Classical MechanicsAn introductory courseRichard FitzpatrickAssociate Professor of PhysicsThe University of Texas at AustinContents1 Introduction 71.1 Major sources: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 What is classical mechanics?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 mks units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Standard prexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.5 Other units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.6 Precision and signicant gures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.7 Dimensional analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 Motion in 1 dimension 182.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2 Displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.4 Acceleration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.5 Motion with constant velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.6 Motion with constant acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.7 Free-fall under gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Motion in 3 dimensions 323.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.2 Cartesian coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.3 Vector displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.4 Vector addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.5 Vector magnitude. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3523.6 Scalar multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.7 Diagonals of a parallelogram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.8 Vector velocity and vector acceleration . . . . . . . . . . . . . . . . . . . . . . . . . 373.9 Motion with constant velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.10 Motion with constant acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.11 Projectile motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.12 Relative velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444 Newtons laws of motion 534.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.2 Newtons rst law of motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.3 Newtons second law of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.4 Hookes law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.5 Newtons third law of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.6 Mass and weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.7 Strings, pulleys, and inclines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.8 Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.9 Frames of reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705 Conservation of energy 785.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 785.2 Energy conservation during free-fall . . . . . . . . . . . . . . . . . . . . . . . . . . 785.3 Work. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815.4 Conservative and non-conservative force-elds . . . . . . . . . . . . . . . . . . . . . 885.5 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9235.6 Hookes law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935.7 Motion in a general 1-dimensional potential . . . . . . . . . . . . . . . . . . . . . . 965.8 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 996 Conservation of momentum 1076.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.2 Two-component systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.3 Multi-component systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1126.4 Rocket science . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.5 Impulses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1186.6 Collisions in 1-dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1216.7 Collisions in 2-dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1277 Circular motion 1367.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1367.2 Uniform circular motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1367.3 Centripetal acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1387.4 The conical pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1417.5 Non-uniform circular motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1427.6 The vertical pendulum. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1487.7 Motion on curved surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1508 Rotational motion 1608.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1608.2 Rigid body rotation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1608.3 Is rotation a vector? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16248.4 The vector product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1668.5 Centre of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1688.6 Moment of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1728.7 Torque. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1798.8 Power and work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1848.9 Translational motion versus rotational motion . . . . . . . . . . . . . . . . . . . . . 1868.10 The physics of baseball . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1868.11 Combined translational and rotational motion. . . . . . . . . . . . . . . . . . . . . 1909 Angular momentum 2049.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2049.2 Angular momentum of a point particle . . . . . . . . . . . . . . . . . . . . . . . . . 2049.3 Angular momentum of an extended object . . . . . . . . . . . . . . . . . . . . . . . 2069.4 Angular momentum of a multi-component system. . . . . . . . . . . . . . . . . . . 20910 Statics 21710.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21710.2 The principles of statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21710.3 Equilibrium of a laminar object in a gravitational eld . . . . . . . . . . . . . . . . 22010.4 Rods and cables. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22310.5 Ladders and walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22610.6 Jointed rods. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22811 Oscillatory motion 23711.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23711.2 Simple harmonic motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237511.3 The torsion pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24111.4 The simple pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24211.5 The compound pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24511.6 Uniform circular motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24612 Orbital motion 25312.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25312.2 Historical background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25312.3 Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26212.4 Gravitational potential energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26512.5 Satellite orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26812.6 Planetary orbits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26913 Wave motion 27913.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27913.2 Waves on a stretched string . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27913.3 General waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28413.4 Wave-pulses. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28513.5 Standing waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28913.6 The Doppler effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29161 INTRODUCTION1 Introduction1.1 Major sources:The sources which I consulted most frequently whilst developing this course are:AnalyticalMechanics: G.R. Fowles, Third edition (Holt, Rinehart, & Winston, NewYork NY, 1977).Physics: R. Resnick, D. Halliday, and K.S. Krane, Fourth edition, Vol. 1 (John Wiley& Sons, New York NY, 1992).EncyclopdiaBrittanica: Fifteenthedition(EncyclopdiaBrittanica, ChicagoIL,1994).Physicsforscientistsandengineers: R.A.Serway, andR.J.Beichner, Fifthedition,Vol. 1 (Saunders College Publishing, Orlando FL, 2000).1.2 What is classical mechanics?Classical mechanicsisthestudyofthemotionofbodies(includingthespecialcaseinwhichbodiesremainatrest)inaccordancewiththegeneral principlesrst enunciated by Sir Isaac Newton in his Philosophiae Naturalis Principia Math-ematica (1687), commonly known as the Principia. Classical mechanics was therstbranchofPhysicstobediscovered, andisthefoundationuponwhichallother branches of Physics are built. Moreover, classical mechanics has many im-portant applications in other areas of science, such as Astronomy (e.g., celestialmechanics), Chemistry (e.g., the dynamics of molecular collisions), Geology (e.g.,the propagation of seismic waves, generated by earthquakes, through the Earthscrust), and Engineering (e.g., the equilibrium and stability of structures).Classi-cal mechanics is also of great signicance outside the realm of science.After all,the sequence of events leading to the discovery of classical mechanicsstartingwith the ground-breaking work of Copernicus, continuing with the researches ofGalileo, Kepler, and Descartes, and culminating in the monumental achievements71 INTRODUCTION 1.2 What is classical mechanics?of Newtoninvolved the complete overthrow of the Aristotelian picture of theUniverse, whichhadpreviouslyprevailedformorethanamillennium, anditsreplacementbyarecognizablymodernpictureinwhichhumankindnolongerplayed a privileged role.In our investigation of classical mechanics we shall study many different typesof motion, including:Translational motionmotion by which a body shifts from one point in space toanother (e.g., the motion of a bullet red from a gun).Rotational motionmotion by which an extended body changes orientation, withrespect to other bodies in space, without changing position (e.g., the motionof a spinning top).Oscillatory motionmotion which continually repeats in time with a xed period(e.g., the motion of a pendulum in a grandfather clock).Circular motionmotion by which a body executes a circular orbit about anotherxed body [e.g., the (approximate) motion of the Earth about the Sun].Ofcourse, thesedifferenttypesofmotioncanbecombined: forinstance, themotionof aproperlybowledbowlingball consistsof acombinationof trans-lationalandrotationalmotion, whereaswavepropagationisacombinationoftranslational and oscillatory motion. Furthermore, the above mentioned types ofmotion are not entirely distinct: e.g., circular motion contains elements of bothrotational and oscillatory motion. We shall also study statics: i.e., the subdivisionof mechanics which is concerned with the forces that act on bodies at rest andin equilibrium. Statics is obviously of great importance in civil engineering: forinstance, the principles of statics were used to design the building in which thislecture is taking place, so as to ensure that it does not collapse.81 INTRODUCTION 1.3 mks units1.3 mks unitsThe rst principle of any exact science is measurement. In mechanics there arethree fundamental quantities which are subject to measurement:1. Intervals in space: i.e., lengths.2. Quantities of inertia, or mass, possessed by various bodies.3. Intervals in time.Any other type of measurement in mechanics can be reduced to some combina-tion of measurements of these three quantities.Eachofthethreefundamental quantitieslength, mass, andtimeismea-sured with respect to some convenient standard. The system of units currentlyused by all scientists, and most engineers, is called the mks systemafter the rstinitials of the names of the units of length, mass, and time, respectively, in thissystem: i.e., the meter, the kilogram, and the second.The mks unit of length is the meter (symbol m), which was formerly the dis-tance between two scratches on a platinum-iridium alloy bar kept at the Inter-national Bureau of Metric Standard in S`evres, France, but is now dened as thedistance occupied by 1, 650, 763.73 wavelengths of light of the orange-red spectralline of the isotope Krypton 86 in vacuum.The mks unit of mass is the kilogram (symbol kg), which is