Some Perspective. Forces and Newton s Laws

Transcription

1 Soe Perspective The language of Kineatics provides us with an efficient ethod for describing the otion of aterial objects, and we ll continue to ake refineents to it as we introduce additional types of coordinate systes later on. However, kineatics tells us nothing about how or why aterial objects ove in the way they do - in order to answer those questions, we need soe sort of physical principle which tells us how aterials bodies interact with each other and how this affects their otion. Even today we don t understand the full extent of these physical principles - science is always a work in progress, and physics is no exception. Every day, physics experients (including ones happening right here at UCSB!) ake further and further refineents to our understanding of these principles. Because of this, the laws of physics which we will discuss in our course, in particular Newtonian echanics, will necessarily be approxiations. However, they are in fact incredibly good approxiations. For hundreds of years, it was believed that Newtonian echanics was THE correct odel of the universe, capable of describing everything in the aterial world around us. While ore efficient versions of Newtonian echanics were later invented (Lagrangian and Hailtonian echanics, which you will learn about next quarter), these were only atheatical odifications which ade no fundaental changes to the underlying physics. Of course, the reason people believed that Newtonian echanics was correct was because the predictions it ade were so accurate, no one could perfor a sensitive enough experient which disagreed with the! The extent to which Newtonian echanics akes incorrect predictions could only be deterined once experients were sensitive enough to be able to detect the detailed structure of the ato, at which point it becae clear that Quantu echanics was ultiately a ore accurate description of the universe. The punchline is that so long as we re not trying to ake detailed predictions about the behaviour of things like atos, we can safely use Newtonian echanics as if it were correct, because we will never know the difference anyways. And there is a huge advantage to doing just that - the laws of Newtonian Mechanics are significantly less coplicated than those of Quantu Mechanics! There are also still plenty of applications for Newtonian echanics - everyday echanical objects are still an iportant part of our lives, and we need efficient tools for describing how they behave. So with these caveats in ind, let s start learning about Newtonian Mechanics. Forces and Newton s Laws As we know fro freshan echanics, Newton s Laws tell us how an object s otion is deterined by the forces acting on it. Intuitively, a force on an object is a sort of push or pull that an object experiences, and based on what is currently happening to the object, we have rules that tell us what the 1

2 force acting on the object is. For exaple, if the object is in a gravitational field, we have rules which tell us what the force acting on the object should be (Newton s Law of gravitation), or if the object happens to be an electron oving in a agnetic field, we have an equation describing the resulting force in that case (the Lorenz force law). Ultiately, all of the forces that act on a body are a result of interactions it experiences with other bodies, and so the idea of forces is really just a succinct atheatical way to describe how bodies interact with each other. Fro your freshan echanics class, you probably have a pretty decent intuitive grasp of what a force is, and for the ost part, this intuitive notion will be sufficient for our purposes. Entire books could (and have been) written on the philosophy of exactly what a force is, but this would be overkill for our course, so we won t dwell on it now (although Taylor discusses this atter in slightly ore detail than I have, for those of you interested in reading ore). In particular, the force acting on a body is a vector quantity. It has a agnitude (roughly speaking, how uch we re pushing on the body), along with a direction (roughly speaking, where we are trying to push it). When ore than one type of force acts on a body, we say that the net force acting on it is the vector su of all of the individual forces. Newton s Second Law tells us that the total force acting on a body causes that body to accelerate according to F = a, (1) where is the ass of the body. The ass of an object is another quantity which you probably have an intuitive sense of fro freshan echanics, and we know it is a rough easure of how uch an object resists acceleration. It turns out there are also soe philosophical subtleties in defining the ass of an object, which we won t dwell on either, but for those that are interested, Kibble s textbook has a good discussion of how to handle these issues (section 1.3). Reeber that the ass of an object is different fro its weight. The weight of an object is the force that object experiences due to the effects of gravity. If I take a ball on Earth and ove it to the Moon, where the effects of gravity are weaker, then the weight of the ball will be reduced. But its ass will stay the sae. If I were to take a agnetic ball out into epty space far fro any other bodies, where any gravitational effects are negligible, and study its reaction to a agnetic force, I would be learning soething about its ass. When there is no force acting on an object, the above equation tells us that its acceleration is zero, and therefore it will travel with a constant velocity. This fact, despite following as an obvious consequence of Newton s Second Law, is given its own nae - Newton s First Law. Newton s second law isn t very useful unless I start telling you soething about what types of forces can act on an object and how they behave. But before I start giving exaples of forces, Newton s third law tells us that there are certain conditions that any valid force ust obey. In particular, whenever two bodies interact, the agnitude of the force exerted on each body is the sae, while the directions of the forces are opposite. This is often stated by 2

