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Physics 430: Lecture 4 Quadratic Air Resistance Dale E. Gary NJIT Physics Department Linear Air Resistance Recap-1 When a projectile moves through the air (or other medium—such as gas or liquid), it experiences a drag force, which depends on velocity and acts in the direction opposite the motion (i.e. it always acts to slow the projectile). ˆ, where the function Quite generally, we can write this force as f f (v ) v f(v) can in general be any function of velocity. At relatively slow speeds, it is often a good approximation to write f (v) bv cv 2 f lin f quad where flin and fquad stand for the linear and quadratic terms, respectively: f lin bv and f quad cv 2 For linear air resistance, the equation of motion is mr mg bv or in terms of velocity, it is a first-order differential equation mv mg bv, which has component equations: mvx bvx mv y mg bv y v y vx b b ; Equations of this form can be written: vx m v y vter m mg where vter is the terminal velocity. b September 10, 2009 Linear Air Resistance Recap-2 Such equations are said to be in separable form (terms involving v on one side, and no dependence on v on the other side). Solutions of this dv y particular form, e.g. dv x b b dt ; dt vx m v y vter m have exponential solutions: v x v xo e t / v y v yoet / vter (1 et / ) m / b which we then integrate to get x and y positions: t / x(t ) x 1 e t / y(t ) vtert v yo vter (1 e ) We can then combine these equations by eliminating t, to get a single equation for the trajectory: v yo vter x y x vter ln 1 vxo v xo Finally, we solved this for the range R, i.e. the value of x for which y = 0, valid for low air resistance: 4 v yo R Rvac 1 3 v ter Linear air resistance applies only to tiny projectiles or viscous fluids. September 10, 2009 2.4 Quadratic Air Resistance For more normal size projectiles (baseball, cannon ball), it is the quadratic drag force that applies. We are now going to follow exactly the same procedure, but starting with the quadratic form of the drag force: f quad cv 2 The equation of motion (in terms of v) then becomes: mv mg cv 2 vˆ with component equations: mvx cv 2 c vx2 v y2 vx mv y mg cv 2 mg c vx2 v y2 v y As we noted last time, these two equations are coupled, and are generally not solvable analytically (in terms of equations), although they can be solved numerically. However, we can solve these equations for special cases of either solely horizontal motion (vy = 0), or solely vertical motion (vx = 0), in which case the equations become mv cv 2 [horizonta l motion] mv mg cv 2 [vertical motion] Let’s look at these one at a time. September 10, 2009 Horizontal Motion with Quadratic Drag-1 As before, we write the equation in separable form (move the terms involving v to one side). For the horizontal equation, it is trivial: dv c dt v2 m This equation is called a non-linear differential equation because one of the derivatives (the zeroth one, in this case) has a non-linear dependence. Such equations are significantly harder to solve, in general. In this case, however, the separable form allows us to integrate both sides directly v dv c t vo v2 m 0 dt 1 1 vo ct to get or v(t ) , where I have introduced the 1 t / m vo v m characteristic time, , in terms of constants: . cvo To find the position, we again integrate the velocity equation to get v0 dt vo ln( 1 t / ) 0 1 t / x(t ) xo t [if xo 0] September 10, 2009 Horizontal Motion with Quadratic Drag-2 The final solutions for v(t) and x(t) are: vo m / cvo v(t ) 1 t / x(t ) vo ln( 1 t / ) [for quadratic drag] Graphs of these functions are: x vx vo t t They may look similar at first to the linear case, but now the velocity as t approaches zero much more slowly, like 1/t, so the position does not approach some limiting value x like in the linear case, but rather continues to increase forever. If this sounds impossible, you are right. What really happens is that as the speed drops, quadratic drag gets swamped by linear drag. September 10, 2009 Vertical Motion with Quadratic Drag We now consider motion solely in the vertical direction, governed by the equation of motion: mv mg cv 2 [vertical motion] Before we write the vertical equation in separated form, however, we notice as before that the gravity force mg is balanced by the drag force cvy2 at terminal velocity mg vter c after which v y 0 , i.e. the velocity becomes constant. In terms of vter, the separated form for the vertical equation is: dv g dt 2 1 v 2 / vter In this separated form, we can integrate both sides directly (assuming vo = 0). v t dv g 0 1 v2 / vter2 0 dt dx Looking at the inside front cover of the book we find 1 x 2 arctanh x which is what we have if we write x = v/vter. What the heck is arctanh? September 10, 2009 Hyperbolic