Fluid in a rotating cylinder

I have been wondering why a fluid in a rotating container has a parabola shape? Is it possible to prove this mathematically?

You will have two forces that act on an elementary mass element $dm$ on the surface. The force in the $x$-direction will be $dF_{x}=\omega^{2}xdm$ and in the $y$-direction $dF_{y}=gdm$. Also, we know that the slope of a curve is $\tan{\alpha}=dy/dx$. However, the tangent is equal also to $\tan{\alpha}=dF_{x}/dF_{y}$. So from this you have that

$$\frac{dy}{dx}=\frac{\omega^{2}x}{g}$$

After integration you get $$y=\frac{\omega^{2}}{2g}x^{2}$$

Which is just the equation for a parabola.

This is a two-dimensional derivation based on the stagnant interface. A more general solution would be as follows. Consider the axis $Oz$ along the cylinders axis. In this case, the velocity components will be $v_{x}=-\omega y$, $v_{y}=\omega x$ ,$v_{z}=0$. Taking Euler's equation

$$\frac{\partial\vec{v}}{\partial t}+(\vec{v}\cdot\nabla)\vec{v}=-\frac{1}{\rho}\mathrm{grad}p$$

Considering that $\partial\vec{v}/\partial t=0$, the projections on the three axis on Euler's equation are

$$x\omega^{2}=\frac{1}{\rho}\frac{dp}{dx}$$

$$y\omega^{2}=\frac{1}{\rho}\frac{dp}{dy}$$

$$\frac{1}{\rho}\frac{dp}{dz}+g=0$$

The general solution of these equations is

$$\frac{p}{\rho}=\frac{1}{2}\omega^{2}(x^{2}+y^{2})-gz+C$$

On the free surface, where the pressure is constant, the surface will have the shape of a paraboloid.

$$z=\frac{\omega^2}{2g}(x^{2}+y^{2})$$

• "However, the tangent is equal also to $\tan{\alpha}=dF_{x}/dF_{y}$" - it might be worth pointing out that this is the condition for the gravitation + centrifugal force to be normal to the surface, which is the only direction that the water other than the element can oppose (leaving aside surface tension). It took me a little while to work this out. Apr 8, 2015 at 1:16