3 saying that bodies exert equal and opposite forces on each other. For exaple, two bodies with ass will be attracted to each other through gravity, and the force they exert on each other will be the sae in agnitude. The direction of the force on one body is such that the force points towards the other body, so that the two forces are opposite in direction. This is sketched in Figure 1. We ll have plenty ore to say about Newton s law of gravitation as the course progresses. To clarify the difference between the two forces in a force pair, we often develop a subscript notation. If I have two bodies which I will call A and B, then the force fro body A acting on body B is written F A on B. Newton s third law then reads F A on B = F B on A. (2) Fro a practical standpoint, Newton s second law is the one we will ake use of the ost in this course, although fro a fundaental standpoint, Newton s third law is absolutely crucial. It turns out that (as discussed in section 1.3 of Kibble s textbook) Newton s third law is necessary in order to be able to define the ass of an object, and later in the course, Newton s third law will play an iportant role in the concept of center of ass otion and conservation of oentu. Projectile Motion with Air Resistance To see a concrete exaple of Newton s laws in action, let s revisit the projectile otion proble we recently studied, but with a slight odification - this tie we will include the effects of air resistance. We re all failiar with the fact that if we ove very quickly through the air, we feel a force pushing on our bodies. While this force is typically a result of coplicated icroscopic interactions between the olecules in the air and our bodies, it is usually possible to odel the force in a siple for. In any realistic situations, it is accurate to odel the force acting on a body due to air resistance as F R = kv nˆv, (3) where v is the agnitude of the velocity, ˆv is a unit vector pointing in the direction of the velocity, n is soe integer power, and k is soe nuerical constant, referred to as the drag coefficient. The constant k is soething which in principle could be calculated fro first principles, and typically depends on the size and shape of the body in question. As a practical atter, it is often easiest to siply easure the value of k by doing an experient on the body. In either case, we will assue that the paraeter k is soething that we know the value of. A free-body diagra for this odified situation is shown in Figure 2. As we usually do, we re going to ake the approxiation that the cup is a point-like object, which, despite sounding soewhat silly, turns out to be a surprisingly reasonable assuption, provided that the shape and overall 3

4 Figure 1: Two assive bodies will exert a gravitational force on each other, and it will obey Newton s third law. Newton s law of gravitation is shown at the botto. structure of the block doesn t change uch (later in the course we ll understand why this approxiation works so well when we discuss the notion of center of ass). Let s consider an exaple with n = 1, when the drag force can be written 4

5 Figure 2: The various forces acting on a projectile in otion through the air. Iage credit: Kristen Moore as F R = kvˆv = k v. (4) If we revisit our projectile otion proble fro our discussion of kineatics, where the aterial body in question is now subject to the force of gravity and air resistance, the total force will be F = F g + F R = g k v, (5) where is the ass of the body, and g is the local acceleration due to gravity. Again, we ll set up a coordinate syste whose x axis is parallel with the ground and whose y axis points vertically upwards away fro the ground. We ll place the origin so that it is on the ground, directly below the uzzle of the gun. This is shown in Figure 3. In our coordinates, the initial location of the bullet is specified by ( 0 r 0 =, (6) h) where h is the height of the nozzle of the gun off of the ground. If the gun akes 5

6 an angle of θ with respect to the horizontal, our initial velocity will be ( ) v0 cos θ v 0 =, (7) v 0 sin θ where v 0 is the initial speed of the bullet. The acceleration due to gravity is given by ( ) 0 g =, (8) g where g is approxiately 9.8 eters per second squared. Figure 3: The initial set up of our projectile otion proble, the sae as it was in the previous lecture. Iage credit: Kristen Moore If we now write Newton s Second Law, F = a, in ters of coponents, we find ( ) ( ) ( ) ax 0 kvx = +, (9) a y g kv y or, ( ) ( ) rx kṙx =. (10) r y g kṙ y The above vector equation is a syste of differential equations - the two coponent equations define the coordinates r x (t) and r y (t) in ters of their own derivatives. In particular, the above syste of differential equations is decoupled 6

