Classical Mechanics PHYS 3610
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Notes for Phys 3610 Last revised 121808 These notes are just intended to give an overview of the major equations covered in class 1 Newton s Laws We present Newton s law as they apply to point masses or particles Such a particle can have translational motion but no internal degrees of freedom Later on we will nd out how the apply the law of mechanics to extended objects Newt0n7s First Law In the absence of forces a particle moves with constant velocity v Newt0n7s Second Law For any particle of mass m the net force F on the particle is always equal to the mass m times the particles acceleration F ma Here a is the particles acceleration dv dzr l a i 7 V dt 172 The second law can also be expressed in terms of the momentum of the particle de ned as p mv In classical mechanics the mass of our idealized particle does not change so that p mv ma so that the second law can be written as F p which is fin spite of what some basic physics books sayi completely equivalent to F ma 11 Reference Frames 12 Discussion of Newton7s 1st and 2nd Laws Newton s First Law is much more profound than the way it is usually expressed in rst year physics courses One might think that it trivially follows from the second law but in fact it really states that there is a reference frame in which no acceleration takes place except when real concrete effects known as forces are present This is not true for many reference frames that one might choose for example the surface of the earth Such a reference frame is an inertial frame For non inertial frames we do get accelerations when there are no forces acting The second law applies to an inertial frame It says F ma where F is the total force acting on a point mass m and a is its acceleration 13 Newton7s 3rd Law Newtons s Third Law states that a force on one object is due to the presence of some other object We will let F12 mean the force on object 1 due to object 2 and F21 mean the force on object 2 due to object 1 Newton7s Third Law If object 1 exerts a force F21 on object 2 then object 2 always exerts a force F12 on object 1 given by F12 F21 2 Projectile Motion with Air Resistance Motion in a Magnetic Field Air exerts a force on a projectile which is opposite to the velocity and depends on the speed f 7fvv The simplest form of u has linear and quadratic parts bUCU2fnnfquad 1 The coef cients 1 and c for spherical projectiles of diameter D are given by 1251 and 0 where B and 39y for air at STP are 3 16 x10 4 N sm2 and 39y 025 N s2m4 It often happens that we can ignore one or the other of the terms in Eq 1 Alas as we ll see for the usual conditions of projectiles moving in air it is fquad which is the largest which does not have an analytic solution 21 Solution for Linear Air Resistance When we point the y axis downward Newton s 2nd law for a projectile gives mv mg 7 bv where g gSI Separating this into components mom 71wz mi y mg 7 buy b 15 7kvm where k 7 m and we also de ne 739 1k mb Consider the z equation rst It is b 15 7kvm where k a and we also de ne 739 1k mb Solving this differential equation gives Umt Umoe tT and Mt x0017 sitT where zoo is the limiting value of x as t 7 00 The y equation can be written as by 7bUy 7 Uter where m Uter 79 For linear drag It has the solution If vyoe tT Uter17 sitT A 1 21 from which we can get yt 3905 Utert 7quot UyO 7 Uter7 1 7 gitT It is possible to combine the z and y equations to get an equation for the trajectory of the projectile 71130 Uterm UterTlHlt1i m gt U10 UmOT but this equation must be solved numerically For zero air resistance the range R of a projectile red from ground level is 7 2123 sin0 cos0 7 2vm0vy0 9 9 R Rvac For low air resistance the range is given apporximately by Ptsz 17 1 0 3 Uter 22 Solutions for Quadratic Air Resistance mom7C4Ugvgvm mvymgic4Ugvgvy 2 for which both equations contain both of the variables Um and 12y Separating we get If the motion is purely horizontal the equation of motion 2 mom icvm has the solution W05 amp where 739 m d t d 1 tT CW0 qua ra 1c rag and Mt UmOTlH1tT If the motion is purely vertical we can also solve the equation If a baseball is dropped from a high place and again the y axis goes downward the equation of motion is mi y mg 7 Cl There will be a terminal velocity which is my Uter 7 which has the solutions vyt Utertanh lt it gt Mt 1ng In COSh wit U er er When we need to have both horizontal and vertical motion the equations in 2 must be solved numerically with a computer The popular packages Mathematica Maple and Matlab have canned routines which will do this you just need to learn how to set up the DES according to their syntax 23 Motion of a Charge in a Uniform Magnetic Field If a particle of charge q and velocity V moves in a magnetic eld B it experiences a force F mv qV X B Consider motion in a uniform magnetic eld B b2 From these we get the equations of motion mom qva 771 iqB mi z 0 The 2 equation just gives vz const so we can focus on Um and W With w E qBm we solve 1 11 wvy by iwvw While these can be solved by conventional means you would combine the z and y equations and use the appropriate initial conditions Taylor goes for a creative approach with the use of complex variables He de nes 7 Um my so that 7 iiwn The solution is 7 Ae m where A is determined from the initial conditions De ning Eiy wehave ndt which with a convenient choice of coordinates gives 5 m iy 057th where C 0 iyo with 0 yo being the initial coordinates From this form one sees that the particle travels in a helical path with the radius of the orbit being r vw mvqB Thus we have derived the elementary result with some rigorl 3 Linear Momentum From the discussion of Newton s Third Law we found P FEXt where P 2 pa Dd and FEXt is the total external force on the system 31 The Rocket Problem77 Rockets accelerate by chucking mass out the back continuously at some speed vex relative to the rocket We consider the change in speed of a rocket in a time interval dt Initially the rocket has mass m and moves with velocity 1 At a time dt later its mass has changes by a negative amount dm The mass dm is traveling at velocity v 7 vex and the rocket now has velocity v dv If no external force acts on the rocket conservation of momentum gives m d1 idm vex which implies thrust mi 7mg where m is the rate at which the rocket expels the mass It s not simply a decrease in its mass This differential equation can be solved to give 1 7 U0 vexlnm0m 32 Center of Mass For a system of particles the center of mass is de ned by 1 R M Z mara Dd From this de nition it follows that P 2 man MP Dd so that Fext MR 4 Angular Momentum The angular momentum for a single particle relative to some origin 0 is l r x p The rate of change of the angular momentum is l r x F P where P is the torque on the particle Not to be confused yet with the external torque acting on a rigid body 41 Kepler7s Second Law Kepler s second law states that the rate at which a planet sweeps