In the monsoon, rain falls from clouds a kilometre or two above the ground. A stone dropped from a kilometre would hit the ground at about \(140\,\text{m/s}\), faster than a cricket ball from any bowler. Yet a raindrop only taps your face, and a fine drizzle seems to float down. The clouds themselves, full of water, hang in the sky.
Drizzle and cloud droplets
A drop of water falling through air feels its weight, the buoyant force of the air and the drag. Air is less than a thousandth as dense as water, so the buoyant force is negligible. The terminal velocity of Lesson 3 becomes \(v_t = 2r^2\rho g/9\eta\), with \(\rho\) the density of water and \(\eta\) the viscosity of air.
A drizzle drop has a radius of \(0.03\,\text{mm}\). Air has viscosity \(2 \times 10^{-5}\,\text{Pa s}\); neglect its density. Take \(g = 10\,\text{m/s}^2\). (a) Find the terminal velocity of the drop. (b) How long does it take to fall \(1\,\text{km}\)?
(a) \(v_t = \dfrac{2 \times (3 \times 10^{-5})^2 \times 1000 \times 10}{9 \times 2 \times 10^{-5}} = \dfrac{1.8 \times 10^{-5}}{1.8 \times 10^{-4}} = 0.1\,\text{m/s}\).
(b) It reaches this speed within a few hundredths of a second, so \(t = 1000/0.1 = 10^4\,\text{s}\), nearly three hours.
The droplets of a cloud are smaller still, about \(0.005\,\text{mm}\) in radius. Since \(v_t\) grows as \(r^2\), they fall \(36\) times slower than drizzle, about \(3\,\text{mm/s}\). The air inside a cloud rises and swirls at centimetres or metres per second. So a cloud droplet is carried along by the air far more than it falls through it. This is why clouds float.
Raindrops are not Stokes drops
Try the formula on a real raindrop of radius \(1\,\text{mm}\). It gives \(v_t = 111\,\text{m/s}\), which is absurd. At such speeds the flow round the drop is far from streamline, and the drag grows nearly as \(v^2\) instead of \(v\). The drag catches up with the weight much sooner. Measured raindrops \(2\,\text{mm}\) across fall at about \(6.5\,\text{m/s}\), and the largest drops at about \(9\,\text{m/s}\). That is why rain does not hurt.
A droplet of water has a radius of \(0.02\,\text{mm}\). Air has viscosity \(2 \times 10^{-5}\,\text{Pa s}\); neglect its density, and take \(g = 10\,\text{m/s}^2\). Find its terminal velocity, in cm/s.
A cloud droplet of radius \(0.005\,\text{mm}\) falls at its terminal velocity through still air. Estimate how many hours it would take to fall \(1\,\text{km}\).
A bubble rises at a steady speed
An air bubble in a liquid is far less dense than the liquid around it. The buoyant force is larger than its weight, so it rises. As it speeds up, the drag grows, and soon the bubble rises at a steady terminal velocity. The formula of Lesson 3 still works. Put \(\rho \approx 0\) for the air, and \(\rho - \sigma\) is negative. The minus sign in \(v_t\) means that the motion is upward.
The drag always opposes the motion. On the falling ball of Fig 4.1 it acts upward, together with the buoyant force. On the rising bubble it acts downward, together with the tiny weight of the air, against the buoyant force. At the terminal velocity the upward and downward forces balance in both.
An air bubble of radius \(1\,\text{mm}\) rises through oil of density \(900\,\text{kg/m}^3\) and viscosity \(0.1\,\text{Pa s}\). Neglect the density of air, and take \(g = 10\,\text{m/s}^2\). Find its terminal velocity. In which direction does the viscous force on it act?
\(v_t = \dfrac{2r^2(0 - \sigma)g}{9\eta} = -\dfrac{2 \times 10^{-6} \times 900 \times 10}{9 \times 0.1} = -0.02\,\text{m/s}\). The minus sign means upward: the bubble rises at \(2\,\text{cm/s}\).
The viscous force opposes the motion, so it acts downward on the rising bubble.
