Chapter7 Lagrangian And Hamiltonian Mechanics

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Chapter 7Lagrangian and Hamiltonian MechanicsAbstract Chapter 7 is devoted to problems solved by Lagrangian and Hamiltonianmechanics.7.1 Basic Concepts and FormulaeNewtonian mechanics deals with force which is a vector quantity and therefore difficult to handle. On the other hand, Lagrangian mechanics deals with kinetic andpotential energies which are scalar quantities while Hamilton’s equations involvegeneralized momenta, both are easy to handle. While Lagrangian mechanics contains n differential equations corresponding to n generalized coordinates, Hamiltonian mechanics contains 2n equation, that is, double the number. However, theequations for Hamiltonian mechanics are linear.The symbol q is a generalized coordinate used to represent an arbitrary coordinate x, θ , ϕ, etc.If T is the kinetic energy, V the potential energy then the Lagrangian L isgiven byL T V(7.1)Lagrangian Equation:ddt dLdq̇ K L 0 q K(K 1, 2 . . .)where it is assumed that V is not a function of the velocities, i.e.(7.2) v 0. Eqn (2) q̇ Kis applicable to all the conservative systems.When n independent coordinates are required to specify the positions of themasses of a system, the system is of n degrees of freedom.Hamiltonrps q̇s LH s 1(7.3)where ps is the generalized momentum and q̇ K is the generalized velocity.287

2887 Lagrangian and Hamiltonian MechanicsHamiltonian’s Canonical Equations H q̇r , pr H ṗr qr(7.4)7.2 Problems7.1 Consider a particle of mass m moving in a plane under the attractive forceμm/r 2 directed to the origin of polar coordinates r , θ . Determine the equationsof motion.7.2 (a) Write down the Lagrangian for a simple pendulum constrained to move ina single vertical plane. Find from it the equation of motion and show thatfor small displacements from equilibrium the pendulum performs simpleharmonic motion.(b) Consider a particle of mass m moving in one dimension under a force withthe potential U (x) k(2x 3 5x 2 4x), where the constant k 0. Showthat the point x 1 corresponds to a stable equilibrium position of theparticle. Find the frequency of a small amplitude oscillation of the particleabout this equilibrium position.[University of Manchester 2007]7.3 Determine the equations of motion of the masses of Atwood machine by theLagrangian method.7.4 Determine the equations of motion of Double Atwood machine which consistsof one of the pulleys replaced by an Atwood machine. Neglect the masses ofpulleys.7.5 A particular mechanical system depending on two coordinates u and v haskinetic energy T v 2 u̇ 2 2v̇ 2 , and potential energy V u 2 v 2 . Writedown the Lagrangian for the system and deduce its equations of motion (do notattempt to solve them).[University of Manchester 2008]7.6 Write down the Lagrangian for a simple harmonic oscillator and obtain theexpression for the time period.7.7 A particle of mass m slides on a smooth incline at an angle α. The incline is notpermitted to move. Determine the acceleration of the block.7.8 A block of mass m and negligible size slides on a frictionless inclined plane ofmass M at an angle θ with the horizontal. The plane itself rests on a smoothhorizontal table. Determine the acceleration of the block and the inclined plane.7.9 A bead of mass m is free to slide on a smooth straight wire of negligible masswhich is constrained to rotate in a vertical plane with constant angular speed ωabout a fixed point. Determine the equation of motion and find the distance xfrom the fixed point at time t. Assume that at t 0 the wire is horizontal.

7.2Problems2897.10 Consider a pendulum consisting of a small mass m attached to one end of aninextensible cord of length l rotating about the other end which is fixed. Thependulum moves on a spherical surface. Hence the name spherical pendulum.The inclination angle ϕ in the xy-plane can change independently.(a) Obtain the equations of motion for the spherical pendulum.(b) Discuss the conditions for which the motion of a spherical pendulum isconverted into that of (i) simple pendulum and (ii) conical pendulum.7.11 Two blocks of mass m and M connected by a massless spring of spring constant k are placed on a smooth horizontal table. Determine the equations ofmotion using Lagrangian mechanics.7.12 A double pendulum consists of two simple pendulums of lengths l1 and l2and masses m 1 and m 2 , with the cord of one pendulum attached to the bobof another pendulum whose cord is fixed to a pivot, Fig. 7.1. Determine theequations of motion for small angle oscillations using Lagrange’s equations.Fig. 7.17.13 Use Hamilton’s equations to obtain the equations of motion of a uniformheavy rod of mass M and length 2a turning about one end which is fixed.7.14 A one-dimensional harmonic oscillator has Hamiltonian H 12 p 2 12 ω2 q 2 .Write down Hamiltonian’s equation and find the general solution.7.15 Determine the equations for planetary motion using Hamilton’s equations.7.16 Two blocks of mass m 1 and m 2 coupled by a spring of force constant k areplaced on a smooth horizontal surface, Fig. 7.2. Determine the natural frequencies of the system.Fig. 7.2