7 - one equation involves ters that only depend on the x coordinate, while the other equation involves ters that only depend on the y coordinate. This type of syste is especially easy to solve, since we can treat each equation separately, without worrying about the other. If we had chosen a drag force with n 1, this would not have been the case - the higher power of the velocity would have ixed the two coordinates together in a ore coplicated way. At this point in your physics career, you should have encountered soe basic differential equations at least once before, but we ll review the basic steps here. Let s start with the equation for the x coordinate, which reads r x = k ṙx. (11) If we ake use of the basic definitions of velocity as the derivative of position, and acceleration as the derivative of velocity, then we can rewrite this equation in ters of the velocity and its derivative, v x = k v x. (12) In this for, the differential equation in the x direction is a first-order differential equation for the x coordinate of the velocity. We say that it is a first-order differential equation because the highest derivative that appears is a first-order one. In order to proceed in solving the differential equation, we use a technique known as separation of variables. The idea is to first rearrange the equation so that it reads v x = k v x, (13) with the velocity and its derivative appearing on the left, and the constant ters appearing on the right. Now, we integrate each side of the differential equation with respect to tie, t v x dt = k t dt, (14) v x t 0 t 0 where t 0 is the initial tie (often taken to be zero), and t is soe later tie at which we want to know the location (and velocity) of the projectile. The integral on the right side is trivial to perfor, and we find t t 0 v x v x dt = k (t t 0). (15) Now, without prior knowledge of what v x (t) is, we can t iediately copute the integral on the left side. However, we have a valuable tool at our disposal - the substitution rule for definite integrals. The substitution rule says that v x dt = dv x dt dt = dv x, (16) 7

8 which allows us to perfor a change of variables in an integral. Using this fact, we find vx(t) dv x = k v x (t t 0). (17) v x(t 0) We can now perfor the integral over the left side, to find ( ) vx (t) ln = k v x (t 0 ) (t t 0). (18) If we now take the exponential of both sides, and do a little bit of algebraic rearrangeent, we find that [ v x (t) = v x (t 0 ) exp k ] (t t 0). (19) This equation tells us that the velocity in the x direction decays exponentially over tie, which akes sense, since we expect the retarding force to decrease the speed of the projectile. Notice that when k = 0, the velocity is constant for all tie, which is exactly what we would expect when there is no drag. We can rewrite the equation slightly, where v x (t) = v x (t 0 ) exp [ (t t 0 ) /τ], (20) τ /k (21) is known as the characteristic tie. The characteristic tie gives a rough sense of how long it takes for the velocity to decay appreciably (ore precisely, it is the tie it takes for the agnitude of the velocity to decrease by a factor of e 2.718). Notice that when k = 0, the characteristic tie is infinity. The above solution of the differential equation for the x coordinate should be failiar to you fro previous courses. If any of this sees unfailiar, you should review ASAP! Notice that if the ters on the right had depended on tie in soe specified way, the integral on the right still would have been straightforward to perfor. For exaple, if the drag coefficient k were a function of tie, so that k k (t), the integral over tie could have still been coputed, so long as we knew the functional for for k (t). This is why this ethod is known as separation of variables - all of the explicit dependence on tie is separated to one side, where all of the explicit dependence on the velocity (and its derivative) is separated to the other side. While this solution technique is a valuable tool, it unfortunately only works for first-order equations. We ll learn how to solve ore general equations later on in the course. The differential equation for the y coordinate can siilarly be written in ters of the velocity, and, after soe rearrangeent, takes the for v y g + k v y = 1. (22) 8