out area as it orbits the sun is constant dAdt is constant In fact7 it follows from the conservation of angular momentum One can show that 1A 7 1 X i t dt 2m r p 2m and since the force on a planet is parallel to fquot giving zero torque then 6 is constant and thus so is dAdt 42 Angular Momentum of a System of Particles The total momentum of a system of particles is N N LZ aZraXpa 041 041 One can show that if the particles obey Newton s third law7 and if the forces between them act along the line joining each pair that is7 they are central force then LZra x F33 re Dd ie the rate of change of the total momentum is just the total external torque acting on the system While we hold off on the serious rotation problems until a leter chapter7 we recall some elemen tary facts bout rotation dynamics so that we can do a few examples Recall that if a rigid object turns about an axis along the z axis7 its angular momentum is Lz 1w where the moment of inertia for a set of mass points is de ned as 12mapi p being the distance of each mass point from the axis of rotation An under appreciated theorem states that we can use the equation L 1 5 even in an accel erating frame as long as the origin stays with the center of mass7 that is d Euabout CM P5Xtabout CM It is only because of this theorem that we can analyze the motion of an accelerating rolling object 5 Energy 51 Kinetic Energy and Work The kinetic energy of a particle is T mv2 which gives g dt For the change in T over a nite displacement over a particular path we have mvvFv gt dTFdr 2 AT FdrEWlt1gt2gt 1 where we ve introduced W the work done on the particle as it moves from 1 to 2 along the path of its motion lf several individual forces act on the particle we add up the work done by each ATZ2FidrZWZ1H2 i 1 i 52 Potential Energy Conservative Forces We say that a force F is consematiue if the work done by F does not depend on the path chosen between points 1 and 2 If that is the case we can choose a reference point re and de ne the potential energy function Ur as Ur 7Wr0 a r E rFr dr ro With this de nition of Ur we than have WI391 H r2 UI392 7 UI 1 AU and after de ning the total mechanical energy as E TU we nd that if only a single conservative force acts on the particle then AE AT U 0 When several different conservative forces act we can add the potential energies from the different forces to get the total mechanical energy ETU1rU2r If there are nonconservative forces in the problem we must leave them in the workienergy theorem as the work which they do ATWWCODSWDC AEATUWDC 53 More About Conservative Forces The de nition given above for a conservative force is equivalent to the requirement that F is derivable from a potential energy that is there is some function Ur such that F 7VU It is also equivalent to the condition that F has zero curl VXFO 54 If F Depends On t We can consider the case that the force F depends explicitly on time Fr7 t For such a force7 a particle could stay in one spot and the force would change with time If such a force satis es V x F 0 we can de ne a timeedependent potential by F 7VU but that is still not good enough to make it useful because with such a potential7 mechanical energy would not be conserved 8U dT U 7 Edt 55 OneiDimensional Systems If a particle can move in just one linear dimension we can get some extra understanding of the motion of the particle In this case the work done by F is m2 Wz1 gt x2 dm 1 All we need for a conservative force in this case is that is only a function of z With the potential energy function given by Um 7 may dm a plot of Um will show uphill and downhill parts The maxima and minima where dUdz 0 are places where the force F idUdm is zero The minima are points of stable equilibrium and the maxima are points of unstable equilibrium When we draw a horizontal line on graph at the value of the total energy E then motion can only take place where the values of Um are below this line The points where Um E are points where the kinetic energy and the particle must reverse its motion for these values of z they are called the turning points of the motion 56 Complete Solution for the Motion If a particle moves in one dimension7 acted on by a conservative force it turns out that in principle we can always solve for We only have to be given the constant value of the total energy E7 then E T Uz mz Um an Hng 7 Uz The ambiguity in signs comes from the fact that the energy value E won t tell you which way the particle is initially movingl Choosing 7 we have dt 7 m dm 2 E 7 Um and integrating the lhs from 0 to t and the rhs from 0 to m we get t 7 m x dx TV 2 m0 E7Ux which fin principlei can be worked out andthen inverted to get x as a afunction of t 57 Central Forces Though we will have more to say about this special type of force later for now we note that some forces are of the form Fltrgt f0quot If the potentially is spherically symmetric then 1 depends only on r In that case U depends only on r and we have 8U F iii 1quot Iquot a We will nd later that symmetric central forces like that of gravity the motion can be expressed as an equivalent sort on onedimensional motion which will simplify the solution considerably 58 Energy of Multiparticle Systems We need to pay special attention to multiparticle systems because here we have coordinates for all the particles to deal with r1 r2 rN The particles will exert forces on one another and also may move under the in uence of external forces The previous work of this chapter only dealt with one particle and we need to see precisely what changes are needed in the theory First off while the force between two particles is some function of the form Fr1 r2 in fact the interparticle forces have to be translationally invariant which means that it must be a function of the di erence of the coordinates F12 F121quot171quot2 Then if such force is conservative we need to have V1 gtlt F12 0 where V1 involves derivatives with respect to the components of r1 The important point here is that we can de ne one potential energy function Ur1 7 r2 such that F12 7V1UI391 7 1392 F21 7V2UI391 7 1392 a result which we will generalize to a system of many particles With this de nition the worki energy theorem is with T T1 1 T2 and only our conservative force acting between the particles dTUdE0 ln collisions between atomic particles the kinetic energy is often perfectly conserved because the collision does not cause any internal excitation in the particles such a collision is elastic Collisions of macroscopic objects can often cause very little change in internal energy so they can be very nearly elastic The condition of elasticity gives us another constraint on the nal velocities on the colliding particles To understand the right way to compute energy we can consider four particles easily gener alizable to many which interact amongst themselves via conservative forces but may also be in uenced by external forces which we ll also take to be conservative There will still be a principle