A bubble of radius \(2\,\text{mm}\) rises through the same oil (\(900\,\text{kg/m}^3\), \(0.1\,\text{Pa s}\)). Neglect the density of air. At what speed does it rise, in cm/s?
A bubble rises at its terminal velocity through a glass of soda water. The viscous force on it
When drops merge
In a cloud, droplets collide and merge. Rain begins when merged drops grow big enough to fall faster than the air rises. So what happens to the speed of a drop when drops merge? The figure below lets eight drizzle drops fall beside the single drop they make.
Eight drizzle drops, each falling at its terminal velocity of \(0.1\,\text{m/s}\), merge into one drop. At what terminal velocity does the merged drop fall?
The volume of water does not change when the drops merge. Eight drops of radius \(r\) make one drop of radius \(R\) with \(\tfrac43\pi R^3 = 8 \times \tfrac43\pi r^3\). So \(R^3 = 8r^3\), and \(R = 2r\). The terminal velocity grows as the square of the radius, so the merged drop falls \(4\) times as fast as each small one. For \(n\) drops, \(R = n^{1/3}r\) and \(v = n^{2/3}v_0\).
Twenty-seven identical small drops, each with terminal velocity \(v_0\), merge into one drop. Its terminal velocity is how many times \(v_0\)?
Measuring a viscosity
Turn the formula for \(v_t\) round, and it measures a viscosity. Fill a tall tube with the liquid, and mark two lines on it well below the surface. Drop a small steel ball of known radius and density down the axis. By the first mark it is already at its terminal velocity. Time it between the marks, and divide the distance by the time. Then \(\eta = 2r^2(\rho - \sigma)g/9v_t\). This is the falling-ball viscometer.
In Fig 4.2 the ball has radius \(1.5\,\text{mm}\) and density \(7500\,\text{kg/m}^3\). The liquid has density \(1500\,\text{kg/m}^3\). The ball takes \(5.0\,\text{s}\) to fall the \(10\,\text{cm}\) from A to B. Take \(g = 10\,\text{m/s}^2\). Find the viscosity of the liquid.
\(v_t = 0.10/5.0 = 0.02\,\text{m/s}\).
\(\eta = \dfrac{2 \times (1.5 \times 10^{-3})^2 \times 6000 \times 10}{9 \times 0.02} = \dfrac{0.27}{0.18} = 1.5\,\text{Pa s}\), close to glycerine.
Check: the ball reaches its terminal velocity with a time constant \(2r^2\rho/9\eta = 2.5\,\text{ms}\), long before mark A.
The tube must be wide compared with the ball, or the walls add their own drag. The ball must be small and slow, so that the flow round it stays streamline. And the first mark must be far enough down for the ball to have reached its terminal velocity. The liquid's temperature must be noted, because its viscosity depends strongly on it.
At Chicago, Robert Millikan sprayed tiny oil drops between two metal plates and watched one through a microscope. With the plates uncharged, the drop fell at its terminal velocity, about a tenth of a millimetre per second. From that speed, Stokes' law and the viscosity of air, he found the drop's radius and so its weight. Switching on the plates, he balanced that weight with an electric force. Every drop carried a whole number of one small charge, the charge of the electron.
A skydiver falling belly-down reaches a terminal velocity of about \(55\,\text{m/s}\), about \(200\,\text{km/h}\). The drag here grows as \(v^2\), not as in Stokes' law. But the idea is the same: the speed grows until the drag equals the weight. An open parachute presents a huge area to the air. The drag then balances the weight at only about \(5\,\text{m/s}\), the speed of a jump from a wall a little over a metre high.
- Using Stokes' law for large raindrops.It holds only for tiny drops such as drizzle. Large drops meet a drag that grows as \(v^2\), and fall at a few metres per second.
- Thinking the drag on a rising bubble points up.Drag points against the velocity. On a rising bubble it points down.
- Taking eight merged drops to fall eight times as fast.The radius doubles, and \(v_t\) grows as \(r^2\), so the speed grows four times.