9 We can again proceed by integrating both sides with respect to tie, and then perforing a change of variables on the left. This results in the equation vy(t) v y(t 0) dv y g + k v y = (t t 0 ). (23) Solving the integral on the left, we find [ ] g + k ln kvy (t) g + kv y (t 0 ) = (t t 0 ). (24) With a little bit of algebraic rearrangeent, this becoes ( ) [ v y (t) = k g + v y (t 0 ) exp k ] (t t 0) g, (25) k or, in ters of the characteristic tie, v y (t) = (gτ + v y (t 0 )) exp [ (t t 0 ) /τ] gτ. (26) The above equation for the y coponent of the velocity tells us that at the initial tie, when t = t 0, v y (t = t 0 ) = (gτ + v y (t 0 )) e 0 gτ = gτ + v y (t 0 ) gτ = v y (t 0 ), (27) exactly as it should. The ore interesting case occurs at infinitely long ties, when v y (t = ) = (gτ + v y (t 0 )) e gτ = gτ. (28) This tells us that in the liit of infinitely long tie, the velocity approaches a constant value, which is known as the terinal velocity. The terinal velocity is the velocity which is reached when the gravitational force on the falling projectile precisely balances out the force due to air resistance. To see that this is so, siply plug the expression for the terinal velocity into the equation for the force due to drag, F Ry = kv y = k k g = g = F gy. (29) Now that we have the two velocity coponents, one property of the projectile s otion that we can copute is its acceleration. The acceleration is found by differentiating the velocity, and we find a x (t) = v x (t 0 ) τ exp [ (t t 0 ) /τ] = k v x (t 0 ) exp [ (t t 0) /τ] (30) for the x coponent. Notice that at t = t 0, this expression tells us that a x (t = t 0 ) = k v x (t 0 ), (31) 9

10 which is siply the force due to drag right at the oent that the projectile is fired. At infinitely long ties, the acceleration decays to zero, as it should, since the velocity is approaching the constant value of zero. For the y coponent, we find a y (t) = (gτ + v y (t 0 )) τ At the initial tie, we find ( exp [ (t t 0 ) /τ] = g + k ) v y (t 0 ) exp [ (t t 0 ) /τ]. (32) a y (t = t 0 ) = g k v y (t 0 ), (33) which is siply the acceleration due to gravity and the acceleration due to the initial drag force. At infinitely long ties, the acceleration in the y direction also decays to zero, since the gravitational and drag forces eventually balance each other, leading to zero net acceleration. The agnitude of the acceleration is also a siple calculation, and after a little algebra, we find a (t) = a 2 x (t) + a 2 y (t) = ( ) 2 ( k v x (t 0 ) + g + k 2 v y (t 0 )) e (t t0)/τ. (34) The agnitude of the acceleration also decays exponentially, approaching zero at long ties. Projectile Range with Air Resistance Now that we know the velocity of the projectile as a function of tie, we can find the position of the projectile as a function of tie siply by integrating the velocity. For the x coponent, we find r x (t) = r x (t 0 ) + t v x (t ) dt t 0 = v x (t 0 ) t exp [ (t t 0 ) /τ] dt, t 0 (35) since the initial value of the x coponent is zero. Coputing the integral, we have [ ] r x (t) = v x (t 0 ) τ 1 e (t t0)/τ. (36) Written in ters of the angle at which our projectile is initially pointed, [ ] r x (t) = v 0 τ cos (θ) 1 e (t t0)/τ. (37) At large ties, this becoes r x (t = ) = v 0 τ cos (θ) [ 1 e ] = v 0 τ cos (θ). (38) This tells us that even if the projectile were to travel through the air for infinitely long ties, it would never reach a horizontal position larger than the above 10

