= Solution
In steady rigid <rotation> of the liquid about its vertical axis, <centrifugal acceleration> is $\Omega^2r$ outward. <Hydrostatic pressure> balance in the rotating frame gives
$$
\frac{\partial p}{\partial r}=\rho\Omega^2r,\qquad
\frac{\partial p}{\partial z}=-\rho g.
$$
The <free surface> is at constant atmospheric <pressure>, so its slope obeys
$$
0=dp=\rho\Omega^2r\,dr-\rho g\,dz,
\qquad z-z_0=\frac{\Omega^2r^2}{2g}.
$$
Compare this <parabolic free surface of a rigidly rotating liquid> with $z-z_0=r^2/(4f)$. For <focal length> $f=1\ \mathrm m$,
$$
\boxed{\Omega=\sqrt{\frac{g}{2f}}
\simeq2.21\ \mathrm{rad\,s^{-1}}
\simeq21.1\ \mathrm{rev\,min^{-1}}.}
$$
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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