1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
|
---
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.
|
