Mechanics Chapter 1-7 PDF

Mechanics 2024–25 Autumn term Jon Fenton Room 313B, Blackett Laboratory [email protected] Contents Contents i 1 Introduction...

Mechanics 2024–25 Autumn term Jon Fenton Room 313B, Blackett Laboratory [email protected] Contents Contents i 1 Introduction 1 2 Newton’s laws of motion and one-dimensional motion 11 2.1 Statement............................. 11 2.2 Newton’s first law......................... 11 2.3 Newton’s second law....................... 13 2.4 Inertial and gravitational mass.................. 15 2.5 Motion due to forces....................... 16 2.6 Motion in one dimension..................... 17 2.7 Impulse.............................. 19 2.8 One-dimensional motion with a velocity-dependent force.... 20 2.9 The work-energy theorem..................... 27 2.10 Power............................... 30 2.11 One-dimensional motion with a position-dependent force.... 32 2.12 Potential functions and equilibrium................ 38 2.13 Small oscillations about equilibrium............... 41 3 Two-body dynamics 43 3.1 Newton’s third law........................ 43 3.2 The centre of mass and the reduced mass............. 44 3.3 Collisions............................. 49 3.4 Elastic collisions in one dimension................ 50 3.5 Inelastic collisions in one dimension............... 52 3.6 Introduction to collisions in three dimensions.......... 53 3.7 The centre-of-mass frame..................... 56 3.8 Elastic scattering in the centre-of-mass frame.......... 58 3.9 The rocket equation........................ 60 4 Three-dimensional motion 63 i C ONTENTS 4.1 Work in three dimensions..................... 63 4.2 Potentials in three dimensions................... 64 4.3 Central forces........................... 65 4.4 Rotating vectors in two dimensions................ 67 4.5 Angular momentum and torque in three dimensions....... 68 5 Orbits 73 5.1 Conservation laws in orbital mechanics.............. 73 5.2 Orbital energy........................... 74 5.3 Circular orbits........................... 76 5.4 Elliptical orbits.......................... 77 5.5 Hyperbolic orbits......................... 81 5.6 Last words............................. 83 6 Rigid-body dynamics 87 6.1 Newton’s laws for extended bodies................ 87 6.2 Centre-of-mass motion...................... 89 6.3 Separating the centre-of-mass and internal physics........ 90 6.4 Rigid bodies............................ 95 6.5 The angular momentum of a rigid body.............. 95 6.6 Rotations about axles....................... 98 6.7 The moment of inertia....................... 100 6.8 Rotational work.......................... 104 6.9 The parallel-axis theorem..................... 106 6.10 Linear and rotational analogues.................. 108 6.11 Into the continuum........................ 109 6.12 Applications............................ 115 7 Rotating frames 123 7.1 Our spinning world........................ 123 7.2 Rotating vectors.......................... 124 7.3 Inertial and rotating coordinate systems............. 127 7.4 Centrifugal and Coriolis forces.................. 130 7.5 The Earth as a rotating frame................... 134 ii  Contents Acknowledgements These notes are an evolution of the textbook-style notes carefully prepared by the previous lecturer, Matthew Foulkes, and are in large part unchanged from the version he prepared, so if you find them useful (which I hope you will), most of the credit belongs to him! I have though made some modifications since taking on the course. Please let me know if you find any errors! The course as a whole has evolved over a number of years and as such (almost certainly) portions of the content of these notes and the course content more broadly were originally developed by earlier lecturers, who include also Caroline Clewley, Tim Horbury, Vijay Tymms and Martin McCall. iii Chapter 1 Introduction Mechanics was the field in which the ideas, approaches and techniques of mod- ern science were invented and remains the platform on which most of physics is built. It was in mechanics that we first realised the mysterious link between mathematics — the manipulation of symbols written on pieces of paper — and the real world; and it was to understand mechanics that we first formulated the laws of nature mathematically. It was also in mechanics that we developed cal- culus as the mathematical language for describing things that change, that we became aware of the existence of conservation laws for energy, momentum and angular momentum, and that we first encountered the concept of (Galilean) rel- ativity. All of these tools and ideas remain central to physics today. Almost everything you learn this term will help you throughout your degree and the rest of your career in physics. Although we all have intuition about mechanics, aspects of this intuition are often quite wrong. For example, it was not until the 16th century that it was widely realised that objects did not naturally come to rest when nothing was pushing them, or that the planets in our solar system and the flight of a tennis ball are described by the same laws of motion. During this course we will derive some results that seem obvious, but also some that are surprising. It pays to be careful: do not assume that answers to mechanics problems are always obvious. Limitations of classical mechanics Classical mechanics is the theory of how bodies (humans, bricks, planets, molecules,...) move in response to forces. Like most classical mechanics courses, this one ignores two important areas of physics. These have almost no direct impact on the motion of bodies as experienced in everyday life, but they matter at sub- atomic and galactic scales. 1 1. I NTRODUCTION No quantum effects: This is a spectacularly good approximation for objects larger than atoms. At temperatures of 1 K or lower, it is sometimes possi- ble to see individual atoms behaving quantum mechanically, but only just. At room temperature, although H and He atoms are very slightly affected by quantum mechanics, the motions of heavier atoms and larger objects are (almost) exactly as Newton described. Most chemical reactions, for example, can be described very well using Newton’s laws. The forces felt by the atoms depend on chemical bonds made of electrons, which are very quantum mechanical, but Newton’s second law tells you how those forces accelerate the atoms. No special or general relativity: This is also a very good approximation ev- erywhere on Earth for all objects moving at speeds v much slower than the speed of light c. Special relativity is essential if you want to under- stand how the fast-moving particles at CERN respond to forces. The core electrons of heavy atoms can have v/c ≈ 0.1, so special relativity also matters there. General relativity (Einstein’s theory of gravity) is important on galactic scales and near very massive objects such as black holes, but Newton’s laws are more than good enough for use on Earth. Tiny general (and also special) relativistic effects have to be taken into account in global positioning systems, which require absurd accuracy, but it is hard to think of other terrestrial examples. The fact that we neglect relativity and quantum theory may strike you as cheat- ing, but all theories of physics turn out to have limitations that restrict their range of validity. Most particle physicists now regard the standard model, once viewed as an essential part of the fundamental theory of everything, as an approxima- tion valid only at long enough length scales and low enough energy scales. Just as quantum theory lurks below classical mechanics, something even more fun- damental lurks below the standard model. For most aspects of the physics of everyday life, Newtonian mechanics is the best theory we have: an engineer try- ing to use quantum mechanics to design a bridge would be making their lives unnecessarily difficult, to say the least! Limitations of this Mechanics course This course does not cover every aspect of classical mechanics. Some of the more difficult areas are left for later in your degree. No fluid mechanics: The motion of (non-relativistic) fluids is governed by New- ton’s laws but is mathematically challenging. There is a fluid mechanics 2  option in the final year of your degree. No Lagrangian or Hamiltonian mechanics Although they look very different mathematically, Lagrangian and Hamiltonian mechanics are just reformu- lations of Newton’s laws: they describe exactly the same physics and pro- duce identical results. A few famous problems are easier to solve when reformulated, but the main use of Hamiltonian and Lagrangian mechanics is in quantum theory: Hamiltonian mechanics is central to non-relativistic quantum mechanics; and Lagrangian mechanics is at the heart of relativis- tic quantum field theory. Non-relativistic quantum mechanics is taught from the second year onwards; Lagrangian and Hamiltonian mechanics come in the third year; and quantum field theory is a final-year option. Assumptions The theory of classical mechanics rests on two assumptions so fundamental that they are almost philosophical. Time is universal: All observers, no matter where they are or how fast they are moving, agree on the time that elapses between any two events. Space is homogeneous, isotropic and Euclidean: A homogeneous space is one that looks the same wherever you are; an isotropic space looks the same in all directions; and a Euclidean space is the flat space familiar from school. Distances in a Euclidean space are given by Pythagoras’s theo- rem: r2 = x2 + y 2 + z 2. These assumptions seem so obviously true that it took thousands of years for people to realise that they were assumptions at all, and a few hundred more years to realise that they are wrong. Time is not universal in special relativity, and space-time is not homogeneous, isotropic and Euclidean in general relativity. Fortunately, although false, both assumptions are very accurate when v ≪ c, length scales are less than galactic, and gravity is not enormously strong. Definitions and notation A particle is a point-like object on which forces may act. A body is a particle or assembly of particles bound together. When trying to work out how bodies respond to forces, they can often be treated as particles. Off-centre forces cause bodies to start rotating, which point-like particles cannot do, but the centre of mass still moves like a particle. 3 1. I NTRODUCTION A reference frame (or ‘frame of reference’) is a set of coordinate axes used to measure positions. Particle positions, velocities and accelerations are vectors. In these notes, I will indicate a vector in bold face and will also add an arrow on top, to remind you that it is important for you to indicate in handwritten work that you are writing a vector. (In most books (and online materials etc), vectors are written in bold face without including the arrow.) In handwritten work, you should indicate the you are writing a vector quantity by an underline, r , or by an arrow on top, ⃗r. Both are fine. Unit vectors also have a hat (caret) on top. The unit vectors along the x, y and z axes are denoted ı̂, ȷ̂ and k̂. (Another common notation is x̂, ŷ, ẑ.) When writing unit vectors by hand using arrows to indicate vectors, the arrow may be omitted. Any vector ⃗ r can be expressed as the sum of its components along the three coordinate axes: ⃗ r = xı̂ + y ȷ̂ + z k̂.   x Often, as a kind of shorthand, we write ⃗ r =  y . r = (x, y, z) or ⃗ z Position, velocity and acceleration As I’m sure you already know, the velocity of a particle is the derivative of its position with respect to time. 4  In one dimension, the position is a scalar: a positive or negative number. You can, if you want, think of the position in one dimension as a one- component vector and underline it, but this is not necessary and rarely useful. The velocity dx v= dt is also a scalar. If v is positive, the particle is moving in the +x direction; if v is negative, it is moving in the −x direction. The speed, |v|, is never negative. (The vertical bars, called modulus signs, mean the absolute value of a number: |3| = | − 3| = 3.) In three dimensions, where the position ⃗ r (t) is a vector, the velocity d⃗ r ⃗ v= dt is also a vector. The absolute value of a three-dimensional vector such as ⃗ v is (by definition) its length: q v | = vx2 + vy2 + vz2. |⃗ 5 1. I NTRODUCTION Aside: How to differentiate vectors Vectors are differentiated component by component:   d⃗ r dx dy dz = , , dt dt dt dt dx dy dz = ı̂ + ȷ̂ + k̂. dt dt dt (You cannot differentiate a vector with respect to another vector.) One way to understand why differentiating component by component is the right thing to do is to start from the definition of the derivative as a limit:   d⃗ r r (t + δt) − ⃗ ⃗ r (t) = lim dt δt→0 δt x(t + δt)ı̂ + y(t + δt)ȷ̂ + z(t + δt))k̂ = lim δt→0 δt ! x(t)ı̂ + y(t)ȷ̂ + z(t)k̂ − δt  x(t + δt) − x(t) y(t + δt) − y(t) = lim ı̂ + ȷ̂ δt→0 δt δt  z(t + δt) − z(t) + k̂ δt     x(t + δt) − x(t) y(t + δt) − y(t) = lim ı̂ + lim ȷ̂ δt→0 δt δt→0 δt   z(t + δt) − z(t) + lim k̂ δt→0 δt dx dy dz = ı̂ + ȷ̂ + k̂. dt dt dt If an object is at ⃗ r1 at time t1 and ⃗ r2 at time t2 , its average velocity during that time interval is r (t2 ) − ⃗ ¯=⃗ r (t1 ) ∆⃗ r ⃗ v =. t2 − t1 ∆t In the limit as ∆t → 0 the average velocity tends to the instantaneous velocity, but the two can be very different for longer time intervals. Think, 6  for example, about a particle orbiting in a circle. Its velocity is never zero and its average speed is constant, but the average of its velocity around a complete orbit is exactly zero. Acceleration is the rate of change of velocity with respect to time. dv d2 x a= = 2 in 1D, dt dt d⃗ v d2⃗r ⃗ a= = 2 in 3D. dt dt Newton denoted time derivatives by dots, d⃗ r r˙ ≡ ⃗ (the ≡ sign means “is equivalent to”), dt and space derivatives by primes, dV V′ ≡. dx We often use Newton’s notation in this course, mainly because it is more concise than Leibniz’s d⃗ r /dt and dV /dx notation. Leibniz’s approach is normally more powerful, but not always clearer. Suppose, for example, that you want to write down the time derivative of ⃗r (t) at some special time t0 : Newton: r˙ (t0 ), ⃗   d⃗r d⃗ r Leibniz: or perhaps (t0 ). dt t=t0 dt The Newtonian version is clearer here. The SUVAT equations The SUVAT equations, v = u + at, 1 s = ut + at2 , 2 describe the motion of a body accelerating at a constant rate. The constant acceleration is a, the initial velocity is u, the time elapsed is t, the final velocity 7 1. I NTRODUCTION is v, and the distance travelled is s. The other SUVAT equations, such as v 2 = u2 + 2as, are easily derived from the two above. The SUVAT equations assume only that the acceleration is constant and stand independent of Newton’s laws, but Newton’s laws tell us when the SUVAT equa- tions are likely to apply. According to Newton’s second law, bodies accelerate at a constant rate if and only if the applied force is constant. If you are used to getting out the SUVAT equations without thinking in any and every situation involving a time, distance and an initial and final velocity, now is the time to “un- learn” that reflex — in this course, many or most of the situations you encounter will not correspond to constant acceleration. Derivation of v = u + at Start by stating that a is constant: dv = a = const. dt Now integrate both sides with respect to time: Z t Z t  dv(t)  dt = a dt.  t=0 dt t=0 The idea of this is OK but the notation is confused: the same symbol t cannot be used for the upper limit of integration, which has a specific value, and for the variable of integration, which varies from 0 to the upper limit. The simplest solution is to change the name of the variable of integration and rewrite the equation above as Z t Z t dv(t′ ) ′ ′ dt = a dt′. ′ t =0 dt ′ t =0 (This is still a bit confusing because Newton used primes to indicate derivatives with respect to x. The symbol called t′ here is just a new name for the variable of integration.) Since a is a constant, the integrals are easy to do: t t [v(t′ )]t′ =0 = [at′ ]t′ =0 ⇒ v(t) − v(0) = at ⇒ v(t) = u + at (as v(0) = u). 8 Derivation of s = ut + 12 at2 This time we start with v = u + at and integrate both sides to get: Z t Z t dx(t′ ) ′ ′ dt = (u + at′ ) dt′ t′ =0 dt t′ =0  t ′ t ′ 1 ′2 ⇒ [x(t )]t′ =0 = ut + at 2 t′ =0 1 ⇒ x(t) − x(0) = ut + at2 2 1 ⇒ s = ut + at2. 2 The SUVAT equations in three dimensions The vector versions of these derivations are no harder (try them) and give the SUVAT equations in three dimensions: ⃗ v=u ⃗ +⃗at, 1 2 ⃗ s=u ⃗t + ⃗ at , 2 where ⃗ r (t) − ⃗ s(t) = ⃗ r (0) is the vector displacement. 