dened as the massof a platinum-iridium alloy cylinder kept at the InternationalBureau of MetricStandard in S`evres, France.The mks unit of time is the second (symbol s), which was formerly dened interms of the Earths rotation,but is now dened as the time for9, 192, 631, 770oscillations associated with the transition between the two hyperne levels of theground state of the isotope Cesium 133.In addition to the three fundamental quantities, classical mechanics also dealswith derived quantities,such as velocity,acceleration,momentum,angular mo-91 INTRODUCTION 1.4 Standard prexesmentum, etc. Each of these derived quantities can be reduced to some particularcombination of length, mass, and time. The mks units of these derived quantitiesare, therefore, the corresponding combinations of the mks units of length, mass,and time. For instance, a velocity can be reduced to a length divided by a time.Hence, the mks units of velocity are meters per second:[v] =[L][T]= ms1. (1.1)Here, v stands for a velocity, L for a length, and T for a time, whereas the operator[ ] representstheunits, ordimensions, of thequantitycontainedwithinthebrackets.Momentum can be reduced to a mass times a velocity. Hence, the mksunits of momentum are kilogram-meters per second:[p] = [M][v] =[M][L][T]= kg ms1. (1.2)Here, p stands for a momentum, and M for a mass. In this manner, the mks unitsof all derived quantities appearing in classical dynamics can easily be obtained.1.4 Standard prexesmks units are specically designed to conveniently describe those motions whichoccur in everyday life.Unfortunately, mks units tend to become rather unwieldywhen dealing with motions on very small scales (e.g., the motions of molecules)orverylargescales(e.g., themotionofstarsintheGalaxy). Inordertohelpcope with this problem, a set of standard prexes has been devised, which allowthe mks units of length, mass, and time to be modied so as to deal more easilywith very small and very large quantities: these prexes are specied in Tab. 1.Thus, akilometer(km)represents103m, ananometer(nm)represents109m,and a femtosecond (fs) represents 1015s. The standard prexes can also be usedto modify the units of derived quantities.101 INTRODUCTION 1.5 Other unitsFactor Prex Symbol Factor Prex Symbol1018exa- E 101deci- d1015peta- P 102centi- c1012tera- T 103milli- m109giga- G 106micro- 106mega- M 109nano- n103kilo- k 1012pico- p102hecto- h 1015femto- f101deka- da 1018atto- aTable 1: Standard prexes1.5 Other unitsThe mks system is not the only system of units in existence. Unfortunately, theobsoletecgs(centimeter-gram-second)systemandtheevenmoreobsoletefps(foot-pound-second) system are still in use today, although their continued em-ployment is now strongly discouraged in science and engineering (except in theUS!). Conversionbetweendifferentsystemsofunitsis, inprinciple, perfectlystraightforward, but, inpractice, afrequent sourceof error. Witness, forex-ample, therecentlossoftheMarsClimateOrbiterbecausetheengineerswhodesigned its rocket engine used fps units whereas the NASA mission controllersemployedmksunits. Table2speciesthevariousconversionfactorsbetweenmks, cgs, and fps units. Note that, rather confusingly (unless you are an engineerin the US!), a pound is a unit of force, rather than mass. Additional non-standardunitsoflengthincludetheinch(1 ft =12 in), theyard(1 ya=3 ft), andthemile(1 mi =5, 280 ft). Additionalnon-standardunitsofmassincludetheton(intheUS, 1 ton=2, 000 lb; intheUK, 1 ton=2, 240 lb), andthemetricton(1 tonne=1, 000 kg). Finally, additional non-standard units of time include theminute (1 min=60 s), the hour (1 hr=3, 600 s), the day (1 da=86, 400 s), andthe year (1 yr = 365.26 da = 31, 558, 464 s).111 INTRODUCTION 1.6 Precision and signicant gures1cm = 102m1g = 103kg1ft = 0.3048 m1lb = 4.448 N1slug = 14.59 kgTable 2: Conversion factors1.6 Precision and signicant guresInthiscourse, youareexpectedtoperformcalculationstoarelativeaccuracyof 1%:i.e., to three signicant gures. Since rounding errors tend to accumulateduring lengthy calculations, the easiest way in which to achieve this accuracy is toperform all intermediate calculations to four signicant gures, and then to roundthe nal result down to three signicant gures. If one of the quantities in yourcalculation turns out to the the small difference between two much larger num-bers, then you may need to keep more than four signicant gures. Incidentally,you are strongly urged to use scientic notation in all of your calculations: theuse of non-scientic notation is generally a major source of error in this course.If your calculators are capable of operating in a mode in which all numbers (notjust very small or very large numbers) are displayed in scientic form then youare advised to perform your calculations in this mode.1.7 Dimensional analysisAs we have already mentioned, length, mass, and time are three fundamentallydifferent quantities which are measured in three completely independent units. It,therefore, makes no sense for a prospective law of physics to express an equalitybetween (say) a length and a mass. In other words, the example lawm = l, (1.3)where m is a mass and l is a length, cannot possibly be correct. One easy way ofseeing that Eq. (1.3) is invalid (as a law of physics), is to note that this equation isdependent on the adopted system of units: i.e., if m = l in mks units, then m = l121 INTRODUCTION 1.7 Dimensional analysisin fps units, because the conversion factors which must be applied to the left- andright-hand sides differ. Physicists hold very strongly to the assumption that thelaws of physics possess objective reality: in other words, the laws of physics arethe same for all observers. One immediate consequence of this assumption is thata law of physics must take the same form in all possible systems of units that aprospective observer might choose to employ. The only way in which this can bethe case is if all laws of physics are dimensionally consistent: i.e., the quantitieson the left- and right-hand sides of the equality sign in any given law of physicsmust have the same dimensions (i.e., the same combinations of length, mass, andtime). A dimensionally consistent equation naturally takes the same form in allpossible systems of units, since the same conversion factors are applied to bothsides of the equation when transforming from one system to another.As an example, let us consider what is probably the most famous equation inphysics:E = mc2. (1.4)Here, Eistheenergyof abody, misitsmass, andcisthevelocityof lightinvacuum. Thedimensionsofenergyare[M][L2]/[T2], andthedimensionsofvelocity are [L]/[T].Hence, the dimensions of the left-hand side are [M][L2]/[T2],whereas the dimensions of the right-hand side are[M] ([L]/[T])2=[M][L2]/[T2].It followsthat Eq. (1.4)isindeeddimensionallyconsistent. Thus, E=mc2holds good in mks units, in cgs units, in fps units, and in any other sensible setofunits. HadEinsteinproposedE=mc, orE=mc3, thenhiserrorwouldhave been immediately apparent to other physicists, since these prospective lawsare not dimensionally consistent. In fact, E=mc2represents the only simple,dimensionally consistent way of combining an energy, a mass, and the velocity oflight in a law of physics.The last comment leads naturally to the subject of dimensional analysis: i.e.,the use of the idea of dimensional consistency to guess the forms of simple lawsof physics.It should be noted that dimensional analysis is of fairly limited appli-cability, and is a poor substitute for analysis employing the actual laws of physics;nevertheless, it is occasionally useful. Suppose that a special effects studio wantstolmasceneinwhichtheLeaningTowerofPisatopplestotheground. Inorder to achieve this, the studio might make a scale model of the tower, which131 INTRODUCTION 1.7 Dimensional analysishmgFigure 1: The Leaning Tower of Pisais (say) 1m tall, and then lm the model falling over. The only problem is thatthe resulting footage would look completely unrealistic, because the model towerwould fall over too quickly. The studio could easily x this problem by slowingthe lm down. The question is by what factor should the lm be slowed down inorder to make it look realistic?Although, at this stage, we do not know how to apply the laws of physics tothe problem of a tower falling over, we can, at least, make some educated guessesas to what factors the timetfrequired for this process to occur depends on. Infact,it seems reasonable to suppose thattfdepends principally on the mass ofthe tower,m, the height of the tower,h, and the acceleration due to gravity, g.See Fig. 1. In other words,tf= Cmxhygz, (1.5)whereC is a dimensionless constant,andx, y,andz are unknown exponents.The exponentsx,y, andz can be determined by the requirement that the aboveequation be dimensionally consistent. Incidentally, the dimensions of an acceler-ation are [L]/[T2].Hence, equating the dimensions of both sides of Eq. (1.5), weobtain[T] = [M]x[L]y__ [L][T2]__z. (1.6)141 INTRODUCTION 1.7 Dimensional analysisWecannowcomparetheexponentsof [L], [M], and[T] oneithersideoftheaboveexpression: theseexponentsmust all match inorder forEq.(1.5)tobedimensionally consistent. Thus,0 = y +z, (1.7)0 = x, (1.8)1 = 2 z. (1.9)It immediately follows that x = 0, y = 1/2, and z = 1/2. Hence,tf= C_hg. (1.10)Now, the actual tower of Pisa is approximately 100m tall. It follows that sincetf h(gisthesameforboththerealandthemodeltower)thenthe1mhigh model tower falls over a factor of _100/1=10 times faster than the realtower. Thus,the lm must be slowed down by a factor 10 in order to make itlook realistic.Worked example 1.1: Conversion of unitsQuestion: Farmer Jones has recently brought a 40 acre eld and wishes to replacethe fence surrounding it. Given that the eld is square, what length of fencing (inmeters) should Farmer Jones purchase? Incidentally, 1 acre equals 43,560 squarefeet.Answer: If 1 acre equals 43,560 ft2and 1 ft equals 0.3048 m (see Tab. 2) then1 acre = 43560 (0.3048)2= 4.047 103m2.Thus, the area of the eld in mks units isA = 40 4.047 103= 1.619 105m2.Now, a square eld with sides of length l has an area A = l2and a circumferenceD = 4l. Hence, D = 4A. It follows that the length of the fence isD = 4 _1.619 105= 1.609 103m.151 INTRODUCTION 1.7 Dimensional analysisWorked example 1.2: Tire pressureQuestion: The recommended tire pressure in a Honda Civic is 28 psi (pounds persquare inch). What is this pressure in atmospheres (1 atmosphere is 105Nm2)?Answer: First, 28 pounds per square inch is the same as 28(12)2= 4032 poundsper square foot (the standard fps unit of pressure). Now, 1 pound equals4.448Newtons (the standard SI unit of force), and 1 foot equals 0.3048 m (see Tab. 2).Hence,P= 4032 (4.448)/(0.3048)2= 1.93 105Nm2.It follows that 28 psi is equivalent to 1.93 atmospheres.Worked example 1.3: Dimensional analysisQuestion: The speed of sound v in a gas might plausibly depend on the pressure p,the density , and the volume V of the gas. Use dimensional analysis to determinethe exponents x, y, and z in the formulav = CpxyVz,whereC is a dimensionless constant. Incidentally, the mks units of pressure arekilograms per meter per second squared.Answer: Equatingthedimensionsofbothsidesoftheaboveequation, weob-tain[L][T]=__[M][T2][L]__x__[M][L3]__y[L3]z.A comparison of the exponents of[L], [M],and[T] on either side of the aboveexpression yields1 = x 3y +3z,0 = x +y,1 = 2x.161 INTRODUCTION 1.7 Dimensional analysisThe third equation immediately givesx=1/2; the second equation then yieldsy = 1/2; nally, the rst equation gives z = 0. Hence,v = C_p.172 MOTION IN 1 DIMENSION2 Motion in 1 dimension2.1 IntroductionThe purpose of this section is to introduce the concepts of displacement, velocity,and acceleration. For the sake of simplicity, we shall restrict our attention to 1-dimensional motion.2.2 DisplacementConsider a body moving in 1 dimension: e.g., a train traveling down a straightrailroad track, or a truck driving down an interstate in Kansas. Suppose that wehave