How Does a Homogeneous Fluid Behave in a Rotating Cylinder?

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A homogeneous self-gravitating fluid in a rotating cylinder exhibits specific behaviors under gravitational and rotational forces. The gravitational force per unit mass is directed radially inward, as derived in part (a). In part (b), the discussion centers on the condition for equilibrium without pressure gradients, leading to the conclusion that the fluid's velocity must be set to zero. This is because, in a rotating frame at equilibrium, the fluid does not experience any net motion, implying that the forces balance out without generating pressure gradients. Understanding this equilibrium concept is crucial for analyzing the fluid's dynamics in a rotating system.
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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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