A horizontally oriented tube AB of length l rotates with a constant angular velocity `omega` about a stationary vertical axis `OO^'` passing through the end A (figure). 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.

Explanation

In a rotating frame (with constant angular velocity) the Eulerian equation is
`-vecVp+rhovecg+2rho(overset(rarr')vxxvecomega)+rhoomega^2vecr=rho(doverset(rarr')v)/(dt)`
In the frame of rotating tube the liquid in the "column" is practically static because the orifice is sufficiently small. Thus the Eulerian Eq. in projection form along `vecr` (which is the position vector of an arbitrary liquid element of length `dr` relative to the rotation axis) reduces to
`(-dp)/(dr)+rhoomega^2r=0`
or, `dp=rhoomega^2rdr`
so, `underset(p_0)overset(p)intdp=rhoomega^2underset((l-h))overset(r)intrdr`
Thus `p(r)=p_0+(rhoomega^2)/(2)[r^2-(l-h)^2]` (1) Hence the pressure at the end B just before the orifice i.e.
`p(l)=p_0+(rhoomega^2)/(2)(2lh-h^2)` (2)
Then applying Bernoull's theorem at the orifice for the points just inside and outside of the end B
`p_0+1/2rhoomega^2(2lh-h^2)=p_0+1/2rhov^2` (where v is the sought velocity)
So, `v=omegahsqrt((2l)/(h)-1)`