a team of observers who continually report the location of this body to usas time progresses. To be more exact, our observers report the distancex of thebody from some arbitrarily chosen reference point located on the track on whichitisconstrainedtomove. Thispointisknownastheoriginofourcoordinatesystem. A positive x value implies that the body is located x meters to the right ofthe origin, whereas a negativex value implies that the body is located|x| metersto the left of the origin. Here,x is termed the displacement of the body from theorigin. See Fig. 2. Of course, if the body is extended then our observers will haveto report the displacement x of some conveniently chosen reference point on thebody (e.g., its centre of mass) from the origin.Our information regarding the bodys motion consists of a set of data points,eachspecifyingthedisplacement xof thebodyat sometimet. It isusuallyilluminating to graph these points. Figure 3 shows an example of such a graph.As is often the case,it is possible to t the data points appearing in this graphusing a relatively simple analytic curve.Indeed, the curve associated with Fig. 3isx = 1 +t +t22t44 . (2.1)182 MOTION IN 1 DIMENSION 2.3 Velocity2.3 VelocityBoth Fig. 3 and formula (2.1) effectively specify the location of the body whosemotion we are studying as time progresses. Let us now consider how we can usethis information to determine the bodys instantaneous velocity as a function oftime. The conventional denition of velocity is as follows:Velocityistherateofchangeofdisplacementwithtime.This denition implies thatv =xt, (2.2)wherev is the bodys velocity at timet, andx is the change in displacement ofthe body between times t and t +t.How should we choose the time interval t appearing in Eq. (2.2)? Obviously,inthesimplecaseinwhichthebodyismovingwithconstantvelocity, wecanmake t as large or small as we like, and it will not affect the value of v. Suppose,however, thatv is constantly changing in time, as is generally the case. In thissituation, tmustbekeptsufcientlysmall thatthebodysvelocitydoesnotchangeappreciablybetweentimestandt +t. If tismadetoolargethenformula (2.2) becomes invalid.Suppose that we require a general expression for instantaneous velocity whichis valid irrespective of how rapidly or slowly the bodys velocity changes in time.We can achieve this goal by taking the limit of Eq. (2.2) ast approaches zero.This ensures that no matter how rapidlyv varies with time,the velocity of thex = 0xorigin track bodydisplacementFigure 2: Motion in 1 dimension192 MOTION IN 1 DIMENSION 2.3 VelocityFigure 3: Graph of displacement versus timebody is always approximately constant in the interval t to t +t. Thus,v =limt0xt=dxdt, (2.3)where dx/dt represents the derivative of x with respect to t. The above denitionis particularly useful if we can representx(t) as an analytic function, because itallows us to immediately evaluate the instantaneous velocity v(t) via the rules ofcalculus. Thus, if x(t) is given by formula (2.1) thenv =dxdt= 1 +t t3. (2.4)Figure 4 shows the graph of v versus t obtained from the above expression. Notethatwhenvispositivethebodyismovingtotheright(i.e., xisincreasingintime). Likewise, whenvisnegativethebodyismovingtotheleft(i.e., xisdecreasing in time). Finally, when v = 0 the body is instantaneously at rest.The terms velocity and speed are often confused with one another. A velocitycanbeeitherpositiveornegative, dependingonthedirectionofmotion. Theconventional denition of speed is that it is the magnitude of velocity (i.e., it isvwith the sign stripped off). It follows that a body can never possess a negativespeed.202 MOTION IN 1 DIMENSION 2.4 AccelerationFigure 4: Graph of instantaneous velocity versus time associated with the motion specied in Fig. 32.4 AccelerationThe conventional denition of acceleration is as follows:Accelerationistherateofchangeofvelocitywithtime.This denition implies thata =vt, (2.5)wherea is the bodys acceleration at timet, andv is the change in velocity ofthe body between times t and t +t.Howshouldwechoosethetimeinterval tappearinginEq.(2.5)? Again,inthesimplecaseinwhichthebodyismovingwith constantacceleration, wecan maket as large or small as we like,and it will not affect the value ofa.Suppose, however, that a is constantly changing in time, as is generally the case.In this situation, t must be kept sufciently small that the bodys accelerationdoes not change appreciably between times t and t +t.A general expression for instantaneous acceleration, which is valid irrespectiveof how rapidly or slowly the bodys acceleration changes in time, can be obtained212 MOTION IN 1 DIMENSION 2.4 AccelerationFigure 5: Graph of instantaneous acceleration versus time associated with the motion specied inFig. 3by taking the limit of Eq. (2.5) as t approaches zero:a =limt0vt=dvdt=d2xdt2. (2.6)The above denition is particularly useful if we can representx(t) as an analyticfunction, because it allows us to immediately evaluate the instantaneous acceler-ation a(t) via the rules of calculus. Thus, if x(t) is given by formula (2.1) thena =d2xdt2= 1 3t2. (2.7)Figure 5 shows the graph ofa versus time obtained from the above expression.Notethatwhenaispositivethebodyisacceleratingtotheright(i.e., visin-creasing in time).Likewise, whena is negative the body is decelerating (i.e.,v isdecreasing in time).Fortunately, it is generally not necessary to evaluate the rate of change of ac-celerationwithtime, sincethisquantitydoesnotappearinNewtonslawsofmotion.222 MOTION IN 1 DIMENSION 2.5 Motion with constant velocityxx0tx00t Figure 6: Graph of displacement versus time for a body moving with constant velocity2.5 Motion with constant velocityThe simplest type of motion (excluding the trivial case in which the body underinvestigation remains at rest) consists of motion with constant velocity. This typeof motion occurs in everyday life whenever an object slides over a horizontal, lowfriction surface: e.g., a puck sliding across a hockey rink.Fig. 6 shows the graph of displacement versus time for a body moving withconstant velocity. It can be seen that the graph consists of a straight-line. Thisline can be represented algebraically asx = x0 +v t. (2.8)Here, x0 is the displacement at time t =0: this quantity can be determined fromthe graph as the intercept of the straight-line with the x-axis. Likewise, v = dx/dtistheconstantvelocityofthebody: thisquantitycanbedeterminedfromthegraph as the gradient of the straight-line (i.e., the ratiox/t, as shown). Notethat a = d2x/dt2= 0, as expected.Fig. 7 shows a displacement versus time graph for a slightly more complicatedcase of motion with constant velocity. The body in question moves to the right232 MOTION IN 1 DIMENSION 2.6 Motion with constant accelerationtx0C D E A B Figure 7: Graph of displacement versus time(since x is clearly increasing with t) with a constant velocity (since the graph is astraight-line) between times Aand B. The body then moves to the right (since x isstill increasing in time) with a somewhat larger constant velocity (since the graphis again a straight line, but possesses a larger gradient than before) between timesB and C. The body remains at rest (since the graph is horizontal) between timesC and D. Finally, the body moves to the left (since x is decreasing with t) with aconstant velocity (since the graph is a straight-line) between times D and E.2.6 Motion with constant accelerationMotion with constant acceleration occurs in everyday life whenever an object isdropped:the object moves downward with the constant acceleration9.81 ms2,under the inuence of gravity.Fig. 8 shows the graphs of displacement versus time and velocity versus timefor a body moving with constant acceleration. It can be seen that the displacement-time graph consists of a curved-line whose gradient (slope) is increasing in time.242 MOTION IN 1 DIMENSION 2.6 Motion with constant acceleration0t00xx0t00t vvvFigure 8:Graphs of displacement versus time and velocity versus time for a body moving with con-stant accelerationThis line can be represented algebraically asx = x0 +v0t +12 at2. (2.9)Here, x0 is the displacement at time t =0: this quantity can be determined fromthe graph as the intercept of the curved-line with thex-axis. Likewise, v0is thebodys instantaneous velocity at time t = 0.Thevelocity-timegraphconsistsofastraight-linewhichcanberepresentedalgebraically asv =dxdt= v0 +at. (2.10)252 MOTION IN 1 DIMENSION 2.7 Free-fall under gravityThequantityv0isdeterminedfromthegraphastheinterceptofthestraight-linewiththex-axis. Thequantityaistheconstantacceleration: thiscanbedetermined graphically as the gradient of the straight-line (i.e., the ratiov/t,as shown). Note that dv/dt = a, as expected.Equations (2.9) and (2.10) can be rearranged to give the following set of threeuseful formulae which characterize motion with constant acceleration:s = v0t +12 at2, (2.11)v = v0 +at, (2.12)v2= v20+2 as. (2.13)Here, s = x x0 is the net distance traveled after t seconds.Fig. 9 shows a displacement versus time graph for a slightly more complicatedcase of accelerated motion. The body in question accelerates to the right [sincethe gradient (slope) of the graph is increasing in time] between timesA andB.The body then moves to the right (sincex is increasing in time) with a constantvelocity (since the graph is a straight line) between timesB andC. Finally, thebody decelerates [since the gradient (slope) of the graph is decreasing in time]between times C and D.2.7 Free-fall under gravityGalileo Galilei was the rst scientist to appreciate that, neglecting the effect of airresistance, all bodies in free-fall close to the Earths surface accelerate verticallydownwards with the same acceleration:namely,g=9.81 ms2.1The neglect ofair resistance is a fairly good approximation for large objects which travel rela-tively slowly (e.g., a shot-putt, or a basketball), but becomes a poor approxima-tion for small objects which travel relatively rapidly (e.g., a golf-ball, or a bulletred from a pistol).1Actually, the acceleration due to gravity varies slightly over the Earths surface because of the combined effectsof the Earths rotation and the Earths slightly attened shape. The acceleration at the poles is about9.834 ms2,whereas the acceleration at the equator is only 9.780 ms2. The average acceleration is 9.81 ms2.262 MOTION IN 1 DIMENSION 2.7 Free-fall under gravityx0A BC DtFigure 9: Graph of displacement versus timeEquations (2.11)(2.13) can easily be modied to deal with the special caseof an object free-falling under gravity:s = v0t 12 gt2, (2.14)v = v0 gt, (2.15)v2= v202 gs. (2.16)Here, g = 9.81 ms2is the downward acceleration due to gravity, s is the distancethe object has moved vertically between times t = 0 and t (if s > 0 then the objecthas risens meters, else ifs 0) that v = u.In other words, the ball hits the ground with an equal and opposite velocity tothat with which it was thrown into the air. Since the ascent and decent phases ofthe balls trajectory are clearly symmetric, the balls time of ight is simply twicethe time required for the ball to attain its maximum height: i.e.,t =2 ug. (2.19)Worked example 2.1: Velocity-time grapht (s)v (m/s)4 8 12 16 0048282 MOTION IN 1 DIMENSION 2.7 Free-fall under gravityQuestion: Consider the motion of the object whose velocity-time graph is givenin the diagram.1. What is the acceleration of the object between times t = 0 and t = 2?2. What is the acceleration of the object between times t = 10 and t = 12?3. What is the net displacement of the object between times t = 0 and t = 16?Answer:1. The v-t graph is a straight-line between t = 0 and t = 2, indicating constantacceleration during this time period. Hence,a =vt=v(t = 2) v(t = 0)2 0=8 02= 4 ms2.2. Thev-t graph is a straight-line betweent=10 andt=12, indicating con-stant acceleration during this time period. Hence,a =vt=v(t = 12) v(t = 10)12 10=4 82= 2 ms2.The negative sign indicates that the object is decelerating.3. Now, v = dx/dt, sox(16) x(0) =