2907 Lagrangian and Hamiltonian Mechanics7.17 A simple pendulum of length l and mass m is pivoted to the block of mass Mwhich slides on a smooth horizontal plane, Fig. 7.3. Obtain the equations ofmotion of the system using Lagrange’s equations.Fig. 7.37.18 Determine the equations of motion of an insect of mass m crawling at a uniform speed v on a uniform heavy rod of mass M and length 2a which isturning about a fixed end. Assume that at t 0 the insect is at the middlepoint of the rod and it is crawling downwards.7.19 A uniform rod of mass M and length 2a is attached at one end by a cord oflength l to a fixed point. Calculate the inclination of the string and the rodwhen the string plus rod system revolves about the vertical through the pivotwith constant angular velocity ω.7.20 A particle moves in a horizontal plane in a central force potential U (r ). Derivethe Lagrangian in terms of the polar coordinates (r, θ ). Find the correspondingmomenta pr and pθ and the Hamiltonian. Hence show that the energy andangular momentum of the particle are conserved.[University of Manchester 2007]7.21 Consider the system consisting of two identical masses that can move horizontally, joined with springs as shown in Fig. 7.4. Let x, y be the horizontaldisplacements of the two masses from their equilibrium positions.(a) Find the kinetic and potential energies of the system and deduce theLagrangian.(b) Show that Lagrange’s equation gives the coupled linear differential equations m ẍ 4kx 3kym ÿ 3kx 4kyFig. 7.4

7.2Problems291(c) Find the normal modes of oscillation of this system and their period ofoscillation.7.22 Two identical beads of mass m each can move without friction along a horizontal wire and are connected to a fixed wall with two identical springs ofspring constant k as shown in Fig. 7.5.Fig. 7.5(a) Find the Lagrangian for this system and derive from it the equations ofmotion.(b) Find the eigenfrequencies of small amplitude oscillations.(c) For each normal mode, sketch the system when it is at maximumdisplacement.Note: Your sketch should indicate the relative sizes as well as the directions ofthe displacements.[University of Manchester 2007]7.23 Two beads of mass 2m and m can move without friction along a horizontalwire. They are connected to a fixed wall with two springs of spring constants2k and k as shown in Fig. 7.6:(a) Find the Lagrangian for this system and derive from it the equations ofmotion for the beads.(b) Find the eigenfrequencies of small amplitude oscillations.(c) For each normal mode, sketch the system when it is at the maximum displacement.Fig. 7.6

2927 Lagrangian and Hamiltonian Mechanics7.24 Three identical particles of mass m, M and m with M in the middle are connected by two identical massless springs with a spring constant k. Find thenormal modes of oscillation and the associated frequencies.7.25 (a) A bead of mass m is constrained to move under gravity along a planarrigid wire that has a parabolic shape y x 2 /l, where x and y are,respectively, the horizontal and the vertical coordinates. Show that theLagrangian for the system ism(ẋ)2L 2 mgx 24x 21 2 ll(b) Derive the Hamiltonian for a single particle of mass m moving in onedimension subject to a conservative force with a potential U (x).[University of Manchester 2006]7.26 A pendulum of length l and mass m is mounted on a block of mass M. Theblock can move freely without friction on a horizontal surface as shown inFig. 7.7.(a) Show that the Lagrangian for the system is L M m2 (ẋ)2 ml cos θ ẋ θ̇ m 2 2l (θ̇ ) mgl cos θ2(b) Show that the approximate form for this Lagrangian, which is applicablefor a small amplitude swinging of the pendulum, is L M m2 m 2 2θ2(ẋ) ml ẋ θ̇ l (θ̇ ) mgl 1 222(c) Find the equations of motion that follow from the simplified Lagrangianobtained in part (b),(d) Find the frequency of a small amplitude oscillation of the system.[University of Manchester 2006]Fig. 7.7

