How Does a Homogeneous Fluid Behave in a Rotating Cylinder?

In summary, the problem is to consider a homogeneous self-gravitating fluid of uniform density contained within a rotating cylinder, and find the condition for the fluid to be in equilibrium with no pressure gradients. Part a shows that the gravitational force per unit mass is directed radially towards the axis and perpendicular to it. Part b involves setting the pressure gradient to zero and finding the condition for the fluid to be in equilibrium, which is achieved by also setting the fluid's velocity to zero. This means that the fluid is not experiencing any acceleration or pressure changes, and is essentially standing still in the rotating frame.
  • #1
xnd996
2
0
Here is the problem: (Question 5-2 in Binney&Tremaine Galactic Dynamics)
Consider a homogeneous self-gravitating fluid of uniform density [tex]\rho_0[/tex] contained within a rotating cylinder of radius [tex]R_0[/tex]. The cylinder and the fluid rotate at angular speed [tex]\Omega[/tex] about the axis of the cylinder, which we take to be the [tex]z[/tex]-axis, so [tex]{\bf\Omega} = \Omega\hat{\bf z}[/tex].

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

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

directed radially toward the axis and perpendicular to it.

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

[tex]\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}[/tex]

Find the condition on [tex]\Omega[/tex] 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 [tex] \nabla P [/tex] which leaves you with
[tex]\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}[/tex]

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

My question:
What part of this question implies that [tex]v = 0 [/tex]? 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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  • #2
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.
 
  • #3
May be the fluid in the rotating frame is standing still.
v=0 in this frame.
 

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What is the relationship between the Coriolis effect and the conservation of angular momentum?

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