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AT Physics II. Air Resistance The motion of falling objects is usually described with constant acceleration. This is only approximately true: air resistance, a friction which increases with increasing speed, acts agains gravity, so the speed of falling objects tends toward a limit (called terminal velocity; terminal speed would be better.) The physics of air resistance applies to motion through any sort of fluid: gas or liquid. To a reasonable approximation, fluid resistance tends to depend on either the first power of the speed (a linear resistance) or the second power (a quadratic resistance). At small speeds, fluid resistance is linear; Fresist = −bv (low speeds) where b has the units of N/(m/s) = kg/s. The force −bv is sometimes called the viscous drag. It arises mainly from the cohesion of the layers of the fluid. For large speeds, fluid resistance is better described by a quadratic dependence on speed; Fresist = −cv 2 (higher speeds) where the units of c are N/(m/s)2 = kg/m. This term is usually called drag. It is related to the momentum transfer between the moving object and the fluid through which it travels. You may recall the arguments used to derive Boyle’s Law from the kinetic theory of gases, that the pressure was related to the momentum transfer between gas atoms or molecules and the walls of the surrounding container. In much the same way the drag coefficient c can be related to other quantities; c = 12 CD ρA where CD is a geometric factor depending on the shape of the object (equal to 21 for a sphere), ρ is the density of the fluid, and A the cross-sectional area of the object. The factor 12 is put in for convenience. In general, both terms (and others, proportional to higher powers of v) may be present; Fresist = −bv − cv 2 + · · · If v 1, drop the second term; ignore the first term if v 1. The speeds of everyday objects are large enough that the effects of fluid resistance are due entirely to quadratic drag. Skydivers are slowed appreciably by the quadratic term; viscosity does nothing. The same is true for Hollywood heroes evading gunfire by swimming underwater. Two meters of water will slow bullets to a few percent of their original speed. Sadly, the realistic quadratic drag needs more intricate math than linear, viscous drag. Students are typically tested on the unrealistic viscous drag. We’ll do both, but let’s start with the easier case of viscous drag. Viscous drag: linear dependence on speed Viscous drag typically appears in one of two situations: with a driving (constant) force, usually gravity, and on its own. First, on its own. Say an object has an initial velocity vo and encounters a resistive force of the form Fresist = −bv directed opposite to the motion. We assume all other forces balance out. For example we might have a car driving along and encountering air resistance (though in reality, the quadratic resistance would be the dominant term.) Alternatively, we might fire a bullet horizontally though a resistive medium like industrial strength jello: Then Fnet = Fresist ⇒ ma = −bv so that dv b = − v, v(0) = vo dt m This is an example of a separable differential equation. Rewrite as a= (V1) b dv = − dt v m and integrate both sides; Z dv =− v Z b dt m or ln v = − b t+C m where C is a constant. There is no reason to add a constant on both sides; if we had added, say, C1 on the left side and C2 on the right, we could combine them by subtracting C1 on both sides; and set C = C2 − C1 . Then exponentiating both sides gives eln v = v = e−(b/m)t+C = eC · e−(b/m)t Determine the value of eC by using the initial condition; v(0) = vo ⇒ vo = e C · e 0 = e C and consequently If dv b = − v, v(0) = vo then v(t) = vo e−(b/m)t dt m The graph is a standard exponential decay: Besides tough jello, the other way viscous drag appears is in free fall : The net force is now the difference between the gravitational force and the viscous drag: Fnet = mg − bv = ma (we are taking down to be positive.) The differential equation from Newton’s Law is a= b dv = g − v, dt m v(0) = vo (V2) Very often, a problem involving free fall begins with an object dropped from rest, i.e. with vo = 0. But it’s easy enough to allow other values of vo . Again the differential equation is separable; dv = dt g − (b/m)v Factor out (b/m) in the denominator, multiply both sides by −(b/m) and integrate: Z Z dv b =− dt v − (mg/b) m The right hand side is easy. For the left hand side, use a “u-substitution”; let u = v − (mg/b). Then du = dv, so Z b du = ln u = ln(v − (mg/b)) = − t + C u m We do exactly what we did before. Exponentiate both sides to obtain v − (mg/b) = eC · e−(b/m)t ⇒ v= mg + eC · e−(b/m)t b Use the initial condition to determine eC : v(0) = vo , or v(0) = mg + eC = v o b so that v= ⇒ e C = vo − mg b mg mg −(b/m)t + vo − e b b or rearranging, mg 1 − e−(b/m)t b Note that in