receives traction and can be spun?

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take a good look at a decanter: youtube.com/watch?v=FhS5vN4r5LA and understand the concepts used. – mart Jul 2 '13 at 10:58

The parabolic shape is due to gravity. Without it, you'd have a much more boring cylindrical shape.

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But even with gravity, given higher speeds, it takes a cylindrical shape, not so? – Dorian Lyder Sep 15 '12 at 6:48
@DorianLyder In the limit that the centrifugal force greatly exceeds gravity, yes. You make the approximation of imagining gravity doesn't exist. – Benjamin Hodgson Sep 17 '12 at 10:01

When a bucket of water is spun, it experiences:

(i) Centrifugal force
(ii) Gravity

Consider a bucket spinning with a constant angular velocity $\omega$.
We give each point on the surface $(x,y)$ coordinates.

Let us assume a particle in the liquid at with coordinates $(x,y)$ and with a mass $dm$. Such a particle will experience a centrifugal force $(dm)\omega^2x$ and a gravitational force $(dm)g$. Let the resultant force be $F_{res}$.

Liquids can NOT withstand tangential stress. If you've noticed, when you move your finger across a water surface, the water will move along with your hand. This is unlike solids, where tangential stress is tolerable.
Therefore the liquid surface will align itself perpendicular to $F_{res}$ so that there is no component of $F_{res}$ which is tangential.

Now, if you can see, as $x$ increases, the centrifugal force keeps increasing ($\because$ it is proprtional to $x$: $(dm)\omega^2x$) while the force of gravity is the same. So as $x$ increases, i.e. we go outwards, $\theta$ will keep increasing and $F_{res}$ will keep getting more horizontal.
That means that as we go outwards, the surface will keep getting more vertical! (Since the liquid surface will align itself perpendicular to $F_{res}$.)

Thus near the axis of rotation, the surface will be horizontal, and it will keep increasing in slope as it goes outwards, forming a sort of3-D parabolic shape.

(If this doesn't convince you that a parabola will be formed, you can add a comment, I have a proof but I thought it would get to complex an answer)

BUT! It will never be completely vertical because there is always a downward component of force as gravity!

When $\omega$ is VERY HIGH, the effects of gravity are less in comparison to that of centrifugal force. Thus we get a shape which almost resembles a hollow cylinder. It is NOT, though, and is slightly angled, as explained above.

Coming to your 0G situation, there is no downward component of gravity! So we'll simply get a hollow cylinder whose height will keep increasing and thickness will keep decreasing as the angular velocity of rotation increases.

Hope I helped clear it up! :D

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At the end of the day all you are doing is theorizing, none of you know what is correct. It could turn out that gravity acts in the same way as it does on a spherical object.

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So, what does this have to do with the question? – Waffle's Crazy Peanut Jul 2 '13 at 10:55