Question

A horizontally oriented tube $A B$ of length $l$ rotates with a constant angular velocity $ω$ , about a stationary vertical axis $O O'$ passing through the end $A$ . The tube is filled with an ideal fluid. The end $A$ of the tube is open and 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 $h$ .

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Accepted Answer

Consider a fluid element of width $d r$ , at a distance $r$ from the axis of rotation $O O'$ , as shown in the following figure.

Clearly, mass $A$ of this element is given by

$h$

The required centripetal force on the considered element is,

$d F = d m ω^{2} r$

$\Rightarrow d F = (ρ S d r) ω^{2} r$

Hence, the net centripetal force on the whole fluid is given by,

$F = ρ S ω^{2} \int_{l - h}^{l} r d r \Rightarrow F = ρ S ω^{2} {[\frac{r^{2}}{2}]}_{l - h}^{l} \Rightarrow F = ρ S ω^{2} [\frac{l^{2} - {(l - h)}^{2}}{2}] \Rightarrow F = ρ S ω^{2} (\frac{2 l h - h^{2}}{2}) . . . (1)$

Now, force $F_{1}$ , at the left end of the fluid, is

$F_{1} = P_{0} S$

where $P_{0}$ , is the atmospheric pressure.

Let force $F_{2}$ be just inside $B$ , such that, $F_{2} = P_{2} S$ , where $P_{2}$ is the pressure just inside $B$ .

Now, we have,

$F_{2} - F_{1} = F \Rightarrow F_{2} = F_{1} + F \Rightarrow P_{2} S = P_{0} S + ρ S ω^{2} (\frac{2 l h - h^{2}}{2}) \Rightarrow P_{2} = P_{0} + ρ ω^{2} (\frac{2 l h - h^{2}}{2}) . . . (2)$

Now, let us apply Bernoulli's theorem just inside and outside the orifice at point $B$ , with respect to the tube.

$\frac{P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 0 = \frac{P_{0}}{ρ g} + \frac{v_{2}^{2}}{2 g} + 0 . . . (3)$

As the height of both the points are at the same horizontal level, $h = 0$ .

Here $v_{1} = 0$ , as the fluid in the tube is in rest with respect to tube.

Let $v_{2} = v$ be the velocity of the fluid, relative to tube, just outside the tube.

Putting the values in equation $(3)$ , we obtain,

$\frac{1}{ρ g} [P_{0} + ρ ω^{2} (\frac{2 l h - h^{2}}{2})] + 0 = \frac{P_{0}}{ρ g} + \frac{v^{2}}{2 g} \Rightarrow v^{2} = ω^{2} (\frac{2 l h - h^{2}}{2})$

$∴ v = ω h \sqrt{\frac{2 l}{h} - 1}$

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