160v(t) dt.In other words, the net displacement between times t = 0 and t = 16 equalsthe area under thev-t curve, evaluated between these two times. Recallingthat the area of a triangle is half its width times its height,the number ofgrid-squares under the v-t curve is 25. The area of each grid-square is 22 =4 m. Hence,x(16) x(0) = 4 25 = 100 m.292 MOTION IN 1 DIMENSION 2.7 Free-fall under gravityWorked example 2.2: Speed trapQuestion: In a speed trap, two pressure-activated strips are placed 120 m apart ona highway on which the speed limit is85 km/h.A driver going110 km/h noticesa police car just as he/she activates the rst strip, and immediately slows down.What deceleration is needed so that the cars average speed is within the speedlimit when the car crosses the second strip?Answer: Letv1=110 km/h be the speed of the car at the rst strip. Letx=120 m be the distance between the two strips, and let t be the time taken by thecar to travel from one strip to the other. The average velocity of the car isv =xt.We need this velocity to be 85 km/h. Hence, we requiret =xv=12085 (1000/3600)= 5.082 s.Here, we have changed units from km/h to m/s. Now, assuming that the accel-eration a of the car is uniform, we havex = v1t +12 a(t)2,which can be rearranged to givea =2 (x v1t)(t)2=2 (120 110 (1000/3600) 5.082)(5.082)2= 2.73 ms2.Hence, the required deceleration is 2.73 ms2.Worked example 2.3: The Brooklyn bridgeQuestion:In 1886, Steve Brodie achieved notoriety by allegedly jumping off therecentlycompletedBrooklynbridge, forabet, andsurviving. Giventhatthe302 MOTION IN 1 DIMENSION 2.7 Free-fall under gravitybridge rises 135ft over the East River, how long would Mr. Brodie have been inthe air, and with what speed would he have struck the water?Give all answersin mks units. You may neglect air resistance.Answer: Mr. Brodies net vertical displacement was h = 1350.3048 = 41.15 m.Assuming that his initial velocity was zero,h = 12 gt2,where t was his time of ight. Hence,t =_2 hg=_2 41.159.81= 2.896 s.His nal velocity wasv = gt = 9.81 2.896 = 28.41 ms1.Thus,the speed with which he plunged into the East River was28.41 ms1,or63.6 mi/h! Clearly, Mr. Brodies story should be taken with a pinch of salt.313 MOTION IN 3 DIMENSIONS3 Motion in 3 dimensions3.1 IntroductionThe purpose of this section is to generalize the previously introduced concepts ofdisplacement, velocity, and acceleration in order to deal with motion in 3 dimen-sions.3.2 Cartesian coordinatesOur rst task,when dealing with 3-dimensional motion,is to set up a suitablecoordinate system. The most straight-forward type of coordinate system is calleda Cartesian system, after Rene Descartes. A Cartesian coordinate system consistsof three mutually perpendicular axes, the x-, y-, and z-axes (say). By convention,the orientation of these axes is such that when the index nger, the middle nger,and the thumb of the right-hand are congured so as to be mutually perpendic-ular, the index nger, the middle nger, and the thumb can be aligned along thex-, y-, and z-axes, respectively. Such a coordinate system is termed right-handed.See Fig. 10. The point of intersection of the three coordinate axes is termed theorigin of the coordinate system.yxz(middle finger)(index finger)(thumb)Figure 10: A right-handed Cartesian coordinate system323 MOTION IN 3 DIMENSIONS 3.3 Vector displacementyxzROrFigure 11: A vector displacement3.3 Vector displacementConsider the motion of a body moving in 3 dimensions. The bodys instantaneousposition is most conveniently specied by giving its displacement from the originof our coordinate system. Note, however, that in 3 dimensions such a displace-ment possesses both magnitude and direction. In other words, we not only haveto specify how far the body is situated from the origin, we also have to specifyin which direction it lies. A quantity which possesses both magnitude and direc-tion is termed a vector. By contrast, a quantity which possesses only magnitudeistermedascalar. Massandtimearescalarquantities. However, ingeneral,displacement is a vector.The vector displacement r of some point R from the origin O can be visualizedas an arrow running from pointO to pointR. See Fig. 11. Note that in typesetdocuments vector quantities are conventionally written in a bold-faced font (e.g.,r) to distinguish them from scalar quantities. In free-hand notation, vectors areusually under-lined (e.g., r).The vector displacement r can also be specied in terms of its coordinates:r = (x, y, z). (3.1)The above expression is interpreted as follows: in order to get from pointO topointR, rst movex meters along thex-axis (perpendicular to both they- andz-axes), then movey meters along they-axis (perpendicular to both thex- and333 MOTION IN 3 DIMENSIONS 3.4 Vector additionSRr1r2rOFigure 12: Vector additionz-axes), nally move z meters along the z-axis (perpendicular to both the x- andy-axes). Note that a positivex value is interpreted as an instruction to movexmeters along the x-axis in the direction of increasing x, whereas a negative x valueis interpreted as an instruction to move |x| meters along the x-axis in the oppositedirection, and so on.3.4 Vector additionSupposethatthevectordisplacementrofsomepoint RfromtheoriginOisspecied as follows:r = r1 +r2. (3.2)Figure 12 illustrates how this expression is interpreted diagrammatically: in orderto get from point O to point R, we rst move from point O to point S along vectorr1, and we then move from point S to point R along vector r2. The net result is thesame as if we had moved from pointO directly to pointR along vector r.Vectorr is termed the resultant of adding vectors r1 and r2.NotethatwehavetwowaysofspecifyingthevectordisplacementofpointSfromtheorigin: wecaneitherwriter1orr r2. Theexpressionr r2isinterpreted as follows: starting at the origin, move along vector r in the directionofthearrow, thenmovealongvectorr2inthe oppositedirectiontothearrow.In other words, a minus sign in front of a vector indicates that we should movealong that vector in the opposite direction to its arrow.343 MOTION IN 3 DIMENSIONS 3.5 Vector magnitudeSuppose that the components of vectors r1 and r2 are (x1, y1, z1) and (x2, y2, z2),respectively. As is easily demonstrated, the components(x, y, z) of the resultantvector r = r1 +r2 arex = x1 +x2, (3.3)y = y1 +y2, (3.4)z = z1 +z2. (3.5)In other words, the components of the sum of two vectors are simply the algebraicsums of the components of the individual vectors.3.5 Vector magnitudeIf r = (x, y, z) represents the vector displacement of point R from the origin, whatis the distance between these two points?In other words, what is the length, ormagnitude,r=|r|, of vector r. It follows from a 3-dimensional generalization ofPythagoras theorem thatr =_x2+y2+z2. (3.6)Note that if r = r1 +r2 then|r| |r1| + |r2|. (3.7)Inotherwords, themagnitudesofvectorscannot, ingeneral, beaddedalge-braically.The only exception to this rule (represented by the equality sign in theabove expression) occurs when the vectors in question all point in the same di-rection. Accordingtoinequality(3.7), ifwemove1mtotheNorth(say)andnext move 1m to the West (say) then, although we have moved a total distanceof 2m, our net distance from the starting point is less than 2mof course, thisis just common sense.3.6 Scalar multiplicationSuppose that s= r. This expression is interpreted as follows: vector s pointsin the same direction as vector r, but the length of the former vector is times353 MOTION IN 3 DIMENSIONS 3.7 Diagonals of a parallelogramABXDC ba abcdFigure 13: A parallelogramthat of the latter. Note that if is negative then vector s points in the oppositedirection to vector r, and the length of the former vector is|| times that of thelatter. In terms of components:s = (x, y, z) = ( x, y, z). (3.8)In other words, when we multiply a vector by a scalar then the components ofthe resultant vector are obtained by multiplying all the components of the originalvector by the scalar.3.7 Diagonals of a parallelogramThe use of vectors is very well illustrated by the following rather famous proofthat the diagonals of a parallelogram mutually bisect one another.Suppose that the quadrilateral ABCD in Fig. 13 is a parallelogram. It followsthat the opposite sides of ABCD can be represented by the same vectors, a andb: this merely indicates that these sides are of equal length and are parallel (i.e.,they point in the same direction). Note that Fig. 13 illustrates an important pointregardingvectors. Althoughvectorspossessbothamagnitude(length)andadirection, theypossessnointrinsicpositioninformation. Thus, sincesidesABandDC are parallel and of equal length,they can be represented by the samevector a, despite the fact that they are in different places on the diagram.The diagonalBD in Fig. 13 can be represented vectorially as d = b a.Like-wise, the diagonal AC can be written c = a +b. The displacement x (say) of the363 MOTION IN 3 DIMENSIONS 3.8 Vector velocity and vector accelerationcentroid X from point A can be written in one of two different ways:x = a + d, (3.9)x = b +a c. (3.10)Equation (3.9) is interpreted as follows: in order to get from pointA to pointX, rst move to pointB (along vector a), then move along diagonalBD (alongvector d) for an unknown fraction of its length.Equation (3.10) is interpretedas follows:in order to get from pointA to pointX, rst move to pointD (alongvector b), then move to point C (along vector a), nally move along diagonal CA(along vector c) for an unknown fraction of its length. Since X represents thesame point in Eqs. (3.9) and (3.10), we can equate these two expressions to givea + (b a) = b +a (a +b). (3.11)Now vectors a and b point in different directions, so the only way in which theaboveexpressioncanbesatised, ingeneral, isifthecoefcientsofaandbmatch on either side of the equality sign.Thus, equating coefcients of a and b,we obtain1 = 1 , (3.12) = 1 . (3.13)Itfollowsthat ==1/2. Inotherwords, thecentroidXislocatedatthehalfway points of diagonalsBD andAC: i.e., the diagonals mutually bisect oneanother.3.8 Vector velocity and vector accelerationConsider a body moving in 3 dimensions. Suppose that we know the Cartesiancoordinates,x,y, andz, of this body as time,t, progresses. Let us consider howwe can use this information to determine the bodys instantaneous velocity andacceleration as functions of time.The vector displacement of the body is given byr(t) = [x(t), y(t), z(t)]. (3.14)373 MOTION IN 3 DIMENSIONS 3.8 Vector velocity and vector accelerationBy analogy with the 1-dimensional equation (2.3), the bodys vector velocity v =(vx, vy, vz) is simply the derivative of r with respect to t. In other words,v(t) =limt0r(t +t) r(t)t=drdt. (3.15)When written in component form, the above denition yieldsvx=dxdt, (3.16)vy=dydt, (3.17)vz=dzdt. (3.18)Thus, the x-component of velocity is simply the time derivative of the x-coordinate,and so on.By analogy with the 1-dimensional equation (2.6), the bodys vector acceler-ationa=(ax, ay, az)issimplythederivativeofvwithrespecttot. Inotherwords,a(t) =limt0v(t +t) v(t)t=dvdt=d2rdt2. (3.19)When written in component form, the above denition yieldsax=dvxdt=d2xdt2, (3.20)ay=dvydt=d2ydt2, (3.21)az=dvzdt=d2zdt2. (3.22)Thus, thex-componentof accelerationissimplythetimederivativeof thex-component of velocity, and so on.As an example, suppose that the coordinates of the body are given byx = sint, (3.23)y = cos t, (3.24)z = 3 t. (3.25)383 MOTION IN 3 DIMENSIONS 3.9 Motion with constant velocityThe corresponding components of the bodys velocity are then simplyvx=dxdt= cos t, (3.26)vy=dydt= sint, (3.27)vz=dzdt= 3, (3.28)whilst the components of the bodys acceleration are given byax=dvxdt= sint, (3.29)ay=dvydt= cos t, (3.30)az=dvzdt= 0. (3.31)3.9 Motion with constant velocityAnobjectmovingin3dimensionswithconstantvelocityvpossessesavectordisplacement of the formr(t) = r0 +v t, (3.32)where the constant vector r0 is the displacement at time t = 0. Note that dr/dt =v andd2r/dt2= 0, as expected. As illustrated in Fig. 14, the objects trajectoryis a straight-line which passes through point r0 at timet=0 and runs parallel tovector v.3.10 Motion with constant accelerationAn object moving in 3 dimensions with constant acceleration a possesses a vectordisplacement of the formr(t) = r0 +v0t +12 a t2. (3.33)393 MOTION IN 3 DIMENSIONS 3.10 Motion with constant accelerationr0rvtrajectoryt = 0t = tFigure 14: Motion with constant velocityHence, the objects velocity is given byv(t) =drdt= v0 +a t. (3.34)Note thatdv/dt=a, as expected. In the above, the constant vectors r0and v0are the objects displacement and velocity at time t = 0, respectively.As is easily demonstrated, the vector equivalents of Eqs. (2.11)(2.13) are:s = v0t +12 a t2, (3.35)v = v0 +a t, (3.36)v2= v20+2 as. (3.37)These equation fully characterize 3-dimensional motion with constant accelera-tion. Here, s= r r0 is the net displacement of the object between timest=0and t.The quantity as, appearing in Eq. (3.37), is termed the scalar product of vectorsa and s, and is denedas = axsx +aysy +azsz. (3.38)The above formula has a simple geometric interpretation, which is illustrated inFig.15. If |a| isthemagnitude(orlength)ofvectora, |s| isthemagnitudeof403 MOTION IN 3 DIMENSIONS 3.11 Projectile motionsa . s = |a| |s| cos |a|acos |s| |s|Figure 15: The scalar productvector s, and is the angle subtended between these two vectors, thenas = |a| |s| cos . (3.39)In other words, the scalar product of vectors a and s equals the product of thelength of vector a times the length of that component of vector s which lies in thesame direction as vector a. It immediately follows that if two vectors are mutuallyperpendicular (i.e.,=90) then their scalar product is zero. Furthermore, thescalarproductofavectorwithitselfissimplythemagnitudesquaredofthatvector [this is immediately apparent from Eq. (3.38)]:aa = |a|2= a2. (3.40)It is also apparent from Eq. (3.38) that a s = s a, and a(b +c) = ab +ac, anda(s) = (a s).Incidentally, Eq. (3.37) is obtained by taking the scalar product of Eq. (3.36)with itself, taking the scalar product of Eq. (3.35) with a, and then eliminating t.3.11 Projectile motionAs a simple illustration of the concepts introduced in the previous subsections, letus examine the following problem. Suppose that a projectile is launched upwardfrom ground level, with speed v0, making an angle with the horizontal. Neglect-ing the effect of air resistance, what is the subsequent trajectory of the projectile?413 MOTION IN 3 DIMENSIONS 3.11 Projectile motionzxv00vv0cos v 0sinFigure 16: Coordinates for the projectile problemOurrsttaskistosetupasuitableCartesiancoordinatesystem. Aconve-nient system is illustrated in Fig. 16. Thez-axis points vertically upwards (thisis a standard convention), whereas thex-axis points along the projectiles initialdirection of horizontal motion. Furthermore, the origin of our coordinate systemcorresponds to the launch point. Thus, z = 0 corresponds to ground level.Neglectingairresistance, theprojectileissubjecttoaconstantaccelerationg=9.81 ms1, due to gravity, which is directed vertically downwards. Thus, theprojectiles vector acceleration is writtena = (0, 0, g). (3.41)Here,the minus sign indicates that the acceleration is in the minusz-direction(i.e., downwards), as opposed to the plus z-direction (i.e., upwards).What is the initial vector velocity v0 with which the projectile is launched intotheairat(say)t =0? AsillustratedinFig. 