7.2Problems2937.27 (a) A particle of mass m slides down a smooth spherical bowl, as in Fig. 7.8.The particle remains in a vertical plane (the xz-plane). First, assume thatthe bowl does not move. Write down the Lagrangian, taking the angle ϑwith respect to the vertical direction as the generalized coordinate. Hence,derive the equation of motion for the particle.Fig. 7.8(b) Assume now that the bowl rests on a smooth horizontal table and has amass M, the bowl can slide freely along the x-direction.(i) Write down the Lagrangian in terms of the angle θ and the xcoordinate of the bowl, x.(ii) Starting from the corresponding Lagrange’s equations, obtain anequation giving ẍ in terms of θ , θ̇ and θ̈ and an equation giving θ̈in terms of ẍ and θ .(iii) Hence, and assuming that M m, show that for small displacements about equilibrium the period of oscillation of the particle issmaller by a factor [M/(M m)]1/2 as compared to the case wherethe bowl is fixed. [You may neglect the terms in θ 2 θ̈ or θ θ̇ 2 comparedto terms in θ̈ or θ .][University of Durham 2004]7.28 A system is described by the single (generalized) coordinate q and theLagrangian L(q, q̇). Define the generalized momentum associated with q andthe corresponding Hamiltonian, H (q, p). Derive Hamilton’s equations fromLagrange’s equations of the system. For the remainder of the question, consider the system whose Lagrangian, L(q, q̇). Find the corresponding Hamiltonian and write down Hamilton’s equations.7.29 Briefly explain what is the “generalized (or canonical) momentum conjugate to a generalized coordinate”. What characteristic feature should theLagrangian function have for a generalized momentum to be a constant ofmotion? A particle P can slide on a frictionless horizontal table with a smallopening at O. It is attached, by a string of length l passing through the opening, to a particle Q hanging vertically under the table (see Fig. 7.9). The twoparticles have equal mass, m. Let τ denote the distance of P to the opening, θthe angle between OP and some fixed line through O and g the acceleration of

2947 Lagrangian and Hamiltonian Mechanicsgravity. Initially, r a, Q does not move, and P is given an initial velocity ofmagnitude (ag)1/2 at right angles to OP.Fig. 7.9(a) Write the Lagrangian in terms of the coordinates r and θ and derive thecorresponding equations of motion.(b) Using these equations of motion and the initial conditions, show that r̈ a 3 g/r 3 g.(c) Hence, (i) show that the trajectory of P is the circle r a, (ii) show thatP describes small oscillations about this circle if it is slightly displacedfrom it and (iii) calculate the period of these oscillations:[v 2p ṙ 2 r 2 θ̇ 2 , where vp is the speed of P]7.30 A particle of mass m is constrained to move on an ellipse E in a vertical plane,parametrized by x a cos θ , y b sin θ , where a, b 0 and a b and thepositive y-direction is the upward vertical. The particle is connected to theorigin by a spring, as shown in the diagram, and is subject to gravity. Thepotential energy in the spring is 12 kr 2 where r is the distance of the point massfrom the origin (Fig. 7.10).Fig. 7.10

7.2Problems295(i) Using θ as a coordinate, find the kinetic and potential energies of theparticle when moving on the ellipse. Write down the Lagrangian andshow that Lagrange’s equation becomes m(a 2 sin2 θ b2 cos2 θ )θ̈ (a 2 b2 )(k m θ̇ 2 ) sin θ cos θ mgb cos θ .(ii) Show that θ π/2 are two equilibrium points and find any other equilibrium points, giving carefully the conditions under which they exist.You may either use Lagrange’s equation or proceed directly from thepotential energy.(iii) Determine the stabilities of each of the two equilibrium points θ π/2(it may help to consider the cases a b and a b separately).(iv) When the equilibrium point at θ π/2 is stable, determine the periodof small oscillations.[University of Manchester 2008]7.31 In prob. (7.12) on double pendulum if m 1 m 2 m and l1 l2 l, obtainthe frequencies of oscillation.7.32 Use Lagrange’s equations to obtain the natural frequencies of oscillation of acoupled pendulum described in prob. (6.46).7.33 A bead of mass m slides freely on a smooth circular wire of radius r whichrotates with constant angular velocity ω. On a horizontal plane about a pointfixed on its circumference, show that the bead performs simple harmonicmotion about the diameter passing through the fixed point as a pendulum oflength r g/ω2 .[with permission from Robert A. Becker, Introduction to theoreticalmechanics, McGraw-Hill Book Co., Inc., 1954]7.34 A block of mass m is attached to a wedge of mass M by a spring with springconstant k. The inclined frictionless surface of the wedge makes an angle α tothe horizontal. The wedge is free to slide on a horizontal frictionless surfaceas shown in Fig. 7.11.Fig. 7.11(a) Show that the Lagrangian of the system isL (M m) 2 1 2kẋ m ṡ m ẋ ṡ cos α (s l)2 mg(h s sin α),222

