A farmer sprays his cotton crop with a hand pump on his back. When he screws the nozzle down to a finer mist, he has to pump harder for the same amount of liquid. The liquid is the same and so is its volume. Something else has grown: the surface of all the tiny drops put together.
Four shapes that physics uses
Nearly every body in a physics problem is drawn as one of four shapes, or as a few of them joined together. Their areas and volumes are worth knowing by heart.
- A circle of radius \(r\) has the circumference \(2\pi r\) and the area \(\pi r^2\). It is the section of a wire, a pipe or a beam of light.
- A sphere of radius \(r\) has the surface \(4\pi r^2\) and the volume \(\tfrac43\pi r^3\). It is a drop, a bubble, a nucleus, a planet or a Gaussian surface round a charge.
- A cylinder of radius \(r\) and length \(L\) has the curved surface \(2\pi rL\), two ends of \(\pi r^2\) each, and the volume \(\pi r^2L\). It is a wire, a pipe or a Gaussian surface round a line of charge.
- A cone of base radius \(r\) and height \(h\) has the volume \(\tfrac13\pi r^2h\), one third of the cylinder that holds it.
Two of these results can be owned in a few lines. Cut a disc into many thin slices from its centre, like a pizza. Each slice is almost a triangle of height \(r\). Cut a ball into many thin cones from its centre in the same way. Each cone has height \(r\) and a tiny base on the surface.
\[ A = \tfrac12 b_1 r + \tfrac12 b_2 r + \dots \]
Why. each thin slice of the disc is almost a triangle with the height \(r\) and a small base \(b\) on the rim.
\[ A = \tfrac12 r\,(b_1 + b_2 + \dots) = \tfrac12 r \times 2\pi r = \pi r^2 \]
Why. the small bases together make the whole circumference, \(2\pi r\).
\[ V = \tfrac13 a_1 r + \tfrac13 a_2 r + \dots \]
Why. each thin cone of the ball has the height \(r\) and a small base \(a\) on the surface. A cone holds one third of its base times its height.
\[ V = \tfrac13 r\,(a_1 + a_2 + \dots) = \tfrac13 r \times 4\pi r^2 \]
Why. the small bases together make the whole surface, \(4\pi r^2\), which Archimedes proved.
\[ V = \tfrac43\pi r^3 \]
Read it. multiply out.
A ball of radius \(10\,\text{cm}\) is held completely under water. The buoyant force is the weight of the water it pushes aside, \(\rho V g\). Take \(\rho = 1000\,\text{kg/m}^3\) and \(g = 10\,\text{m/s}^2\). Find the buoyant force, in newtons.
Areas grow as the square, volumes as the cube
Double the radius of a drop. Its surface \(4\pi r^2\) becomes four times larger, and its volume \(\tfrac43\pi r^3\) eight times larger. The same is true of any shape at all. An area is a length times a length, and a volume is a length times a length times a length.
Multiply every length of a body by the same factor \(k\). Then every area of the body is multiplied by \(k^2\) and every volume by \(k^3\). The shape stays the same.
This rule settles every question of drops that merge or split. Let \(n\) equal drops of radius \(r\) merge into one drop of radius \(R\). The water is the same, so the volume is the same: \(\tfrac43\pi R^3 = n \times \tfrac43\pi r^3\). So \(R = n^{1/3}\,r\). Eight drops make one of twice the radius, and a thousand make one of ten times the radius.
The surface is not kept. Before the merger the total surface is \(n \times 4\pi r^2\). After it, the surface is \(4\pi R^2 = n^{2/3} \times 4\pi r^2\). So merging divides the total surface by \(n^{1/3}\). Splitting one drop into \(n\) multiplies the surface by \(n^{1/3}\). That is the farmer's harder pumping: a mist of a thousand times as many drops has ten times the surface.
Eight equal raindrops merge into one drop. What happens to the radius and to the total surface of the water?
The volume of a nucleus is proportional to its number of nucleons \(A\), so its radius is \(R = R_0A^{1/3}\), with \(R_0 = 1.2\,\text{fm}\). What is the radius of a nucleus with \(A = 64\), in fm?
Surface for each unit of volume
Divide the surface of a sphere by its volume. The result is its surface-to-volume ratio, \(4\pi r^2 / \tfrac43\pi r^3 = 3/r\). A cube of side \(a\) has \(6a^2/a^3 = 6/a\). For every shape the ratio falls as the body grows, because the surface grows as the square and the volume as the cube.
A small body is almost all surface. So whatever acts at a surface rules it. Surface tension pulls a tiny drop into a near-perfect sphere, while its weight, which acts on the volume, can hardly flatten it. A large drop on a table sags into a puddle. A hot small body cools quickly, because it loses heat through its surface but holds heat in its volume. So a tiny drop of hot tea cools in seconds, while the cup takes many minutes.