9 Chapter 2 Newton’s laws of motion and one-dimensional motion 2.1 Statement Isaac Newton formulated his laws of mechanics in the Philosophiæ Naturalis Principia Mathematica published in 1687. From his point of view they were empirical laws, verified by comparison with experiment. Newton’s laws can now be derived from quantum theory, but that is no less empirical in nature. At heart, all scientific theories are empirical. Newton’s first law: A body on which no forces act remains at rest or moves at a constant velocity. Newton’s second law: The rate of change of momentum of a body is propor- tional to the force applied to it. Newton’s third law: When two bodies interact, they exert equal and opposite forces on each other. We’ll consider Newton’s first two laws now and return to Newton’s third law later. 2.2 Newton’s first law Newton’s first law (N1) can be viewed as a special case of Newton’s second law (N2). If the applied force is zero, N2 says that the momentum is constant. Since 11 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION the momentum p is equal to mv in classical mechanics (this is a definition), it follows that the velocity is also constant. N1 follows from N2. It seems unlikely that Isaac Newton missed such a simple observation. Why, then, did he decide to state the first law separately? The main reason may well be historical. From the time of Aristotle (384–322 BC), the prevailing view had been that bodies slow down and come to rest unless a force is applied to keep them moving. Newton wanted to make sure his readers understood that this was wrong. Inertial frames A reinterpreted version of N1 remains useful today for another reason, which Newton himself may or may not have understood. The first thing to realise is that N1 does not always hold. If, for example, you draw your coordinate axes on the floor of an accelerating train carriage and drop a ball, you will see the ball accelerate backwards, just as if pushed by a force. We understand that this force is fictitious — its apparent existence is a side effect of the acceleration of the coordinate axes — but in order to make that deduction we have to know that the frame of reference is accelerating. How can we tell? We could look at the landscape passing the train window, but perhaps the rest of the world is accelerating backwards and the train is stationary? Suppose the train is in outer space, far from any visible landmarks? How can we work out whether it is accelerating then? We could check whether objects accelerate when none of the real forces we know about are acting on them, but perhaps there are forces we have not yet discovered? Frames of reference in which N1 holds are called inertial frames, and N1 is nowadays often reformulated as a statement about their existence. Reformulation of Newton’s first law: There exist frames of reference, known as inertial frames, in which bodies on which no forces act remain at rest or move at a constant velocity. Although the Earth’s surface is not an inertial frame, the rotation of the coor- dinate system is slow and the corresponding acceleration is small. It is usually small enough to ignore. Galilean relativity If you can find one inertial frame, any other frame moving at constant velocity with respect to the first is also inertial. The Galilean version of the principle of 12  2.3. Newton’s second law relativity states that any two such frames are completely equivalent. Neither can be considered “special” and the laws of physics must be the same in both. This does not mean that everything you can measure (velocity, position,...) has to be the same in both frames, but that the laws of physics relating the measured quantities to each other must have the same form. It is interesting to check whether Newton’s second law obeys this principle. Sup- pose the coordinate axes of inertial frame 2 move at constant velocity u⃗ relative to the coordinate axes of inertial frame 1. A body with velocity ⃗ v1 (t) as mea- sured in frame 1 has velocity ⃗ v1 (t) − u v2 (t) = ⃗ ⃗ as measured in frame 2. If a ⃗ force F (t) acts on the body, N2 in frame 1 reads: d⃗ v1 ⃗ (t). m =F dt Since u u/dt = ⃗0. (Note that a zero vector is still a vector, ⃗0 = ⃗ is constant, d⃗ (0, 0, 0), and should still be written in bold face or with an arrow/underline.) This means that d⃗ v2 d⃗ v1 d⃗ u d⃗ v1 ⃗ (t), m =m −m =m =F dt dt dt dt showing that N2 has exactly the same form in both frames. N2 does obey the Galilean principle of relativity. 2.3 Newton’s second law Why did we write N2 as F ⃗ = d⃗ p/dt instead of F ⃗ = md⃗ v /dt = m⃗ a? In classical mechanics, where the momentum p ⃗ is just another name for m⃗ v , the two equations are equivalent and the choice is irrelevant. In the Relativity course in term 3 you will see that, in special relativity, the equation F⃗ = d⃗ p/dt still 1 holds but the momentum is defined differently : m⃗ v ⃗≡ p p. (≡ means “is defined to be equal to”) 1 − v 2 /c2 Because of the more complicated ⃗ v dependence of the relativistic momentum, the F⃗ = md⃗v /dt = m⃗ a version of N2 no longer holds when relativistic effects are important. The essence of N2 in classical physics is the statement that the acceleration of an object is proportional to the force applied. If you double the force, the accelera- tion also doubles. There is a bit more to it than that, though. Although Newton 1 This equation is non-examinable for the Mechanics part of the course 13 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION did not state this explicitly, N2 also assumes that the mass is a property of the body itself; the value of F/a does not depend on where the body is located, what sort of force is applied, or whether it happens to be tea time. Measuring mass One can use N2 to establish a mass scale. First choose an object (until very re- cently this was a platinum cylinder in Paris) and say that it has mass m1 = 1. This defines the unit of mass. To measure the mass m2 of any other object, all you have to do is apply the same force to m1 and m2 and measure the accelera- tions ⃗ a1 and ⃗ a2. Since ⃗ = m1⃗ F a1 = m2⃗ a2 and m1 = 1, m1 |⃗ a1 | |⃗ a1 | m2 = =. |⃗ a2 | |⃗ a2 | If you can measure accelerations, you can also measure masses. How can you be sure that you have applied exactly the same force to both masses? One way would be to use the same spring. That might work, but a neater way is to allow the two masses to collide and exert forces on each other. By Newton’s third law (N3), the force exerted on mass 1 by mass 2 is equal and opposite to the force exerted on mass 2 by mass 1. Equal and opposite is not the same as equal, of course, but because m2 = |⃗ a1 |/|⃗ a2 | only depends on the absolute values of the two accelerations it is good enough. Groups of particles N2 holds (bar small quantum corrections) for each of the more than 1027 atoms in my body. Why does it also work for my body as a whole? The mystery is that, although every atom exerts forces on every other atom, only the externally applied forces appear in N2 for my body as a whole. What happened to all the internal forces? Start by thinking about a body made of only two particles and write down N2 for each: d⃗ p1 d⃗ p2 = f1ext + f2 on 1 , = f2ext + f1 on 2. dt dt Adding these two equations gives d⃗ p1 d⃗p2 + = f1ext + f2ext + f2 on 1 + f1 on 2 = f1ext + f2ext , dt dt 14  2.4. Inertial and gravitational mass ⃗ =p where the last step made use of N3. Since P ⃗1 + p ⃗2 is the total momentum ⃗ ext ext ext of the body and F = f1 + f2 is the total applied force, this is just N2 for the body as a whole: dP ⃗ ⃗ ext. =F dt The argument is easy to extend to bodies made of arbitrarily large numbers of particles. (Try it.) Notice that we would not be able to apply N2 to bodies consisting of many particles if N3 did not hold. Without N3, N2 would be more or less useless. We have just shown that the total momentum of an isolated system of interacting particles is always conserved, no matter how complicated the internal interac- tions. Nothing is truly isolated because gravitational and electromagnetic forces are so long ranged, but the external forces on many objects — the solar system, an astronaut in free fall, a free-wheeling bicycle — are often small enough to be ignored. It is important to understand that, although the total momentum of an isolated system is conserved, its total kinetic energy need not be. The total momentum of the two astronauts playing tug of war in problem sheet 2 remains constant before, during and after the collision, but their kinetic energy increases as they pull in on the rope and decreases when they run into each other. As they pull on the rope, potential energy stored in chemical bonds in their muscles is turned into kinetic energy. As their bodies crunch together, that kinetic energy becomes a mixture of elastic energy and heat energy. Sadly for the astronauts, people are not very springy, so the energy stored elastically soon turns to heat as well. Some of the kinetic energy might also be stored as potential energy associated with broken chemical bonds in fractured bones! One can argue that most of the heat energy left after the collision exists as vibra- tions of atoms and that this is also a kind of kinetic energy (or, more accurately, a rapidly interchanging mixture of kinetic energy and potential energy, as in a vibrating mass on a spring). This is true, but the kinetic energy associated with disordered atomic motion is not the sort of kinetic energy we talk about in classical mechanics. Only the kinetic energy associated with macroscopic (“human-scale”) motions of the body as a whole counts. 