Question

A horizontally oriented tube $AB$ of length $I$ rotates with a constant angular velocity $\omega$ about a stationary vertical axis $O{O}^{\prime }$ passing through the end $A.$ The tube is filled with an ideal fluid. The end $A$ of the tube is open, the closed end $B$ has a very small orifice. Find the velocity of the fluid relative to the tube as a function of the column 'height' h.

Answer 1

In a rotating frame (with constant angular velocity) the Bulerian equation is

$-\overrightarrow{V}p+\rho \overrightarrow{g}+2\rho (\overrightarrow{v}\times \overrightarrow{\omega })+\rho {\omega }^2\overrightarrow{r}=\rho \frac{d\overrightarrow{v}}{dt}$

In the frame of rotating tube the liquid in the column in partically static because the orific is sufficient small. Thus the Eulerian Eq. In projection form along $\overrightarrow{r}$ (which is the position vector of an arbitray liquid element of length $dr$ relative to the rotation axis)

re duce to

$\frac{-dp}{dr}+\rho {\omega }^{2}r=0$

or, $dp=\rho {\omega }^{2}rdr$

so, ${\int }_{p_0}^{p}dp=\rho {\omega }^2{\int }_{(l-k)}^{r}rdr$

Thus $p\left(r\right)=p_0+\frac{\rho {\omega }^2}{2}[r^2-(l-h{)}^2]$

Hence the pressure at the end $B$ just before the orifice i.e.

$p\left(l\right)=p_0+\frac{\rho {\omega }^2}{2}(2lh-h^2)$

Then applying Bernoull's theorem at the orifice for the point just inside and outside of the end $B$

$p_0+\frac{1}{2}\rho {\omega }^2(2lh-h^2)=p_0+\frac{1}{2}\rho v^2$ (where $v$ is the sought velocity)

So, $v=\omega g\sqrt{\frac{2l}{h}-1}$