Rotational velocity of water in spinning bucket

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.