2.4 Inertial and gravitational mass The mass in F ⃗ = m⃗a measures the inertia of a body: it tells you how much force is required to make that body accelerate at 1 m · s−2. The mass in Newton’s law 15 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION of gravitation, F = GM m/r2 , tells you how hard gravity pulls on a body: it is a gravitational “charge”, analogous to the electrical charge in F = q1 q2 /4πε0 r2. Why should the inertial and gravitational masses, mI and mG , be the same? The two concepts appear unrelated. The electrical charge of a body is not propor- tional to its inertia. Why should gravity be different? The equivalence of inertial and gravitational mass was first investigated by Galileo in his famous (and probably apocryphal) Leaning Tower of Pisa experiment showing that different masses drop at the same rate. It has since been verified experimentally to about 1 part in 1012 , but that does not make the reason for the equivalence any more obvious. It is surprising that we had to wait for Albert Einstein to point out the strangeness of this equivalence and investigate its consequences. He noticed that the ficti- tious forces seen by observers living in accelerated frames of reference are pro- portional to the inertial mass and wondered whether gravity might also be a kind of fictitious force. Perhaps some of the frames we think of as inertial are not inertial after all? Einstein reasoned that the frame of reference attached to a free-falling lift, in which the unfortunate riders experience zero gravity, is (lo- cally) inertial, but that the frame associated with the Earth’s surface into which the lift plummets is not. This was the first step of the chain of reasoning that ultimately led him to the General Theory of Relativity. 2.5 Motion due to forces The force acting on an object may depend on its position (perhaps it is moving in a non-uniform electric or gravitational field), its velocity (if it is experiencing v × B forces), and time: drag, friction or magnetic q⃗ d2 ⃗ r m ⃗ (⃗ =F r˙ , t). r, ⃗ dt2 This vector differential equation is shorthand for three coupled differential equa- tions, d2 x m = Fx (x, y, z, ẋ, ẏ, ż, t), dt2 d2 y m 2 = Fy (x, y, z, ẋ, ẏ, ż, t), dt d2 z m 2 = Fz (x, y, z, ẋ, ẏ, ż, t), dt 16  2.6. Motion in one dimension which already look a bit daunting to solve. If there are lots of particles (the 6 × 1023 C atoms in a mole of diamond, for example), the force on one parti- cle depends on the coordinates of all of the others and the problem gets really difficult to solve. Newton’s equations of motion can only be solved analytically (that is, by using a pencil and paper to find a formula for ⃗ r (t)) in a few very simple cases. Even solving them on a computer is often challenging. Many dynamical systems are chaotic, which means that tiny differences in the initial positions and veloci- ties of the particles or bodies grow exponentially with time. To get accurate numerical results at long times, the initial conditions have to be specified to an exponentially large number of digits. Unfortunately, standard computers (using 8-byte real numbers) keep only about 14 significant figures. Predicting the fu- ture accurately using Newton’s laws is impossible in practice in chaotic systems. When trying to work out the weather a week from now, the best that one can do is run simulations starting from lots of slightly different initial conditions and use the set of results to estimate probabilities. Because of these difficulties, the best way to learn about classical mechanics is to study simple examples motivated by the real world. The knowledge we gain will help us derive general principles, especially conservation laws, which are often more useful than Newton’s laws themselves. 2.6 Motion in one dimension The position x, velocity v, and acceleration a of an object moving in one dimen- sion are one-dimensional vectors: the only directions they can have are + (right) or − (left). There is usually no need to write one-dimensional vectors in bold or underline them or put arrows above them, but you can if you want to and it sometimes helps to distinguish, say, between speed (which is always positive) and velocity (which can be positive or negative). One-dimensional vector spaces are valid mathematical objects. The point of working in one dimension is to keep the mathematics simple, but from in this course we move beyond the world of school-level physics in which almost all forces are constant, independent of time, position and velocity. The 17 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION SUVAT equations, which only apply when the acceleration is constant, will only rarely be useful. There are more examples of one-dimensional motion than you might imagine. Objects falling under gravity move along lines, as do trains on tracks, particles in head-on collisions, and bodies in circular motion. The circumference of the circle is admittedly not straight, but positions on it can be specified using a single variable, the angle of rotation θ, so the coordinate space is still one dimensional. Another reason for studying one-dimensional motion is that any three-dimensional motion can be resolved along any axis. The x-component of the force is easily obtained using a dot product, Fx = |F ⃗ | cos θ = |F ⃗ | |ı̂| cos θ = F ⃗ · ı̂, and only the x component causes acceleration in the x direction: mẍ = Fx. This means that the x-component of the motion is governed by the familiar one- dimensional version of Newton’s second law, the only proviso being that Fx may now depend on y and z (and perhaps also vy and vz ). 18  2.7. Impulse 2.7 Impulse You probably already know that impulse = force × time. This section shows how to work out the impulse when the force depends on time, as it does, for example, during a collision between two objects. You might reasonably ask whether assuming that the force depends on time is general enough? What about a spaceship moving through the solar system? The gravitational forces it feels depend on its position, not time. Fortunately, as the location of the spaceship ⃗ r (t) is a function of time, we can think of the force it experiences as a function of time too. We just have to understand that the force is to be measured along the track of the spaceship. If F ⃗ (⃗ r ) is the force as a ˜ ⃗ function of position, the corresponding time-dependent force F (t) is given by: ˜ ⃗ ⃗ (⃗ F (t) ≡ F r (t)). ˜ ⃗ ⃗ (⃗ ˜ ⃗ The mathematical forms of F (t) and F r ) are of course very different: F maps a single variable, t, to a force, whereas F⃗ (⃗r ) maps a three-dimensional vector, ⃗˜ ⃗ , for the ⃗ r , to a force. This explains why I have used different symbols, F and F two functions. To a mathematician this is the right thing to do. From the point of view of a physicist, however, both functions give the same result — the force on the spaceship — and assigning them different names is a bit fussy. From now on I will call both functions by the same name: F ⃗ (⃗ ⃗ (t). r ) and F Return now to working in one dimension and suppose that a time-dependent force F (t) acts from time ti to time tf (> ti ) on a mass m. Newton’s second law tells us that dv m = F (t) dt Z Z tf tf dv(t) ⇒ m dt = F (t)dt. ti dt ti We cannot do anything more with the right-hand side without knowing more about the form of F (t), which we are trying to keep general, but we can integrate the left-hand side to get: Z tf tf m [v(t)]ti = F (t)dt, ti and so 19 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION Z tf F (t)dt = mvf − mvi = change in momentum ti Rt The integral tif