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+---
+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.