Functions—Problem 2.33(a) Statement of the Problem: The hyperbolic functions cosh z and sinh z are defined as follows: e z e z e z e z cosh z and sinh z 2 2 for any z, real or complex. (a) Sketch the behavior of both functions over a suitable range of real values of z. cosh z e z 2 sinh z ez 2 1 1 2 ez 2 1 2 z 12 z z e 2 September 10, 2009 Hyperbolic Functions—Problem 2.33(b) Statement of the Problem, cont’d: (b) Show that cosh z = cos(iz). What is the corresponding relation for sinh z? Solution: To do this part, you have to know the relations: eix e ix cos x 2 and eix e ix sin x 2i Then the solution is very easy: ei (iz ) e i (iz ) e z e z cos iz cosh z 2 2 ei (iz ) e i (iz ) e z e z e z e z sin iz i i sinh z 2i 2i 2 So sinh z sin iz i sin iz i September 10, 2009 Hyperbolic Functions—Problem 2.33(c) Statement of the Problem, cont’d: (c) What are the derivatives of cosh z and sinh z? What about their integrals? Solution: The derivatives are: d cosh z 1 d z 2 (e e z ) 12 (e z e z ) sinh z dz dz d sinh z 1 d z 2 (e e z ) 12 (e z e z ) cosh z dz dz The integrals are equally straightforward: cosh z dz sinh z sinh z dz cosh z September 10, 2009 Hyperbolic Functions—Problem 2.33(d) Statement of the Problem, cont’d: (d) Show that cosh2 z – sinh2 z = 1. Solution: 2 e z ez 2 Since cosh z 2 14 (e 2 z 2e z e z e 2 z ) 14 (e 2 z 2 e 2 z ) e z ez 2 sinh z 2 14 (e 2 z 2e z e z e 2 z ) 14 (e 2 z 2 e 2 z ) and the difference is 2 cosh 2 z sinh 2 z 12 12 1 September 10, 2009 Hyperbolic Functions—Problem 2.33(e) Statement of the Problem, cont’d: dx (e) Show that 1 x 2 arcsinh x . [Hint: One way to do this is to make the substitution x = sinh z.] Solution: Making that substitution, we have dx = cosh z dz, so: cosh z dz 1 sinh z 2 cosh z dz cosh z sinh z sinh z 2 2 2 dz z but sinh z x z arcsinh x so we have shown that dx 1 x 2 arcsinh x September 10, 2009 Vertical Motion with Quadratic Drag Likewise, you can do Problem 2.34, which gives the definition: tanh z and leads you through the steps needed to show Now back to our equation: The left side is dv 0 1 v2 / vter2 g v sinh z cosh z dx 1 x 2 arctanh x . t dt 0 v / vter dv dx v 0 1 v2 / vter2 ter 0 1 x 2 v vter arctanh x 0 v / vter v vter arctanh vter gt v(t ) vter tanh while the right side is just gt, so solving for v, we get v ter To get the position, integrate (see Prob. 2.34) to get 2 gt vter y (t ) ln cosh g vter September 10, 2009 Example 2.5 A Baseball Dropped from a High Tower Find the terminal speed of a baseball (diameter D = 7 cm , mass m = 0.15 kg). Make plots of its velocity and position for the first six seconds after it is dropped from a tall tower. Solution Recall that the constant c can be written c = gD2., where g = 0.25 Ns2/m2. So mg (0.15 kg)(9.8 m/s 2 ) vter 35 m/s 2 2 2 c (0.25 Nm /s )(0.07 m) which is nearly 80 mph. You can sketch the velocity and position, or you can calculate it in Matlab. Here are the plots. As expected, the velocity increases more slowly than it would in a vacuum under gravity (dashed line), and approaches vter = 35 m/s (dotted line). As a consequence, the position is less than the parabolic dependence in vacuum. September 10, 2009 Quadratic Drag with Horizontal and Vertical Motion As we said before, the general problem of combined horizontal and vertical motion yields a set of coupled equations mvx c vx2 v y2 vx mv y mg c vx2 v y2 v y where now we take y positive upward. The projectile does not follow the same x and y equations we just derived, because the drag in the x direction slows the projectile and changes the drag in the y direction, and vice versa. In fact, these equations cannot be solved analytically at all! The best we can do is a numerical solution, but that requires specifying initial conditions. That means we cannot find the general solution—we have to solve them numerically on a case-by-case basis. Let’s take a look at one such numerical solution. September 10, 2009 Example 2.6 Trajectory of a baseball The baseball of the previous example is thrown with velocity 30 m/s (about 70 mph) at 50o above the horizontal from a high cliff. Find its trajectory for the first 8 s of flight and compare with the trajectory in a vacuum. If the same baseball were thrown on the same trajectory on horizontal ground, how far would it travel (i.e. what is its horizontal range)? Solution First, what are the initial conditions for the position and velocity? For the position, we are free to choose our coordinate system, so we certainly would choose xo = 0 and yo = 0 at t = 0. For the velocity, the statement of the problem gives the initial conditions vxo = vocos q = 19.3 m/s, vyo = vosin q = 23.0 m/s. Using these values, we need to write a program in Matlab that performs a numerical solution to the equations 2 2 mvx c vx v