2967 Lagrangian and Hamiltonian Mechanicswhere l is the natural length of the spring, x is the coordinate of the wedgeand s is the length of the spring.(b) By using the Lagrangian derived in (a), show that the equations of motionare as follows:(m M)ẍ m s̈ cos α 0,m ẍ cos α m s̈ k(s s0 ) 0,where s0 l (mg sin α)/k.(c) By using the equations of motion in (b), derive the frequency for a smallamplitude oscillation of this system.[University of Manchester 2008]7.35 A uniform spherical ball of mass m rolls without slipping down a wedge ofmass M and angle α, which itself can slide without friction on a horizontaltable. The system moves in the plane shown in Fig. 7.12. Here g denotes thegravitational acceleration.Fig. 7.12(a) Find the Lagrangian and the equations of motion for this system.(b) For the special case of M m and α π/4 find(i) the acceleration of the wedge and(ii) the acceleration of the ball relative to the wedge.[Useful information: Moment of inertia of a uniform sphere of mass m2and radius R is I m R 2 .]5[University of Manchester 2007]7.3 Solutions7.1 This is obviously a two degree of freedom dynamical system. The square of theparticle velocity can be written asv 2 ṙ 2 (r θ̇)2(1)Formula (1) can be derived from Cartesian coordinatesx r cos θ,y r sin θẋ ṙ cos θ r θ̇ sin θ,ẏ ṙ sin θ r θ̇ cos θ

7.3Solutions297We thus obtainẋ 2 ẏ 2 ṙ 2 r 2 θ̇ 2The kinetic energy, the potential energy and the Lagrangian are as follows:1 21mv m(ṙ 2 r 2 θ̇ 2 )22μmV r1μmL T V m(ṙ 2 r 2 θ̇ 2 ) 2rT (2)(3)(4)We take r , θ as the generalized coordinates q1 , q2 . Since the potential energyV is independent of q̇i , Lagrangian equations take the formd L L 0 (i 1, 2)dt q̇i qi L Lμm m ṙ, mr θ̇ 2 2Now ṙ rr L L2 0 mr θ̇ , θ θ̇(5)(6)(7)Equation (5) can be explicitly written asddtddt L ṙ L θ̇ L 0 r(8) L 0 θ(9)Using (6) in (8) and (7) in (9), we getm r̈ mr θ̇ 2 md(r 2 θ̇ ) 0dtμm 0r2(10)(11)Equations (10) and (11) are identical with those obtained for Kepler’s problemby Newtonian mechanics. In particular (11) signifies the constancy of arealvelocity or equivalently angular momentum (Kepler’s second law of planetarymotion). The solution of (10) leads to the first law which asserts that the pathof a planet describes an ellipse.This example shows the simplicity and power of Lagrangian method whichinvolves energy, a scalar quantity, rather than force, a vector quantity in Newton’s mechanics.

2987 Lagrangian and Hamiltonian Mechanics7.2 The position of the pendulum is determined by a single coordinate θ and so wetake q θ . Then (Fig. 7.13)Fig. 7.131 211mv mω2l 2 ml 2 θ̇ 2222V mgl(1 cos θ )1L T V ml 2 θ̇ 2 mgl(1 cos θ )2 T T 0 ml 2 θ̇ , θ θ̇ V mgl sin θ 0, θ θ̇ Ld L 0dt θ̇ θ d (T V ) 0(T V ) dt θ̇ θd(ml 2 θ̇ ) mgl sin θ 0dtor l θ̈ g sin θ 0(equation of motion)T For small oscillation angles sin θ θgθ(equation for angular SHM)θ̈ l g2πω or time period T 2πlω2 (1)(2)(3)(4)(5)(6)lg

