I need to show that the surface of water in a bucket rotating with constant angular velocity will have parabolic shape. I'm quite confused by this problem, but here's what I did:

$$\vec{F}_{cf} + \vec{F}_{grav} = -\vec{\nabla} U = m(\vec{\Omega}\times\vec{r})-mg\hat{z}$$ where $F_{cf}$ is the centrifugal force, $F_{grav}$ is the force of gravity, $U$ is potential energy, $\vec{r}:=(x,y,z)$. So $\nabla U(z) = mg-m\Omega^2 z$, hence $$U(z) = gmz-\frac{1}{2}\Omega^2 z^2+C$$ which is a parabola.

In this approach I was trying to use the fact that the surface is equipotential for $F_{cf}+F_{grav}$. But apparently my approach doesn't quite hold any water. Please give some suggestion on approaching this problem, as I have no other idea.

  • 9
    $\begingroup$ Fun fact: The fact that a rotating bucket of liquid takes the shape of a parabola has very real applications Liquid Mirror. $\endgroup$
    – PGnome
    Nov 16, 2016 at 22:58
  • $\begingroup$ I once worked with a technician that previously worked for a company that fabricated parabolic reflectors in this manner using epoxy as the fluid. The hardened surface was afterwards removed from the spinning fixture, polished and coated with aluminum. $\endgroup$
    – docscience
    Nov 16, 2016 at 23:09
  • $\begingroup$ I think my approach wasn't bad at all. I had simply forgotten that it is $\vec{\nabla} U$ which is orthogonal to the surface, but not $U$ itself. $\endgroup$
    – sequence
    Nov 16, 2016 at 23:10
  • $\begingroup$ related question here $\endgroup$
    – Graviton
    Nov 17, 2016 at 5:26
  • $\begingroup$ BTW, there is typo in your question. $-\frac{\Omega^2 r^2}{2}$ instead of the same with $z$. $\endgroup$
    Jan 10, 2017 at 21:33

4 Answers 4


Consider the following cylindrical container with liquid to be rotating at uniform $\omega$:

Rotating liquid surface

Consider an infinitesimal liquid element $\mathrm{d}m$ at height $h$ above the minimum of the parabola. The forces acting on it are:

1) gravity: $$g\mathrm{d}m$$ 2) the centripetal force: $$\mathrm{d}F_c=\omega^2r\mathrm{d}m$$ Consider the angle $\alpha$: $$\tan\alpha=\frac{\mathrm{d}h}{\mathrm{d}r}=\frac{F_c}{g\mathrm{d}m}=\frac{\omega^2r\mathrm{d}m}{g\mathrm{d}m}=\frac{\omega^2r}{g}$$ This means that: $$\omega^2r\mathrm{d}r-g\mathrm{d}h=0$$ Integrate this differential equation: $$\int_0^r\omega^2r\mathrm{d}r=\int_0^hg\mathrm{d}h$$ $$\frac{\omega^2r^2}{2}=gh$$ $$\implies h=\frac{\omega^2r^2}{2g}$$

This is a quadratic parabola.

  • 1
    $\begingroup$ The crucial thing used here is that $dR$ is normal to the surface of water. $\endgroup$
    Sep 21, 2021 at 14:39

Your potential energy function $U(z)$ doesn't show at all that the water surface is parabolic. What you need to find is the functional form of the rotating water surface, i.e. the surface height $z$ as a function of $r$ in cylindrical coordinates $r$ and $z$. (Because of rotational symmetry $\phi$ is not necessary.) The centrifugal force is $$F_{cf}=m\omega^2 r$$ and the gravitational force is $$F_{grav}=-mg$$ The water surface is orthogonal to the direction of the resultant force $$\vec F=\vec F_{cf}+\vec F_{grav}$$ Thus the slope of the water surface is $$\frac {dz(r)}{dr}=\frac{|F_{cf}|}{|F_{grav}|}=\frac {m\omega^2 r}{mg}$$ From this we get by integration $$z-z_0=\frac{1}{2g}\omega^2 r^2$$ Thus we get indeed a parabolic surface in the rotating water in the bucket.