of energy conservation but how to treat kinetic and especially potential energy The total kinetic energy is TT1T2T3T4 while for each pair of particles there is a potential energy function U which depends on the difference of the coordinates of the pairs for example U34 U34Iquot3 F4 and for an external conservative force there is potential energy U for each particle so that the total potential energy of the system is UU12U13U34UfmUth and then with E T U and no nonconservative forces acting the total energy is conserved dE dT dU 0 So for N particles the de nitions are T 0 TD and U Uint Uext Z Z Ua Z ngt 0 gtuz 0 6 Oscillations 61 Hookels Law and Simple Harmonic Motion Hooke s law77 is the same given to the linear restoring force of an ideal spring Fm 7km which gives a potential energy function of The law applies to more than springs because near the minimum of any onedimensional potential Um can be approximated as a quadratic function with some appropriate constant k From this force we get the equation of motion i77m7wx 3 where k m While we know perfectly well that the solution to 3 is mt B1 coswt B2 sinwt 4 it is a good idea to think about the solution in terms of complex exponential functions There are two ways conceptually to approach the solution We can not that the functions mt em and 57m 5 are solutions to 3 as is any linear combination of them When we insist that the linear combination be real we get 4 That solution can also be written in the form mt Acoswt 7 6 6 10 Another approach is the write down x as a linear combination of the the terms in 5 and then specify that z is the real part of the solution We get the same result7 and we will take this point of View when we have to solve the harder problem of a driven oscillator With the potential and kinetic energies given by U W2 and T m2 substitution of solution 6 gives E T U kA2cos2wt 7 6 kA2 sin2wt 7 6 kA2 which is constant7 as it should be for the conservative force we have here 62 Oscillators in 2D an 3D A central force of the form F ikf gt Fm 7km Fy 7kg gives the equations of motion and so has solution Mt Am COSwt 7 6m yt Ay coswt 7 6y A particle s motion in such a potential mapped out in the my plane can be a straight line if there is no difference in the phases 6 or else an ellipse If the potential has different force constants for the z and y motion Fm 7kmm Fy ikyy then there are different frequencies for the z and y coordinates 2 2 i iwmm y iwyy For this kind of force the particle s trajectory in the my plane can be quite complicated7 resulted in the famous Lissajous gures or a spacek lling curve 63 Damped Oscillations Next we consider a onedimensional oscillator for which there is a linear resistive force ie as studied in Chapter 2 of the form f ibz then the equation of motion is mi7biikx gt mibikz0 This equation has the same form as one we might write for an LRC circuit with no driving voltage7 yet7 where qt7 the charge on the capacitor has the same role as De ning 23 E and wo 4 the equation to be solved is i23iwgm0 11 For this DE we need any two independent solutions which work and then the general solution is a linear combination of them If we try a solution of the form Mt e we nd that it is a solution if T has either of the values n 76 mng r2 7 767 mng To get the behavior of zt it is most instructive to consider three separate cases Undamped B 0 In this case we just have our old solution for the simple harmonic oscillator Mt Clem 0257th Weak Damping B lt we Also called underdampmg things are clearer if we rst write l z7wgiw1 Mt e t Claim Cge wlt where w1 is real Then the solution is which can also be written in the form Mt Ae t cosw1t 7 6 Clearly this is a wiggly trig function which falls off in amplitude due to the decaying exponential out in front The falloff is governed by B that is Decay parameter B underdamped Strong Damping B gt we Also called ouerdampmg the solution is now 7 7 7 2 7 7 2 W 016 9 x Z me 025 rm192 we which is a sum of two real exponential functions which both decrease with time Comparing the coef cients in the exponents we see that the rst coef cient is smallest so the rst rst term decays least slowly so the long7term motion is governed by the rst term So here we want to de ne Decay parameter B 7 32 7 wg overdamped Critical Damping B we This singular case must be treated separtely because the two independent solutions are the same solution so the original DE must be solved again As it so happens two independent solutions are t 57m and Mt te t so the general solution is t Claim Cgtei t so that both terms have the exponential falloff with Decay parameter B we critical damping 12 64 Driven Damped Oscillations We now include the possibility of a time7dependent external force Ft that acts on the mass as it oscillates on the spring still with linear damping The total force is now 71912 7 ks Ft so that the equation of motion can be written mibikmFt Here there is a corresponding equation for the LRC circuit when there is a driving EMF t De ne 2B bm km tag and ft Ftm Then the equation of motion and the DE we need to solve is i 2322 wgz ft 7 This equation has linear operators on the lhs but the rhs is not zero it is inhomogeneous the latter fact makes things a little harder For such equations the procedure is to nd the general solution of the corresponding homogeneous DE that is 7 with the rhs replaced by zero and then nd any particular solution to the full DE The general solution to the full DE is the sum of the two We would like to solve 7 for a sinusoidal driving force ft f0 coswt Note that this function is the real part of the function foem We will use that on the rhs when we make it into a DE for a complex function We already have the general solution to the homogeneous equation to get a particular solution for the sinusoidal driving force we can try a complex solution of the form 2t Gem with the intention of taking the real part afterwards and after a bit of algebra we nd that it works with f0 0 wg7w22l w It is best to have C in the form C Ae i s and we nd 2 2 7 f0 7 26w A 7 W and 6 7 arctan Of course we just want the real part of this solution so with A and 6 as given above the particular solution is mt Acoswt 7 6 8 Now get the general solution of the driven damped harmonic oscillator by adding the general homogeneous solution to the particular solution in 8 We get mt Acoswt 7 6 Clem 025m which again contains a little too much to be enlightening for the case of weak damping B lt tag it is mt Acoswt 7 6 Atre t cosw1t 7 6n Here the tr is short for transient since it goes with an oscillation which will die off while the rst term is an oscillation which goes on forever The motion of the oscillator eventually settles down to the same oscillatory motion regardless of the initial conditions This is not always the case for nonlinear oscillators the behavior can be more complicated 13 6 5 Resonance Now we focus on the long term motion of the driven damped oscillator In particular we look at the