Sometimes a body catches something on one area and loses it from another. Sunlight comes to the Earth as a parallel beam. Only the rays within a distance \(R\) of the centre line strike the Earth. So the Earth catches the sunlight that crosses a disc of area \(\pi R^2\). This is its projected area. The Earth radiates heat from its whole surface, \(4\pi R^2\), four times as much.
Two steel balls, of radius \(1\,\text{cm}\) and \(2\,\text{cm}\), are taken from the same furnace and left in the same room. How does the rate at which the small ball's temperature falls compare with the large ball's, at first?
The temperature of the Earth, if it reflected no sunlight, is found from \(\pi R^2 S = 4\pi R^2 \sigma T^4\). Here \(S\) is the sunlight falling on each square metre that faces the Sun. Why are the two areas different?
Back to physics: drops that merge
Now the farmer's sprayer, run backwards. When many small drops merge into one, surface disappears. Each square metre of water surface stores an energy equal to the surface tension \(T\), so the lost surface sets energy free as heat. The chapter Surface Tension works this for a thousand droplets.
A thousand water droplets, each of radius \(0.1\,\text{mm}\), merge into one drop. Take \(T = 0.07\,\text{N/m}\). Find the radius of the new drop and the energy set free.
The volume is kept, so the new radius is \(R = 1000^{1/3} \times 0.1\,\text{mm} = 1\,\text{mm}\).
The surface before is \(1000 \times 4\pi(0.1)^2 = 40\pi\,\text{mm}^2\). After, it is \(4\pi(1)^2 = 4\pi\,\text{mm}^2\).
The surface lost is \(36\pi\,\text{mm}^2 = 113\,\text{mm}^2 = 1.13 \times 10^{-4}\,\text{m}^2\). The energy is \(0.07 \times 1.13 \times 10^{-4} \approx 7.9 \times 10^{-6}\,\text{J}\).
One drop of \(1\,\text{mm}\) forms, and \(7.9 \times 10^{-6}\,\text{J}\) is set free as heat, the answer of the physics chapter. The total surface has fallen ten times, from \(40\pi\) to \(4\pi\,\text{mm}^2\).
The farmer's question is answered too. Splitting drops is the same sum in reverse. A mist of a thousand times as many drops has ten times the surface. Every square metre of it costs the pump the work \(T\). The smallest drops are the roundest, because they are nearly all surface. They also cool the fastest, for the same reason.
Sixty-four water droplets, each of radius \(0.5\,\text{mm}\), merge into one drop. Take \(T = 0.07\,\text{N/m}\). How much energy is set free, in microjoules?
These formulas are for exact shapes. A drop sitting on a table is not a sphere, and a pipe with thick walls is not a thin cylinder. The shell rule \(V \approx 4\pi r^2t\) needs the thickness \(t\) much smaller than \(r\). For a shell of outer radius \(10\,\text{cm}\) and thickness \(1\,\text{cm}\) it gives \(1257\,\text{cm}^3\), but the true volume is \(\tfrac43\pi(10^3 - 9^3) = 1135\,\text{cm}^3\). Using the middle radius, \(9.5\,\text{cm}\), gives \(1134\,\text{cm}^3\), much closer.
Archimedes proved that a sphere has two thirds of the volume of the cylinder that just holds it. He also proved that its surface is four times the area of its largest circle, \(4\pi r^2\). He was so proud of these results that he asked for a sphere inside a cylinder to be carved on his tomb. More than a century later, in 75 BCE, the Roman writer Cicero found the tomb overgrown with bushes near Syracuse. He knew it by that carving.
Archimedes' Principle finds the buoyant force from the volume a body displaces, and asks why a hollow steel ship or a balloon can float. Surface Tension counts the surface of drops and bubbles, the work to blow a bubble and the energy of merging drops. Viscosity follows merged drops that fall faster, their terminal speed growing as \(n^{2/3}\). The Nucleus, Binding Energy and Nuclear Reactions finds the nuclear radius and a density that is the same for every nucleus. Radiation, Convection and Cooling finds the temperatures of the Sun and the Earth from a sphere and a disc. Centre of Mass and Rigid Bodies and Moment of Inertia find the mass of a cavity from its volume. Gravitational Force and Field finds the mass inside a sphere, and Electric Flux and Gauss's Law finds the charge inside spheres and cylinders.
- Using the surface \(4\pi r^2\) where the volume \(\tfrac43\pi r^3\) is needed.A mass, a buoyant force or a count of nucleons goes with the volume. A surface energy or a rate of cooling goes with the area.
- Keeping the total surface when drops merge.Only the volume is kept. Find the new radius from the volume, then find the new surface.
- Taking the radius \(n\) times larger when \(n\) drops merge.The volume is \(n\) times larger, so the radius is only \(n^{1/3}\) times larger.