16, giventhatthemagnitudeofthis velocity isv0, its horizontal component is directed along thex-axis, and itsdirection subtends an angle with this axis, the components of v0 take the formv0= (v0 cos ,0,v0 sin). (3.42)Note that v0 has zero component along the y-axis, which points into the paper inFig. 16.Since the projectile moves with constant acceleration, its vector displacement423 MOTION IN 3 DIMENSIONS 3.11 Projectile motions = (x, y, z) from its launch point satises [see Eq. (3.35)]s = v0t +12 a t2. (3.43)Making use of Eqs. (3.41) and (3.42), the x-,y-, and z-components of the aboveequation are writtenx = v0 cos t, (3.44)y = 0, (3.45)z = v0 sint 12 gt2, (3.46)respectively. Note that the projectile moves with constant velocity,vx=dx/dt=v0 cos , inthex-direction(i.e., horizontally). Thisishardlysurprising, sincethere is zero component of the projectiles acceleration along thex-axis. Note,further, that since there is zero component of the projectiles acceleration alongthe y-axis, and the projectiles initial velocity also has zero component along thisaxis, the projectile never moves in the y-direction. In other words, the projectilestrajectory is 2-dimensional, lying entirely within thex-z plane. Note, nally, thatthe projectiles vertical motion is entirely decoupled from its horizontal motion.In other words, the projectiles vertical motion is identical to that of a second pro-jectile launched vertically upwards, at t = 0, with the initial velocity v0 sin (i.e.,the initial vertical velocity component of the rst projectile)both projectiles willreach the same maximum altitude at the same time, and will subsequently strikethe ground simultaneously.Equations (3.44) and (3.46) can be rearranged to givez = x tan 12gx2v20sec2. (3.47)As was rst pointed out by Galileo, and is illustrated in Fig. 17, this is the equa-tionofaparabola. ThehorizontalrangeRoftheprojectilecorrespondstoitsx-coordinate when it strikes the ground (i.e., whenz=0). It follows from theabove expression (neglecting the trivial result x = 0) thatR =2 v20gsin cos =v20gsin2. (3.48)433 MOTION IN 3 DIMENSIONS 3.12 Relative velocityxzhRFigure 17: The parabolic trajectory of a projectileNote that the range attains its maximum value,Rmax=v20g, (3.49)when =45. Inotherwords, neglectingairresistance, aprojectiletravelsfurthest when it is launched into the air at 45 to the horizontal.The maximum altitudeh of the projectile is attained whenvz=dz/dt=0(i.e., when the projectile has just stopped rising and is about to start falling). Itfollows from Eq. (3.46) that the maximum altitude occurs at time t0= v0 sin/g.Hence,h = z(t0) =v202 gsin2. (3.50)Obviously, the largest value of h,hmax=v202 g, (3.51)is obtained when the projectile is launched vertically upwards (i.e., = 90).3.12 Relative velocitySuppose that, on a windy day, an airplane moves with constant velocity va withrespect to the air,and that the air moves with constant velocity u with respect443 MOTION IN 3 DIMENSIONS 3.12 Relative velocityvavguFigure 18: Relative velocitytotheground. Whatisthevectorvelocityvgoftheplanewithrespecttotheground? In principle, the answer to this question is very simple:vg= va +u. (3.52)In other words, the velocity of the plane with respect to the ground is the vectorsum of the planes velocity relative to the air and the airs velocity relative to theground. See Fig. 18. Note that, in general, vg is parallel to neither va nor u. Letus now consider how we might implement Eq. (3.52) in practice.As always, our rst task is to set up a suitable Cartesian coordinate system.Aconvenient system for dealing with 2-dimensional motion parallel to the Earthssurface is illustrated in Fig. 19. Thex-axis points northward, whereas they-axispoints eastward. In this coordinate system, it is conventional to specify a vector rin term of its magnitude, r, and its compass bearing, . As illustrated in Fig. 20, acompass bearing is the angle subtended between the direction of a vector and thedirection to the North pole: i.e., the x-direction. By convention, compass bearingsrun from0to360. Furthermore, the compass bearings of North, East, South,and West are 0, 90, 180, and 270, respectively.According to Fig. 20, the components of a general vector r, whose magnitudeis r and whose compass bearing is , are simplyr = (x,y) = (r cos ,r sin). (3.53)Note that we have suppressed thez-component of r (which is zero), for ease ofnotation. Although, strictly speaking, Fig. 20 only justies the above expressionfor in the range0to90, it turns out that this expression is generally valid:i.e., it is valid for in the full range 0 to 360.453 MOTION IN 3 DIMENSIONS 3.12 Relative velocityW ESNxyFigure 19: Coordinates for relative velocity problemr cosrrxyr sinNEFigure 20: A compass bearing463 MOTION IN 3 DIMENSIONS 3.12 Relative velocityAs an illustration, suppose that the planes velocity relative to the air is 300 km/h,at acompassbearingof 120, andtheairsvelocityrelativetothegroundis85 km/h, atacompassbearingof 225. Itfollowsthatthecomponentsofvaand u (measured in units of km/h) areva= (300 cos 120,300 sin120) = (1.500 102,2.598 102), (3.54)u = (85 cos 225,85 sin225) = (6.010 101,6.010 101). (3.55)According to Eq. (3.52), the components of the planes velocity vg relative to theground are simply the algebraic sums of the corresponding components of va andu. Hence,vg= (1.500 1026.010 101,2.598 1026.010 101)= (2.101 102,1.997 102). (3.56)Our nal task is to reconstruct the magnitude and compass bearing of vectorvg, given its components (vgx, vgy). The magnitude of vg follows from Pythagorastheorem [see Eq. (3.6)]:vg=_(vgx)2+ (vgy)2=_(2.101 102)2+ (1.997 102)2= 289.9 km/h. (3.57)In principle, the compass bearing of vg is given by the following formula: = tan1__vgyvgx__. (3.58)Thisfollowsbecausevgx=vg cos andvgy=vg sin[seeEq.(3.53)]. Un-fortunately, theaboveexpressionbecomesalittledifculttointerpretif vgxisnegative. An unambiguous pair of expressions for is given below: = tan1__vgyvgx__, (3.59)if vgx 0; or = 180 tan1__ vgy|vgx|__, (3.60)473 MOTION IN 3 DIMENSIONS 3.12 Relative velocityifvgxm1, and viceversa. Note that the acceleration of the system is less than the full accelerationdue to gravity, g, since both masses contribute to the inertia of the system, but664 NEWTONS LAWS OF MOTION 4.8 FrictionmTTm1mg1mg22..Figure 31: An Atwood machinetheir weights partially cancel one another out. In particular, if the two masses arealmost equal then the acceleration of the system becomes very much less than g.Incidentally, the device pictured in Fig. 31 is called an Atwood machine, aftertheeighteenthCenturyEnglishscientistGeorgeAtwood, whousedittoslowdown free-fall sufciently to make accurate observations of this phenomena us-ing the primitive time-keeping devices available in his day.4.8 FrictionWhenabodyslidesoveraroughsurfaceafrictional forcegenerallydevelopswhich acts to impede the motion. Friction, when viewed at the microscopic level,is actually a very complicated phenomenon. Nevertheless,physicists and engi-neers have managed to develop a relatively simple empirical law of force whichallows the effects of friction to be incorporated into their calculations. This law offorce was rst proposed by Leonardo da Vinci (14521519), and later extendedby Charles Augustin de Coulomb (17361806) (who is more famous for discov-674 NEWTONS LAWS OF MOTION 4.8 Frictionering the law of electrostatic attraction). The frictional force exerted on a bodysliding over a rough surface is proportional to the normal reaction Rn at that sur-face, the constant of proportionality depending on the nature of the surface. Inother words,f = Rn, (4.22)where is termed the coefcient of (dynamical) friction.For ordinary surfaces,is generally of order unity.Consider a block of massm being dragged over a horizontal surface,whosecoefcient of frictionis , byahorizontal forceF. SeeFig. 32. TheweightW= mg of the block acts vertically downwards, giving rise to a reaction R = mgacting vertically upwards. The magnitude of the frictional force f, which impedesthe motion of the block, is simply times the normal reactionR=mg. Hence,f = mg. The acceleration of the block is, therefore,a =F fm=Fmg, (4.23)assuming that F > f. What happens if F < f: i.e., if the applied force F is less thanthe frictional forcef?In this case, common sense suggests that the block simplyremains at rest (it certainly does not accelerate backwards!). Hence, f=mgisactuallythemaximumforcewhichfrictioncangenerateinordertoimpedethe motion of the block. If the applied forceF is less than this maximum valuethen the applied force is canceled out by an equal and opposite frictional force,and the block remains stationary. Only if the applied force exceeds the maximumfrictional force does the block start to move.Consider a block of mass m sliding down a rough incline (coefcient of friction) which subtends an angle to the horizontal, as shown in Fig 33. The weightmg of the block can be resolved into components mg cos , acting normal to theincline, andmg sin, acting parallel to the incline. The reaction of the inclinetotheweightoftheblockactsnormallyoutwardsfromtheincline, andisofmagnitude mg cos . Parallel to the incline, the block is subject to the downwardgravitational forcemg sin, andtheupwardfrictional forcef(whichactstoprevent the block sliding down the incline). In order for the block to move, themagnitudeoftheformerforcemustexceedthemaximumvalueofthelatter,684 NEWTONS LAWS OF MOTION 4.8 Frictionm gWRfFFigure 32: Frictionwhich is time the magnitude of the normal reaction, or mg cos . Hence, thecondition for the weight of the block to overcome friction, and, thus, to cause theblock to slide down the incline, ismg sin > mg cos , (4.24)ortan > . (4.25)In other words, if the slope of the incline exceeds a certain critical value, whichdepends on, then the block will start to slide. Incidentally, the above formulasuggests a fairly simple way of determining the coefcient of friction for a givenobject sliding over a particular surface. Simply tilt the surface gradually until theobject just starts to move: the coefcient of friction is simply the tangent of thecritical tilt angle (measured with respect to the horizontal).Up to now, we have implicitly suggested that the coefcient of friction betweenan object and a surface is the same whether the object remains stationary or slidesover the surface.In fact, this is generally not the case.Usually, the coefcient offriction when the object is stationary is slightly larger than the coefcient whenthe object is sliding. We call the former coefcient the coefcient of static friction,s, whereasthelattercoefcientisusuallytermedthecoefcientofkinetic(ordynamical) friction,k. The fact thats>k simply implies that objects have atendency to stick to rough surfaces when placed upon them. The force requiredto unstick a given object,and,thereby,set it in motion,isstimes the normal694 NEWTONS LAWS OF MOTION 4.9 Frames of referencemg mg cosmg cos mg sinmfFigure 33: Block sliding down a rough slopereaction at the surface. Once the object has been set in motion, the frictional forceacting to impede this motion falls somewhat to k times the normal reaction.4.9 Frames of referenceAs discussed in Sect. 1, the laws of physics are assumed to possess objective real-ity. In other words, it is assumed that two independent observers, studying thesame physical phenomenon, would eventually formulate identical laws of physicsinordertoaccountfortheirobservations. Now, twocompletelyindependentobservers are likely to choose different systems of units with which to quantifyphysical measurements. However,as we have seen in Sect.1,the dimensionalconsistency of valid laws of physics renders them invariant under transformationfromonesystemofunitstoanother. Independentobserversarealsolikelytochoose different coordinate systems. For instance,the origins of their separatecoordinate systems might differ, as well as the orientation of the various coordi-nate axes. Are the laws of physics also invariant under transformation betweencoordinate systems possessing different origins, or a different orientation of thevarious coordinate axes?Consider the vector equationr = r1 +r2, (4.26)which is represented diagrammatically in Fig. 12. Suppose that we shift the origin704 NEWTONS LAWS OF MOTION 4.9 Frames of referenceof our coordinate system, or rotate the coordinate axes. Clearly, in general, thecomponentsofvectorsr, r1, andr2aregoingtobemodiedbythischangeinour coordinate scheme. However, Fig. 12 still remains valid. Hence, we concludethatthevectorequation(4.26)alsoremainsvalid. Inotherwords, althoughtheindividualcomponentsofvectorsr, r1, andr2aremodiedbythechangein coordinate scheme, the interrelation between these components expressed inEq. (4.26) remains invariant. This observation suggests that the independenceof the laws of physics from the arbitrary choice of the location of the underlyingcoordinate systems origin,or the equally arbitrary choice of the orientation ofthe various coordinate axes, can be made manifest by simply writing these lawsas interrelations between vectors. In particular, Newtons second law of motion,f = ma, (4.27)is clearly invariant under shifts in the origin of our coordinate system, or changesin the orientation of the various coordinate axes. Note that the quantitym (i.e.,themassof thebodywhosemotionisunderinvestigation), appearingintheabove equation, is invariant under any changes in the coordinate system, sincemeasurements of mass are completely independent of measurements of distance.We refer to such a quantity as a scalar (this is an improved denition). We con-clude that valid laws of physics must consist of combinations of scalars and vec-tors, otherwise they would retain an unphysical dependence on the details of thechosen coordinate system.Up to now, we have implicitly assumed that all of our observers are stationary(i.e.,they are all standing still on the surface of the Earth). Let us,now,relaxthis assumption. Consider two observers, O andO

