Irodov 1.324: Draining Uphill Against Its Own Spin – Amit & Deepali Sinha Foundation

Problem Irodov 1.324

A narrow tube $AB$ of length $l$ lies horizontally and spins at a constant angular velocity $\omega$ about a fixed vertical axis passing through its own end $A$ . The tube is filled with an ideal (incompressible, inviscid) fluid. The end $A$ is open to the atmosphere; the opposite end $B$ is closed except for a very small orifice, through which fluid escapes as the tube spins. At some instant a column of fluid of length $h$ remains inside the tube, occupying the stretch adjacent to the closed end $B$ (so the fluid’s own free surface, facing the open end $A$ , sits a distance $l-h$ from the axis). Find the fluid’s speed relative to the tube, as a function of $h$ .

Side elevation of the tube AB, spinning at ω about the dashed axis through A. The shaded length h next to the closed end B is the fluid that remains, spraying from the orifice at B. — Side elevation of the tube $AB$ , spinning at $\omega$ about the dashed axis through $A$ . The shaded length $h$ next to the closed end $B$ is the fluid that remains, spraying from the orifice at $B$ .

Given

$l$
$\omega$
$h$ (remaining fluid column, measured from $B$ )

Find

The fluid’s speed

v

relative to the tube

Solution

Step 1Work in the tube’s own rotating frame

In the frame that spins with the tube, every fluid element also feels a centrifugal force per unit volume $\rho\omega^2 r$ directed outward along the tube, where $r$ is the distance from the axis at $A$ . This is a conservative force with potential energy per unit volume $-\rho\omega^2r^2/2$ (minus its gradient is exactly that outward force), so Bernoulli’s equation applies along the tube in this frame with that potential added in.

p+\dfrac{1}{2}\rho v_{\text{rel}}^2-\dfrac{1}{2}\rho\omega^2 r^2 = \text{const}

(1)

Step 2Identify the two ends of the streamline

The fluid’s own free surface sits at $r=l-h$ (the boundary between fluid and the open air stretching back to $A$ ), where the pressure is atmospheric, $p=p_0$ , and the relative speed is negligible: the orifice at $B$ is very small compared with the tube’s own bore, so the free surface retreats far more slowly than the fluid escapes through it, exactly as in a draining vessel with $s\ll S$ . At the orifice, $r=l$ , the fluid discharges directly into the atmosphere, so $p=p_0$ there too, with relative speed $v$ , the quantity sought.

Step 3Apply the rotating-frame Bernoulli equation between the two ends

Evaluating (1) at $r=l-h$ (surface) and at $r=l$ (orifice) and equating, the atmospheric pressure $p_0$ cancels from both sides:

\begin{align*}p_0+0-\frac{1}{2}\rho\omega^2(l-h)^2 &= p_0+\frac{1}{2}\rho v^2-\frac{1}{2}\rho\omega^2 l^2 \\ \frac{1}{2}\omega^2 l^2-\frac{1}{2}\omega^2(l-h)^2 &= \frac{1}{2}v^2\end{align*}

v^2 = \omega^2\left[l^2-(l-h)^2\right]

(2)

Step 4Simplify the bracket

Expanding the difference of squares in (2),

l^2-(l-h)^2 = l^2-\left(l^2-2lh+h^2\right) = h(2l-h),

v^2 = \omega^2 h(2l-h) = \omega^2 h^2\left(\frac{2l}{h}-1\right).

v = \omega h\sqrt{\dfrac{2l}{h}-1}

(3)

Check

At $h=l$ (the tube still full) the free surface sits on the axis itself, where the centrifugal potential vanishes, so the fluid falls through the whole centrifugal head $\omega^2l^2/2$ and the bracket duly gives $v=\omega l\sqrt{2-1}=\omega l$ . That this matches the tube’s tip speed is a coincidence of magnitude: $v$ is measured relative to the tube and points along it, whereas the tip speed is absolute and tangential (in the laboratory the fluid leaving at $h=l$ moves at $\sqrt{2}\,\omega l$ , the resultant of the two). As $h\to0$ (the tube nearly empty), $v\to0$ : with almost nothing left near $B$ , there is almost no column length for the centrifugal effect to accelerate.

Remark

The result depends on $h$ through $l^2-(l-h)^2$ , the difference of the squared distances of the two ends of the fluid column from the axis, not through $h$ alone: draining is driven by centrifugal potential, which goes as $r^2$ , so a short column near the rim ( $h\ll l$ ) feels a much stronger effective gravity than the same length $h$ would near the axis. This is the same $r^2$ structure that makes a centrifuge so much more effective at separation than ordinary settling under gravity.