dependence of the amplitude A given by A2 1 02 6137 LUZ 432612 and the phase shift 6 The fact that the amplitude is proportional to the strength of the driving force f0 is clear The dependence on the driving frequency w is a little more complicated but it is clear that A gets big when w is close to the natural frequency of the oscillator 4 If we vary wo then clearly the maximum occurs at we w If we are varying w the maximum occurs at wEw24w372 z The Q factor is a measure of the width of the resonance peak The full width at half maximum of the curve of A2 versus an can be shown to be FWHMz B The sharpness of the peak is the ratio of the value of w to the FWHM The Q factor is 7 Q Q 7 26 It can be shown that the Q factor is 7139 times the number of cycles the oscillator makes in one decay time The phase of the oscillator at resonance is 6 arctan 410 7 LU At low w we nd 6 is small ie the oscillations are in phase with the driving force As an approaches we 6 approaches 7r2 so that at resonance the oscillations are 90 behind the driving force At very large w 6 goes to 7139 and the oscillations are nearly totally out of phase with the driving force To deal with a driving force which is not sinusoidal we can still use the results from this section if the driving force is periodic In that case we can use the mathematical machinery of Fourier analysis to express the driving force as a sum of sinusoidal driving forces and then put our particular solutions for the sinusoidal cse to work Though Taylor goes through this in the book I think I want to go on to the other important topics in classical mechanics Of which there are lots 7 The Calculus of Variations Newton s laws of motion really are complete but they can be recast in a form which has practical and aesthetic advantages The new form is are known as the Lagrange equations They are useful for working in general and sometimes peculiar coordinate systems such as may occur when a mass s motion is constrained in some way The Lagrange equations arise naturally from a study of the calculus of variations and how it is applied to mechanics This branch of mathematics has important applications in almost every branch of physics It s important 71 TWO Examples One example is to prove that the shortest path between two points is a straight line How do we properly state this problem From basic calculus if a curve connects the two given points the total length of path between points 1 and 2 is 2 m2 ds 1y mdx 1 m1 and we want the path function such that L is a minimum Another problem is the nd that path that light will travel between two points which is inter esting when the index of refraction n and the speed of light depends on the coordinates In one version of the problem we ought to recover Snell s law for refraction Since the time of travel over a small distance ds is given by dt dsv with v cn the time of travel over a full path is 2 2 2 d 1 timeoftravel dt 3 nds 1 1 U c 1 which is minimized when we minimize the integral 2 m2 nwiyds Way 1y 2d90 1 11 The principle that light will follow the path that takes the least time is called Fermat s Principle but in fact in some cases light will follow the path where the time of travel is locally a maximum so that the real rule is the path in one which makes the travel time stationary 72 The EulerLagrange Equation The general problem we presently want to solve in one where we have an integral of the form 2 s ax y z zidz 1 where the curve must join the two given points 1 yl and 2 yg and where we want S to be a minimum or maximum The subject in which we study these problems is known as the calculus of variations One can show that the path which solves the problem satis es 8f 7 1 8f 7 0 8y dm 83 7 An interesting example solved in the test is the brachistochrone problem the problem where we nd the path on which a particle moving under gravity but without friction will get from a speci ed upper point to a speci ed lower point in the least time The interesting result is that the path is a cycloid We can pose the problem of extremum values of integrals where there are two or more functions which depend on a parameter as will be the usual case in our mechanics problems where the independent variable is the time t We nd that we get Euler Lagrange equations for each of the dependent variables The material of this section is relevant for mechanics in that if we describe the motion of a particle by the coordinates q1t q2t we will solve for these functions of time because some function of these functions to be called E will satisfy a stationary principle that is the integral t2 S q17 17q27 27qmqmtdt 1 is stationary This condition will give the equations 8 7d8 8 7d8 8 7d8 aql 7 dt 8amp1 7 6 qi 7 dt 8amp1 7 7 9 7 dt 9 So what is this function E 8 Lagrange s Equations 81 Unconstrained Motion First we consider the motion in 3 dimensions of a single particle At rst we describe the motion with Cartesian coordinates Recall T mvz mi2 y 22 and U Uhquot UW y 2 Choosing the x coordinate one can show that SE 7 1 8E 8x 7 dt 832 and likewise for y and 2 Since these equations have the same form as the BK equations when we are minimizing the integral S f dt we can state Hamilton s principle The actual path which a particle follows between points 1 and 2 in a given time interval 251 to 252 is such that the integral t2 S Edt 1 is stationary when taken along the actual path This integral is called the action integral and so we have three equivalent ways to get the path of a particle rt 1 Use F ma 2 Determine the path from the Lagrange equations using Cartesian coordinates 3 Determine the path from Hamilton s principle This demonstrates an alternate way to do mechanics when Cartesian coordinates are used but the beauty of the way Hamilton s principle is that S is minimized for the right path regardless of the coordinates in which it is expressed lf instead of z y 2 we use the set q1q2q3 and express the Lagrangian in terms of them then the action integral is 2 S q17q27q37Q17QZ7QSdt t1 and the condition that is is stationary give EiL equations in the q s 8118 6218 an 6218 aqlidt8q17 aqgidt8q27 aqgidtaqig 16 When we apply this to motion described with plane polar coordinates r7 gt7 the r equation gives 8U mr gt2 7 E m7quot and the 1 equation gives 8U d 2 g 07 1 The latter can be written as the old relation between torque and angular momentum dL P 7 dt The latter result is of course Newton s second law for rotations77 but we see that an equation much like F ma has dropped out of the EL equation for a generalized coordinate 1 Such an equation could have dropped out for any generalized coordinate qi7 and when we write out 8E 7 1 8E 8 7 dt 9 we can identify the left side with a force associated with the coordinate qi and on the right side we can associate a momentum for qi with araq i then we will have a force equal to a rate of change of a momentum The discussion generalizes to N particles Then there are 3N Cartesian coordinates which can also