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I'm trying to solve for the shape of the free surface of an incompressible, perfect fluid in a bucket which is spinning with (uniform) angular velocity $\Omega$. I know the solution is a paraboloid, and I know how to obtain it assuming the velocity profile $\mathbf{v}=v_{\phi}(r)\hat{\phi}$ where $v_{\phi}(r)=\Omega r$.

I'm wondering how to derive this velocity profile. It should just come out of the equations, I would think. I would use symmetries to determine that $\mathbf{v}=v_{\phi}(r)\hat{\phi}$ and $P=P(r,z)$. Then there's the continuity equation and Euler's equation for a rotating, steady-state fluid

$$ \nabla \cdot \mathbf{v}=0$$ $$ \mathbf{\Omega}\times\mathbf{v}+(\mathbf{v}\cdot \nabla)\mathbf{v}=-\frac{1}{\rho}\nabla P + \mathbf{g}.$$

I find that the continuity equation and the $\phi$ component of the Euler equation give no extra information. The $r$ component of the Euler equation gives $$ \Omega v_{\phi}(r) = \frac{1}{\rho}\frac{\partial P(r,z)}{\partial r} $$ and the $z$ component gives $$ \frac{\partial P(r,z)}{\partial z} = -\rho g. $$

Clearly, given $v_{\phi}(r)$ I can solve these equations. But, it's also clear I can't find $v_{\phi}(r)$ from these equations. Is there a way to obtain the fluid velocity profile?

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  • $\begingroup$ possible duplicate: fluid in a rotating cylinder $\endgroup$ – Cleonis Nov 15 '20 at 7:46
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    $\begingroup$ It's not a duplicate. The other question just postulates that the velocity field is as they say. I want to know if/how that can be derived from the Euler equation and continuity equation. $\endgroup$ – Physics_Plasma Nov 15 '20 at 8:06
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    $\begingroup$ If it is a perfect fluid, (no shear stresses), the spinning movement of the bucket is not transmitted to the fluid. It is a theoretical situation where its surface remains unchanged. $\endgroup$ – Claudio Saspinski Nov 15 '20 at 23:02
  • $\begingroup$ @ClaudioSaspinski I believe that even for perfect fluids, "no-slip" boundary conditions are usually assumed, such that the fluid has the same velocity as the solid at the boundary. $\endgroup$ – Physics_Plasma Nov 15 '20 at 23:06
  • $\begingroup$ See also the discussion under Bernhard's answer, where we discuss this further. You are right, Claudio. $\endgroup$ – Physics_Plasma Nov 16 '20 at 19:01
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I don't think the Euler equations will be of to much use here. In case of doubt, I always return to the Navier-Stokes equations. In this case, I would directly use the cylindrical formulation.

The $z$ dimension does not play a role, so I can say $v_z=0$ and $\frac{\partial}{\partial z}=0$. Furthermore, from rotational symmetry is also know there is no $\phi$-dependency in the problem. Because of the above assumptions we can also tell that $v_r=0$. Another solution would violate the continuity equations (this is also your assumption).

Then we can take a look at the azimuthal component of the Navier-Stokes equations. Many of the terms vanish (left as an exercise to the reader), and we end up with

$$\frac{1}{r}\frac{d}{dr}\left(r\frac{du_\phi}{dr}\right)-\frac{u_\phi}{r^2}=0$$

You can use your differential equation courses to solve this equation for $u_\phi$. I skipped this step and verified that $u_\phi=cr$ solves this equation and is hence the solution. Obviously $c$ follows from the boundary condition.

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  • $\begingroup$ Can you explain where this equation comes from? The Euler equation in fluid dynamics is just NS but with zero viscosity. I'm not sure how you get a term with a second order derivative from the Euler equation. $\endgroup$ – Physics_Plasma Nov 16 '20 at 3:40
  • $\begingroup$ The assumption of zero viscosity is wrong. Viscosity and hence friction is what makes your fluid rotate in a laminar way. The equation I wrote is just the viscous term. In your answer it looks like you solve the Euler equation, but you don't really do that. Where does the assumption of $v=0$ from? $\endgroup$ – Bernhard Nov 16 '20 at 16:57
  • $\begingroup$ Oh that's interesting, I see what you did. Yes, in my answer I didn't really need the Euler equation to get the fluid velocity, that's right. It does give the pressure $P=P(r,z)$, however. I just went into the rotating frame of the bucket, where the fluid velocity would be zero (because the fluid is rotating along with the bucket - that's the type of solutions we're looking for). Then I transformed that velocity vector field back to the inertial lab frame and find the same answer as you. $\endgroup$ – Physics_Plasma Nov 16 '20 at 18:51
  • $\begingroup$ And I guess your point is that an ideal fluid wouldn't start rotating along with the bucket, right? Which is why you would want to use the Navier-Stokes equation. $\endgroup$ – Physics_Plasma Nov 16 '20 at 18:53
  • $\begingroup$ @Physics_Plasma Exactly. And if you use solid body motion anyhow (like in the other answer), you also don't really need to do a coordinate transformation. $\endgroup$ – Bernhard Nov 16 '20 at 18:55
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I'm wondering how to derive this velocity profile.

Implicit in the problem is that the state of the fluid is one of solid body rotation.

It appears that the problem was presented to you in a form such that the state of solid body rotation wasn't mentioned.

If a fluid is not in solid body rotation then there is friction, and the fluid is dissipating energy. The state of solid body rotation is the state that is reached when there is no more opportunity to dissipate energy. When all parts of the fluid have zero velocity relative to each other there is no more opportunity for friction, hence no more opportunity for energy dissipation.

Solid body rotation is a form of static equilibrium in the sense that the state of solid body rotation is a state with zero conversion of energy. The state of solid body rotation is so simple that it cannot be derived; it is rockbottom simple.

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  • $\begingroup$ Thanks, I never thought of it that way. I posted an answer to the problem, where I found the lab-frame velocity by transforming from the rotating frame to the lab frame. I'm not sure if I had to make the assumption of rigid-body motion of the fluid at all. $\endgroup$ – Physics_Plasma Nov 16 '20 at 3:42
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I've figured out the answer to my own question. It's easiest to go into the noninertial frame of the rotating bucket. In that case, the fluid is static, $\mathbf{v}=0$. The Euler equation in the rotating frame $$ \frac{\partial \mathbf{v}}{\partial t} + (\mathbf{v}\cdot\nabla)\mathbf{v}+2\Omega\times\mathbf{v}+\mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{r})=-\frac{1}{\rho}\nabla P + \mathbf{g} $$ becomes $$ \mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{r})=-\frac{1}{\rho}\nabla P + \mathbf{g} $$.

Because of rotational symmetry, the pressure is $P=P(r,z)$. From here, Euler's equation ($r$ and $z$ components) give you the pressure and allow you to determine the paraboloidal shape of the fluid surface.

Then, to get the fluid velocity in the lab frame, you can use the relation $$\mathbf{v}_{lab} = \mathbf{v}_{rotating}+\mathbf{\Omega}\times\mathbf{r}$$ where $\mathbf{r}=r\hat{r}+z\hat{z}$, to get that the velocity of the fluid in the lab is $\mathbf{v}_{lab} = \Omega r \hat{\phi}$.

In transforming the fluid velocity in the frame of the rotating bucket to the lab frame, I may have assumed rigid-body motion for the fluid, as discussed by "Cleonis" in his answer - I don't know.

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