7.3Solutions2997.3 In this system there is only one degree of freedom. The instantaneous configuration is specified by q x. Assuming that the cord does not slip, the angularvelocity of the pulley is ẋ/R, Fig. 7.14.Fig. 7.14The kinetic energy of the system is given byT 111 ẋ 2m 1 ẋ 2 m 2 ẋ 2 I 2222 R(1)The potential energy of the system isV m 1 gx m 2 g(l x)(2)And the Lagrangian is 1Im 1 m 2 2 ẋ 2 (m 1 m 2 )gx m 2 glL 2R(3)The equation of motionddt L ẋ L 0 x(4)

3007 Lagrangian and Hamiltonian Mechanicsyields Im 1 m 2 2 ẍ g(m 1 m 2 ) 0R(m 1 m 2 )gor ẍ m 1 m 2 RI2(5)which is identical with the one obtained by Newton’s mechanics.7.4 By problem the masses of pulleys are negligible. The double machine is anAtwood machine in which one of the weights is replaced by a second Atwoodmachine, Fig. 7.15. The system now has two degrees of freedom, and its instantaneous configuration is specified by two coordinates x and x . l and l denotethe length of the vertical parts of the two strings. Mass m 1 is at depth x belowthe centre of pulley A, m 2 at depth l x x and m 3 at depth l l x x .The kinetic energy of the system is given byT 111m 1 ẋ 2 m 2 ( ẋ ẋ )2 m 3 ( ẋ ẋ )2222(1)while the potential energy is given byV m 1 gx m 2 g(l x x ) m 3 g(l x l x )Fig. 7.15(2)

7.3Solutions301The Lagrangian of the system takes the formL T V 111m 1 ẋ 2 m 2 ( ẋ ẋ )2 m 3 ( ẋ ẋ )2222 (m 1 m 2 m 3 )gx (m 2 m 3 )gx m 2 gl m 3 g(l l )(3)The equations of motion are thenddtddt L ẋ L ẋ L 0 x(4) L 0 x (5) which yield(m 1 m 2 m 3 )ẍ (m 3 m 2 )ẍ (m 1 m 2 m 3 )g (m 3 m 2 )ẍ (m 2 m 3 )ẍ (m 2 m 3 )g(6)(7)Solving (6) and (7) we obtain the equations of motion.7.5 T v 2 u̇ 2 2v̇ 22V u v(1)2(2)2 222L T V v u̇ 2v̇ u v L L 2v 2 u̇, 2u u̇ u L L 4v̇, 2v(u̇ 2 1) v̇ v2(3)(4)(5)The equations of motionddtddt L u̇ L v̇ L 0 u(6) L 0 v(7) yieldd 2(v u̇) 2u 0dtor v 2 ü 2u̇ v̇ 2u 0222v̈ v(u̇ 1) 0(8)(9)

3027 Lagrangian and Hamiltonian Mechanics7.6 Here we need a single coordinate q x:11 2m ẋ , V kx 2221 2 1 2L m ẋ kx22 L L m ẋ, kx ẋ x Ld L 0dt ẋ xT (1)(2)(3)(4)m ẍ kx 0(5)Let x A sin ωt(6)2ẍ Aω sin ωt(7)Inserting (6) and (7) and simplifying mω2 k 0 km2πor T0 2πω mωkwhere T0 is the time period.7.7 Only one coordinate q x (distance on the surface of the incline) is adequateto describe the motion:T 1 2m ẋ ,2 L m ẋ, ẋV mgx sin α, L mg sin α xL 1 2m ẋ mgx sin α2Equation of motionddt L q̇ L 0 qyieldsd (m ẋ) (mgx sin α) 0dt xor ẍ g sin α7.8 This is a two degree of freedom system because both mass m and M are moving. The coordinate on the horizontal axis is described by x for the inclinedplane and x for the block of mass m on the incline. The origin of the coordinate

7.3Solutions303system is fixed on the smooth table, Fig. 7.16. The x- and y-components of thevelocity of the block are given byFig. 7.16vx ẋ ẋ cos θv y ẋ sin θv 2 vx2 v 2y ẋ 2 2ẋ ẋ cos θ ẋ 2 (1)(2)(3)Hence, the kinetic energy of the system will beT 11M ẋ 2 m(ẋ 2 2ẋ ẋ cos θ ẋ 2 )22(4)while the potential energy takes the formV mgx sin θ(5)and the Lagrangian is given byL 11M ẋ 2 m(ẋ 2 2ẋ ẋ cos θ ẋ 2 ) mgx sin θ22(6)The equations of motionddtddt L ẋ L ẋ L 0 x(7) L 0 x (8)yieldm(ẍ cos θ ẍ) mg sin θ 0M ẍ m(ẍ ẍ cos θ ) 0(9)(10)

