Physics 430 Lecture 4 Quadratic Air Resistance Dale E Gary NJIT Physics Department Linear Air Resistance Recap1 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 f f v v Quite generally we can write this force as where the function f v can in general be any function of velocity At relatively slow speeds it is often f v bv cv 2 f lin a good f quad approximation to write 2 where flin and fquad stand and terms f lin bvfor the andlinearf quad cvquadratic respectively m r mg bv mv mg bv mv x bv x For linear air resistance the equation of motion is or in v y mg bvdifferential terms of velocity it is amfirst order y v y v x b equations b equation which has component vx m v y vter m mg vter b Equations of this form can be written September 10 2009 where is the terminal velocity Linear Air Resistance Recap2 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 dv y of this particular form dv x e g b b dt dt vx m v y vter m have exponential solutions v x v xo e t v y v yo e t vter 1 e t m b which we then integrate to get x and y positions t x t x 1 e t y t vter t v yo vter 1 e We can then combine these equations by eliminating t to get a v yo vter single equation for the trajectory x y x vter ln 1 v xo 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 2 with the quadratic form of the fdrag force quad cv mv mg cv 2 v The equation of motion in terms of v then becomes with component equations mv x cv 2 c v x2 v y2 v x mv y mg cv 2 mg c v x2 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 mv become cv 2 horizontal motion which case the equations 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 dv it isc trivial involving v to one side For the horizontal equation 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 v dv us to general In this case however the separable form allows c t vo v 2 m 0 dt integrate both sides directly 1 1 v ct v t o 1 t where I have introduced the m or to get vo v m cvo characteristic time in terms of constants To find the position we again integrate the velocity equation to get t v0 x t xo dt vo ln 1 t if xo 0 0 1 t September 10 2009 Horizontal Motion with Quadratic Drag 2 The final solutions for v t and x t are v m cvo v t o 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 velocityt as approaches zero much more slowly like 1 t so the position does not approach some limiting value xlike 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 whichv 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 v t dv assuming vo 0 g 0 1 v 2 vter2 0 dt dx 1 x 2 arctanh x Looking at the inside front cover of the book we find September which is what we have if we write x v vter What the heck10 is 2009 arctanh 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 ez 2 1 2 1 2 z 1 2 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 e ix e ix and sin x 2i Then the solution is very easy e i iz e i iz e z e z cos iz cosh z 2 2 e i 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 d coshare z The integrals are equally straightforward d z 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 12 …