the limit as t → ∞, the object tends toward the terminal velocity vT ; i mg h mg 1 − e−(b/m)t = vT = lim vo e−(b/m)t + t→∞ b b v = vo e−(b/m)t + You can also obtain vT easily from the original differential equation (V2) by setting a = 0: a=0=g− b vT m ⇒ vT = mg b There are three possible graphs, depending as whether vo is greater than, equal to, or less than vT : In every case, the velocity tends toward the limiting value of vT . Nearly always, though, the initial value of the velocity is taken to be zero, corresponding to the bottom-most curve above: If dv b mg = g − v, v(0) = 0 then v(t) = 1 − e−(b/m)t dt m b Distance traveled with viscous resistance By definition, v = dx/dt. It’s easy enough to write down an expression for the distance ∆x = x − xo traveled as a function of time in terms of an integral; Z t ∆x = v dt 0 For the “tough jello”, (V1), Z t ∆x = vo e−(b/m)t dt = 0 mvo 1 − e−(b/m)t b The object keeps moving till v = 0, which happens only in the limit t → ∞. Then mv mvo o ∆x = lim 1 − e−(b/m)t = t→∞ b b There is a second, elegant way to obtain this result. We know a = −(b/m)v. Write, using the Chain Rule, a= dv dv dx dv b = · =v =− v dt dx dt dx m Divide both sides by v to obtain the simple b dv =− dx m which is a constant. That means the graph of v vs x is a straight line; b v=− x + vo m In particular, when v = 0, we obtain ∆x = mvo /b, exactly as before. For the case of free fall, (V2), Z t ∆x = 0 mg m −(b/m)t mg 1 − e−(b/m)t = t+ e −1 b b b It’s clear that this increases without limit; as t → ∞, this becomes (mg/b)t, which is just vT t. No surprise there. It’s worth looking at the special case when b 1; then e−(b/m)t = 1 − (b/m)t + 12 (b/m)2 t2 + · · · and consequently . m2 g mg t + 2 1 − (b/m)t + 21 (b/m)2 t2 + · · · − 1 = 21 gt2 b b which is exactly what you’d expect, in agreement with Galileo. This is about the limit of what you might see on an AP examination, but here’s the quadratic resistance for completeness. ∆x = Drag: Quadratic dependence on speed Again, say that either an object is passing horizontally through a resistive medium or else is freely falling. Take the “tough jello” case first: Fresist = −cv 2 = Fnet = ma so a=− c 2 dv v = m dt and v(0) = vo Again this differential equation is separable; Z dv = v2 Z − c dt m and both sides are easily integrated; 1 c =− t+C v m Use the initial condition to determine C; when t = 0, v = vo ; − − 1 =C vo (D1) so 1 1 c − = t v vo m which can be manipulated into v= vo 1 + (cvo /m)t Here’s the graph: The bullet does stop in the limit as t → ∞, but in that limit, its acceleration is proportional to 1/t2 : a= cvo vo m dv =− · →− 2 dt m (1 + (cvo /m)t)2 ct which becomes very small very rapidly. For free fall, the math is quite a bit more complex. Then Fnet = ma = −cv 2 + mg or dv c 2 =g− v dt m This equation is separable, but the left hand side looks tough to integrate: Z Z dv c c = − dt = − t + B v 2 − (mg/c) m m (D2) where B is a constant. For ease of manipulation let α2 = mg/c, and let I be the integral of the left hand side; Z dv I= 2 v − α2 Were the sign of α2 positive, I would be the inverse tangent. This is a known integral, but probably unfamiliar: Z dx 1 −1 x = tanh x2 − a2 a a where tanh−1 x is the inverse hyperbolic tangent of x. The hyperbolic trigonometric functions are not generally known, even among calculus students. It turns out that all of the inverse hyperbolic trig functions are expressible in terms of natural logs. And that’s how we’ll do this integral: by factoring, and using partial fractions: 1 1 1 1 1 = = − v 2 − α2 (v − α)(v + α) 2α v − α v + α Then Z I= so that dv 1 = v 2 − α2 2α Z 1 ln 2α dv 1 − v − α 2α v−α v+α Z =− dv 1 = ln v+α 2α c t+B m v−α v+α Multiply both sides by 2α and exponentiate both sides to get v−α = e−2α (c/m)t · e2αB v+α If, as is usually the case, v(0) = 0, then e2αB = −1, and v−α = −e−2α (c/m)t v+α which can be solved for v; v=α Resubstitute α = p 1 − e−2α (c/m)t 1 + e−2α (c/m)t =α eα (c/m)t − e−α (c/m)t eα (c/m)t + e−α (c/m)t = α tanh αc t m mg/c and find r v= mg tanh c r gc t m Here’s the graph: As x → ∞, tanh x → 1, so terminal velocity in this case is r r r mg gc mg tanh t = vT = lim t→∞ c m c which is also obtained easily from the original equation of motion when a = 0: r mg c 2 a=0=g− vT ⇒ vT = m c Distance traveled with quadratic resistance For free fall with quadratic resistance (D2), it’s clear that the distance traveled will again be without limit. Here it is formally (cosh x is the hyperbolic cosine, and cosh(0) = 1): r r r r Z tr mg gc mg m gc tanh t = · ln cosh t − ln 1) ∆x = c m c gc m 0 r m gc = ln cosh t c m By definition, cosh x = ex + e−x 2 so that lim cosh x = 12 ex x→∞ and consequently ∆x → m r gc c m r t= mg t = vT t c again as would be expected from Galileo (and much as we found with linear drag). This increases without limit as t → ∞. For the quadratic resistant “tough jello” (D1), we