be expressed as 3N generalized coordinates q17 7q3N7 and for each of these we have a Lagrange equation7 8E 1 8E 7 87 7 7 2 7 7 7 82 Constrained Systems As interesting as the Lagrange equations are in having the same form for generalized coordinates7 their real advantage is in how they can be used to treat systems where the motion of the particles is constrained Examples of constrained motion are the simple pendulum position of bob needs one coordinate there is one degree of freedom the double pendulum two coordinates needed7 a bead sliding on a frictionless wire7 a block sliding down a smooth plane7 and even a rigid body In each7 an N particle system which could possibly require the use of 3N coordinates needs only 71 coordinates to specify its con guration It might be supposed from considering a few simple examples that the number of degrees of freedom is always equal to the number of coordinates needed but in fact that is not the case If it 23957 we say that the constraints are holonomic7 but one can concoct examples where it is not the case Taylor gives the example of a ball rolling with no slipping7 but also no spinning on a horizontal table While there are 2 degrees of freedom for the motion7 the con guration of the system needs 5 generalized coordinates The study of the mathematics of such systems and their constraints is covered extensively in stuffy old English books which no one reads anymore7 because we all have better things to study In the study of such systems7 one encounters such impressive biologicalisounding words like scleronomous and rheonomous It is very tedious For our purposes we will only consider holonomic systems in our problemisolving Do the rest on your own time We will nd that for constrained system with their reduced number of coordinates we will still get the Lagrange equations for the equations of motion but we should note that at this point in our study we have not proven that this is the case This is because is our derivation using Newton s laws and the variational principle to go over to generalized coordinates we considered all the forces acting on the particles But when we deal with constrained systems we want to ignore the forces which con ne the particles to the reduced number of dimensions So more work is needed 83 Lagrangels Equations with Constraints To accomplish the proof we need we divide up the forces acting on our particles into those which constrain the particles and all the rest The rest of the forces need to be conservative so that they are derivable from a potential U The proof uses the fact that the constraint forces are normal to the path over which the particles move and nally the obvious fact that the generalized coordinates qg qg qn we wish to use are consistent with the constraints In the end we get an equation which looks like what we ve written down before but now we know that we can use it for the n coordinates of our constrained systems or 7 d or 8 7 dt 9 At this point there is no substitute for doing lots of problems for constrained systems nding their equations of motion with the Lagrange equations Examples are given in your textbook Read them 84 Generalized Momenta Ignorable Coordinates If for the coordinate qi we de ne 8E8qi as a generalized force associated with qi E and 8E3q i as a generalized momentum then the Lagrange equation can be written as 11 F 7 1 dt In particular if E is independent of q then F is zero and the momentum p is constant In such a case we say that the coordinate q is ignorable or cyclic Having ignorable coordinates in a problem simpli es the solution 85 Conservation Laws and the Lagrangian Thinking back on the law of conservation of momentum and energy since in our previous approach we arrived at those laws from considering forces which are not made explicit in the Lagrangian approach we might ask how we get those laws from the present variational Lagrangian point of view One can show that momentum conservation arises from the translational invariance of the Lagrangian that is if we can replace rat by ra e for all 04 This would happen if we had a system of isolated particles such a substitution would change the form of an external force but not the forces between the particles In this way we get a conservation law from a mathematical symmetry of the system Likewise one can show that energy conservation follows if E has no explicit dependence on the time t In that case one can show that the quantity Zipiqi 7 E is constant This quantity is very 18 important in mechanics and is called the Hamiltonian of the system 71 H 21m 7 E i1 and for the case that the Cartesian and generalized coordinates are related by time7mdependent relations r0 raq17 7 then the Hamiltonian is the same as the total energy H T U and so the total energy is conserved 86 Magnetic Forces Magnetic forces are not dissipative they don t do any work and they are certainly fundamental so it would be a shame if we could not incorporate them into the Lagrangian formalism But they do depend on the velocity of the particle and that was why we did not consider them We can handle magnetic forces but we need to expand the de nition of the Lagrangian we will set up a function E but it won t be simply T 7 U Nevertheless this function is intended to used in the Lagrange equation to give the equations of motion First a review of magnetic forces Generally on a charge q in an EM eld the force is FEMqEVgtltB and in the general case where the elds can depend on time the elds follow from the scalar and vector potentials E7VV77 and BVXA 8t We claim that we will get the right force equation for a particle of mass m and charge q if the use the Lagrangian rrt mr27qV7r A 9 One can note that q2A2 2m mr2 qr A iw 39 qA2 7 and the last term on the rhs can be ignored because in the Lagrange equation either partial derivative would give zero We see that if we de ne a kind of extended momentum as General momentum miquot qA then the new terms in Eq 9 come from replacing miquot gt miquot qA When going over to quantum mechanics one nds that the momentum operator is replaced by V But now that we know a bit more about momentum which does it correspond to the old momentum or our new general momentum It turns out it is the new momentum and so if we have need of an operator for the old momentum miquot in quantum mechanics we need to correct for it h my0p 7V 7 qA Z 19 9 TW07B0dy Central Force Problem Now we turn to the case where two masses exert a central force on each other with no external forces This includes the case of two masses interacting through gravitation but also the problem where we might model a diatomic molecule with classical motion and forces Of course molecules really require quantum mechanics but the work that we do in working out the general problem will be very useful in the QM solution The basic problem is that the particles need 6 coordinate to describe the motion they have locations r1 and r2 The only forces are F12 and F21 with a potential energy