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


1 Answer 1


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


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




The general solution of these equations is


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


  • 3
    $\begingroup$ "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. $\endgroup$ Apr 8, 2015 at 1:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.