y vx mv y mg c vx2 v y2 v y for the time range 0 < t < 8 s. We will use the routine ode45 (ode stands for ‘ordinary differential equation’). September 10, 2009 Example 2.6, cont’d Solution, cont’d First we have to write a function that will be called by ode45. The heart of that routine is quite simple, just write expressions for the two equations: Vdot_x = -(c/m)*sqrt(v(1)^2+v(2)^2)*v(1); = [block 2] Vdot_y = -g-(c/m)*sqrt(v(1)^2+v(2)^2)*v(2); Here, v is the velocity vector, so v(1) is the horizontal velocity vx and v(2) is the vertical velocity vy. Before these equations will work, we have to define the constants, g, c, and m. Recall that c = gD2. m = 0.15; % Mass of baseball, in kg g = 9.8; % Acceleration of gravity, in m/s diam = 0.07; % Diameter of baseball, in m gamma = 0.25; % Coefficient of drag in air at STP, in Ns^2/m^2 c = gamma*diam^2; = [block 1] The last step is to name the function and indicate the inputs and outputs. ODE45 specifies that the function must have two inputs—the limits of the independent variable (time in this case), and the array of initial conditions (start velocity in x and y in this case). function vdot = quad_drag(t,v) … [block 1] [block 2] … vdot = [vdot_x; vdot_y]; After saving this function as quad_drag.m, we call ODE45 with [T,V] = ode45('quad_drag',[0 8],[19.3; 23.0]); September 10, 2009 Example 2.6, cont’d Solution, cont’d The arrays T and V that are returned are the times and x and y velocities, but what we need is the trajectory, i.e. the x and y positions. For those, we have to integrate the velocities. There is probably an elegant way to do this in Matlab, but I wrote a simple (and rather inaccurate) routine to do that, given the T and V arrays: function y = int_yp(t,yp) n = length(t); y = yp; y(1,:) = [0 0]; for i=1:n-1 dt = t(i+1)-t(i); dy = yp(i,:)*dt; y(i+1,:) = y(i,:)+dy; end y = y(1:n-1,:); Save this as int_yp.m, and then call it by pos = int_yp(T,V); which returns the position array [pos(1,:) is x, pos(2,:) is y]. All that remains is to plot the trajectory ( i.e. pos(1,:) vs. pos(2,:) ). plot(19.3*T,23.0*T-4.9*T.^2); % Plot vacuum case hold on plot(pos(:,1),pos(:,2),'color','red'); % Overplot quadratic drag case hold off September 10, 2009 Example 2.6, cont’d Solution, cont’d Here is the resulting plot (somewhat improved by labels). Note that the range is about 60 m, much shorter than the equivalent trajectory in a vacuum. Note also that the baseball does not reach quite as high as in a vacuum, and reaches its peak earlier. You will be given homework problems in which I will ask you to try your hand at such numerical solutions and plotting. I will help you learn these very useful skills, or you can make use of the Matlab helpers. September 10, 2009 2.5 Motion of a Charge in a Uniform Magnetic Field You may recall from Physics 121 that the force on a charge moving in a magnetic field is F qv B where q is the charge and B is the magnetic field strength. The equation of motion then becomes mv qv B which is a first-order differential equation in v. In this type of problem, we are often free to choose our coordinate system so that the magnetic field is along one axis, say the z-axis: B (0,0, B ) and the velocity can in general have any direction v (vx , v y , vz ). Hence, v B (v y B, vx B, 0) and the three components of the equation of motion are: mvx qv y B mv y qvx B mvz 0 September 10, 2009 Motion of a Charge in a Uniform Magnetic Field-2 This last equation simply says that the component of velocity along B, vz = const. Let’s now focus on the other two components, and ignore the motion along B. We can then consider the velocity as a two-dimensional vector (vx, vy) = transverse velocity. To simplify, we define the parameter w = qB/m:, so the equations of motion become: vx w v y v y w vx We will take the opportunity provided by these two coupled equations to introduce a solution based on complex numbers. As you should know, a complex number is a number like z = x + iy, where imaginary i is the square root of 1. Let us define: part h = vx + ivy and then plot the value of h as a vector in the vy complex plane whose components are vx and vy. real part vx September 10, 2009 Motion of a Charge in a Uniform Magnetic Field-3 Next, we take the time derivative of h: h vx iv y wv y iwvx iw (vx iv y ) or h iwh So the equation in terms of this new relation has the same form we saw in the previous lecture for linear air resistance, with the familiar solution h Aeiwt The only difference is that this time the argument of the exponential is imaginary, but it turns out that this makes a huge difference. Before we can discuss the solution in detail, however, we need to introduce some properties of complex exponentials, which we will do next time. September 10, 2009