Liquid rise in rotating U-tube

Data: U-tube of uniform bore with base length $L$, rotating about the left vertical arm at $\omega=10$ rad/s. The relative density 0.85 of the liquid does not affect the result.

At rest the liquid fills only the base. When the tube spins, the liquid is thrown outwards: it rises a height $h$ in the outer arm, and the same volume leaves the inner end of the base. With a uniform bore, the liquid now fills the base only from $r=h$ to $r=L$, with a free surface (at atmospheric pressure, since air enters through the inner arm) at $r=h$ on the base level.

For rigid-body rotation the pressure satisfies

$$p+\rho g z-\tfrac{1}{2}\rho\omega^{2}r^{2}=\text{constant}$$

Apply this between the two free surfaces, both at $p=0$: the inner surface in the base ($r=h$, $z=0$) and the surface in the outer arm ($r=L$, $z=h$):

$$\rho g h-\tfrac{1}{2}\rho\omega^{2}L^{2}=-\tfrac{1}{2}\rho\omega^{2}h^{2} \quad\Rightarrow\quad h=\frac{\omega^{2}\left(L^{2}-h^{2}\right)}{2g}$$

With $k=\dfrac{\omega^{2}}{2g}=\dfrac{100}{19.62}=5.097\ \text{m}^{-1}$, this is $k h^{2}+h-kL^{2}=0$, so

$$h=\frac{\sqrt{1+4k^{2}L^{2}}-1}{2k}=\frac{\sqrt{1+103.9\,L^{2}}-1}{10.19}\ \text{m}\quad(L\text{ in metres})$$

For example, $L=0.1$ m gives $h=0.042$ m (42 mm), and $L=0.3$ m gives $h=0.218$ m. If the loss of liquid from the inner end of the base were ignored, we would get $h=\omega^{2}L^{2}/2g=5.1\,L^{2}$, which overestimates the rise (51 mm for $L=0.1$ m).

Answer: the rise in the outer arm satisfies $h=\dfrac{\omega^{2}(L^{2}-h^{2})}{2g}$, i.e. $h=\dfrac{\sqrt{1+103.9L^{2}}-1}{10.19}$ m (about 42 mm for $L=0.1$ m), independent of the liquid density.