11 constant. Intuitively, this akes sense - just like a block sliding on a table eventually coes to rest, a projectile s horizontal otion is eventually ipeded by air resistance, and since there are no other forces in the horizontal direction, it will stay at rest along this direction. Of course, in a realistic situation, a projectile will generally run into the ground in an aount of tie which is less than infinity, so the horizontal range of the projectile will be less than the above constant. To find out exactly how far the projectile will travel, we need to know how long it takes for the vertical coponent of the position to reach zero, which indicates that the projectile has hit the ground. The y coponent of the position is also found by integrating, and we have t t [ ] r y (t) = r y (t 0 ) + v y (t ) dt = h + (gτ + v y (t 0 )) e (t t0)/τ gτ dt, t 0 which, after integration, yields t 0 (39) r y (t) = h + [ gτ 2 + v y (t 0 ) τ ] [ 1 e (t t0)/τ ] gτ (t t 0 ). (40) Notice that as the tie becoes large and the exponential ter becoes sall, we have r y (t ) = [ h + gτ 2 + v y (t 0 ) τ ] gτ (t t 0 ), (41) which is just otion with a constant velocity of gτ, as we found earlier. In order to find the tie at which the projectile hits the ground, we need to set the vertical coordinate of the position equal to zero. Taking t 0 = 0 for siplicity, and writing the initial velocity in ters of the angle of the projectile, this eans that we need to solve the equation r y (t R ) = 0 h + [ gτ 2 + v 0 τ sin (θ) ] [ 1 e t R/τ ] = gτt R. (42) Unfortunately, it is not possible to solve the above equation in a siple closed for. No clever cobination of taking logariths or exponentials will result in a siple expression for t R. Because of the presence of the exponential ter, this type of equation is known as a transcendental equation, since the exponential is an exaple of a transcendental function (a transcendental function is a function which is not a siple algebraic function of its arguents). However, there are several ways to find the value of t R, depending on what our needs are. If we are trying to calculate the range of a specific projectile in a real-world application (say, for calculating the range of a issile on soe navy boat), then a nuerical approach is often the best approach. If we specify a particular nuerical value for all of the paraeters in the proble (the ass of the projectile, the drag coefficient, the acceleration due to gravity, and the initial position and velocity), then there are any coputer progras which can copute an approxiate value for t R. If we only need to know the range of one specific projectile, but we need it to a high level of precision, this is often the best approach - odern coputers can perfor billions of calculations per second, 11

12 and so can attain very accurate nuerical solutions in a short period of tie. You ll explore how soe of these algoriths work in the hoework (although I proise I won t ake you do a billion calculations). However, in any cases we want to gain a better intuitive understanding of how the details of the proble vary as we odify or change its paraeters. This ight be iportant, for exaple, if we were trying to understand how to odify the paraeters of a projectile to ensure that it had the best possible perforance under different circustances. In this case, a perturbative approach is typically a better route. In fact, perturbative approaches are so coonplace in physics that it pays to set aside soe tie and understand how exactly they work. We ll use a perturbative ethod here to solve our current projectile proble, although we will see perturbative approaches appear several ore ties before the course is through. When we use a perturbative approach, we first identify soe paraeter in our proble which we can think of as being sufficiently sall. Usually we choose this paraeter so that when the paraeter is equal to zero, we know how to solve the proble in closed for. In our case, the drag coefficient k is a natural choice for this paraeter, since we ve already seen how to calculate t R when there is no drag. Next, we assue that the quantity we are trying to find, in this case t R, is a function of k, t R t R (k), which has a well-defined Taylor series expansion in ters of k, t R (k) = t Rn k n = t R0 + t R1 k + t R2 k (43) This is alost always the case in real-world probles, and we expect it to be true here - as we slowly add air resistance to the projectile, the range of the projectile should decrease slightly, due to the effects of drag. The goal of the perturbative approach is then to identify each of the coefficients which appear in the Taylor series expansion, one at a tie. To see how this ethod works, let s assue that the drag coefficient, k, is indeed suitably sall in soe sense. Exactly what constitutes sall can soeties be a subtle question, the answer to which is soeties only clear after perforing the perturbative ethod. However, it is often the case that the approach works very well even for fairly large paraeter values. So let s start by taking our original equation and rewriting it slightly, h + [g 2 k 2 + v 0 sin (θ) ] [ ] 1 e kt R/ = g k k t R, (44) so that the dependence on k is explicit. We ll also define and ultiply both sides by k 2, in order to get φ v 0 sin (θ), (45) hk 2 + [ g 2 + φk ] [ 1 e kt R/ ] = gkt R. (46) 12