F (t)dt is called the impulse, often denoted J, and this equation tells R tf us that the impulse is equal to the change in momentum. The units of −2 ti F (t)dt are N·s. Since F = ma, 1 Newton-second = 1 kg· m· s · s = 1 kg· m· s−1. This is also the unit of momentum, mv. We have shown that the equation “impulse = change in momentum” works even when the applied force is not constant. This is no surprise but can be useful, especially in collisions. The forces acting during a collision are complicated and hard to work out, but their time integral, the impulse, is just the difference between the momentum before and after the collision, which is easy to measure. In “inelastic” collisions such as car crashes, some of the kinetic energy turns into heat and kinetic energy is not conserved, but Newton’s second law always holds and the change in momentum is still equal to the impulse. You already know how to differentiate a time-dependent vector component by component. Integrating vectors works the same way: Z Z  Z  Z  ⃗ F (t)dt ≡ Fx (t)dt ı̂ + Fy (t)dt ȷ̂ + Fz (t)dt k̂ Z Z Z  = Fx (t)dt, Fy (t)dt, Fz (t)dt. Once you know this, it is easy to derive the three-dimensional version of the impulse equation: Z tf ⃗ (t)dt = m⃗ F vf − m⃗ vi = change in momentum ti Try the derivation yourself! 2.8 One-dimensional motion with a velocity-dependent force Our aim here is to find the trajectory x(t) of an object moving in response to an externally applied force that depends on velocity. The SUVAT equations do 20  2.8. One-dimensional motion with a velocity-dependent force not hold, but we can still obtain analytic solutions in simple cases. The parti- cle velocity v(t) depends on time, so a velocity-dependent force is also a time- dependent force in the sense discussed in Sec. 2.7. This means that impulse is still a useful concept (although we are not going to use it in this example). Objects moving through fluids (gases or liquids) feel velocity-dependent drag forces of the form: ⃗ = −α(v)⃗ F v. The minus sign indicates that the force opposes the velocity. In detail, the speed- dependent function α(v) is complicated and depends on the type of fluid and the shape of the moving object, but it is more or less constant at low speed, when the flow is laminar, and more or less proportional to v at higher speed, when the flow is turbulent. The simplest model of friction says that the forces between solid objects in rela- tive motion are independent of their relative speed, although the direction of the force depends on the direction of the relative velocity. If N is the normal force acting between the two objects, µk is the kinetic friction coefficient, and ⃗ v is the relative velocity of the two objects, the frictional force is: ⃗ = −µk N v̂. F Note that v̂ is a unit vector in the ⃗ v direction. Example: Fluid drag proportional to speed At low speeds it is often a good approximation to assume that F = −αv, with α a constant. Suppose the wind pushing the sailing boat pictured on the next page drops at time t = 0, after which the boat gradually drifts to a stop. Newton’s second law 21 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION The velocity-dependent drag force on a sailing boat. reads: dv m = −αv dt 1 dv α ⇒ =− v dt m dv α ⇒ = − dt (“physicist’s” notation) v Z v ′ m α t ′ Z dv ⇒ ′ =− dt v0 v m 0 v α t ⇒ [ln v ′ ]v0 = − [t′ ]0 m α ⇒ ln v − ln v0 = − t   m v α ⇒ ln =− t v0 m ⇒ v = v0 e−αt/m  −t/τ m characteristic time ⇒ v = v0 e , where τ ≡ = α 1/e time The speed drops exponentially with time, decreasing by a factor of e every τ = m/α seconds. A graph of speed against time can be found near the end of this section. 22  2.8. One-dimensional motion with a velocity-dependent force Aside: Physicist’s Notation If you speak to some physicists (or mathematicians), they may tell you that the “physicist’s” notation (not standard terminology) we saw just now is incorrect and we “should” write something like this instead: 1 dv α =− vZ dt m t 1 dv(t′ ) ′ α t ′ Z ⇒ ′ ′ dt = − dt. 0 v(t ) dt m 0 Changing the variable of integration, setting u = v(t′ ), then gives Z v α t ′ Z 1 du = − dt , v0 u m 0 which is the same as line 4 of the version we just saw (although with v ′ renamed as u). The same notation is used when changing variables in integrals. Asked to work out Z t 1 dv(t′ ) ′ ′ ′ dt , 0 v(t ) dt dv(t′ ) ′ let u = v(t′ ). Then du = dt′ dt and Z t Z v(t) 1 dv ′ 1 dt = du. 0 v dt′ v(0) u Despite supposedly being incorrect, almost all working physicists and mathematicians are happy with physicist’s notation and use it all the time. In essence, in order to “multiply out” terms like dt, what we are really doing is treating dt as if it is δt, that is, a small but finite term, so that δt is a normal algebraic quantity which can be manipulated as any other x, y or whatever.a At the point of turning this into an integral we are then taking the limit that δt tends to zero, so that the ≈ becomes an exact = again, since in the limit of δt → 0, δv δt = dvdt. a In terms of notation, writing dt is really indicating an infinitesimal quantity, whereas δt indicates a small but finite change in t. ∆t would indicate a change in t which is not necessarily particularly small. Later in your Functions course, you will see the further notation ∂t; this is similar to dt but indicates a partial derivative. It indicates where, for a function of several variables, you are only considering variations in that one variable. 23 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION Aside: Physicist’s Notation (continued) If you really want to be a bit more explicit when using physicist’s notation you could write this out as for example: dv m = −αv dt 1 δv α ⇒ ≈− v δt m δv α ⇒ ≈ − δt v m In the limit dt → 0: Z Z dv α =− dt v m Z v ′ α t ′ Z dv ⇒ ′ =− dt , v0 v m 0 again the same as line 4 in the original working. 24  2.8. One-dimensional motion with a velocity-dependent force What is the range of the boat? To find that we need to integrate again: dx = v0 e−t/τ dt ⇒ dx = v0 e−t/τ dt Z x Z t ′ ′ ⇒ dx = v0 e−t /τ dt′ x0 0 h it −t′ /τ ⇒ x − x0 = −v0 τ e 0 −t/τ ⇒ x − x0 = −v0 τ e + v0 τ −t/τ ⇒ x − x0 = (1 − e )v0 τ. It is worth investigating how the position x depends on time when t ≪ τ and t ≫ τ: When t ≪ τ ,  2 −t/τ t 1 t e =1− + −... τ 2! τ t ≈1− τ and so x(τ ) ≈ x0 + v0 t. The boat starts at x0 and is initially moving at speed v0 , exactly as ex- pected. When t ≫ τ , e−t/τ ≈ 0 and 1 − e−t/τ ≈ 1, so x − x0 ≈ v0 τ , which is constant. The range is v0 τ. There is a graph of position versus time on the next page. 25 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION Boat speed versus time. Boat position versus time. 26  2.9. The work-energy theorem 2.9 The work-energy theorem We normally think of the velocity of a body and the force it experiences as functions of t. We might ask, for example: “how fast and in what direction was the car moving at 3pm?” If we choose to, however, we can equally well think of the velocity and applied force as functions of the position of the body. We could instead ask: “how fast and in what direction was the car moving when it passed the front door of the Blackett Laboratory?” This viewpoint is useful in the derivation of the work energy theorem. A body starts at position xi at time ti and moves to xf at time tf under the influ- ence of an applied force F that may depend on position, velocity and time. We are going to write the velocity and applied force as functions of the position x of the body, but this does not mean that they may not be different on different jour- neys from xi to xf. Once all the details of the journey (starting time, velocity profile) have been specified, we can work out the force and velocity as func- tions of the position x of the body, but the results may be different on different journeys. Although we are going to treat the velocity and applied force as functions of the position of the body, the position depends on t:  v = v(x) where x = x(t). F = F (x) This means that we can use the chain rule: dv dv dx dv 1 d v2.
= = v= dt dx dt dx 2 dx Substituting into Newton’s second law then gives 1 d v 2 = F (x),
