diff options
Diffstat (limited to 'source/know/concept/newtons-bucket/index.md')
-rw-r--r-- | source/know/concept/newtons-bucket/index.md | 91 |
1 files changed, 91 insertions, 0 deletions
diff --git a/source/know/concept/newtons-bucket/index.md b/source/know/concept/newtons-bucket/index.md new file mode 100644 index 0000000..30fe079 --- /dev/null +++ b/source/know/concept/newtons-bucket/index.md @@ -0,0 +1,91 @@ +--- +title: "Newton's bucket" +date: 2021-05-13 +categories: +- Physics +- Fluid mechanics +- Fluid statics +layout: "concept" +--- + +**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. |