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 Incoming links: Parabolic free surface of a rigidly rotating liquid

Past exam of the physics B course of the University of Cambridge / 2017 / Paper 1 / Section A / A2 / Solution 2026-10-18

In steady rigid rotation of the liquid about its vertical axis, centrifugal acceleration is

Ω^{2} r

outward. Hydrostatic pressure balance in the rotating frame gives

\frac{\partial p}{\partial r} = ρ Ω^{2} r, \frac{\partial p}{\partial z} = - ρ g .

The free surface is at constant atmospheric pressure, so its slope obeys

0 = d p = ρ Ω^{2} r d r - ρ g d z, z - z_{0} = \frac{Ω ^{2} r ^{2}}{2 g} .

Compare this parabolic free surface of a rigidly rotating liquid with

z - z_{0} = r^{2} / (4 f)

. For focal length

f = 1 m

,

Ω = \frac{g}{2 f} ≃ 2.21 rad s^{- 1} ≃ 21.1 rev mi n^{- 1} .

Either sense of rotation gives the same shape. The result assumes the liquid has reached rigid rotation before solidifying and that surface-tension and solidification distortions can be neglected.

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