Homework 5 Physics 14 Methods in Theoretical Physics Due: Wednesday February 16, 4 Reading Assignment: Thornton and Marion, Ch..4. 1. è1d motion, Newtonian resistance.è In class, we considered the motion of an object in 1D subjected to gravity and Newtonian air resistance F r = ækv, where the negative sign corresponds to the case in which the object is rising and the positive sign when the object is falling. We found vètè for the rising case. Write down and solve Newton's second law forvètè in case in which the object is falling. Try to put the solution explicitly in the following form: t, c vètè =,v t tanh èdownward motionè è1è where the terminal speed v t = p mg=k, the characteristic time = m=kv t, and c is a constant of integration. Newton's second law for this falling object is m dt =, mg! m kv kv, mg = t èè The terminial speed occurs when =dt =, so it's easy to see that v t = p mg=k. We want tointegrate equation èè and solve forvètè. To do so the following integral is helpful: dx ax + bx + c = p, ax + b tanh,1 p,æ,æ è3è which holds when æ é. Here x = v; a = k; b = ;c =,mg; æ = 4ac, b = p,4kmg;,æ = p kmg =k p mg=k =kv t. Unfortunately, in class when I gave you this integral I forgot to include the negative sign in front of the on the right hand side. I hope you discovered this for yourself, but if you didn't you won't be penalized. It's a good habit not to believe formulas you get from other people, but to check them out for yourself. Using these substitutions, we can now integrate equation èè: t = m kv, mg =,m kv tanh,1 + c è4è kv t kv t v,kvt = tanh v t m èt, cè = tanh, 1 1 èt, cè =, tanh èt, cè è5è from which equation è1è follows directly. Note that the last step in equation è5è was to use the fact that tanh is an odd function, so that tanhè,xè =, tanhèxè.. è1d motion, Newtonian resistance.è This problem is similar to è1, but we want to ænd the vèxè rather than vètè. Assume the projectile is æred straight up èor downè with an initial speed v. Again, let's assume that the drag on the projectile varies as the square of its speed, but let's write the Newtonian resistive force as F r = æmkv, where k is a positive constant. Note that here F r has a factor of m in it explicitly, unlike in è1. Show that the speed varies with height according to the equations: v uèxè = Ae,kx, v t èupward motionè; è6è v dèxè = v t, Bekx èdownward motionè; è7è 1
where A and B are constants of integration, v t = p g=k is the terminal velocity, g is the acceleration of gravity, and m is the mass of the projectile. Position x is measured positive upward, and the gravitational force is assumed to be constant. èhint: You need to solve this with an integral in x rather than t, soyou should let a = v=dx. This will recast the problem for you.è This problem is formally mis-stated. It should read that the projectile is æred straight up and not include the possibility that it could be æred straight down. Again, you will not be penalized for this mis-statement of the problem. This is really a problem in the application of initial conditions. In this problem the force applied depends on the direction of motion. Choosing a coordinate system such that +x is up, gravity points in the downward direction. Thus, air drag and gravity point in the same direction when the bullet is moving upward, but point in opposite directions when the bullet is coming back down to the ground: F u = ma u =,mg, mkv! a u = v dx =,g, kv ; F d = ma d =,mg + mkv! a d = v dx =,g + kv : In the upward direction, the initial speed is v and the ænal speed when it reaches its highest point is. The initial location is x =. These quantities need to be substituted into the limits on the integrals. From equation è8è: x vu v dx =, : è1è v g + kv x =, 1 k lnèg + èj vu kv v =, 1 h i lnèg + kvuè, lnèg + kvè è11è k =, 1 k ln g + kv u : è1è g + kv e,kx = g + kv u : è13è g + kv vuèxè = 1 g + kv e,kx, g k k = Ae,kx, g k = Ae,kx, v t ; è14è where A =èg + kv è=k = v t + v. In the downward direction, the initial velocity is at the top of the projectile's trajectory which is =. It's initial height isx, which is the top of its trajectory. From equation è9è: v x where B = ge,kx =k = v t e,kx. vd v dx =, : è15è x g, kv x, x = 1 k lnèg, èj v kv d = 1 h i lnèg, kv dè, lnègè è16è k = 1 k ln g, kv d : è17è g e kèx,x è = g, kv d : è18è g v d = g g k, k e,kx e kx = g k, Bekx = v t, Be kx ; è19è Again, notice the diæerence in the limits on the integrals èinitial and ænal conditionsè in the up and down directions. In the up direction, the bullet starts at ground level èx =èwith velocity v, but the bullet begins to drop at rest èv =èfromanelevation x. è8è è9è