13 Now, we insert the expansion of t R in ters of k into this equation, to find hk 2 + ( g 2 + φk ) ( [ { 1 exp k }]) { } t Rn k n = gk t Rn k n. (47) This expression ay not iediately see like an iproveent, since we still need to solve for a bunch of ters which are in the exponential. However, the trick is to realize that the exponential function itself also has a Taylor series expansion, which can be written as e x = p=0 x p p! = 1 + x + x2 2 + x (48) 6 Because the sall paraeter k also shows up inside of the exponential, we can use this expansion in our equation, to find hk 2 + ( g 2 + φk ) ( [ { 1 1 k }] p ) { } t Rn k n = gk t Rn k n. p! p=0 (49) We now have an equation which involves an expansion only in powers of k, which thus eliinates the issue of the exponential function. Of course, this expression is still soewhat intiidating looking, especially with one infinite su nested inside of the other. The trick to being able to actually do soething with this equation is to reeber a theore about Taylor series: if two power series expansions are equal to each other, then it ust be true that the individual ters are equal. In other words, if we have then it ust be the case that a 0 + a 1 x + a 2 x = b 0 + b 1 x + b 2 x (50) a 0 = b 0, a 1 = b 1, a 2 = b 2,... (51) This eans that if we expand out both sides of our equation, we can atch powers of k to deterine the coefficients. If we only want a sall nuber of coefficients, we only need to expand out a few ters. Let s see exactly how expanding out both sides of our equation in this way tells us soething about the coefficients. The right side of our equation is easy to expand - it siply becoes { } gk t Rn k n = gt R0 k + gt R1 k 2 + gt R2 k (52) For now, let s expand each side out to third order in k. We ll see later why this was a good choice. As for the left side of our equation, a little ore care is required. We want to expand the sus on the left so that we keep all of the 13

14 powers of k that appear up through k 3, so that we can atch powers on both sides of the equation. To begin this expansion, let s notice that ) 1 e x = 1 (1 + x + x2 2 + x = p=1 x p p!, (53) since the first ter in the su cancels. Using this, our equation becoes hk 2 ( g 2 + φk ) ( [ { 1 k }] p ) t Rn k n = gt R0 k+gt R1 k 2 +gt R2 k p! p=1 (54) This eans that the quantity on the left we need to expand is [ { 1 k }] p { t Rn k n ( 1) p k p } p = p! p! p t Rn k n (55) p=1 keeping ters as high as k 3. In order to perfor the expansion, we ll consider the su over p, one ter at a tie. Let s start with the p = 1 ter, which gives { k } t Rn k n p=1 = t R0 k t R1 k2 t R2 k (56) Despite the fact that the inside su over n has infinitely any ters, we only need to keep the first three, since we are only expanding both sides through k 3. Any additional ters contribute a factor of k 4 or higher, which contributes to a power that we are not interested in atching. Continuing with the expansion over p, the p = 2 ter is 1 2 k 2 2 { } 2 t Rn k n = 1 k ( t R0 + t R1 k +... ) ( t R0 + t R1 k +... ). (57) The sus inside the parentheses contain infinitely any ters, which then ust be ultiplied. While it ay see like carrying out this ultiplication is ipossible, the fact that we only want an expansion through k 3 eans that this is actually a relatively siple task. First, we notice that because the expression already contains an overall factor of k 2, we only need to expand ultiplication of the parentheses out to order one. If we start to perfor the ultiplication by expanding one ter at a tie, we find ( t R0 + t R1 k +... ) ( t R0 + t R1 k +... ) = (58) t R0 ( t R0 + t R1 k +... ) + t R1 k ( t R0 + t R1 k +... ) +... Despite the fact that there are infinitely any ters in this su, each ter contributes increasingly higher powers of k. If we only want to expand this 14

15 ultiplication to first order in k, then we siply take ( t R0 + t R1 k +... ) ( t R0 + t R1 k +... ) = (59) t R0 ( t R0 + t R1 k +... ) + t R1 k ( t R ) +... = t 2 R0 + 2t R0 t R1 k +... Therefore, if we only want to expand our expression through k 3, the only part of the p = 2 ter that we need is 1 2 k 2 2 { } 2 t Rn k n = 1 k 2 ( t R0 + 2t R0 t R1 k +... ) = 1 2 t 2 R0 2 k2 + t R0t R1 2 k (60) Every additional ter is at least as large as k 4. The p = 3 ter in the su is given by { 1 k 3 } t Rn k n = 1 k ( t R0 + t R1 k +... ) 3. (61) Because the overall pre-factor on this ter is already third order in k, the ter in parentheses only needs to be expanded to zero order in k, which siply gives the constant value t 3 R0. Therefore, the relevant contribution fro the p = 3 ter is 1 k ( t R0 + t R1 k +... ) 3 = 1 t 3 R0 6 3 k (62) Continuing further, the p = 4 ter in the su is given by { 1 k 4 } t Rn k n = 1 k ( t R0 + t R1 k +... ) 4. (63) However, this ter does not contribute any factors of k which are less than fourth order, due to the overall factor of k 4. Of course, this will also be the case for every higher power of p. Thus, the only two powers of p which are iportant are the first three, which is uch less than infinity! Taking our expressions for the p = 1, p = 2 and p = 3 ters and cobining the together, we find that the infinite su on the left side of our equation can be expressed as { ( 1) p k p } p p! p t Rn k n (64) p=1 t 2 R0 = t R0 k t R1 k2 t R2 k k2 + t R0t R1 2 k k = t ( R0 1 k + t 2 R0 2 2 t ) ( R1 k 2 tr0 t R t 3 R0 6 3 t ) R2 k All of the ters that appear after the ellipses are at least fourth order in k. Despite the appearance of an infinite su, the nuber of ters we need to t 3 R0 15

