20Lecture 3. Tuesday 27 Jan 2015Case (e) continued, velocity dependent forces rF =rF(rv )A complication: Linear drag + gravity. F(v) = -mg – cv ( displacement and velocity upwards isdefined as positive. ) Thus (still defining κ= c/m which has units of inverse time):m dv / dt = −mg − cv sodt = −mdv / (mg + cv) sodttot∫=vov∫mF(v)dv = −vov∫mmg + cvdv so ⇒t − to= −dvg +κvvov∫= −1κlng +κvg +κvo(where we have defined κ = c/m which has units of inverse time)We take to = 0 and vo to be its velocity at to and solve for v(t):v(t) = voexp(−κt) −gκ[1 − exp(−κt)]This describes how the velocity starts at time zero at vo and then is affected by gravity anddiminished by drag. It is worth checking to confirm it agrees with the expression foundpreviously when g = 0.The expression may be re-arranged as follows:v(t) = −gκ+ exp(−κt)[vo+ g /κ]in which form you can see that the behavioris just a constant plus an exponentiallydiminishing term. If plotted, v(t) must looklike one of these two curves: The solid curvecorresponds to a particle that is originallymoving up (vo > 0 ) and is brought to rest andthen moves downwards, asymptoting at theterminal velocity –g/κ. The dashed curvecorresponds to a particle that starts movingdown at a speed greater than terminalvelocity (v < -g/κ) but is then is slowed bydrag to terminal velocity.21{Sanity check: Does g/κ have units of velocity? YES. Does this expression have the right valueat t= 0? YES. How does this behave as t→ ∞ ? We see that v → v∞ = -g/κ. Thus the speedapproaches some limiting terminal downward value that scales with the strength of gravity andinversely with the resistance. It does this regardless of initial condition. This makes sense. Thespeed v=-g/κ =mg/c corresponds to a drag force cv = mg that neutralizes the weight. }We can integrate again (with respect to t from 0 to t) to get the height z(t), should we want thatz − zo= v dt0t∫= [voexp(−κt)]dt0t∫−0t∫gκ[1− exp(−κt)]dt= −gtκ+g +κvoκ2[1− exp(−κt)]In the special case vo = 0 ( i.e., just dropping the object from a height zo) we findz = zo−gtκ+gκ2[1 − exp(−κt)]You are invited to check to make sure that, at short times, this reduces to the familiar z = zo – gt2/2. You can do this by expanding the exponential exp(-ε) ~ 1 - ε + ε2/2 +…z = zo−gtκ+gκ2−gκ2[1−κt +κ2t2/ 2 + ....]Nonlinear Drag Linear drag F= -cv is realistic only for the low speeds and small objects and thick fluids ( thinkof a marble in honey ) . Drag of an airplane or car in air is more commonly modeled asproportional to the square of the speed. F(v) = - C v2 sgn(v). (in Discussion you examined thecase F ∝ -v3) The coefficient C is equal to the cross sectional area times the mass density of airtimes a dimensionless factor of order unity that depends on shape. Making C as small aspossible is important for fuel economy. Let us revisit the falling object. Let us take v to benegative, i.e downwards throughout the interval of interest, so that F(v) = -mg + Cv2. (If we don'tdo that, the sgn function may cause complications.) Then, including gravity and this drag force(and let to = 0)….mF(v)vov∫dv = −mmg − Cv2∫dv = ∫ dt = to= t ⇒t − to= −dvg −σv2∫= −1σdvg /σ− v2vov∫( We have defined σ = C/m; you are invited to check to confirm that it has units of inverse length1/[L] ) The integral can be done by partial fractions, or more easily, by looking it up in a table.22σt = −dvg /σ− v2vov∫= −12 g /σlng /σ+ vg /σ− vvovLet's make it simpler by choosing vo = 0. I.e. the object is dropped from rest. Then the ln vanishesat the lower limit and,σt = −12 g /σlng /σ+ vg /σ− vNow solve for v….v =gσexp{−2 gσt} −1exp{−2 gσt} +1As t→∞, we see that the object achieves a terminal velocity. v∞ = -√(g/σ) , at which weight equalsdrag force mg = Cv2. We can also check if our formula satisfies v = 0 at t= 0. This is readilyconfirmed. Also we can check its short time behavior ( where v ought be –gt ) At small time, theexponentials areexp{−2 gσt} ≈ 1 − 2 gσtso the numerator becomes –2√gσ t and the denominator becomes ~ 2. It checks.If you wanted x(t), you could integrate the above ∫ v(t) dt. You can find it in tables of integrals.The answer is (if x(0) = 0)x(t) =1σln(cosh( gσt))Case f) There are a couple of other ways that F could depend on v and x and t such that we mightbe able to solve the ODE analytically. (In general it is not possible.)f1.) One such is F = f(v) g(t); (cf HW1B.1) thenm dv/dt = f(v) g(t)som ∫ dv/ f(v) = ∫ g(t) dtf2.) Another is F = h(x) f(v). Thenm dv/dt = h(x) f(v)23Multiply by v = dx/dt:m v dv/dt = h(x) (dx/dt) f(v)som ∫ v dv / f(v) = ∫ h(x) dxOther cases must be handled numerically.Some 2 and 3-d cases: ( still Cartesian coordinates)Linear drag in 2-dHaving solved the linear drag problem both with and without the gravitational force, wecan now put the two together into a two-dimensional problem.Consider an object falling under the influence of gravity and subjected to linear drag forceF(D)=-cv. The object is projected horizontally at time zero at some speed vo . How does itmove?We note that the total force is F = – mgk - cvSo its x-component is Fx = -cvx and its z-component is Fz = -mg-cvzThe components of F = ma are:m d2z/dt2 = -mg - cvzm d2x/dt2 = - cvxWe have ignored the motion in the y-direction (into or out of the paper) because we can orient ourcoordinate system so that the initial horizontal velocity is purely in the x-direction. In that case,the forces are all in the x and z directions, and y is constant. The problem is therefore 2-d.These two differential equations are uncoupled. By which we mean that we can solve either ofthem without thinking about the other. The solution for x is (as seen above, with κ = c/m)24x(t ) =vxoκ[1 − exp(−κt )] + xoThe solution for z is, as was seen above,z = zo−gtκ+gκ2[1 − exp(−κt)]Again one should check these expressions for dimensional consistency and for reasonableness ofbehavior. It may be worthwhile to sketch x(t) and z(t) , and also the parametric plot x(z) that givesthe object's trajectory.-------------------------An ODE with a very different character occurs for gravity plus nonlinear drag in 2-d. Consider adrag force on a falling