Why a tap stream breaks into drops
Tonight's idea
Turn a tap down until the stream is thin. Near the tap it is a smooth rod of water. Lower down it has become separate drops, though nothing cut it.
The surface of water acts like a stretched skin. This surface tension pulls the water towards shapes with less surface area. The volume cannot change. A flow is stable if tiny wobbles die away, and unstable if they grow into something new.
A long thin cylinder of water is unstable. A small wobble in its thickness can grow until the stream is a row of drops.
Tonight you compare a cylinder of radius \(r\) and length \(L\) (side area \(2\pi r L\), volume \(\pi r^2 L\)) with a sphere of radius \(R\) (area \(4\pi R^2\), volume \(\tfrac43 \pi R^3\)).
Model: take one piece of the stream. Its ends join more water, so only its side is surface. Ignore gravity and air.
Watch one: a cylinder against a ball
A piece of stream has radius \(r = 1\) mm and length \(L = 10\) mm. If the same water became one ball, would it have less surface?
Cylinder. Side area and volume:
\[2\pi r L = 2\pi \times 1 \times 10 = 20\pi \approx 62.8\ \text{mm}^2, \qquad \pi r^2 L = 10\pi\ \text{mm}^3.\]Ball. It holds the same volume, so \(\tfrac43 \pi R^3 = \pi r^2 L\). That gives \(R^3 = \tfrac34 r^2 L\) every time. Here
\[R^3 = \tfrac34 \times 1^2 \times 10 = 7.5, \qquad R \approx 1.957\ \text{mm}.\]Its area is \(4\pi R^2 \approx 4\pi \times 3.832 \approx 48.1\ \text{mm}^2\). The ball has less surface: 48.1 against 62.8, a saving of about 23%. Surface tension favours the ball.
This compares the start with one possible finish, not the route between them. The stretch comes back to that.
Your turn
- Keep \(r = 1\) mm. For each length, find the side area of the cylinder and the area of the ball with the same volume. Which has less surface? (a) \(L = 2\) mm. (b) \(L = 4\) mm. (c) \(L = 6\) mm.
- Somewhere between \(L = 4\) and \(L = 6\) the winner changes. (a) Show that the ball has less surface when \(2R^2 \lt rL\). (b) Cube both sides and use \(R^3 = \tfrac34 r^2 L\) to show that this happens exactly when \(L \gt 4.5\,r\). (c) What is \(R\) when \(L = 4.5\,r\)?
- A stream of radius \(r\) breaks into pieces of length \(9r\), and each becomes one drop. (a) Find the drop radius as a multiple of \(r\). (b) Find it for streams of radius 1 mm and 0.5 mm. How many times more water is in the larger drop? (c) What fraction of its surface does a piece lose when it becomes a drop?
- Made-up but sensible numbers: where a stream breaks up it has radius 1.5 mm and falls at 1.2 m/s. The pieces are \(9r\) long. (a) How far apart are the centres of the drops as they form? (b) How many drops form each second? (c) Further down they fall at 2.4 m/s, and the same number pass each second. How far apart are they now?
Stretch: the hill in the way
In the 1800s Joseph Plateau and Lord Rayleigh studied this breakup. In the model, with stickiness ignored too, a slight wobble in thickness with wavelength \(\lambda\) (the distance from one bulge to the next) lowers the surface area exactly when \(\lambda \gt 2\pi r\), the distance round the stream. Such wobbles grow. Shorter ones raise the area and do not grow. The wobble that grows fastest has \(\lambda\) about \(9r\). (a) Find \(2\pi r\) for \(r = 1\) mm and compare it with your \(4.5\,r\) from exercise 2. (b) A piece of length \(5r\) would have less surface as a ball, yet a wobble of wavelength \(5r\) does not grow. Explain how both can be true.
