Imagine a still pond at the edge of a forest. Make it a beautiful autumn day because that’s a nice thought. The surface is like smooth glass despite the rugged shape of the ground beneath it. This is because the surface of a fluid defines a zero potential height. Essentially the only forces acting on it are gravity and gravity is acting perpendicular to the surface.

But what if you could include another force on the pond? How would the surface change?

One easy method to add a new force to the fluid is to spin it. A fluid parcel in rotating frame will experience centrifugal and Coriolis acceleration.

So, let’s consider a tank rotating at a constant angular velocity \Omega .

Like all good fluid dynamics problems, it starts by examining the system using the Navier-Stokes equation. In this example, we will take advantage of the symmetry around the axis of rotation (z) and use cylindrical coordinates. In this example, r defines the horizontal distance from the axis of rotation and z defines height of the fluid from a zero height defined by fluid before it is rotating.

Because the system has been spinning a long time, we can say that the system is at steady state, so the sum of the forces equals zero.

0 = -\nabla P + \rho \vec{g} + \rho \Omega^2 \vec{r}

Such that

\hat{z}: 0 = -\frac{dP}{dz} - \rho g

\hat{r}: 0 = -\frac{dP}{dr} + \rho \Omega^2 r

Where by separation of variables we get

\hat{z}: \int_{P(z = 0)}^{P(z = h)} dP  = \int_{0}^{h} \rho g dz

P(z) - P(0) = -\rho gh

The pressure at point z and point 0 are both just atmospheric pressure so

0 = -\rho gh

Similarly for the r-coordinate we get

\hat{r}: \int_{P(r = 0)}^{P(r = r)} dP  = \int_{0}^{r} -\rho \Omega^2 dr

P(r) - P(0) = -\frac{\rho \Omega^2 r^2}{2}

where again the pressure at point r and 0 is just atmospheric pressure so

0 = -\frac{\rho \Omega^2 r^2}{2}

So by subtracting the z-component from the r-component (because they are both equal to zero) we get,

0 =  -\rho gh + \frac{\rho \Omega^2 r^2}{2}

which describes a height h which follows a parabolic shape where the height increases further from the axis of rotation.

h = \frac{\Omega^2r^2}{2g}

This height is what we refer to as the parabolic free surface.

In this experiment, the fluid was in a cylindrical tank and spun up using a record player. The green light is used to emphasize the position of the surface so you can clearly see the resultant paraboloid.