16 worry about is actually fairly sall, all things considered. Using this expression for the su, our original equation now reads hk 2 ( g 2 + φk ) ( t ( R0 1 k + t 2 R0 2 2 t ) ( R1 k 2 tr0 t R t 3 R0 6 3 t ) ) R2 k (65) = gt R0 k + gt R1 k 2 + gt R2 k Finally, our original equation is starting to look sipler. The last thing we need to do is perfor the ultiplication on the left side, and then atch powers of k. When we ultiply out the ters on the left side, there will be soe powers of k 4 which appear. However, we can also ignore these, since we are only atching the first few powers of k. If we expand out these ters and ignore the k 4 contributions, we find ( hk 2 g 2 t R0 k + ( φ t R0 k2 + ( 1 t 2 R0 2 2 t ) ( R1 k 2 tr0 t R t 3 R0 6 3 t ) ) R2 k 3 ( 1 t 2 R0 2 2 t ) ) R1 k = gt R0 k + gt R1 k 2 + gt R2 k (66) At last, we now have an expansion of our equation, on both sides, which is valid through k 3. Matching powers of k now gives us three equations. The first one coes fro atching first-order powers on both sides, and it reads gt R0 = gt R0. (67) This equation is certainly consistent, although it isn t exactly very interesting. So let s ove on to the second equation, which is ore interesting. Matching powers of k 2, we find h + gt R1 1 2 gt2 R0 + φt R0 = gt R1, (68) or siply h 1 2 gt2 R0 + φt R0 = 0. (69) This is a quadratic equation for t R0, and in fact it is exactly the sae equation that we found when we neglected air resistance. This is exactly what we want, because our expansion in powers of k tells us that when there is no air resistance, we should have t R (k = 0) = t Rn (0) n = t R0. (70) 16

17 In other words, t R0 is precisely the result we get when there is no drag. Solving this quadratic equation gives the sae result as before, t R0 = φ + φ 2 + 2gh. (71) g The third equation is the one which yields new inforation about the effects of air resistance. Matching powers of k 3, we find ( g 2 tr0 t R1 2 1 t 3 R0 6 3 t ) ( R2 1 t 2 R0 φ 2 2 t ) R1 = gt R2, (72) or, after a little rearrangeent, (φ gt R0 ) t R1 + g 6 t3 R0 φ 2 t2 R0 = 0. (73) Since the ter t R1 only appears linearly in this equation, soe siple algebra iediately yields t R1 = (3φ gt R0) t 2 R0 (φ gt R0 ) 6. (74) We now have an explicit expression for t R1 in ters of the paraeters of the proble (reeber that t R0 has its own expression in ters of the paraeters of the proble, which we found previously). With this knowledge, we can now write where t R0 is given by t R (k) = t R0 + t R1 k +... = t R0 + (3φ gt R0) (φ gt R0 ) t 2 R0 k +... (75) 6 t R0 = φ + φ 2 + 2gh. (76) g Despite looking slightly coplicated, we have now achieved our goal of finding an explicit expression for t R, which is valid to first order in the drag coefficient k. If the drag coefficient is sall enough, so that the force due to air resistance is not too strong, this should be a reasonably good approxiation for the value of t R. Of course, with this value for t R, we can finally return to our original goal of finding the range of the projectile. The horizontal range of the projectile is found by evaluating the x coordinate at the tie t R, [ ] R = r x (t = t R ) = v 0 τ cos (θ) 1 e t R/τ. (77) Now, before we rush to plug in our value of t R, let s reeber that our original constraint equation was given by r y (t R ) = 0 h + [ gτ 2 + φτ ] [ 1 e t R/τ ] = gτt R. (78) 17

