A U-tube of narrow, uniform bore has a horizontal part 30 cm long between its two vertical limbs…

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A U-tube of narrow, uniform bore has a horizontal part $30\ \text{cm}$ long between its two vertical limbs. The left limb is open at the top, and the right limb is closed at the top. At rest, mercury fills the horizontal part and stands $10\ \text{cm}$ high in each limb. Above the mercury in the closed limb there is a column of air $20\ \text{cm}$ long, at the air's pressure of $76\ \text{cm}$ of mercury. The tube is now turned about the axis of the open limb, at a steady angular speed $\omega$ , as shown in the figure. The air stays at a steady temperature. Take $g = 10\ \text{m/s}^2$ and ignore the bends at the corners.

A U-tube with a horizontal part 30 cm long. The left limb is open, and a dashed vertical axis runs through it with a curved arrow showing rotation at angular speed omega. The right limb is closed at the top. Mercury fills the horizontal part and stands 10 cm high in each limb, and a 20 cm column of air is trapped above the mercury in the closed limb.

In the horizontal part of the tube, how does the pressure change from the axis out to the closed limb? Write the difference in terms of $\omega$ , the length $L$ of the horizontal part and the density $\rho$ of mercury.
For what $\omega$ does the mercury rise by $4.0\ \text{cm}$ in the closed limb? What is then the pressure of the trapped air?
For what $\omega$ does the mercury level in the open limb come right down to the horizontal part?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Measure pressures in centimetres of mercury, so that $\rho g$ times a height is just that height. Follow the pressure round the tube from one free surface to the other.

Step 1. Along the horizontal part (a). A short piece of mercury of length $dr$ at a distance $r$ from the axis moves in a circle. The pressure difference across it gives its centripetal force: $A\,dp = (\rho A\,dr)\,\omega^2 r$ . Adding up from $0$ to $L$ :

\displaystyle \Delta p = \tfrac12\rho\,\omega^2 L^2

or, in centimetres of mercury, $\dfrac{\omega^2 L^2}{2g}$ .

Step 2. Round the tube. Let the mercury rise by $x$ in the closed limb, so it falls by $x$ in the open limb. Start at the open surface, at the air's pressure $P_0 = 76$ . Down the open limb, $+\,(h - x)$ . Along the bottom, $+\,\omega^2L^2/2g$ . Up the closed limb, $-\,(h + x)$ . So the pressure at the closed surface is

\displaystyle P_0 - 2x + \frac{\omega^2L^2}{2g}

This must equal the pressure of the trapped air, which by Boyle's law is $P_0\,\dfrac{l_0}{l_0 - x}$ .

Step 3. A rise of $4.0\ \text{cm}$ (b). The air is squeezed from $20\ \text{cm}$ to $16\ \text{cm}$ , so its pressure is $76 \times \dfrac{20}{16} = 95.0\ \text{cm}$ of mercury. Then

\displaystyle \frac{\omega^2L^2}{2g} = 95.0 - 76 + 8 = 27\ \text{cm} = 0.27\ \text{m}

\displaystyle \omega^2 = \frac{2 \times 10 \times 0.27}{0.09}\ \text{s}^{-2} = 60\ \text{s}^{-2}, \qquad \omega = 7.75\ \text{rad/s}

Step 4. The open limb empties (c). Now $x = h = 10\ \text{cm}$ . The air is squeezed to $10\ \text{cm}$ , so its pressure is $152\ \text{cm}$ of mercury, and

\displaystyle \frac{\omega^2L^2}{2g} = 152 - 76 + 20 = 96\ \text{cm}, \qquad \omega^2 = 213.3\ \text{s}^{-2}, \qquad \omega = 14.6\ \text{rad/s}

Answer: (a) $\tfrac12\rho\omega^2L^2$ , higher at the closed limb. (b) $7.75\ \text{rad/s}$ ; the air is at $95.0\ \text{cm}$ of mercury. (c) $14.6\ \text{rad/s}$ .

Check. Without the trapped air (both limbs open), the levels would differ by $\omega^2L^2/2g$ , the usual result for a spinning U-tube. Here the air pushes back, so a much larger $\omega$ is needed for the same rise.

Watch out. The mercury in the open limb is on the axis, so it has no centripetal acceleration. All the pressure change due to the spin happens in the horizontal part.

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