Ur1 r2 Examples are the gravitational force and the Coulomb force Gmlmg kqqu and Ucoul m U 7 grav r1 7 r2 We will consider only central forces in which the potential energy depends only on r1 7 r2 Ur1 7 r2 U rl 7 rgl and for simplicity we will denote the relative position as r r1 7 r2 And so U Ur The Lagrangian is thus s 7 imlr m21 quot Um 91 CM and Relative Coordinates The Reduced Mass We begin the task of reducing the number of coordinates that we really need to worry about The relative position r can be used as one of the variables of the problem and the other can be taken as the CM position miri mzrz 7711 m2 R but as we saw before with no external force on the system P with M m1 m2 is constant so R is constant so later we can choose an inertial reference frame where the CM is at rest However in the current reference frame one can show that 1 2 1 2 T EM R Eur where u is an important combination of the two masses called the reduced mass mlmz mlmz Id 7 m1 m2 7 M and so the Lagrangian has a very simple form r T 7 U MR2 gm 7 Ur Em Em From the fact that the Lagrangian splits into to pieces we can solve for the CM motion R and the relative motion r separately Using the Lagrange equation for components of the the CM coordinate R gives 0 or Rconstant But since this is an isolated system we ve known that for a long time The Lagrange equation for the relative coordinate r gives u39r39 7VUT which is equivalent to a force problem for a single particle of mass 1 moving in a potential Ur It is at this point that we jump into the CM reference frame it is also inertial so we can do that so that the CM is now at rest Then the Lagrangian is simply e era m2 7 Um and with it we have reduced a twoibody problem to effectively a oneibody problem But we are solving for r a vector which points from mg to m1 and the mass of the ctional particle is u If one of the masses is very large compared to the other say 7712 gtgt m1 then the CM is very close to particle 2 and the vector r is nearly the same as r1 Then we very nearly have m1 orbiting a xed center of force One can show that the total angular momentum of the system is L r x pr which is conserved which implies that r and r remain in a xed plane So we only need to consider motion in that xed plane and thus we have reduced the problem to two dimension With the vector r now speci ed by the plane polar coordinates r and 1 our previous Lagrangian becomes s W2 M e Um which is independent of 1 Thus 1 is ignorable and its corresponding momentum is constant 8E it wzzj const 6 19 This is of course expresses the conservation of angular momentum The Lagrange equation for r gives A dU W l z OT W T which has the familiar form of F ma 92 Equivalent OneiDimensional Problem A MTZ39 We exploit the idea that we can replace with Doing that we can write the r equation as dU dU ur if pr gt2 if Fcf dr dr where Fcf which will be treated as a ctitious outward force is i 52 Fcf urz z is m Yes you are at last old enough now learn about the word centiifugal and about an outward force77 for circular motion It is only something that acts like a force to make simple sense of the radial equation And it does point outward The centrifugal force can be expressed as a centrifugal potential Fcf dU 52 77 where Ucfr 21 2 and so the genuine and centrifugal potentials combine to give an e ectiue potential at d 52 7EUr Ucf7 iaUeg where U53 Ur I 7 21 2 In the context of U53 and the equivalent one dimensional problem the condition of energy conservation is in U530 const as one might expect When we plot U530 versus r for the gravitational potential Ur GmlmgT we get some thing very illuminating The curve goes to 00 at small r As r increases the potential goes negative reaches a minimum and then approaches zero asymptotically from the negative side If we draw a horizontal line with a yiintercept at the value of the total energy of the system the line will cross the U53 curve at the tuming points of the motion A circular orbit stays at the bottom of the well at the lowest point and a single value of r Comets usually have very elongated orbits so that the turning points are widely separated If E lt 0 the motion has two turning points rmin and rm and the motion is bounded For values of the total energy E gt 0 there is only one turning point namely rum and the motion is unbounded 93 The Equation and Shape of the Orbit Go back to the radial equation and express it in terms of the actual radial force Fr 62 m4 Fr W This contains time derivatives ofr but our purpose here is to nd r as a function of the angle r gt To get to this we use a couple clever tricks one of which is to work with the variable 1 1 147 gt 7 T u then to write the operator ddt in terms of dd gt One can show 1 E1 dt T u d gt After some algebraic work we nd a DE for u gt which is good for any radial force F 94 Gravity The Kepler Orbits Now specialize to the gravitational attraction Fr 7 i39yuz where 39y Gmlmg The solutions is now fairly easy The DE for u gt is U gt W VIM52 which after a little more work can be shown to have the solution ult gt lt1 mow Since we really want r gt this can be written as 2 c 6 where c 7 T 1ecos gt mm For bounded orbits the previous equation gives us the perihelion and aphelion distances C C rmm m and rm 1 76 While to many of us it still isn t clear what shape is represented here one can show that in the Cartesian coordinates the relation has the form d 2 2 w y 1 a b that is it is an ellipse where a C b C d as which also give us I 7 4 1 7 62 a which more clearly shows the importance of the parameter 6 It is the eccentricity of the ellipse One can show using the result for areal velocity77 found earlier in the book that the period and semi major axis a are related by where the last approximation holds if one of the masses is much bigger than the other one such is the case with the Sun and the planets and in that case the relation is known as Kepler s Third Law The case 6 1 E 0 can be shown to give a parabolic trajectory The case 6 gt 1 or E gt 0 can be shown to give a hyperbolic path 23 10 Mechanics in Noninertial Frames Having accepted that we must be in an inertial frame to properly apply F ma we now deal with non inertial frames if for no other reason than the fact that we live in one due to the rotation of the earth We want to nd how the non inertialness 7 affects the use of Newton s 2nd law 101 Accelerated Frame No Rotation We rst look at a relatively simple case that where one frame to be called 5 is accelerating uniformly with respect to an inertial frame which we ll call 50 at some acceleration A One nds that if F is the total true force on a mass m then in frame 5 we have mr F 7 mA so that we can use Newton s 2nd law if we add an extra term the inertial force given by Finertial mA So that lack of suitability of the frame is accounted for by the addition of a ctitious force called in some texts a pseudoforce We will adopt the same strategy for dealing with the more complicated case of a rotating