Describe surfaces of equal pressure in a rotating fluid

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In a rotating fluid within a closed vessel, the surfaces of equal pressure form circular cylinders with their axis positioned at a height of g/Ω² above the rotation axis. The problem can be approached using the conservation of momentum and analyzing the forces acting on the fluid, including centrifugal effects. By simplifying the momentum equation and focusing on the pressure gradient, one can derive the relationship between pressure and radial distance from the axis of rotation. The discussion highlights the importance of understanding inviscid flow and the implications of neglecting the Coriolis effect. This theoretical framework provides insight into fluid dynamics in rotating systems.
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Hi, I am trying to solve a basic question from a Fluid dynamics textbook. Could you help me with the answer? The question is as follows:

A closed vessel full of water is rotating with constant angular velocity \Omega about a horizontal axis. Show that the surfaces of equal pressure are circular cylinders whose common axis is at a height g/\Omega^{2} above the axis of rotation.

I don't know how to tackle this problem. Is there a technique in solving such theoretical questions?

Thanks,

P.S:- This is not a homework or coursework question. I am also new to Physics forum, and hence, my question may not be appropriate for this section. In that case please tell me in which section I should pose fluid dynamics questions.
 
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Interesting question... I only have a partial answer, working from Tritton's 'Physical Fluid Dynamics'. In it, he starts with:

\frac{Du}{Dt} =\frac{1}{\rho}\nabla p -\Omega \times \Omega \times r - 2\Omega \times u + \nu \nabla^{2} u +\rho g

So, assuming conservation of momentum, Du/Dt = 0. Also, the second term on the rhs can be written as
-\nabla (\frac{1}{2}\Omega^{2}r^{2})

and combined to give a reduced pressure

p - \frac{1}{2}\Omega^{2}r^{2}

Then, ignoring the Coriolus term and assuming inviscid flow, I can maybe see how you get the result you mention. Maybe...

hope this helps.
 
Thanks!
 
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