Stir a cup of tea with loose leaves in it, briskly, and take the spoon out. The surface dips into a hollow in the middle and climbs up the sides. As the tea slows down, the leaves on the bottom do something odd. They do not fly out to the rim. They gather in a small heap at the centre of the bottom.
Turning as one body
Set a cylinder of water turning about its vertical axis at a steady angular speed \(\omega\). At first only the water near the walls turns. Friction between the layers soon drags the rest along. After a while the whole of the water turns with the vessel, like a solid body. Each small part of it then moves in a horizontal circle about the axis.
A part at a distance \(r\) from the axis has a centripetal acceleration \(\omega^2 r\), towards the axis. So it needs a net inward force, as in the chapter on Dynamics of Circular Motion. In a liquid, that force can only come from a difference in pressure. The pressure must be larger on the outer side of each part than on the inner side.
\[ dm = \rho A\,dr \]
Why. the element of Fig 6.1 has faces of area \(A\), facing the axis, and a thickness \(dr\).
\[ (P + dP)A - PA = dm\,\omega^2 r \]
Why. the net inward force is the push on the outer face minus the push on the inner face. It gives the mass its centripetal acceleration.
\[ A\,dP = \rho A\,dr\,\omega^2 r \]
Why. put in the mass from step 1.
\[ \dfrac{dP}{dr} = \rho\omega^2 r \]
Read it. divide by \(A\,dr\). The pressure grows away from the axis, faster the farther out.
Take a liquid turning as one body at angular speed \(\omega\). At one level, its pressure grows with the distance \(r\) from the axis at the rate \(\rho\omega^2 r\). Vertically it still grows downward at the rate \(\rho g\).
It is the rule of the last lesson again. At a distance \(r\) the acceleration is \(\omega^2 r\) towards the axis. So the effective gravity \(\vec g - \vec a\) has a part \(\omega^2 r\) pointing outward, as well as \(g\) down. In the frame turning with the water, that outward part is the centrifugal force, \(m\omega^2 r\) on each mass \(m\). The chapter on Dynamics of Circular Motion met it first.
Water turns with its vessel at a steady \(\omega\). Two points A and B are at the same level, A near the axis and B near the wall. How do their pressures compare?
The shape of the surface
At one level, the pressure grows from the axis outward. Integrate \(dP/dr = \rho\omega^2 r\) from the axis to \(r\): \(P(r) - P(0) = \tfrac12 \rho\omega^2 r^2\). Now the extra pressure at \(r\) must come from extra water above that point. So the free surface must stand higher at \(r\) than at the axis by a height \(y\), with \(\rho g y = \tfrac12 \rho\omega^2 r^2\).
\[ P(r) - P(0) = \int_0^r \rho\omega^2 r'\,dr' = \tfrac12 \rho\omega^2 r^2 \]
Why. integrate the radial gradient along a level line, from the axis out to \(r\).
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Your turn. which line turns the extra pressure into a height of water?
\[ y = \dfrac{\omega^2 r^2}{2g} \]
Read it. cancel \(\rho\) and divide by \(g\). The height grows as the square of the distance from the axis.
The curve \(y = \omega^2 r^2/2g\) is a parabola. Turned about the axis, it sweeps out a bowl called a paraboloid. The old route to it goes through the turning frame. There each bit of surface feels \(mg\) down and \(m\omega^2 r\) outward. The surface lies at right angles to their sum, so its slope is \(dy/dr = \omega^2 r/g\), which integrates to the same curve.
At \(10\,\text{rad/s}\), about \(95\) turns a minute, the rim stands \(100 \times 0.01/20 = 0.05\,\text{m}\), or \(5\,\text{cm}\), above the centre. Doubling \(\omega\) to \(20\,\text{rad/s}\) makes it \(20\,\text{cm}\), four times as much. Where does the dip come from? The water's volume does not change. In a cylinder, the centre falls below the level at rest by half the rim height. The rim rises by the other half.
Why exactly half? The water missing from the hollow above the paraboloid is half of the cylinder that just holds it. Archimedes proved that a paraboloid fills half of its surrounding cylinder, more than two thousand years ago. So a paraboloid of rim height \(y\) holds as much water as a flat layer \(y/2\) thick. The level at rest lies halfway between the centre and the rim.
A cylinder of radius \(R = 10\,\text{cm}\) holds water \(20\,\text{cm}\) deep and turns about its axis at \(\omega = 10\,\text{rad/s}\). Take \(g = 10\,\text{m/s}^2\). Find the height of the rim above the centre, and the depth of water at the centre and at the rim. At what \(\omega\) does the centre of the base just become dry?