3. èd motion, no resistance.è Thornton and Marion, è.8. èhint: This problem is easy to do if you ænd the shape of the trajectory or the projectile, which is a parabola. We did this in class.è y v h P Q α x Figure 1: Notation and geometry for Thornton and Marion è.8. Inspection of Figure 1 reveals that the x and y components of the initial velocity vector v are v cos æ and v sin æ, respectively. The projectile suæers a constant acceleration,g in the y-direction and is unaccelerated in the x-direction. Newton's second law can, therefore, be written in component form as follows: míx = èè míy =,mg: è1è Solving these equations for x and y by integration where we use the initial conditions on position x =;y =anelocity discussed above, we ænd: xètè = v t cos æ èè yètè = v t sin æ, 1 gt : è3è We areasked to set up the problem for maximum distance traveled. In the absence of air resistance, that will occur when æ =45 æ, as we willshowbelow. Substituting æ =45 æ into equations èè and è3è we get: xètè = v p t yètè = v p t, 1 gt è4è è5è Eliminating t from equations è4è and è5è by solving è4è for t and substituting into è5è we ænd: x, v g x + v g y =: è6è This is the trajectory equation that we wanted; yèxè. It's a parabola. 3
We want the distance between points P and Q at height y = h. Substituting h for y in equation è6è we get a quadratic equation for x, yielding two solutions for the x locations of points P and Q: x P = v g, v g x Q = v g + v g The horizontal distance between points P and Q is what we seek: d = x Q, x P = v q v, 4gh è7è q v, 4gh: è8è q v, 4gh: è9è Now, we tidy up the loose end and ænd the angle æ for maximum distance traveled. First we need the horizontal distance traveled as a function of the initial angle æ, which we ænd by substituting y = into equation è3è and get t = èv =gè sin æ and then substitute this into equation èè to ænd xèæè =èv =gè cos æ sin æ =gè sin æ. This clearly has its maximum when sin æ = 1 which =èv occurs when æ =45 æ. If you wanted to be fancier, you could take the derivative ofxèæè with respect to æ and show that it is zero when æ =45 æ. 4. èd motion, no resistance.è Thornton and Marion, è.17. Choose a coordinate system such the ball starts at x =and y =:7m. Let the softball initially move inthe+x and +y directions. Breaking the motion of the softball into x and y components we get èfrom Galileo's Kinematical Equations for Falling Bodiesè: x = x + v xt = v t cos ; è3è y = y + v yt, 1 gt = y + v t sin ; è31è where the initial angle with respect to the horizontal is =35 æ. The ball will cross the fence when x = R = 6m which, by equation è3è, will occur at a time = R=èv cos è: è3è The ball must be at least h = m high at this time and place èi.e., y h at t = ;x = Rè. Thus, h, y = v sin, 1 g! v = gr cosër sin, èh, y ècosë ; è33è where the latter equation resulted by solving equation è33è for v after substituting for from equation è3è. Substituting the values of g; R; h; y ; and gives v = 5.4 mès. 5. èd motion, Stokes resistance.è Assume that air resistance is linear in v: F r =,mkv. Consider D motion èx; yè subject to this air resistance. èaè Write down Newton's second law and show that the equation for the trajectory of an object shot from the horizontal plane at point èx = y =èwith initial velocity v = v xæ^x + v yæ^y is given by the equations: xètè = v xæ 1, e,t= è34è yètè = èv yæ + v t è 1, e,t=, v t t; è35è where =1=k and v t = g=k = g is the terminal velocity. 4
From Newton's second law dt = g, kv!,t = kv + g = k æ = 1 lnèv + v tè æ v y v y = 1 v + v t v + v t = 1 è! ln v y + v t v y + v t è36è è37è Multiplying both sides by, exponentiating, and solving for v y,we see that: v y ètè =,v t +èv y + v t è e,t= ; è38è yètè =,v t t + èv y + v t è 1, e,t= ; è39è where the second equation derives by time-integrating the ærst. You can get the equation for xètè from the equation for yètè simply by setting gravity to zero. g appears in the equation through v t = g=k, thus v t = in the x equation. Then equation è35è follows immediately. èbè Show that the trajectory equation èyèxèè is: yèxè = v yæ + v t x + v t ln 1, x v xæ v x æ : è4è Solve equation è34è for t and substitute into equation è35è and you get the result with little eæort. ècè Assume for the moment that there is no air resistance. Go back tothe original equations of motion, set air resistance to zero, and show that the trajectory can be given by: vyæ yèxè = x, 1 x v xæ g è41è v xæ Without air resistance, xètè =v xæ t and yètè =v yæ t, gt =. Solving the ærst for t and substituting into the second gives the desired result. èdè The correct trajectory, including a linear drag force, is given by equation è4è. Show that this reduces to equation è41è when air resistance is switched oæ è!1;v t = g!1è. èhint: Use the Taylor Series expansion for lnè1, æèè. As air resistance is switched oæ,!1, and the second term inside the log term of equation è4è becomes small. Thus we can use the Taylor series for the log, ln 1, x =, x, 1 x,æææ; in equation è4è. For suæciently large, we can neglect all remaining terms in this series and equation è4è becomes è yèxè v yo + v t x x, v t + 1! x v è4è xo è = v yo + g x x, ègè + 1! x v è43è xo vyæ = x, 1 x v xæ g : è44è v xæ èeè èextra Credit, 5 ptsè Set k = s,1, jv j = p mès, and the initial angle is 45 æ. Using a computer, plot the correct trajectories using equations è4è and è41è and compare. 5
trajectories.nb 1 y1 x_ x 5 x^ y x_ 6x.5 Log 1 x graph1 Plot y1 x, x,,., PlotRange,.5, PlotLabel "No Air Resistance" graph Plot y x, x,.1,., PlotRange,.5, PlotLabel "Air Resistance".5.4.3..1 No Air Resistance.5.5.1.15. Air Resistance.4.3..1.5.1.15. Show graph1, graph, PlotLabel "Comparison".5 Comparison.4.3..1.5.1.15. Figure : 6