m 2 dx which integrates to Z xf   Z xf d 1 2 mv dx = F (x)dx xi dx 2 xi  x Z xf 1 2 f ⇒ mv = F (x)dx 2 xi xi Z xf 1 2 1 2 ⇒ mv − mv = F (x)dx. 2 f 2 i xi We have derived: 27 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION The Work-Energy Theorem The change in kinetic energy of a body moving under the influence of an externally applied force is equal to the work done: Kf − Ki = Wif where Kf ≡ 12 mvf2 is the final kinetic energy of the body, Ki ≡ 21 mvi2 is the initial kinetic energy of the body, and Z xf Wif ≡ F (x)dx xi is the work done on the body. Notes Wif is the work done on the body by the external apparatus that applies the forces. By Newton’s third law, the work done by the body on the external apparatus is −Wif. The work done is measured in Joules: 1 J = 1 N · m = 1 kg · m2 · s−2 = 1 W · s. If F = const, Z xf Z xf Wif = F dx = F dx = F × (xf − xi ) = force × distance, xi xi as at school. The three-dimensional version of the work-energy theorem reads: Z ⃗ rf Kf − K i = ⃗ (⃗ F r ) · d⃗ r. ⃗ ri The left-hand side is exactly as in one dimension, but the right-hand side is more interesting. The integral is now along the path ⃗ r (t) taken by the body, which need not be a straight line. You will not need to use or under- stand path integrals for this course, but you will meet them again in your Functions course and use them in electromagnetism next term. 28  2.9. The work-energy theorem Aside: Path Integrals (not examinable) To understand the path integral that gives the work done in three dimen- sions, imagine dividing the path ⃗ r (t) into tiny segments, δ⃗ r , and working ⃗ out the scalar product, F (⃗r ) · δ⃗ r , for each segment. Because the segments are so small, the force can be approximated as a constant within each one, although it varies from segment to segment along the path. The value of the path integral is the limit as the segment length tends to zero of the sum of the tiny scalar products, F ⃗ (⃗r ) · δ⃗ r , from all of the segments along the path. The work δW associated with the segment from ⃗ r to ⃗ r + δ⃗ r is a dot ⃗ product, δW = F · δ⃗ r , because the force need not be parallel to δ⃗ r. Only ⃗ the component of F parallel to δ⃗ r does any work: ⃗ · δ⃗ δW = F ⃗ ||δ⃗ r = |F r | cos θ = F∥ |δ⃗ r |. Example A spring-like restoring force F = −3x N pulls a mss from x = 3 m to x = 2 m. How much work does it do? Z 2  2 3 3 3 15 Wif = (−3x)dx = − x2 = − 22 + 32 = = 7.5 J. 3 2 3 2 2 2 The final velocity vf is measured at x = 2 m; the initial velocity vi is measured at x = 3 m. The work done by the external force, 7.5 J, is equal to the change in kinetic energy of the mass. 29 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION 2.10 Power Power ≡ work done by external force per second Suppose that a body moves a distance δx in time δt. An externally applied force F , which may depend on time, position and velocity, acts during the displace- ment. According to Sec. 2.9, the work done during the infinitesimal move is δW = F δx, where F is the force currently acting on the body. The time interval δt is so short that the applied force hardly changes from beginning to end. Dividing both sides by δt, δW δx =F , δt δt and taking the limit as δt → 0 (so that δx also tends to 0) gives dW dx =F = F v. dt dt The power supplied by the external apparatus that applies the force is given by the formula: Power supplied = F v Given the power dW/dt as a function of time, the total work done between any time ti and any later time tf (> ti ) can be found by integration: Z tf Z tf Z xf dW dx Wif = dt = F dt = F dx. ti dt ti dt xi Section 2.9 explained that only the component of the force parallel to δ⃗ r does work in three dimensions. The displacement of a body moving at velocity ⃗ v for time δt is δ⃗ r =⃗v δt, so the three-dimensional formula for the power is: Power supplied = F ⃗ ·⃗v. 30  2.10. Power Example A mass m attached to a spring of spring constantp s undergoes simple harmonic motion of amplitude A at angular frequency ω = s/m: x(t) = A sin(ωt). Find an expression for the power supplied by the spring. The force applied by the spring is given by Hooke’s law: F (t) = −sx(t) = −sA sin(ωt). Differentiating the expression for x(t) gives the velocity: v(t) = Aω cos(ωt). These are the ingredients we need to work out the power supplied by the spring: Power(t) = F (t)v(t) = −sωA2 sin(ωt) cos(ωt) 1 = − sωA2 sin(2ωt). 2 Power flows from the spring, increasing the kinetic energy of the mass, when the mass is moving towards the origin, and back into the spring, decreasing the kinetic energy of the mass, when the mass is moving away from the origin. No work is done on average because the mass is oscillating backwards and forwards. The displacement x(t) and power supplied to an oscillating mass hanging from a spring. 31 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION Aside: Momentum versus Kinetic Energy Kinetic energy and momentum are both properties of bodies in motion, but there are important differences between them: Momentum Kinetic Energy Proportional to m. Proportional to m. A vector proportional to ⃗ v. A positive scalar proportional to v 2. Can be exchanged with other Can be exchanged with other bodies via forces. bodies via forces. Cannot convert to anything Can convert to other forms else. Always conserved. of energy (potential energy, heat,...). Although total en- ergy is conserved, the kinetic energy alone need not be. 2.11 One-dimensional motion with a position-dependent force This section concerns forces that depend on the position of a body but not on its velocity or on time. Gravitational forces, for which F = GM m/r2 , and electrostatic forces, for which F = q1 q2 /4πε0 r2 , are obvious examples. Almost all of the forces we experience in everyday life are gravitational or electrostatic in origin, so this is a very important special case. It is gravity that accelerates a sledge down a snowy hill, holds the atmosphere to the Earth, and keeps the Earth orbiting the Sun. It is electrostatic forces you feel when the Earth pushes up on your feet or you bump into something. Electrostatic forces also bind electrons to nuclei and atoms to each other, and so determine the chemical properties of matter. The only commonly experienced forces that are not functions of position alone are drag and friction (where the force depends on v or the sign of v) and Lorentz magnetic forces given by F ⃗ = q⃗ v × B. This section introduces the idea of a potential energy function, U (x), which provides an alternative description of any position-dependent force field F (x). (To a physicist, a “force field” is just a function F (x) that maps position to force.) Switching from a force-based description to a potential-based description brings 32  2.11. One-dimensional motion with a position-dependent force limited gains in one dimension, but is very useful in three dimensions. This is because the three-dimensional potential function, U (⃗r ), is a position-dependent ⃗ (⃗ scalar and easier to deal with than the three-dimensional force field, F r ), which is a position-dependent vector. Work done by a position-dependent force If the applied force only depends on position, the work done in moving a body from xi to xf is always the same: Z xf Wif = F (x)dx. xi The form of the function F (x) is given, so it does not matter when the body left xi , how fast it travelled, or whether it went past xf and back again. The work done is just the area under the curve F (x) between x = xi and x = xf. Contrast this with the case of a force that depends on t and/or v. The force experienced at position x may then be different on different journeys, so Wif will be journey- dependent too. The potential energy function The potential energy function U (x) is defined by the equation dU (x) (IMPORTANT) F (x) = − dx Make sure you don’t forget the minus sign! Only the value of −dU/dx matters (it’s the force), so we are free to add any constant to U (x). We say that U (x) is defined “to within an arbitrary constant.” It is usual to pick the arbitrary constant by: (a) Choosing a convenient point x0. (b) Setting U (x0 ) = 0. 