How Does a Homogeneous Fluid Behave in a Rotating Cylinder?

xnd996
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Here is the problem: (Question 5-2 in Binney&Tremaine Galactic Dynamics)
Consider a homogeneous self-gravitating fluid of uniform density \rho_0 contained within a rotating cylinder of radius R_0. The cylinder and the fluid rotate at angular speed \Omega about the axis of the cylinder, which we take to be the z-axis, so {\bf\Omega} = \Omega\hat{\bf z}.

(a) Show that the gravitational force per unit mass at distance r from the axis is

\begin{displaymath} -\nabla\phi = -2\pi G\rho_0 (x\hat{\bf x} + y\hat{\bf y}), \end{displaymath}

directed radially toward the axis and perpendicular to it.

(b) Euler's equation for the fluid in the rotating frame is

\begin{displaymath} \frac{\partial{\bf v}}{\partial t} + ({\bf v}\cdot\nabla) \bf v = - \frac{\nabla P} {\rho}- \nabla \phi - 2 \Omega \times{\bf v} + \Omega^2 (x\hat{\bf x} + y\hat{\bf y})\,. \end{displaymath}

Find the condition on \Omega such that the fluid is in equilibrium with no pressure gradients (i.e. no Jeans swindle).


My problem:
So I understand part a but I am having difficulty with part b.

There are no pressure gradients so you set \nabla P which leaves you with
\begin{displaymath} \frac{\partial{\bf v}}{\partial t} + ({\bf v}\cdot\nabla) \bf v = - \nabla \phi - 2 \Omega \times{\bf v} + \Omega^2 (x\hat{\bf x} + y\hat{\bf y})\,. \end{displaymath}

I know you are also supposed to set v = 0 (I know this because I have an answer key that says set v = 0) but why?

My question:
What part of this question implies that v = 0? Is it zero because it is in equilibrium? What does it mean for the fluid to be in equilibrium? I would like a little help with this concept please.
 
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It seems to me that a rotating fluid creates a force and thus pressure. I imagine that to keep pressure either constant or zero, the velocity should be zero as well. I could be wrong though.
 
May be the fluid in the rotating frame is standing still.
v=0 in this frame.
 
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