have Z ∆x = 0 t m cvo vo ln 1 + = t 1 + (cvo /m)t c m This also increases without limit as t → ∞, because limt→∞ ln t = ∞. With the linear viscosity, the object stopped, but with quadratic viscosity, it does not (or at least, the distance increases without limit). What’s going on? The elegant approach provides more insight. As before, write a= dv dv dx dv c = · =v = − v2 dt dx dt dx m and dividing both sides by v leads to dv c =− v dx m This is a separable differential equation, with v(0) = vo and x(0) = 0; Z Z dv c dx =− v m We’ve seen this before! It leads to v = vo e−(c/m)x It’s clear that v = 0 only in the limit x → ∞. So ∆x becomes unbounded as v approaches zero. Evidently if the acceleration a depends on a power of v greater than or equal to 2, then a → 0 as t → ∞ at least as fast as t−2 , and v decreases too slowly to keep the stopping distance ∆x finite. Example. Water as a Bulletproof Vest. A person pursued by enemies with guns attempts to avoid injury by diving into some body of water. Assuming drag only, at what depth of water will the speed of bullets fired vertically into the water be reduced by 99%? Use the following: ρwater = 1 × 103 kg/m3 , Abullet = πr2 , . rbullet = 0.5 cm, vo = 800 m/s, and m = 10 grams. Assume CD = 0.5, and ignore gravity. Solution. From the earlier empirical formula, . c = 12 CD ρA = 1 2 . 3 × 0.5 × (1000 kg/m )(π × (0.5 × 10−2 m)2 ) = 0.02 kg/m because 1 cm = 1/100 m. Also, 10 grams = 0.01 kg, and assume CD = 0.5. Then (c/m) = 2 m−1 , and using the formula from before, v(x) = 800e−2x If the velocity is to be reduced by 99%, it means that the speed will equal 1% of its original value, namely 8 m/s. (At this speed, the bullets would cause very little damage.) That is, 8 = 800e−2x or what is the same thing, 1 100 = e−2x ⇒ x= 1 ln 100 = ln 10 m = 2.30 m 2 m−1 about what the television show Mythbusters found some years back. More about viscosity When adjacent layers of a fluid move at different speeds, the resulting behavior is called shear. Make a thin sandwich of a fluid between two flat plates of area A. The bottom plate is fixed, the top plate is free to move laterally (parallel to the plate.) Exert a lateral force on the top plate. The fluid layer nearest the top acquires a speed vtop ; deeper layers acquire lesser speed. Let v(y) describe the fluid’s velocity as a function of depth y. The viscosity η (the Greek letter eta, ē) is defined from Newton’s Law of Viscosity in terms of the shear stress τ = F/A, which has the units of pressure, or Pascals (N/m2 ); τ = F ∂v =η A ∂y It might be clearer if we used “viscosity” to refer to the linear resistance −bv itself, and η the coefficient of viscosity, in analogy with dry friction, but this is not the language traditionally used. The units of viscosity are poise, after the French physicist J. L. M. Poiseuille (1797-1869), who studied blood flow in narrow tubes; 1 poise (P) = 1 g/cm-s = 0.1 pascal-second (Pa-s) The viscosity of water at 20◦ is about 0.01 poise, or 10−3 Pa-s; air at STP has a viscosity of about 1.8 × 10−5 Pa-s. In 1851, the Irish mathematical physicist G. G. Stokes (1819-1903) derived the formula for b for a sphere moving slowly in a fluid: b = 6πrη where r is the sphere’s radius. In the same article Stokes introduced a dimensionless ratio which others made much use of later, notably the English engineer Osborne Reynolds (1842-1912). This ratio was named by the German physicist Arnold Sommerfeld (in a 1908 lecture) the Reynolds number. It is usually denoted Re; Re = drag ρvL ∼ η viscous drag where L is a characteristic length for the object moving through a fluid (say the radius or the diameter of a sphere), v its speed, ρ the density of the liquid and η its viscosity. Generally, high Reynolds number (anything much bigger than 1) means that viscosity is negligible; low Reynolds number (less than 1) means viscosity dominates over drag. If we say L ∼ 1 m and v ∼ 1 m/s for a person walking, and use ρ = 1.2 kg/m3 for air at STP, we get Re = 6.7 × 104 . The viscosity of air is not important for everyday objects. The term viscosity derives from the Latin viscum, (mistletoe), used to make a sticky goo called birdlime for catching birds on twigs. The ancient Romans used to eat small songbirds; this custom is still followed in some parts of Italy and France, where it is now illegal. Arnold Sommerfeld (1868-1951) may have been the most spectacularly successful supervisor of physics students in history; four of his Ph.D. students won the Nobel Prize. (No one else is even close.) Among these were Werner Heisenberg, Wolfgang Pauli and Hans Bethe (who gave Richard Feynman his first university job at Cornell.) Nominated a record 81 times for the Nobel, he did not win it himself. Sommerfeld wrote a highly influential series of textbooks in theoretical physics which have been translated into many languages. v.2.0.1 26 November 2011