Chapter7 Lagrangian And Hamiltonian Mechanics

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