, whose coordinate systemscoincidemomentarilyatt =0. SupposethatobserverOisstationary(onthesurface of the Earth), whereas observerO

moves (with respect to observerO)with uniform velocity v0. As illustrated in Fig. 34, if r represents the displacementof some body P in the stationary observers frame of reference, at time t, then thecorresponding displacement in the moving observers frame of reference is simplyr

= r v0t. (4.28)The velocity of bodyP in the stationary observers frame of reference is dened714 NEWTONS LAWS OF MOTION 4.9 Frames of referenceOPOv0trrFigure 34: A moving observerasv =drdt. (4.29)Hence, thecorrespondingvelocityinthemovingobserversframeofreferencetakes the formv

=dr

dt= v v0. (4.30)Finally, the acceleration of bodyP in stationary observers frame of reference isdened asa =dvdt, (4.31)whereas the corresponding acceleration in the moving observers frame of refer-ence takes the forma

=dv

dt= a. (4.32)Hence, the acceleration of body P is identical in both frames of reference.It is clear that if observer O concludes that body P is moving with constant ve-locity, and, therefore, subject to zero net force, then observerO

will agree withthis conclusion. Furthermore, if observer O concludes that body P is accelerating,and, therefore, subject to a force a/m, then observer O

will remain in agreement.It follows that Newtons laws of motion are equally valid in the frames of refer-ence of the moving and the stationary observer. Such frames are termed inertialframes of reference. There are innitely many inertial frames of referencewithin724 NEWTONS LAWS OF MOTION 4.9 Frames of referencewhich Newtons laws of motion are equally validall moving with constant ve-locity with respect to one another. Consequently, there is no universal standardof rest in physics. Observer O might claim to be at rest compared to observer O

,and vice versa: however, both points of view are equally valid. Moreover, thereis absolutely no physical experiment which observer O could perform in order todemonstrate that he/she is at rest whilst observer O

is moving. This, in essence,is the principle of special relativity, rst formulated by Albert Einstein in 1905.Worked example 4.1: In equilibriumQuestion:Consider the diagram. If the system is in equilibrium, and the tensionin string 2 is 50 N, determine the mass M.4021M M3o40oAnswer:It follows from symmetry that the tensions in strings 1 and 3 are equal.Let T1bethetensioninstring1, andT2thetensioninstring2. Considertheequilibriumof theknot abovetheleftmost mass. Asshownbelow, thisknotissubject tothreeforces: thedownwardforceT4=Mgduetothetensioninthestringwhichdirectlysupportstheleftmostmass, therightwardforceT2due to the tension in string 2, and the upward and leftward forceT1 due to thetension in string 1. The resultant of all these forces must be zero, otherwise thesystem would not be in equilibrium. Resolving in the horizontal direction (withrightward forces positive), we obtainT2 T1 sin40= 0.Likewise, resolving in the vertical direction (with upward forces positive) yieldsT1 cos 40 T4= 0.734 NEWTONS LAWS OF MOTION 4.9 Frames of referenceCombining the above two expressions, making use of the fact that T4= Mg, givesM =T2g tan40.Finally, since T2= 50 N and g = 9.81 m/s2, we obtainM =509.81 0.8391= 6.074 kg.40T21T4ToWorked example 4.2: Block accelerating up a slopeQuestion: Considerthediagram. Supposethattheblock, mass m=5 kg, issubject to a horizontal forceF=27 N. What is the acceleration of the block upthe (frictionless) slope?Fm25oAnswer: Only that component of the applied force which is parallel to the inclinehas any inuence on the blocks motion: the normal component of the appliedforce is canceled out by the normal reaction of the incline. The component ofthe applied force acting up the incline isF cos 25. Likewise, the component of744 NEWTONS LAWS OF MOTION 4.9 Frames of referencethe blocks weight acting down the incline ismg sin25.Hence, using Newtonssecond law to determine the acceleration a of the block up the incline, we obtaina =F cos 25 mg sin25m.Since m = 5 kg and F = 27 N, we havea =27 0.9063 5 9.81 0.42265= 0.7483 m/s2.Worked example 4.3: Raising a platformQuestion: Consider the diagram. The platform and the attached frictionless pulleyweigh a total of 34N. With what force F must the (light) rope be pulled in orderto lift the platform at 3.2 m/s2?platformpulleyFAnswer: Let W be the weight of the platform, m = W/g the mass of the platform,andTthe tension in the rope. From Newtons third law, it is clear thatT =F.LetusapplyNewtonssecondlawtotheupwardmotionoftheplatform. Theplatform is subject to two vertical forces: a downward force W due to its weight,and an upward force2 Tdue to the tension in the rope (the force is2 T, ratherthanT, because both the leftmost and rightmost sections of the rope, emergingfrom the pulley, are in tension and exerting an upward force on the pulley). Thus,754 NEWTONS LAWS OF MOTION 4.9 Frames of referencethe upward acceleration a of the platform isa =2 TWm.Since T= F and m = W/g, we obtainF =W(a/g +1)2.Finally, given that W= 34 N and a = 3.2 m/s2, we haveF =34 (3.2/9.81 +1)2= 22.55 N.Worked example 4.4: Suspended blockQuestion: Considerthediagram. ThemassofblockAis 75 kgandthemassofblockBis15 kg. Thecoefcientofstaticfrictionbetweenthetwoblocksis=0.45. The horizontal surface is frictionless. What minimum forceF must beexerted on block A in order to prevent block B from falling?FABAnswer: Suppose that blockA exerts a rightward forceR on blockB. By New-tons third law, blockB exerts an equal and opposite force on blockA. ApplyingNewtons second law of motion to the rightward accelerationa of blockB,weobtaina =RmB,wheremB is the mass of blockB. The normal reaction at the interface betweenthe two blocks isR. Hence,the maximum frictional force that blockA can ex-ertonblockBisR. InordertopreventblockBfromfalling, thismaximum764 NEWTONS LAWS OF MOTION 4.9 Frames of referencefrictional force (which acts upwards) must exceed the downward acting weight,mBg, of the block. Hence, we requireR > mBg,ora >g.ApplyingNewtonssecondlawtotherightwardaccelerationaofbothblocks(rememberingthat theequal andoppositeforcesexertedbetweentheblockscancel one another out), we obtaina =FmA +mB,where mA is the mass of block A. It follows thatF >(mA +mB) g.Since mA= 75 kg, mB= 15 kg, and = 0.45, we haveF >(75 +15) 9.810.45= 1.962 103N.775 CONSERVATION OF ENERGY5 Conservation of energy5.1 IntroductionNowadays, the conservation of energy is undoubtedly the single most importantidea in physics. Strangely enough, although the basic idea of energy conservationwas familiar to scientists from the time of Newton onwards, this crucial conceptonlymovedtocentre-stageinphysicsinabout1850(i.e., whenscientistsrstrealized that heat was a form of energy).According to the ideas of modern physics, energy is the substance from whichall things in the Universe are made up. Energy can take many different forms:e.g., potential energy, kineticenergy, electrical energy, thermal energy, chemi-cal energy, nuclear energy, etc. In fact, everything that we observe in the worldaroundusrepresentsoneofthemultitudinousmanifestationsofenergy. Now,there exist processes in the Universe which transform energy from one form intoanother: e.g., mechanical processes (which are the focus of this course), thermalprocesses, electrical processes, nuclear processes, etc. However, all of these pro-cesses leave the total amount of energy in the Universe invariant. In other words,whenever, andhowever, energyistransformedfromoneformintoanother, itis always conserved. For a closed system (i.e., a system which does not exchangeenergy with the rest of the Universe), the above law of universal energy conserva-tion implies that the total energy of the system in question must remain constantin time.5.2 Energy conservation during free-fallConsider a massm which is falling vertically under the inuence of gravity. Wealreadyknowhowtoanalyzethemotionofsuchamass. Letusemploythisknowledge to search for an expression for the conserved energy during this pro-cess. (N.B., This is clearly an example of a closed system, involving only the massand the gravitational eld.)The physics of free-fall under gravity is summarizedby the three equations (2.14)(2.16). Let us examine the last of these equations:785 CONSERVATION OF ENERGY 5.2 Energy conservation during free-fallv2= v20 2 gs. Suppose that the mass falls from height h1 to h2, its initial velocityisv1, and its nal velocity isv2. It follows that the net vertical displacement ofthemassiss=h2 h1. Moreover, v0=v1andv=v2. Hence, thepreviousexpression can be rearranged to give12 mv21+mgh1=12 mv22+mgh2. (5.1)The above equation clearly represents a conservation law,of some description,since the left-hand side only contains quantities evaluated at the initial height,whereas the right-hand side only contains quantities evaluated at the nal height.In order to clarify the meaning of Eq. (5.1), let us dene the kinetic energy of themass,K =12 mv2, (5.2)and the gravitational potential energy of the mass,U = mgh. (5.3)Notethatkineticenergyrepresentsenergythemasspossessesbyvirtueofitsmotion. Likewise, potential energy represents energy the mass possesses by virtueof its position. It follows that Eq. (5.1) can be writtenE = K +U = constant. (5.4)Here,E is the total energy of the mass: i.e., the sum of its kinetic and potentialenergies. It is clear thatE is a conserved quantity:i.e., although the kinetic andpotential energies of the mass vary as it falls, its total energy remains the same.Incidentally, the expressions (5.2) and (5.3) for kinetic and gravitational po-tential energy,respectively,are quite general,and do not just apply to free-fallunder gravity. The mks unit of energy is called the joule (symbol J). In fact,1joule is equivalent to 1 kilogram meter-squared per second-squared, or 1 newton-meter. Note that all forms of energy are measured in the same units (otherwisethe idea of energy conservation would make no sense).One of the most important lessons which students learn during their studies isthat there are generally many different paths to the same result in physics. Now,795 CONSERVATION OF ENERGY 5.2 Energy conservation