3047 Lagrangian and Hamiltonian MechanicsSolving (9) and (10)(M m)g sin θg sin θ 21 m cos θ/(M m)M m sin2 θmg sin θ cos θg sin θ cos θ ẍ (M m)/m cos2 θM m sin2 θẍ (11)(12)which are in agreement with the equations of prob. (2.14) derived from Newtonian mechanics.7.9 T 1m(ẋ 2 ω2 x 2 )2(1)because the velocity of the bead on the wire is at right angle to the linear velocity of the wire:V mgx sin ωt(2)because ωt θ , where θ is the angle made by the wire with the horizontal attime t, and x sin θ is the height above the horizontal position:1m(ẋ 2 ω2 x 2 ) mgx sin ωt2 L L m ẋ, mω2 x mg sin ωt ẋ xL (3)(4)Lagrange’s equationddt L ẋ L x(5)then becomesẍ ω2 x g sin ωt 0 (equation of motion)(6)which has the solutionx Aeωt Be ωt gsin ωt2ω2(7)where A and B are constants of integration which are determined from initialconditions.At t 0, x 0 and ẋ 0(8)Further ẋ ω( Aeωt(9)gcos ωt Be ωt ) 2ω

7.3Solutions305Using (8) in (7) and (9) we obtain0 A B(10)g0 ω(A B) 2ω(11)gSolving (10) and (11) we get A 2 ,4ωgB 4ω2(12)The complete solution for x isx g ωt(e eωt 2 sin ωt)4ω27.10 Let the length of the cord be l. The Cartesian coordinates can be expressed interms of spherical polar coordinates (Fig. 7.17)Fig. 7.17x l sin θ cos φy l sin θ sin φz l cos θV mgz mgl cos θ22222(1)222v ẋ ẏ ż l (θ̇ sin θ φ̇ )1T ml 2 (θ̇ 2 sin2 θ φ̇ 2 )21L ml 2 (θ̇ 2 sin2 θ φ̇ 2 ) mgl cos θ2 L L ml 2 sin θ cos θ φ̇ 2 mgl sin θ ml 2 θ̇, θ θ̇ L L 0 ml 2 sin2 θ φ̇, φ φ̇(2)(3)(4)(5)

3067 Lagrangian and Hamiltonian MechanicsThe Lagrangian equations of motionddt give L Ld L 0 and 0 θdt φ̇ φgθ̈ sin θ cos θ φ̇ 2 sin θ 0l L θ̇ d(sin2 θ φ̇) 0dtHence sin2 θ φ̇ constant Cml 2(6)(7)(8)(9)and eliminating φ̇ in (7) with the use of (9) we get a differential equation in θonly.gcos θsin θ C 2 3 0lsin θ LThe quantity Pφ ml 2 sin2 θ φ̇ φ̇θ̈ (10)(5)is a constant of motion and is recognized as the angular momentum of thesystem about the z-axis. It is conserved because torque is not produced eitherby gravity or the tension of the cord about the z-axis. Thus conservation ofangular momentum is reflected in (5).Two interesting cases arise. Suppose φ constant. Then φ̇ 0 and C 0.In this case (10) reduces toθ̈ gsin θ 0l(11)which is appropriate for simple pendulum in which the bob oscillates in thevertical plane.Suppose θ constant, then from (9) φ̇ constant. Putting θ̈ 0 in (7)we get ω φ̇ g l cos θ gH(12)and time period2π T ω Hg(13)appropriate for the conical pendulum in which the bob rotates on horizontalplane with uniform angular velocity with the cord inclined at constant angle θwith the vertical axis.

7.3Solutions3077.11 The two generalized coordinates x and y are indicated in Fig. 7.18. The kineticenergy of the system comes from the motion of the blocks and potential energyfrom the coupling spring:Fig. 7.181 2 1m ẋ M ẏ 2221 k(x y)22111 m ẋ 2 M ẏ 2 k(x y)2222 L m ẋ, k(x y) x L M ẏ, k(x y) yT (1)V(2)L L ẋ L ẏ(3)(4)(5)Lagrange’s equations are written asddt L ẋ Ld 0, xdt L ẏ L 0 y(6)Using (4) and (5) in (6) we obtain the equations of motionm ẍ k(x y) 0m ÿ k(y x) 0(7)(8)7.12 This problem involves two degrees of freedom. The coordinates are θ1 and θ2(Fig. 7.19)T 11m 1 v12 m 2 v2222v12 (l1 θ̇1 )2v22 (l1 θ̇1 )2 (l2 θ̇2 )2 2l1l2 θ̇1 θ̇2 cos(θ2 θ1 ) (by parallelogram law)For small angles, cos(θ2 θ1 ) 1(1)(2)(3)