reference frame One should note that if you are in an accelerating frame the phony force will seem very real to you The use of an accelerating frame with a ctitious force can be mroe enlightening than a solution using an inertial frame as Taylor shows with the example of a suspended mass in an accelerating railroad car See Taylor s discussion of the tides The tides are caused by the uneven pull of the moon on the earth but if one thinks about it too simplistically one incorrectly comes up with one bulge of the water surface on the side of the moon Using the idea of an inertial force in the accelerating frame of the earth one can deduce an e ectz39ue force that pulls on the near side and pushes on the far side giving a double bulge to the water layer One can also deduce the shape of the equipotential surface incorporating the inertial force which has a height difference of 54 cm between the high and low parts for the tide due to the moon 102 Angular Velocity We will consider coordinate systems attached to a rigid body again called 5 where the origin coincides with the origin of an inertial frame called 50 The coordinates of S rotate possibly in a very exotic way with respect to the coordinates of 0 One can show using a geometrical theorem known as Euler s theorem77 that the most general motion of the rigid body and the attached coordinates relative to a xed point 0 is a rotation about some axis through 0 It follows that the rate of rotation is given by an angular velocity vector speci ed by a unit vector u and rotation rate w wwu and the sign of w is given by the right hand rule One can show that if r is the position of some point xed on the rotating rigid body then its velocity is V w x r 24 We note that V is the rate of change of the vector r in the inertial frame of course r does not change in the frame 5 And here we arrive at the important idea that even though a vector is a single mathematical entity it has different rates of change depending on the reference frame There is a more general relation for any vector which is xed in the rotating body In particular if e is a unit vector xed in the body then One can show that if we have set of rotating frames the relative angular velocities add just as their relative velocities do in non relativistic mechanicsl We have men wsz w21 if frame 3 rotates with respect to frame 1 with angular velocity 0131 etc 103 Time Derivatives in a Rotating Frame As Newton s 2nd law involves force vectors and time derivatives the fact that observers in two frames can disagree on the time derivatives of vectors will be central to understanding what an observer in a rotating frame will make of the 2nd law Here we have a rotating frame 5 which rotates with angular velocity 9 with respect to the inertial frame 50 A very special angular velocity which we ll use in applying the results of the chapter is that of earth in its rotation o z 73 x 105 1 The Very Important Theorem of this section relates d rate of change of vector Q relative to inertial frame 50 50 d rate of change of vector Q relative to rotating frame 5 5 dt dQ dQ also EL t 9 X Q for any vector Q One can show One can use this theorem twice in fact and Newton s 2nd law to calculate the second time derivative of the position r in the rotating frame From now one the dot notation will stand for the time derivative in the rotating frame since that s where we re interested in doing the calculations and with that we get mrF2mrXQmerx which is an odd version of Newton s 2nd law as it should be because it s not being used in an inertial frame The extra terms here are the Coriolis force Fcor er39 x Q and the centrifugal force Fcfm xrx 25 so that if we want to use physics in a rotating frame we have to add a couple terms to the real forces mi F Fcor Fcf 104 Centrifugal Force If an object has a small velocity in the rotating frame or if the time of observation is small we can ignore the Coriolis force and focus on the centrifugal We can estimate that if the speed of the object is much less than 1000 mih this is safe We can show that on the earth s surface the centrifugal force is Fcf mnzp where 5 is the unit vector pointing outward from the axis Applying this to problems in free fall we write mi Fng FEf mgo mQZR sin 05 which gives us a new effective value of the gravitational acceleration for freefall g g0 QstinQ This has radial and tangential components grad go 7 92Rs1n20 9mg Qstin0cos0 giving a maximum angle between g and the radial direction of about 010 which would seem easy to measure but it is not because we would have to make comparisons with a plumb line but the plumb line points along gl We basically have to give up and call g the new vertical direction 105 Coriolis Force This is 2mv x n where V is the object s velocity in the rotating frame For an object with speed say 50 a fast baseball the maximum acceleration due to the Coriolis force is about 0007 5 which is small compared to 9 but measurable For objects with speeds much larger than this it is more important and if it is allowed to act for a long period of time as with a Foucault pendulum it can also have a large One can show that for objects moving in the Northern Hemisphere the Coriolis force always de ects them to the right and in the Southern Hemisphere is de ects them to the left A cyclone arises from this effect when air rushed into a low pressure region and is simultaneously de ected causing a circulating swirl If we let g be the acceleration in our rotating frame due to the actual force of gravity and the small centrifugal force the equation of motion for a falling object is i g2ix Choosing the axes so that the origin is on the earth s surface with 2 up z East and y North one gets the equations i29ycos0i sin0 372Qm cos0 quot217g2 2m sin0 For an object dropped from a height h we can get an approximate solution by rst approximating x0 y0 zh7gt2 and then putting this result back into the equations We nd as Qgt3 sin0 and that if an object is dropped by 100 m at the equator it is de ected by 22 cm a small but measurable amount 106 The Foucault Pendulum Possibly the most impressive application of the Coriolis effect the Foucault pendulum can be seen in many ne science museums In it a heavy mass is suspended from a light but very long wire and set to swing freely in a plane for a very long time The plane of the oscillations changes with time and the rate of change depends on the latitude of the pendulum s location We solve the problem by solving for the z and y motion of the bob For small oscillations won can approximate the string tension as T x mg and wth this we can nd the components of the tension force as T1 7mgmL Ty 7mgyL and with the Coriolis force also acting on the bob we get the z and y equations of motion i igzL 239 cos0 3quot igyL 7 2i9 cos0 which can be rewritten as i 7 2sz wg 0 y 29232 wg 0 where Lug gL and Dz Qcos0 We can solve these using a trick from back in Chapter 2 Use the complex number n m iy and then the coupled equations