Fluids at Rest — Question 10

Question 10

Rigid rotation produces a curved liquid surface

An open cylindrical tank of radius $R$ contains liquid rotating rigidly at angular speed $\omega$ about its vertical axis. The settled liquid is at rest in the rotating frame. Initially at rest its depth was $h_0$. Ignore capillarity; assume the bottom stays covered and no liquid spills.

1. Use radial balance and vertical hydrostatics to derive the free-surface shape $z(r)$ relative to the tank bottom.

2. Find the center and rim depths for $R=0.20\,\mathrm{m}$, $\omega=10\,\mathrm{rad/s}$, $g=10\,\mathrm{m/s^2}$ and $h_0=0.30\,\mathrm{m}$. State the bottom-coverage condition.

Question 10 - Solution

Rigid rotation produces a curved liquid surface

(a) Derivation and calculation

$$\frac{\partial p}{\partial r}=\rho\omega^2r,\quad\frac{\partial p}{\partial z}=-\rho g,\quad dp=0\Rightarrow\frac{dz}{dr}=\frac{\omega^2r}{g}.$$$$z(r)=z_c+\frac{\omega^2r^2}{2g}.$$$$\pi R^2h_0=\int_0^R2\pi r z(r)\,dr=\pi R^2z_c+\frac{\pi\omega^2R^4}{4g}.$$$$\boxed{z(r)=h_0-\frac{\omega^2R^2}{4g}+\frac{\omega^2r^2}{2g}}.$$

(b) Result and interpretation

$$\omega^2R^2/(4g)=100(.20)^2/40=.100\,\mathrm m.$$$$\boxed{z_c=.200\,\mathrm m},\qquad\boxed{z(R)=.400\,\mathrm m}.$$$$\boxed{h_0>\omega^2R^2/(4g)}\quad\text{for strictly positive center depth}.$$If the bottom becomes exposed, full-disk volume integration no longer describes the liquid domain.

Physical check

The disk-average depth remains $.300\,\mathrm m$, midway between the center and rim depths for this paraboloid.