18 If we rearrange this constraint equation slightly, it becoes 1 e t R/τ = gτt R h gτ 2 + φτ. (79) Using this in the expression for the horizontal range, we find [ ] [ ] gτtr h gtr hk R = v 0 τ cos (θ) gτ 2 = v 0 cos (θ). (80) + φτ g + φk Using our result for t R, we finally arrive at R = v 0 cos (θ) {g + φk} 1 { gt R0 hk + g (3φ gt R0) (φ gt R0 ) t 2 } R0 6 k. (81) We now have an approxiate expression for the range of the projectile, in ters of all of the paraeters of the proble. However, there is actually one ore siplification we can ake. Notice that our final answer contains the ter {g + φk} 1 = 1 g + φk. (82) This is not a siple algebraic function of k, and in fact the Taylor series expansion of this expression contains infinitely any powers of k, 1 g + φk = 1 g φ g 2 2 k + φ2 g 3 3 k (83) This eans that in fact, we can write our expression for the range as { 1 R = v 0 cos (θ) g φ } { g 2 2 k + φ2 g 3 3 k gt R0 hk + g (3φ gt R0) t 2 } R0 (φ gt R0 ) 6 k. (84) Since the second bracket ter, the one involving t R, is only expanded out to first order in k, it doesn t really ake sense to keep the first bracket ter to a higher order either. If we ultiply the two bracket ters together, and ignore any ter which is higher than first order in k, we find { ( (3v0 sin (θ) gt R0 ) t 2 R0 R = v 0 cos (θ) t R0 + (v 0 sin (θ) gt R0 ) 6 v 0 sin (θ) g t R0 h ) } k, g (85) where, as before, t R0 = v 0 sin (θ) + v0 2 sin2 (θ) + 2gh. (86) g This is our final expression for the horizontal range of the projectile, as an approxiation to first order in k. 18

19 The last thing we should do before wrapping up this subject is ake sure that our perturbative answer akes sense. When k = 0, we have R (k = 0) = v 0 cos (θ) t R0 = v 0 cos (θ) v 0 sin (θ) + v0 2 sin2 (θ) + 2gh, (87) g which is exactly what we found when we neglected air resistance. We can also get an intuitive sense for how the drag affects the otion of the projectile by considering the slightly sipler case when h = 0 (when the projectile is fired fro the ground). In this case, and thus t R0 = 2φ g = 2v 0 sin (θ), (88) g t R1 = (3φ gt R0) t 2 R0 (φ gt R0 ) 6 = 2 3 g 2 = 2 3 The expression for the range in this case siplifies to { ( R (h = 0) = 2v2 0 sin (θ) cos (θ) 4v0 sin (θ) 1 g 3g φ 2 v 2 0 sin 2 g 2. (89) ) } k. (90) When there is no drag, the overall ter in brackets is siply equal to one. We see now that the effect of drag is to reduce the range slightly, since when k 0, the ter in brackets is slightly less than one. While this ay have been obvious to us, we can now see in ore detail exactly how the range is reduced. Perhaps the ost striking effect we see is that once air resistance is introduced, the range of the projectile is no longer independent of the ass - the saller the ass, the ore iportant the first-order correction is. Of course, this agrees with our coon intuition that, for a given shape and size, less assive objects are affected by air resistance ore strongly. We also see that the reduction in range increases as the initial velocity increases, which akes sense, as the drag force increases with increasing velocity. Siilarly, the effects of air resistance also becoe ore iportant as gravity becoes weaker. Interestingly, the reduction in the range also depends on the initial firing angle. It ay see as though this was a lot of work to get an answer which is only approxiately correct. However, if you becoe a professional physicist, this type of calculation will becoe very failiar to you. Despite what soe textbook hoework probles ay lead you to believe, the vast ajority of probles which show up in physics cannot be solved in a siple, closed for. In order to ake any kind of progress, it is alost always the case that we need to ake soe sort of approxiation, like the one we have ade here. In fact, this type of approach is so coon in physics that it ight be accurate to say that Taylor series are THE ost iportant tool in all of physics (or, as soe people say, the ost iportant physicist was a atheatician, and his nae was Taylor). With this in ind, it s good to get soe practice with these ethods as soon a possible, because they will show up over and over again in your future physics courses, especially if you go to graduate school. 19