can be written as one equation 2mmw3 0 Trying a solution of the form 7725 5 gives the general solution 7 5494 Cleiwot Czeiiwot Some examination of this solution shows that that plane in which the pendulum oscillates rotates at an angular rate given by Dz At a latitude of 42 0 48 the rate of rotation is g9 or about MOOday If the pendulum can swing for 6 hours without signi cant damping then the effect is easily observable 11 Rotational Motion of Rigid Bodies A rigid body can be viewed as a collection of N particles whose motion is restricted such that the shape of the object cannot change This means that the distance between any two masses is xed Whereas the N particles would normally require 3N coordinates to describe its motion a rigid body needs only 6 coordinates 3 to give the location of its center of mass and 3 mroe to give its orientation 111 The Center of Mass Recall that the position of the center of mass of a set of N particles is 1 N N R 7 Z mara where M 2 m0 M 041 041 and we use the integral form for continuous masses Recall our previous theorems PZpaMR and FmMR 04 We also want to investigate the role of the center of mass in the total angular momentum of the system We de ne the position of the mass ma relative the center of mass r by raRrD Now the total angular momentum of the system of particles relative to the origin 0 is L Zea Era x mara Dd Dd Substituting for rat to get things in terms of coordinates relative to the center of mass one can show L R x P Z r x mar Lmotion of CM Lmotion relative to CM 01 that is the total angular momentum can be broken down into the angular momentum of the center of mass plus the angular momentum relative to the center of mass One can apply this fact to the motion of the planets around the sun For them we would break down the total angular momentum as L Lorbital l Lspin where Lorbital R X Fem from which one can show Lspin Pemabout CM so if the external torque about the CM is zero the spin angular momentum is separately conserved For the earth there is a small torque about the CM which results in the precession of the equinoxes 28 One can show a similar sort of separation for the kinetic energy The result is T Zing3 Tmotion of CM Tmotion relative to CM Dd For a rigid body the second term is the kinetic energy for rotation lt s noteworthy that the past formula applies for any point xed within the body not just the GM with the r s being relative to that point 112 Rotation About a Fixed Axis By way of starting the present grown up discussion of rotating objects we consider an arbitrary rigid body rotating about the z axis eg a block of wood spinning on a xed rod that had stuck through it Here the angular velocity of the object and any mass point is w 0 0 w When we evaluate the total angular momentum L we nd something that may surprise us if we take our elementary physics too seriously The vector L doesn t just have a 2 component In fact LZ zzw L1 w Ly Iyzw where 122 Zmai 1 9 112 7 Z mamaza Iyz 7 27774043404204 Thus the 2 component of L is the old business from Phys 2110 but the z and y components are not and they involve something like the old moment of inertia but where we have a product of different coordinates like We need to solve the general situation of rotation about an arbitrary axis that is the case where the angular velocity of the rigid body is w wmwywz We need to have a form for the general relation between w which is a description of the kinematics of the rigid body and the total angular momentum L which is perhaps the most important physical quantity for the body One can show that b a l Immwm l Izywy l Izzwz Ly 2me Iyywy Iyzwz Lz Izmwz l Izywy l Izzwz where m Z szW l with similar de nitions for W and I and my 7 Z mamaya with similar de nitions for W and so on We then have 3 Li Zlijwj j1 29 which using the notation mm my m2 L1 Wm I 1ym W W L Ly w by 2m 2y 122 L2 LUZ can be written L Iw Here I is the moment of inertia tensor which says a bit more than just noting that it is a matrix which it is L and w are of course the angular momentum and angular velocity vectors The matrix I is symmetric meaning that it is equal to its tranpose Iii ji or I i Calculating the elements of I gives some practice with some funky integrals in three dimensions Often we nd that some of elements of I due to symmetries of the rigid body with respect to the chosen axes 113 Principal Axes 0f Inertia In general L is not parallel to w but one can come up with examples often for objects with lots of symmetry where it is A principal axis of an object is an axis such that if w points along it then L is parallel to w L Aw Then is the simple moment of inertia about the axis in question If there are three mutually perpendicular principal axes then when I is evaluated for these axes it is diagonal 1 0 0 I 0 A2 0 0 0 A3 and then 1 A2 and 3 are the principal moments In fact any rigid body rotating around a point has three principal axes It can be shown that the kinetic energy of a rotating body is L 1 T 7 Ew L and if the axes we use are principal axes this simpli es to T 1w1 2w2 3w3 For a rotation about a principal axis we have Iww gt Ii1w0 This equation is true if u 0 but we don t want that For nonzero w it must be true that detI 7 1 0 an equation often called the characteristic or secular equation for the matrix I Generally it gives a cubic equation for the number there are generally 3 solutions 1 A2 A3 which are then the principal moments Eigenvalues for matrices can be easily found on a computer if you have to resort to that Figure 1 Spinning top is made from rod OP fastened through the center of a uniform disc it is freely pivoted at O 114 Precession of a Top due to Weak Torque We will have to develop some dynamical equations for the motion of rigid bodies but already we have enough to do an important example A symmetric top has an axis of symmetry and as a result7 two of the principal moments are the same With the z axis along the axis of symmetry7 we have 1 0 0 I 0 1 0 0 0 A3 The top is set spinning so that its angular velocity is w meg Then its initial angular momentum is L 3 J 3w83 and it would keep this angular momentum if there were no torque The vector 33 points at an angle 9 for the inertial frame s z axis Now introduce gravity7 which acts as shown in Fig 1 Gravity causes a torque P R x Mg7 which has magnitude RMg sin0 and is perpendicular to both the z axis and the axis of the top 3933 The torque will give a change in Lso that the components of the angular velocity wl and wg will be non zero7 but if the effect of the torque is small enough7 we can expect wl and wg to remain small so that L 3w83 continues to hold approximately Then the equation L F gives Agw g R gtlt Mg and with R Reg and g 792 we can show g 9 x 33 where MgR A Q Asw z and this says that the axis of the top7 I337 rotates with angular velocity 9 about the 2 direction We have all observed this effect in a spinning ans precessing top
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