3087 Lagrangian and Hamiltonian MechanicsFig. 7.1911m 1l12 θ̇12 m 2 l12 θ̇12 l22 θ̇22 2l1 l2 θ̇1 θ̇222V m 1 gl1 (1 cos θ1 ) m 2 gl1 (1 cos θ1 ) m 2 gl2 (1 cos θ2 )θ2m2g 2l1 θ1 l2 θ22 m 1 gl1 1 22θ211L m 1l12 θ̇12 m 2 l12 θ̇12 l22 θ̇22 2l1l2 θ̇1 θ̇2 m 1 gl1 1222m2g22l1 θ1 l2 θ2 2 L m 1l12 θ̇1 m 2l12 θ̇1 m 2l1l2 θ̇2 θ̇1 L m 1 gl1 θ1 m 2 gl1 θ1 (m 1 m 2 )gl1 θ1 θ1 L m 2l22 θ̇2 m 2l1l2 θ̇1 θ̇2 L m 2 gl2 θ2 θ2T (4)(5)(6)(7)(8)(9)(10)Lagrange’s equations areddt L θ̇1 L 0, θ1ddt L θ̇2 L 0, θ2(11)using (7, (8), (9) and 10) in (11) we obtain the equations of motion(m 1 m 2 )l1 θ̈1 m 2l2 θ̈2 (m 1 m 2 )gθ1 0l2 θ̈2 gθ2 l1 θ̈1 0(12)(13)

7.3Solutions3097.13 Writing pθ and pφ for the generalized momenta, by (4) and (5) and V by (1)of prob. (7.10)Pθ L Pθ ml 2 θ̇ , θ̇ ml 2 θ̇pφ L pφ ml 2 sin2 θ φ̇, φ̇ 2 φ̇ml sin2 θ L LH q̇i L or H L 2T q̇i q̇i q̇i L L1p 2 cosec2 θ pφ22T θ̇ φ̇ ml 2 θ θ̇ φ̇1p 2 cos ec2 θ pφ2 mgl cos θ H T V 2ml 2 θ(1)(2)(3)(4)The coordinate φ is ignorable, and therefore pφ is a constant of motion determined by the initial conditions. We are then left with only two canonical equations to be solved. The canonical equations are H H, ṗ j pj q j Hpθθ̇ pθml 2pφ2 cos θ H ṗθ mgl sin θ θml 2 sin3 θq̇ j (5)(6)where pφ is a constant of motion. By eliminating pθ we can immediatelyobtain a second-order differential equation in θ as in prob. (7.10).7.14H 1 2 1 2 2p ω q22 H q̇ p p H ṗ ω2 q q(1)(2)(3)Differentiating (2)q̈ ṗ ω2 q(4)Let q x, then (4) can be written asẍ ω2 x 0(5)

3107 Lagrangian and Hamiltonian MechanicsThis is the equation for one-dimensional harmonic oscillator. The generalsolution isx A sin(ωt ε) B cos(ωt ε)(6)which can be verified by substituting (6) in (5). Here A, B and ε are constantsto be determined from initial conditions.7.15 Let r , θ be the instantaneous polar coordinates of a planet of mass m revolvingaround a parent body of mass M:1m(ṙ 2 r 2 θ̇ 2 )2 11 V G Mm2arT (1)(2)where G is the gravitational constant and 2a is the major axis of the ellipse:pr T m ṙ , ṙ ṙmpθ Tpθ mr 2 θ̇ , θ̇ mr 2 θ̇ pθ21112 H pr 2 G Mm2m2arrpr (3)(4)(5)and the Hamiltonian equations are Hpr ṙ , prm Hpθ , pθmr 2p2 HG Mm θ3 ṗr rmrr2 H 0 ṗθ θ(6)(7)Two equations in (7) show thatpθ constant mr 2 θ̇(8)meaning the constancy of angular momentum or equivalently the constancyof areal velocity of the planet (Kepler’s second law of planetary motion).Two equations in (6) yieldr̈ p2G MmG Mmṗr r θ̇ 2 2θ 3 mm rr2r2(9)