during free-fallwe have already analyzed free-fall under gravity using Newtons laws of motion.However, it is illuminating to re-examine this problem from the point of view ofenergyconservation. Supposethatamassmisdroppedfromrestandfallsadistance h. What is the nal velocity v of the mass? Well, according to Eq. (5.1),if energy is conserved thenK = U : (5.5)i.e., any increase in the kinetic energy of the mass must be offset by a correspond-ing decrease in its potential energy. Now, the change in potential energy of themass is simplyU=mgs=mgh, wheres=h is its net vertical displace-ment. The change in kinetic energy is simplyK=(1/2) mv2,wherev is thenal velocity. This follows because the initial kinetic energy of the mass is zero(since it is initially at rest). Hence, the above expression yields12 mv2= mgh, (5.6)orv =_2 gh. (5.7)Suppose that the same mass is thrown upwards with initial velocityv. Whatis the maximum heighth to which it rises? Well, it is clear from Eq. (5.3) thatas the mass rises its potential energy increases. It, therefore, follows from energyconservationthatitskineticenergymust decreasewithheight. Note, however,from Eq. (5.2), that kinetic energy can never be negative (since it is the productof the two positive denite quantities, m andv2/2). Hence, once the mass hasrisen to a heighth which is such that its kinetic energy is reduced to zero it canrise no further, and must, presumably, start to fall. The change in potential energyof the mass in moving from its initial height to its maximum height is mgh. Thecorrespondingchangeinkineticenergyis(1/2) mv2; since(1/2) mv2istheinitial kinetic energy, and the nal kinetic energy is zero. It follows from Eq. (5.5)that (1/2) mv2= mgh, which can be rearranged to giveh =v22 g. (5.8)It should be noted that the idea of energy conservationalthough extremelyusefulis not a replacement for Newtons laws of motion. For instance,in the805 CONSERVATION OF ENERGY 5.3 Workpreviousexample, thereisnowayinwhichwecandeducehowlongittakesthemasstorisetoitsmaximumheightfromenergyconservationalonethisinformation can only come from the direct application of Newtons laws.5.3 WorkWe have seen that when a mass free-falls under the inuence of gravity some ofits kinetic energy is transformed into potential energy, or vice versa. Let us nowinvestigate, in detail, how this transformation is effected. The mass falls becauseit is subject to a downwards gravitational force of magnitudemg. It stands toreason, therefore, thatthetransformationofkineticintopotential energyisadirect consequence of the action of this force.Thisis, perhaps, anappropriatepointatwhichtonotethattheconceptofgravitational potential energyalthough extremely usefulis, strictly speaking,ctitious. Tobemoreexact, thepotentialenergyofabodyisnotanintrinsicpropertyofthatbody(unlikeitskineticenergy). Infact, thegravitationalpo-tential energy of a given body is stored in the gravitational eld which surroundsit. Thus, when the body rises, and its potential energy consequently increases byan amountU; in reality, it is the energy of the gravitational eld surroundingthe body which increases by this amount. Of course,the increase in energy ofthe gravitational eld is offset by a corresponding decrease in the bodys kineticenergy. Thus, when we speak of a bodys kinetic energy being transformed intopotential energy, we are really talking about a ow of energy from the body to thesurrounding gravitational eld. This energy ow is mediated by the gravitationalforce exerted by the eld on the body in question.Incidentally, accordingtoEinsteinsgeneral theoryofrelativity(1917), thegravitational eld of a mass consists of the local distortion that mass induces inthefabricofspace-time. Fortunately, however, wedonotneedtounderstandgeneral relativity in order to talk about gravitational elds or gravitational po-tentialenergy. Allweneedtoknowisthatagravitationaleldstoresenergywithout loss: i.e., if a given mass rises a certain distance, and, thereby, gives upa certain amount of energy to the surrounding gravitational eld, then that eld815 CONSERVATION OF ENERGY 5.3 Workwill return this energy to the masswithout lossif the mass falls by the samedistance. In physics, we term such a eld a conservative eld (see later).Suppose that a massm falls a distanceh. During this process, the energy ofthe gravitational eld decreases by a certain amount (i.e., the ctitious potentialenergy of the mass decreases by a certain amount), and the bodys kinetic energyincreases by a corresponding amount. This transfer of energy, from the eld tothe mass,is,presumably,mediated by the gravitational forcemg (the minussign indicates that the force is directed downwards) acting on the mass. In fact,given that U = mgh, it follows from Eq. (5.5) thatK = f h. (5.9)In other words, the amount of energy transferred to the mass (i.e., the increase inthe masss kinetic energy) is equal to the product of the force acting on the massand the distance moved by the mass in the direction of that force.In physics, we generally refer to the amount of energy transferred to a body,when a force acts upon it, as the amount of workW performed by that force onthe body in question. It follows from Eq. (5.9) that when a gravitational forcef acts on a body, causing it to displace a distancex in the direction of that force,then the net work done on the body isW= f x. (5.10)It turns out that this equation is quite general, and does not just apply to grav-itational forces. IfWis positive then energy is transferred to the body,and itsintrinsic energy consequently increases by an amountW. This situation occurswhenever a body moves in the same direction as the force acting upon it. Like-wise, ifWis negative then energy is transferred from the body, and its intrinsicenergy consequently decreases by an amount|W|. This situation occurs when-ever a body moves in the opposite direction to the force acting upon it. Since anamount of work is equivalent to a transfer of energy, the mks unit of work is thesame as the mks unit of energy: namely, the joule.Inderivingequation(5.10), wehavemadetwoassumptionswhicharenotuniversallyvalid. Firstly, wehaveassumedthatthemotionofthebodyupon825 CONSERVATION OF ENERGY 5.3 Workwhich the force acts is both 1-dimensional and parallel to the line of action of theforce. Secondly, we have assumed that the force does not vary with position. Letus attempt to relax these two assumptions, so as to obtain an expression for thework W done by a general force f.Let us start by relaxing the rst assumption. Suppose, for the sake of argument,that we have a mass m which moves under gravity in 2-dimensions. Let us adoptthecoordinatesystemshowninFig. 35, withzrepresentingverticaldistance,andxrepresentinghorizontaldistance. Thevectoraccelerationofthemassissimply a = (0, g). Here, we are neglecting the redundant y-component, for thesake of simplicity. The physics of motion under gravity in more than 1-dimensionis summarized by the three equations (3.35)(3.37). Let us examine the last ofthese equations:v2= v20+2 as. (5.11)Here, v0is the speed att=0, v is the speed att=t, and s=(x, z) is thenet displacement of the mass during this time interval. Recalling the denitionof a scalar product [i.e., ab = (axbx +ayby +azbz)], the above equation can berearranged to give12 mv212 mv20= mgz. (5.12)Since the right-hand side of the above expression is manifestly the increase in thekinetic energy of the mass between times t = 0 and t = t, the left-hand side mustequal the decrease in the masss potential energy during the same time interval.Hence, we arrive at the following expression for the gravitational potential energyof the mass:U = mgz. (5.13)Ofcourse, thisexpressionisentirelyequivalenttoourpreviousexpressionforgravitational potential energy, Eq. (5.3). Theaboveexpressionmerelymakesmanifest a point which should have been obvious anyway: namely, that the grav-itational potential energy of a mass only depends on its height above the ground,and is quite independent of its horizontal displacement.Let us now try to relate the ow of energy between the gravitational eld andthe mass to the action of the gravitational force, f=(0, mg). Equation (5.12)835 CONSERVATION OF ENERGY 5.3 WorkzxmFigure 35: Coordinate system for 2-dimensional motion under gravitycan be rewrittenK = W= f s. (5.14)In other words, the work W done by the force f is equal to the scalar product of fand the vector displacement s of the body upon which the force acts. It turns outthat this result is quite general, and does not just apply to gravitational forces.Figure 36 is a visualization of the denition (5.14). The workWperformedby a force f when the object upon which it acts is subject to a displacement s isW= |f| |s| cos . (5.15)where is the angle subtended between the directions of f and s. In other words,theworkperformedistheproductofthemagnitudeoftheforce, |f|, andthedisplacement of the object in the direction of that force, |s| cos . It follows thatany component of the displacement in a direction perpendicular to the force gen-erates zero work. Moreover, if the displacement is entirely perpendicular to thedirection of the force (i.e.,if=90) then no work is performed,irrespectiveofthenatureoftheforce. Asbefore, ifthedisplacementhasacomponentinthe same direction as the force (i.e., if 90) then negative work is performed.Suppose, now, that an object is subject to a force f which varies with position.Whatisthetotal workdonebytheforcewhentheobjectmovesalongsomegeneraltrajectoryinspacebetweenpointsAandB(say)? SeeFig. 37. Well,one way in which we could approach this problem would be to approximate thetrajectory as a series ofN straight-line segments, as shown in Fig. 38. Suppose845 CONSERVATION OF ENERGY 5.3 Workscos |s||f|f|s| Figure 36: Denition of workthat the vector displacement of the ith segment is ri. Suppose, further, that N issufciently large that the force f does not vary much along each segment. In fact,let the average force along theith segment be fi. We shall assume that formula(5.14)which is valid for constant forces and straight-line displacementsholdsgood for each segment. It follows that the net work done on the body, as it movesfrom point A to point B, is approximatelyW N

