61 6161 61532010-05-13 00:43:32 / rev b667c9e4c1f1+3.3 Drag using conservation of energyConservation of energy helps analyze drag – one of the most difficultsubjects in classical physics. To make drag concrete, try the followinghome experiment.3.3.1 Home experiment using falling conesPhotocopy this page at 200% enlargement, cut out the templates, thentape the their edges together to make two cones:r = 1 incutout = 90◦r = 2 incutout = 90◦When you drop the small cone and the big cone, which one falls faster?In particular, what is the ratio of their fall speeds vbig/vsmall? The largecone, having a large area, feels more drag than the small cone does. Onthe other hand, the large cone has a higher driving force (its weight)than the small cone has. To decide whether the extra weight or the extradrag wins requires finding how drag depends on the parameters of thesituation.However, finding the drag force is a very complicated calculation. Thefull calculation requires solving the Navier–Stokes equations:(v·O)v +∂v∂t= −1ρOp + νO2v.And the difficulty does not end with this set of second-order, coupled,nonlinear partial-differential equations. The full description of the situa-tion includes a fourth equation, the continuity equation:62 6262 62542010-05-13 00:43:32 / rev b667c9e4c1f1+O·v = 0.One imposes boundary conditions, which include the motion of the ob-ject and the requirement that no fluid enters the object – and solves forthe pressure p and the velocity gradient at the surface of the object. Inte-grating the pressure force and the shear force gives the drag force.In short, solving the equations analytically is difficult. I could spendhundreds of pages describing the mathematics to solve them. Even then,solutions are known only in a few circumstances, for example a sphereor a cylinder moving slowly in a viscous fluid or a sphere moving at anyspeed in an zero-viscosity fluid. But an inviscid fluid – what Feynmancalls ‘dry water’ [7, Chapter II-40] – is particularly irrelevant to real lifesince viscosity is the reason for drag, so an inviscid solution predictszero drag! Conservation of energy, supplemented with skillful lying, is asimple and quick alternative.The analysis imagines an object of cross-sectional area A moving througha fluid at speed v for a distance d:Avolume ∼ Addistance ∼ dThe drag force is the energy consumed per distance. The energy is con-sumed by imparting kinetic energy to the fluid, which viscosity eventu-ally removes from the fluid. The kinetic energy is mass times velocitysquared. The mass disturbed is ρAd, where ρ is the fluid density (here,the air density). The velocity imparted to the fluid is roughly the velocityof the disturbance, which is v. So the kinetic energy imparted to the fluidis ρAv2d, making the drag forceF ∼ ρAv2.The analysis has a divide-and-conquer tree:63 6363 63552010-05-13 00:43:32 / rev b667c9e4c1f1+force ∼ E/dρAv2energy imparted, ∼ mv2ρAv2dmass disturbedρAddensityρvolumeAdvelocity imparted∼ vdistance dThe result that Fdrag∼ ρv2A is enough to predict the result of the coneexperiment. The cones reach terminal velocity quickly (see Problem 7.10),so the relevant quantity in finding the fall time is the terminal velocity.From the drag-force formula, the terminal velocity isv ∼sFdragρA.The cross-sectional areas are easy to measure with a ruler, and the ratiobetween the small- and large-cone terminal velocities is even easier. Theexperiment is set up to make the drag force easy to measure: Since thecones fall at their respective terminal velocities, the drag force equals theweight. Sov ∼sWρA.Each cone’s weight is proportional to its cross-sectional area, because theyare geometrically similar and made out of the same piece of paper. Sothe terminal velocity v is independent of the area A: so the small andlarge cones should fall at the same speed.To test this prediction, I stood on a table and dropped the two cones. Thefall lasted about two seconds, and they landed within 0.1 s of one another.So, the approximate conservation-of-energy analysis gains in plausibility(all the inaccuracies are hidden within the changing drag coefficient).64 6464 64562010-05-13 00:43:32 / rev b667c9e4c1f1+3.4 CyclingThis section discusses cycling as an example of how drag affects the per-formance of people as well as fleas. Those results will be used in theanalysis of swimming, the example of the next section.What is the world-record cycling speed? Before looking it up, predict itusing armchair proportional reasoning. The first task is to define the kindof world record. Let’s say that the cycling is on a level ground using aregular bicycle, although faster speeds are possible using special bicyclesor going downhill.To estimate the speed, make a model of where the energy goes. It goesinto rolling resistance, into friction in the chain and gears, and into drag.At low speeds, the rolling resistance and chain friction are probably im-portant. But the importance of drag rises rapidly with speed, so at high-enough speeds, drag is the dominant consumer of energy.For simplicity, assume that drag is the only consumer of energy. Themaximum speed happens when the power supplied by the rider equalsthe power consumed by drag. The problem therefore divides into twoestimates: the power consumed by drag and the power that an athletecan supply.The drag power Pdragis related to the drag force:Pdrag= Fdragv ∼ ρv3A.It indeed rises rapidly with velocity, supporting the initial assumptionthat drag is the important effect at world-record speeds.Setting Pdrag= Pathletegivesvmax∼ PathleteρA!1/3To estimate how much power an athlete can supply, I ran up one flightof stairs leading from the MIT Infinite Corridor. The Infinite Corridor,being an old building, has spacious high ceilings, so the vertical climb isperhaps h ∼ 4m (a typical house is 3 m per storey). Leaping up the stairsas fast as I could, I needed t ∼ 5 s for the climb. My mass is 60 kg, so mypower output was65 6565 65572010-05-13 00:43:32 / rev b667c9e4c1f1+Pauthor∼potential energy suppliedtime to deliver it=mght∼60 kg × 10 m s−2× 4 m5 s∼ 500 W.Pathleteshould be higher than this peak power since most authors are notOlympic athletes. Fortunately I’d like to predict the endurance record.An Olympic athlete’s long-term power might well be comparable to mypeak power. So I use Pathlete= 500 W.The remaining item is the cyclist’s cross-sectional area A.