Answers
- (a) Cylinder: \(2\pi \times 1 \times 2 \approx 12.6\ \text{mm}^2\). Ball: \(R^3 = \tfrac34 \times 2 = 1.5\), \(R \approx 1.145\), area \(\approx 16.5\ \text{mm}^2\). The cylinder has less. (b) Cylinder: \(8\pi \approx 25.1\ \text{mm}^2\). Ball: \(R^3 = 3\), \(R \approx 1.442\), area \(\approx 26.1\ \text{mm}^2\). The cylinder has less, but only just. (c) Cylinder: \(12\pi \approx 37.7\ \text{mm}^2\). Ball: \(R^3 = 4.5\), \(R \approx 1.651\), area \(\approx 34.3\ \text{mm}^2\). The ball has less. A short piece of stream gains nothing by becoming a ball. A long one does.
- (a) The ball wins when \(4\pi R^2 \lt 2\pi r L\). Divide both sides by \(2\pi\): \(2R^2 \lt rL\). (b) Both sides are positive, so cubing keeps the order: \(8R^6 \lt r^3 L^3\). Now \(R^6 = (R^3)^2 = \tfrac{9}{16} r^4 L^2\), so \(8R^6 = \tfrac92 r^4 L^2\). The condition is \(\tfrac92 r^4 L^2 \lt r^3 L^3\). Divide by \(r^3 L^2\), which is positive: \(\tfrac92 r \lt L\). Every step can be reversed, so the ball wins exactly when \(L \gt 4.5\,r\). This agrees with exercise 1, since \(4 \lt 4.5 \lt 6\). (c) \(R^3 = \tfrac34 r^2 \times 4.5\,r = 3.375\,r^3\), so \(R = 1.5\,r\). Check: the ball's area is \(4\pi \times 2.25\,r^2 = 9\pi r^2\) and the cylinder's side is \(2\pi r \times 4.5\,r = 9\pi r^2\). Equal, as they should be.
- (a) \(R^3 = \tfrac34 r^2 \times 9r = 6.75\,r^3\), so \(R \approx 1.89\,r\). (b) About 1.89 mm and about 0.94 mm. The drop radius is always the same multiple of \(r\), so a thinner stream makes smaller drops. Volume goes with the cube of the radius, so the larger drop holds \(2^3 = 8\) times as much water. (c) The side area is \(2\pi r \times 9r = 18\pi r^2 \approx 56.5\,r^2\). The drop's area is \(4\pi \times (1.89\,r)^2 \approx 44.9\,r^2\). The ratio is about 0.79, so about 21% of the surface is lost. It is the same for every \(r\).
- (a) Each piece is \(9 \times 1.5 = 13.5\) mm long, so the centres are 13.5 mm apart. (b) The stream moves 1200 mm each second, and \(1200 \div 13.5 \approx 89\) drops a second. That is far too quick to see, so the broken stream looks like a blur. (c) \(2400 \div 89 \approx 27\) mm. The speed doubled and the number passing each second stayed the same, so the gaps doubled. Each drop has radius about \(1.89 \times 1.5 \approx 2.8\) mm. In a real stream the pieces are not all the same length, so treat these as estimates.
Stretch. (a) \(2\pi \times 1 \approx 6.28\) mm, which is longer than 4.5 mm. The true condition is harder to meet than the ball test suggests. (b) The ball test compares only the start and the finish. To get from one to the other the surface must pass through every shape in between, starting with a slight wobble. For a wavelength of \(5r\) that first wobble has more surface than the smooth cylinder, so surface tension flattens it again. Think of a marble in a dip on a hillside. There is lower ground further down, but to reach it the marble must first go uphill, so a small nudge is not enough. Lengths between \(4.5\,r\) and about \(6.28\,r\) are like that. Beyond \(2\pi r\) it is downhill from the first step. Very long wobbles also lower the area, but the water has further to travel, so they grow slowly. The fastest, at about \(9r\), usually wins the race, which is why exercises 3 and 4 used that length. Real streams often leave a tiny extra drop between the main ones, which this model leaves out.