33 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION Integrating F = −dU/dx from x0 to x then gives Z x Z x dU (x′ ) ′ ′ dx = − F (x′ )dx′ x0 dx Z xx0 ⇒ U (x) −  U 0) = − (x  F (x′ )dx′ , x0 where the U (x0 ) term vanished because we chose to set it to zero. We can now write U (x) as an integral: Z x U (x) = − F (x′ )dx′ (IMPORTANT) x0 In words, the potential function at x is minus the work done in moving from the reference point x0 to x. If the force is a function of time or velocity, the work done may depend on when theR x body′ left′ x0 or the speed at which it moved from x0 to x. The inte- gral − x0 F (x )dx from the reference point x0 to the final point x can still be evaluated, but the result depends on details of the journey from x0 to x and dif- ferent journeys give different answers. Since the value of the integral cannot be expressed as a function of x alone, it is impossible to define a potential energy function U (x). There is no equivalent of potential energy for friction and drag forces. Conservation of energy Since F = −dU/dx, the work done in moving from xi to xf can be expressed in terms of U : Z xf Z xf   dU (x) Wif = F (x)dx = − dx = −U (xf ) + U (xi ). xi xi dx Plugging this into the work-energy theorem, Kf − Ki = Wif , gives Kf − Ki = U (xi ) − U (xf ), which rearranges into a statement of the conservation of energy: Conservation of Energy Kf + U (xf ) = Ki + U (xi ) 34  2.11. One-dimensional motion with a position-dependent force Using only Newton’s laws, and assuming that the force F (x) is a function of position alone, we have shown that the sum of the kinetic and potential energies remains constant as the body moves around. The potential energy U (x) is stored in the external system (springs, petrol, grav- itational field, electric field) that applies the forces. As the body moves around, the external system may do work on the body, converting some of its stored potential energy into kinetic energy, or the body may do work on the external system, converting some of its kinetic energy into stored potential energy. The total energy K + U (x) is always conserved. Notes Since K + U (x) = const., K is also a function of x. Whenever the body passes position x, regardless of which way it is moving, it always has the same kinetic energy. Any force that depends only on x is called conservative. – Any conservative force field F (x) has a corresponding potential func- tion U (x). – Motion in any conservative force field conserves energy. Beware: – In classical mechanics, U (x) is the potential energy of the body at x. – In electromagnetism and electronics, U (x) is the potential energy per unit charge at x. The potential energy of the body is qU (x), where q is the body’s charge. Like all energies in classical mechanics, U (x) is a (positive or negative) scalar. Its gradient F (x) = −dU/dx is of course a vector. Over time, we have identified other forms of energy. These include heat energy (the random jiggling of atoms), field energy (stored in electric, magnetic, and gravitational fields), and mass energy (Einstein’s famous E = mc2 equivalence). It is an empirical observation that total energy is conserved once you have taken all stores of energy into account. Physicists believe in energy conservation so strongly that any evidence to the contrary immediately leads them to start looking for new types of energy to balance the books. So far, this has always worked out. A good example was the discovery of the neutrino. After experiments showed 35 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION an apparent failure of energy (and momentum and angular momentum) conservation in beta decay, Wolfgang Pauli proposed that a hitherto unseen particle must be responsible. This was later shown by experiments to be correct! Aside: Conservative Forces in 3D You might expect conservative force fields to be very rare in three dimen- sions. Even if the force depends on position only, there are so many pos- sible paths from the reference point ⃗ r0 to ⃗ r that it seems hard to imagine that Z ⃗r − ⃗ (⃗ F r ′ ) · d⃗ r′ ⃗ r0 could be independent of the path taken. If we cannot express the result of the integral as a simple function of the end point ⃗ r , we cannot define a potential energy function U (⃗r ). This reasoning is correct: if you invent a three-dimensional force field at random, it is very unlikely to correspond to a potential function. The three-dimensional force fields that occur in nature are not chosen at ran- dom, however, and many of them are conservative. Why this should be is not obvious, but perhaps it is because energy conservation is a fundamen- tal aspect of physical reality. The gravitational and electrostatic forces that govern the everyday aspects of the world around us are both conservative. Example: Gravity near the Earth Imagine throwing a ball of mass m upwards. The force acting on it is dU F (h) = − = −mg, dh 36  2.11. One-dimensional motion with a position-dependent force so the potential function U (h) is given by Z h Z h ′ ′ U (h) = − F (h )dh = mg dh′ = mgh − mgh0. h0 h0 If we choose h0 = 0, we get U (h) = mgh. The conservation of energy in a uniform gravitational field reads: 1 2 mv + mgh = constant = E. 2 Example: Gravity far from the Earth What is the gravitational potential energy far from the Earth? The force field is GM m F (r) = − (negative because force is in −r direction) r2 so Z r Z r ′ ′ GM m ′ GM m GM m U (r) = − F (r )dr = dr = − +. r0 r0 r′2 r r0 We now have to pick the position r0 at which we set U (r) to zero. Choosing r0 = 0 is no good because GM m/r0 → ∞ as r0 → 0. The simplest and standard choice is to set r0 to infinity, in which case   GM m dU GM m U (r) = −. Check: F (r) = − =− 2 ✓. r dr r 37 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION The positive number Z r −U (r) = F (r′ )dr′ (notice the minus sign) ∞ is the work done on m by the gravitational field as m moves in from ∞ to r. It is positive because F (r′ ) < 0 and dr′ < 0. The same positive number, −U (r), is the work done by m on the gravita- tional field as the particle moves out from r to ∞. 2.12 Potential functions and equilibrium The force F (x) corresponding to the potential function U (x) is given by dU (x) F (x) = − = − slope of U (x). dx To illustrate this idea, here is a diagram of a complicated potential function and the corresponding forces. 38  2.12. Potential functions and equilibrium The force arrows show that any minimum of U (x) is a point of stable equilib- rium: the force is zero at the minimum and pushes particles towards the mini- mum from either side. Any maximum of U (x) is a point of unstable equilibrium: the force is zero at the maximum but pushes particles away on either side. Since this force field was obtained by differentiating a potential function, it must be conservative, so the total energy must be conserved: E = K + U = const. We can use this and the simple observation that K = 21 mv 2 is always greater than or equal to zero to learn quite a lot about the motion of a particle in a potential U (x) without doing any mathematics at all. Cases E < U1 : Since E = K + U (x) and K is always greater than or equal to zero, the particle is restricted to the viable region shown in pink in the diagram below. 39 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION A particle moving to the right in the viable region accelerates right. A particle moving left decelerates until it runs out of kinetic energy at the point where E = U (x) and K = E − U (x) = 0. It then turns around starts accelerating off to the right, “bouncing off” the potential barrier. U1 < E < U2 : Now there are two viable regions. A particle that starts in the well oscillates back and forth, turning around every time it reaches one of the ends of the viable region. A particle that starts in the right-hand viable region behaves like the particle in the case when E < U1. The particle never has enough kinetic energy to surmount the potential barrier and move from one viable region to the other. E > U2 : The particle now has enough kinetic energy to move past the potential barrier and there is only one viable region again. 40  2.13. Small oscillations about equilibrium The behaviour of the particle is similar to the case when E < U1 , except it slows down and then speeds up again when travelling past the hump. Ball on a hill analogy When you feel uncertain about the behaviour of a body moving in a potential U (x), think what would happen to a ball rolling on a hilly landscape with height h(x) = U (x)/mg. The gravitational potential energy of the ball, mgh(x), is then equal to U (x). The analogy is imperfect because the ball on the hill is moving up or down the hillside instead of along the x axis, but it is close enough to give you the right idea. 2.13 Small oscillations about equilibrium We have shown that a particle bound in a potential well oscillates between the points A and B where E = U (x) (see the diagram on the next page). If A and B are close enough to the minimum of the potential function at x0 , which we can always arrange by reducing the total energy E towards U (x0 ), almost any smooth potential function is approximately quadratic in the region A ≤ x ≤ B explored by the particle. Imagine zooming in on the region near x0. The more you zoom in, the more quadratic the potential function appears to be. This can be shown mathematically by expanding U (x) in a Taylor series about its minimum at x = x0 : ′ 1 ′′ U (x0 )(x − x0 )2 ((− (( U (x) = U (x0 ) + ( U( (x(0 )(x ( x0 ) + 2! ( 1 ′′′ (((( + ( U((x 0 )(x ( (( −(x0 )3 +... 