7.3Solutions311Equation (9) describes the orbit of the planet (Kepler’s first law of planetarymotion)11m 1 ẋ12 m 2 ẋ22221V k(x1 x2 )22111L m 1 ẋ12 m 2 ẋ22 k(x1 x2 )22227.16 T (1)(2)(3)Equations of motion areddtddt L ẋ1 L ẋ2 L 0 x1(4) L 0 x2(5)Using (3) in (4) and (5)m 1 ẍ1 k(x1 x2 ) 0m 2 ẍ2 k(x1 x2 ) 0(6)(7)It is assumed that the motion is periodic and can be considered as superposition of harmonic components of various amplitudes and frequencies. Let oneof these harmonics be represented byx1 A sin ωt,x2 B sin ωt,ẍ1 ω2 A sin ωt2ẍ2 ω B sin ωt(8)(9)Substituting (8) and (9) in (6) and (7) we obtain(k m 1 ω2 ) A k B 0 k A (k m 2 ω2 )B 0The frequency equation is obtained by equating to zero the determinantformed by the coefficients of A and B:(k m 1 ω2 ) k 0 k(k m 2 ω2 )Expansion of the determinant givesm 1 m 2 ω4 k(m 1 m 2 )ω2 0

3127 Lagrangian and Hamiltonian Mechanicsorω2 [m 1 m 2 ω2 k(m 1 m 2 )] 0which yields the natural frequencies of the system: ω1 0 and ω2 k(m 1 m 2 ) m1m2 kμwhere μ is the reduced mass. The frequency ω1 0 implies that there is nogenuine oscillation of the block but mere translatory motion. The second frequency ω2 is what one expects for a simple harmonic oscillator with a reducedmass μ.7.17 Let x(t) be the displacement of the block and θ (t) the angle through whichthe pendulum swings. The kinetic energy of the system comes from themotion of the block and the swing of the bob of the pendulum. The potentialenergy comes from the deformation of the spring and the position of the bob,Fig. 7.20.Fig. 7.20The velocity v of the bob is obtained by combining vectorially its linearvelocity (l θ̇ ) with the velocity of the block (ẋ). The height through which thebob is raised from the equilibrium position is l(1 cos θ ), where l is the lengthof the pendulum:v 2 ẋ 2 l 2 θ̇ 2 2l ẋ θ̇ cos θ11T M ẋ 2 m(ẋ 2 l 2 θ̇ 2 2l ẋ θ̇)22( for θ 0, cos θ 1)1V kx 2 mgl(1 cos θ )21θ2 kx 2 mgl22(1)(2)(3)

7.3Solutions313111θ2M ẋ 2 m(ẋ 2 l 2 θ̇ 2 2l ẋ θ̇ ) kx 2 mgl2222L (4)Applying Lagrange’s equationsddt L ẋ L 0, xddt L θ̇ L 0 θ(5)we obtain(M m)ẍ ml θ̈ kx 0l θ̈ ẍ gθ 0(6)(7)7.18 Considering that at t 0 the insect was in the middle of the rod, the coordinates of the insect x, y, z at time t are given byx (a vt) sin θ cos φy (a vt) sin θ sin φz (a vt) cos θand the square of its velocity isẋ 2 ẏ 2 ż 2 v 2 (a vt)2 (θ̇ 2 φ̇ 2 sin2 θ )2m T Ma 2 (θ̇ 2 φ̇ 2 sin2 θ ) {v 2 (a vt)2 (θ̇ 2 φ̇ 2 sin2 θ )}32V Mga cos θ mg(a vt) cos θ constantL T VThe application of the Lagrangian equations to the coordinates θ and φ yieldsddt 44Ma 2 θ̇ m(a vt)2 θ̇ Ma 2 m(a vt)2 φ̇ 2 sin θ cos θ33 {Ma m(a vt)}g sin θ 4dandMa 2 m(a vt)2 φ̇ sin2 θ 0dt5(1)(2)Equation (2) can be integrated at once as it is free from φ: 422Ma m(a vt) φ̇ sin2 θ constant C5(3) pa

Lagrangian and Hamiltonian Mechanics Abstract Chapter 7 is devoted to problems solved by Lagrangian and Hamiltonian mechanics. 7.1 Basic Concepts and Formulae Newtonian mechanics deals with force which is a vector quantity and therefore dif-ficult to handle. On the other hand, Lag

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