The surface is a paraboloid with \(y_{\text{rim}} = \omega^2 R^2/2g\). Its volume is unchanged, so the centre falls \(y_{\text{rim}}/2\) and the rim rises \(y_{\text{rim}}/2\).
Centre depth \(= H - \omega^2 R^2/4g\), rim depth \(= H + \omega^2 R^2/4g\), with \(H = 0.20\,\text{m}\).
\(y_{\text{rim}} = 100 \times 0.01/20 = 0.05\,\text{m} = 5\,\text{cm}\). The centre is \(20 - 2.5 = 17.5\,\text{cm}\) deep and the rim \(22.5\,\text{cm}\). Dry centre: \(\omega^2 = 4gH/R^2 = 4 \times 10 \times 0.20/0.01 = 800\), so \(\omega = 20\sqrt2 \approx 28\,\text{rad/s}\).
The rim stands \(5\,\text{cm}\) above the centre, with \(17.5\,\text{cm}\) of water at the centre and \(22.5\,\text{cm}\) at the rim. At about \(28\,\text{rad/s}\) the centre of the base is bare, if the vessel is tall enough.
A bucket of radius \(15\,\text{cm}\) holds water and turns about its axis at \(6\,\text{rad/s}\). Take \(g = 10\,\text{m/s}^2\). How high does the rim of the surface stand above its centre, in cm?
The turning speed of a cylinder of water is doubled. What happens to the depth of the hollow, the height of the rim above the centre?
The tea leaves
Now the cup of the opening. While the tea turns, its surface is a paraboloid, and the pressure at each level grows outward. That pressure pattern is set by the fast-turning tea in the body of the cup. Near the bottom, friction with the cup slows a thin layer of tea. This slow layer still feels the full inward push of the pressure. But it turns too slowly to need all of it.
So the slow layer at the bottom is pushed inward, towards the centre. It flows across the bottom to the middle, rises up the axis, and moves out again near the surface. The leaves are denser than the tea and lie on the bottom. The inward flow along the bottom sweeps them to the centre and leaves them there in a heap.
In 1926 Albert Einstein wrote a short paper on why rivers wind. He began with a cup of tea. Friction at the bottom of a stirred cup slows the tea there, he explained. So the inward push of the pressure wins, and the bottom layer flows to the centre with the leaves. The same flow in a bending river scours its outer bank and drops sand on the inner one. The river bends more and more.
The centrifuge
A spinning vessel makes an effective gravity \(\omega^2 r\) outward, and it can be enormous. A centrifuge spins test tubes about a vertical axis, so each tube feels \(\omega^2 r\) along its length. The denser parts of a mixture move outward, to the far end of the tube. Sand settles to the bottom of a glass of water in the same way. They move far faster than under \(g\) alone.
A dairy's cream separator turns at about \(6000\) revolutions a minute, or \(628\,\text{rad/s}\). At \(10\,\text{cm}\) from its axis, \(\omega^2 r = 628^2 \times 0.1 \approx 3.9 \times 10^4\,\text{m/s}^2\), about \(4000\,g\). The heavier skimmed milk is flung to the outside, and the lighter cream collects near the axis. In a pathology laboratory, a centrifuge separates blood cells from plasma in the same way.
Which statements about liquids that turn are true?
Before machines, cream was separated by letting milk stand for a day, so that the lighter cream rose under gravity. A cream separator does the same work in seconds. The milk enters a stack of spinning cones, the skimmed milk is thrown to the outside and the cream flows out near the centre. The physics is the settling of a lighter liquid, with \(\omega^2 r\) in place of \(g\).
- Saying the water is thrown outward by a real outward force.In the ground frame each part needs an inward force, supplied by a pressure that grows outward. The centrifugal force exists only in the turning frame.
- Using \(\rho\omega^2 r^2\) for the extra pressure at the rim.Integrate \(\rho\omega^2 r\,dr\): the extra pressure is \(\tfrac12 \rho\omega^2 r^2\), and the height is \(\omega^2 r^2/2g\).
- Letting only the rim rise in a turning cylinder.The volume is fixed. The centre falls and the rim rises, each by half the rim height.
- Expecting the tea leaves to fly to the rim.Friction slows the bottom layer, the inward pressure wins, and the bottom flow carries the leaves to the centre.