i=1firi. (5.16)We can always improve the level of our approximation by increasing the numberN of the straight-line segments which we use to approximate the bodys trajectorybetween pointsAandB. In fact,if we takethelimitN then theaboveexpression becomes exact:W=limNN

i=1firi=

BAf(r)dr. (5.17)Here, r measures vector displacement from the origin of our coordinate system,and the mathematical construct

BAf(r)dr is termed a line-integral.The meaning of Eq. (5.17) becomes a lot clearer if we restrict our attention to1-dimensional motion. Suppose, therefore, that an object moves in 1-dimension,withdisplacementx, andissubjecttoavaryingforcef(x)(directedalongthex-axis). WhatistheworkdonebythisforcewhentheobjectmovesfromxA855 CONSERVATION OF ENERGY 5.3 WorkABFigure 37: Possible trajectory of an object in a variable force-eldABFigure 38: Approximation to the previous trajectory using straight-line segments865 CONSERVATION OF ENERGY 5.3 Workx x x ->A Bf ->Figure 39: Work performed by a 1-dimensional forcetoxB? Well, a straight-forward application of Eq. (5.17) [with f =(f, 0, 0) anddr = (dx, 0, 0)] yieldsW=

xBxAf(x) dx. (5.18)In other words, the net work done by the force as the object moves from displace-mentxA toxB is simply the area under thef(x) curve between these two points,as illustrated in Fig. 39.Let us, nally, round-off this discussionbyre-derivingtheso-calledwork-energytheorem, Eq.(5.14), in1-dimension, allowingforanon-constantforce.According to Newtons second law of mot