3! ( ( 41 2. N EWTON ’ S LAWS OF MOTION AND ONE - DIMENSIONAL MOTION The U ′ (x0 ) term is zero because dU/dx = 0 at the minimum of U (x). The cubic and higher order terms are not zero, but are small enough to be ignored when x − x0 is very small. If these terms are not small enough, all you have to do is reduce the energy, so reducing the maximum value of |x − x0 | and making them smaller. There are functions that do not become approximately quadratic when you zoom in on the minimum enough times — a V-shaped notch is one example — but any function with a Taylor expansion about its minimum (such functions are said to be “analytic at x0 ”) and a non-zero value of U ′′ (x0 ) is approximately quadratic when viewed on a small enough scale. Let’s assume that our potential well is one such function, so that 1 U (x) ≈ U (x0 ) + U ′′ (x0 )(x − x0 )2 2 when x is close enough to x0. Since F (x) = −dU (x)/dx, this is equivalent to F (x) = −U ′′ (x0 )(x − x0 ), which you should recognise as Hooke’s law, F = −s∆x, with spring constant s = U ′′ (x0 ). It follows that: Almost all small oscillations are simple harmonic! This surprising result has all sorts of interesting applications which you will hear much more about in your Oscillations & Waves course! 42 Chapter 3 Two-body dynamics Until now, most of this course has been about how single bodies move under the influence of externally applied forces. We now move on to look at pairs of bodies interacting with each other. The main topic will be collisions, but we will also look at how rockets use momentum conservation to accelerate by ejecting gas at high speed. The ejected gas is not a single body, so rocket motion is not quite a two-body problem, but it is similar in many respects. However, first of all, let’s return to think about Newton’s third law. 3.1 Newton’s third law Although it looks simple, Newton’s third law is often misunderstood. The first thing to realise is that the action and reaction forces in N3 always affect different bodies. Two forces acting on the same body are never an action-reaction pair. 43 3. T WO - BODY DYNAMICS Figure: Not an N3 action–reaction pair! The force of gravity, mg, pulling down on the brick in the figure is equal and opposite to the force N exerted on the brick by the table, but the two are not an action-reaction pair because both are acting on the brick. The reaction cor- responding to N is the force the brick exerts on the table top. The reaction corresponding to mg is the upward gravitational force the brick exerts on the Earth. The second thing to realise about N3 is that it admits no ifs or buts. If two objects are interacting, the force exerted by the first on the second is always equal and opposite to the force exerted by the second on the first. It does not make any difference whether the objects are moving at constant velocity or accelerating, and it does not matter whether you regard one object as pushing the other. In reality, they are always pushing each other. 3.2 The centre of mass and the reduced mass The idea of the centre of mass, introduced here for two interacting particles, is very general, applying to assemblies of arbitrarily many particles, including people and trees. The idea of reduced mass is much more limited, applying only to two-body systems, but is useful enough to be worth knowing about. Centre-of-mass motion Think about two bodies exerting forces on each other whilst also affected by external forces. 44  3.2. The centre of mass and the reduced mass Starting from Newton’s second law for both bodies, r¨ 1 = F m1⃗ ⃗2 on 1 + F ⃗1ext , r¨ 2 = F m2⃗ ⃗1 on 2 + F ⃗2ext , and adding the two equations gives r¨ 1 + m2⃗ m1⃗ r¨ 2 = F ⃗2 on 1 + F ⃗1 on 2 + F ⃗1ext + F ⃗2ext. ⃗1 on 2 and F The internal forces, F ⃗2 on 1 , cancel out by Newton’s third law, and the two external forces add to give the total external force, F⃗ ext , so we get m1⃗ r¨ 2 = F r¨ 1 + m2⃗ ⃗ ext ! r¨ 1 + m2⃗ m1 ⃗ r¨ 2 ⃗ ext ⇒ (m1 + m2 ) =F m1 + m2 d2 m1⃗   r1 + m2⃗ r2 ⃗ ext ⇒ M 2 =F (M ≡ m1 + m2 ). dt M If we now define The centre-of-mass position ⃗ ≡ m1⃗ R r1 + m2⃗ r2 m1 + m2 the result above can be written in a form that looks like Newton’s second law: ¨ ⃗ ⃗ ext MR =F As long as the internal forces obey Newton’s third law, the centre of mass accel- erates like a point particle of mass M = m1 + m2 , feeling only the total external force. 45 3. T WO - BODY DYNAMICS Notes ¨ ⃗ ext = ⃗0, then M R ⃗ ⃗˙ is If F = ⃗0 and the centre-of-mass momentum, M R, constant. It is easy to show (try it!) that an analogous result holds for N interacting bodies. The centre of mass, defined by PN N i=1 mi ⃗ ri X ⃗ R= , M= mi , M i=1 evolves according to Newton’s second law, ¨ ⃗ ⃗ ext , MR =F feeling only the total external force, N X ⃗ ext = F ⃗iext. F i=1 The centre of mass is the mass-weighted average position. To help under- ⃗ adding and stand this statement, it is useful to rewrite the definition of R, subtracting m2⃗ r1 to and from the numerator: ⃗ = (m1 + m2 )⃗ r1 + m2 (⃗r2 − ⃗r1 ) m2 R =⃗r1 + r2 − ⃗ (⃗ r1 ). M M The diagram below illustrates this vector equation: – R⃗ lies on the line from ⃗ r1 to ⃗r2. – If m1 > m2 , R ⃗ is closer to ⃗ r1. – If m2 > m1 , R ⃗ is closer to ⃗ r2. – ⃗ If m1 = m2 , R lies midway between ⃗ r1 and ⃗ r2. 46  3.2. The centre of mass and the reduced mass Example: Two masses on a spring Since F⃗ ext = m1⃗ g +m2⃗ g = M⃗ g , New- ton’s second law for the centre of mass reads: ¨ ⃗ ¨ ⃗ MR g ⇒ R = M⃗ =⃗ g. The centre of mass free falls, regardless of the internal forces. Internal motion The equation governing the internal motion of a two-body system can also be made to look like Newton’s second law. Return to the two-masses-on-a-spring r2 − ⃗ example above but concentrate on the separation vector ⃗ r1 instead of the ⃗ Newton’s second law for the two masses, centre-of-mass position R. r¨ 1 = F m1 ⃗ ⃗2 on 1 + m1⃗ ⃗1 on 2 + m1⃗ g = −F g, r¨ 2 = F m2 ⃗ ⃗1 on 2 + m2⃗ g, can be rearranged to give −F⃗1 on 2 r¨ 1 = ⃗ +⃗g, m1 ⃗1 on 2 F r¨ 2 = ⃗ +⃗g. m2 Subtracting the first equation from the second then gives d2   ⃗ 1 1 r2 − ⃗ (⃗ r1 ) = F1 on 2 +. dt2 m1 m2 If we re-express this in terms of the position of the second body relative to the first, r≡⃗ ⃗ r2 − ⃗r1 , and the reduced mass µ defined by Reduced mass 1 1 1 m1 m2 ≡ + (⇔ µ = ) µ m1 m2 m1 + m2 it looks like Newton’s second law: d2 ⃗ r F⃗1 on 2 = ⇒ r¨ = F µ⃗ ⃗1 on 2. dt2 µ 47 3. T WO - BODY DYNAMICS Notes The internal motion (of ⃗ r2 − ⃗ r =⃗ r1 ) is governed by the internal forces and the reduced mass µ. Warning: if the external forces cause the two bodies to accelerate at differ- ent rates, the internal motion is more complicated. The concept of reduced mass is only useful in two-body systems experiencing gravitational or zero external forces. Unlike the centre of mass, the concept of reduced mass does not generalise easily to more than two particles. Example: Two masses on a spring The internal coordinate ⃗ r = ⃗ r2 − ⃗ r1 evolves according to µ⃗r¨ = −s(⃗r−⃗ req ), where s is the spring constant and ⃗ req is the end-to-end vector of the spring at its natural length. The motion is one-dimensional, so we drop the vector notation from now on. Introducing a new variable d ≡ r − req , and noting that d¨ = r̈, the equation of motion becomes µd¨ = −sd. This is the equation of simple harmonic motion. If m1 = m2 , µ = m1 /2 and 2s d¨ = − d. m1 The factor of two on the right-hand side arises because the two masses feel equal and opposite internal forces and both accelerate. This makes the acceleration of the length of the spring twice as large as you might at first expect. If m2 ≫ m1 , µ ≈ m1 and s d¨ ≈ − d. m1 The factor of two has vanished because the second body is so massive that it hardly moves. It is like attaching m1 via a spring to a wall. 48  3.3. Collisions Summary External forces determine the motion of the centre of mass. The relevant mass is M = m1 + m2. Internal forces determine the internal motion. The relevant mass is µ = m1 m2 /(m1 + m2 ). 3.3 Collisions Collisions are brief interactions of two or more bodies m