  • $\begingroup$ Why can't Bernoulli's principle be used? External pressure will cancel out. So pressure difference from centre to wall of container will be 2g∆H, where ∆H is the change in height of the surface from centre to wall. This will be equal to change in velocity outwards, given by W^2 times R^2, where w is angular velocity and R is the Radius or x coordinate. This gives ∆h = (w^2/2g)r^2 $\endgroup$
    – AP2261
    Nov 20, 2018 at 6:32

This problem has different solutions. One of them is to use the principal of stationary action. It is convenient to use cylindrical coordinates: $$\vec{R} = \vec{R}(r, \phi, z).$$ Let us consider a water volume rotating like a whole body i.e. each small volume of water (blob) has the same angular velocity $\Omega$. Next step is to reduce this problem to a static problem. In order to attain this we should take a look at the water from the reference frame which also rotates with angular velocity $\Omega$ (around the same axis as the water volume). From that point of view the water is at rest. Then there are two forces acting on a water blob of mass $m$: gravitational $mg$ and centrifugal $m\Omega^2r$. What are the potentials of these forces? So, they are $$V_{gr} = mgz,\quad V_{cf} = -\frac{m\Omega^2r^2}{2}$$ It's up to you to check that the above potentials cause the right forces. Therefore the water possesses only potential energy. Now we can write the action: $$S = -\int_{t_1}^{t_2}dt\int_{vol}d^3r\Bigl\{\rho gz-\frac{\rho\Omega^2r^2}{2}\Bigr\} = \mathrm{const}\cdot \int_{vol}\mathcal{L}d^3r,$$ where $\mathcal{L}$ is Lagrangian or Lagrangian density. But, of course, we must take into account that entire mass of the water $M_w$ is conserved or $$\int_{vol}\rho d^3 r = M_{w} = \mathrm{const}.$$ In cylindrical coordinates the above integrals are ($\rho=1$) $$2\pi\cdot\int_{0}^{R}rdr\int_{0}^{z(r)}\Big\{...\Big\}dz = \pi\cdot\int_{0}^{R}\Big\{gz^2r-\Omega^2zr^3\Big\} dr$$ and $$2\pi\cdot\int_{0}^{R}rdr\int_{0}^{z(r)}dz=2\pi\int_{0}^{R}rzdr$$ So we have a variation problem with additional condition. It is easy to solve this using Lagrange multipliers' method. Necessary condition for $S$ to have extrema with additional condition (about $M_w$) is the existing of a such multiplier $\lambda$ that $$\frac{d}{dr}\frac{\partial G}{\partial \frac{\partial z}{\partial r}} - \frac{\partial G}{\partial z} = 0,$$ where $$G = gz^2r-\Omega^2zr^3 + \lambda rz.$$ Which gives $$2gzr - \Omega^2r^3 + \lambda r = 0$$ or $$z(r) = \frac{\Omega^2r^2}{2g} - \lambda.$$ Voila.


I'm pretty late to the question but I think I know a simpler way. You can solve this really easily with conservation of energy, the kinetic energy of a circle of water at a constant radius is $\frac{m r^2 \Omega^2}{2}$, in a idealised system the energy has nowhere else to go but increasing the waters gravitational potential energy, so $mgz = \frac{m r^2 \Omega^2}{2}$ or $z = \frac{r^2 \Omega^2}{2g}$.

  • $\begingroup$ I don't really understand your equation. You should add some explanation. You can't just write an equation and think it makes sense. Please. $\endgroup$
    Sep 20, 2019 at 11:48
  • 1
    $\begingroup$ @LRDPRDX What don't you understand? It really is very simple. The kinetic energy term is the only part that's not explicitly spelled out but it's just substituting $v=r\Omega$ into the bog standard kinetic energy equation. You use the same equation found in a more roundabout way in your own answer. It's very odd if you can use Lagrangian mechanics yet have trouble with conservation of energy. You can't just say you don't understand something then think I'll know which bit you're having trouble with. Please. $\endgroup$ Sep 22, 2019 at 10:00
  • $\begingroup$ For example what is "a circle of water at constant radius"? $\endgroup$
    Sep 23, 2019 at 6:14

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