Why do tiny drops take so long to settle?
Tonight's idea
Drop a glass bead into a jar of honey. It does not keep speeding up. Almost at once it sinks at one steady speed. Three forces act on it. Its weight pulls down. The honey buoys it up, as water buoys up a swimmer. And the honey drags on it, because a fluid resists being pushed aside. That resistance to flowing is called viscosity. The faster the bead moves, the bigger the drag, so the speed settles where drag plus buoyancy balance the weight.
For small balls of one material moving slowly through one thick fluid, the steady speed goes with the radius squared. Half the radius, a quarter of the speed.
This is called Stokes's law. It is a model: a small smooth ball, moving slowly, far from the walls. It fails for a marble falling fast through water. Drops are small balls too. An oil drop in vinegar is lighter than the vinegar, so it rises instead of sinking. If the drop is tiny, the same rule sets its speed.
Watch one: three beads in honey
A glass bead of radius 2 mm sinks through honey at a steady 4 cm per minute. (The speeds on this sheet are illustrative, not measured.) How fast do glass beads of radius 1 mm and 3 mm sink? Compare each radius with 2 mm, then square.
Radius 1 mm is \(\tfrac12\) of 2 mm, and \(\tfrac12 \times \tfrac12 = \tfrac14\). The speed is \(\tfrac14 \times 4 = 1\) cm per minute.
Radius 3 mm is \(1.5\) times 2 mm, and \(1.5 \times 1.5 = 2.25\). The speed is \(2.25 \times 4 = 9\) cm per minute.
So in this honey, a glass bead of radius \(r\) mm sinks at \(r \times r\) cm per minute. To sink 12 cm the three beads take \(12 \div 1 = 12\) minutes, \(12 \div 4 = 3\) minutes and \(12 \div 9 = 1\tfrac13\) minutes.
Your turn
- Same honey and same glass as the example. (a) Find the steady speed of beads of radius 0.5 mm, 1 mm, 2 mm and 4 mm. (b) How long does each take to sink 8 cm? (c) The radius 4 mm bead is 8 times as wide as the radius 0.5 mm bead. How many times as fast is it? (d) Which radius sinks at 25 cm per minute?
- Oil drops rise through vinegar. Suppose a drop 0.1 mm across rises at a steady 2.5 cm per minute (an illustrative figure). (a) How long does it take to rise 5 cm to the top? (b) Another drop is 10 times smaller across. How fast does it rise? (c) How long does the smaller drop take to rise 5 cm? Give your answer in hours and minutes. (d) How many times as long is that?
- Shaking breaks big drops into smaller ones. The volume of a ball goes with radius \(\times\) radius \(\times\) radius. (a) One drop breaks into 8 equal drops. Show that each has half the radius of the old drop. (b) How many times slower does each new drop rise? (c) Shake harder, and one drop becomes 1000 equal drops. What happens to the radius and to the speed? (d) Now explain why hard-shaken dressing stays mixed for longer.
- A race, with made-up times. The same glass bead, 1 mm across, sinks 4 cm through three liquids: honey in 16 minutes, cooking oil in 4 seconds, maple syrup in 10 seconds. (a) Put the liquids in order from least to most viscous. (b) How many times as long does the bead take in honey as in oil? (c) A glass bead half as wide is dropped into the honey. How long does it take to sink 4 cm?
Stretch: where does the square come from?
The pull on a bead (its weight minus the buoyancy) goes with its volume, so with radius \(\times\) radius \(\times\) radius. For slow, sticky flow the drag goes with radius \(\times\) speed. (a) Double the radius. How many times bigger is the pull? At the same speed, how many times bigger is the drag? So how many times faster must the bead go before drag balances pull again? (b) Repeat for 3 times the radius. (c) For a marble falling fast through water, a rough rule is that drag goes with radius \(\times\) radius \(\times\) speed \(\times\) speed. Double the radius. About how many times faster does the marble fall?
Answers
- (a) Speed is \(r \times r\) cm per minute: \(0.5 \times 0.5 = 0.25\), then 1, 4 and 16 cm per minute. (b) \(8 \div 0.25 = 32\) minutes, \(8 \div 1 = 8\) minutes, \(8 \div 4 = 2\) minutes and \(8 \div 16 = \tfrac12\) minute. (c) \(16 \div 0.25 = 64\) times as fast, which is \(8 \times 8\). (d) \(5 \times 5 = 25\), so radius 5 mm.
- (a) \(5 \div 2.5 = 2\) minutes. (b) A tenth of the size means \(\tfrac1{10} \times \tfrac1{10} = \tfrac1{100}\) of the speed: \(2.5 \div 100 = 0.025\) cm per minute. (c) \(5 \div 0.025 = 200\) minutes, which is 3 hours 20 minutes. (d) \(200 \div 2 = 100\) times as long. Whether you measure across the drop or use its radius makes no difference, because only the ratio of the sizes matters.
- (a) Each new drop has \(\tfrac18\) of the volume, and \(\tfrac12 \times \tfrac12 \times \tfrac12 = \tfrac18\), so half the radius fits. (b) \(\tfrac12 \times \tfrac12 = \tfrac14\) of the speed, so 4 times slower. (c) \(\tfrac1{10} \times \tfrac1{10} \times \tfrac1{10} = \tfrac1{1000}\), so the radius is a tenth and the speed is a hundredth. (d) The amount of oil has not changed, but it is now in far more, far smaller drops. Every one of them rises much more slowly, so the oil takes much longer to gather at the top. This is a model. In a real dressing drops also bump into each other and merge into bigger, faster ones, and crowded drops get in each other's way.
- (a) Cooking oil, maple syrup, honey. The same bead sinks more slowly through a more viscous liquid. (b) 16 minutes is 960 seconds, and \(960 \div 4 = 240\) times as long. That suggests honey is very roughly 240 times as viscous as the oil. Only roughly, because honey is also denser than oil and buoys the bead up more. (c) Half the width means a quarter of the speed, so 4 times as long: \(4 \times 16 = 64\) minutes.
Stretch. (a) The pull is \(2 \times 2 \times 2 = 8\) times bigger. At the same speed the drag is only 2 times bigger. To make the drag 8 times bigger, the speed must be \(8 \div 2 = 4\) times bigger, and \(4 = 2 \times 2\). (b) Pull 27 times bigger, drag 3 times bigger at the same speed, so the speed is \(27 \div 3 = 9 = 3 \times 3\) times bigger. In general the speed goes up by radius cubed divided by radius, which is radius squared. (c) The pull is 8 times bigger and, at the same speed, the drag is \(2 \times 2 = 4\) times bigger. So speed \(\times\) speed must double, and the speed goes up about 1.4 times, since \(1.4 \times 1.4 = 1.96\). Twice the marble is nowhere near 4 times as fast. That is why tonight's rule is only for slow, sticky flow.