---
title: "Newton's bucket"
firstLetter: "N"
publishDate: 2021-05-13
categories:
- Physics
- Fluid mechanics
- Fluid statics

date: 2021-05-13T17:06:45+02:00
draft: false
markup: pandoc
---

# Newton's bucket

**Newton's bucket** is a cylindrical bucket
that rotates at angular velocity $\omega$.
Due to [viscosity](/know/concept/viscosity/),
any liquid in the bucket is affected by the rotation,
eventually achieving the exact same $\omega$.

However, once in equilibrium, the liquid's surface is not flat,
but curved upwards from the center.
This is due to the centrifugal force $\va{F}_\mathrm{f} = m \va{f}$ on a molecule with mass $m$:

$$\begin{aligned}
    \va{f}
    = \omega^2 \va{r}
\end{aligned}$$

Where $\va{r}$ is the molecule's position relative to the axis of rotation.
This (fictitious) force can be written as the gradient
of a potential $\Phi_\mathrm{f}$, such that $\va{f} = - \nabla \Phi_\mathrm{f}$:

$$\begin{aligned}
    \Phi_\mathrm{f}
    = - \frac{\omega^2}{2} r^2
    = - \frac{\omega^2}{2} (x^2 + y^2)
\end{aligned}$$

In addition, each molecule feels a gravitational force $\va{F}_\mathrm{g} = m \va{g}$,
where $\va{g} = - \nabla \Phi_\mathrm{g}$:

$$\begin{aligned}
    \Phi_\mathrm{g}
    = \mathrm{g} z
\end{aligned}$$

Overall, the molecule therefore feels an "effective" force
with a potential $\Phi$ given by:

$$\begin{aligned}
    \Phi
    = \Phi_\mathrm{g} + \Phi_\mathrm{f}
    = \mathrm{g} z - \frac{\omega^2}{2} (x^2 + y^2)
\end{aligned}$$

At equilibrium, the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) $p$
in the liquid is the one that satisfies:

$$\begin{aligned}
    \frac{\nabla p}{\rho}
    = - \nabla \Phi
\end{aligned}$$

Removing the gradients gives integration constants $p_0$ and $\Phi_0$,
so the equilibrium equation is:

$$\begin{aligned}
    p - p_0
    = - \rho (\Phi - \Phi_0)
\end{aligned}$$

We isolate this for $p$ and rewrite $\Phi_0 = \mathrm{g} z_0$,
where $z_0$ is the liquid height at the center:

$$\begin{aligned}
    p
    = p_0 - \rho \mathrm{g} (z - z_0) + \frac{\omega^2}{2} \rho (x^2 + y^2)
\end{aligned}$$

At the surface, we demand that $p = p_0$, where $p_0$ is the air pressure.
The $z$-coordinate at which this is satisfied is as follows,
telling us that the surface is parabolic:

$$\begin{aligned}
    z
    = z_0 + \frac{\omega^2}{2 \mathrm{g}} (x^2 + y^2)
\end{aligned}$$



## References
1.  B. Lautrup,
    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
    CRC Press.