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# [Solution manual] classical mechanics, goldstein

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### [Solution manual] classical mechanics, goldstein

1. 1. Goldstein Classical Mechanics Notes Michael Good May 30, 20041 Chapter 1: Elementary Principles1.1 Mechanics of a Single ParticleClassical mechanics incorporates special relativity. ‘Classical’ refers to the con-tradistinction to ‘quantum’ mechanics. Velocity: dr v= . dt Linear momentum: p = mv. Force: dp F= . dt In most cases, mass is constant and force is simpliﬁed: d dv F= (mv) = m = ma. dt dt Acceleration: d2 r a= . dt2 Newton’s second law of motion holds in a reference frame that is inertial orGalilean. Angular Momentum: L = r × p. Torque: T = r × F. Torque is the time derivative of angular momentum: 1
2. 2. dL T= . dt Work: 2 W12 = F · dr. 1 In most cases, mass is constant and work simpliﬁes to: 2 2 2 dv dv W12 = m · vdt = m v· dt = m v · dv 1 dt 1 dt 1 m 2 2 W12 = (v2 − v1 ) = T2 − T1 2 Kinetic Energy: mv 2 T = 2 The work is the change in kinetic energy. A force is considered conservative if the work is the same for any physicallypossible path. Independence of W12 on the particular path implies that thework done around a closed ciruit is zero: F · dr = 0 If friction is present, a system is non-conservative. Potential Energy: F = − V (r). The capacity to do work that a body or system has by viture of is positionis called its potential energy. V above is the potential energy. To express workin a way that is independent of the path taken, a change in a quantity thatdepends on only the end points is needed. This quantity is potential energy.Work is now V1 − V2 . The change is -V. Energy Conservation Theorem for a Particle: If forces acting on a particleare conservative, then the total energy of the particle, T + V, is conserved. The Conservation Theorem for the Linear Momentum of a Particle statesthat linear momentum, p, is conserved if the total force F, is zero. The Conservation Theorem for the Angular Momentum of a Particle statesthat angular momentum, L, is conserved if the total torque T, is zero. 2
3. 3. 1.2 Mechanics of Many ParticlesNewton’s third law of motion, equal and opposite forces, does not hold for allforces. It is called the weak law of action and reaction. Center of mass: mi ri mi ri R= = . mi M Center of mass moves as if the total external force were acting on the entiremass of the system concentrated at the center of mass. Internal forces that obeyNewton’s third law, have no eﬀect on the motion of the center of mass. d2 R (e) F(e) ≡ M = Fi . dt2 i Motion of center of mass is unaﬀected. This is how rockets work in space. Total linear momentum: dri dR P= mi =M . i dt dt Conservation Theorem for the Linear Momentum of a System of Particles:If the total external force is zero, the total linear momentum is conserved. The strong law of action and reaction is the condition that the internal forcesbetween two particles, in addition to being equal and opposite, also lie alongthe line joining the particles. Then the time derivative of angular momentumis the total external torque: dL = N(e) . dt Torque is also called the moment of the external force about the given point. Conservation Theorem for Total Angular Momentum: L is constant in timeif the applied torque is zero. Linear Momentum Conservation requires weak law of action and reaction. Angular Momentum Conservation requires strong law of action and reaction. Total Angular Momentum: L= ri × pi = R × M v + ri × pi . i i 3
4. 4. Total angular momentum about a point O is the angular momentum of mo-tion concentrated at the center of mass, plus the angular momentum of motionabout the center of mass. If the center of mass is at rest wrt the origin then theangular momentum is independent of the point of reference. Total Work: W12 = T2 − T1 1 2 where T is the total kinetic energy of the system: T = 2 i mi vi . Total kinetic energy: 1 2 1 1 T = mi vi = M v2 + mi vi2 . 2 i 2 2 i Kinetic energy, like angular momentum, has two parts: the K.E. obtained ifall the mass were concentrated at the center of mass, plus the K.E. of motionabout the center of mass. Total potential energy: 1 V = Vi + Vij . i 2 i,j i=j If the external and internal forces are both derivable from potentials it ispossible to deﬁne a total potential energy such that the total energy T + V isconserved. The term on the right is called the internal potential energy. For rigid bodiesthe internal potential energy will be constant. For a rigid body the internalforces do no work and the internal potential energy remains constant.1.3 Constraints • holonomic constraints: think rigid body, think f (r1 , r2 , r3 , ..., t) = 0, think a particle constrained to move along any curve or on a given surface. • nonholonomic constraints: think walls of a gas container, think particle placed on surface of a sphere because it will eventually slide down part of the way but will fall oﬀ, not moving along the curve of the sphere. 1. rheonomous constraints: time is an explicit variable...example: bead on moving wire 2. scleronomous constraints: equations of contraint are NOT explicitly de- pendent on time...example: bead on rigid curved wire ﬁxed in space Diﬃculties with constraints: 4
5. 5. 1. Equations of motion are not all independent, because coordinates are no longer all independent 2. Forces are not known beforehand, and must be obtained from solution. For holonomic constraints introduce generalized coordinates. Degrees offreedom are reduced. Use independent variables, eliminate dependent coordi-nates. This is called a transformation, going from one set of dependent variablesto another set of independent variables. Generalized coordinates are worthwhilein problems even without constraints. Examples of generalized coordinates: 1. Two angles expressing position on the sphere that a particle is constrained to move on. 2. Two angles for a double pendulum moving in a plane. 3. Amplitudes in a Fourier expansion of rj . 4. Quanities with with dimensions of energy or angular momentum. For nonholonomic constraints equations expressing the constraint cannot beused to eliminate the dependent coordinates. Nonholonomic constraints areHARDER TO SOLVE.1.4 D’Alembert’s Principle and Lagrange’s EquationsDeveloped by D’Alembert, and thought of ﬁrst by Bernoulli, the principle that: (a) dpi (Fi − ) · δri = 0 i dt This is valid for systems which virtual work of the forces of constraint van-ishes, like rigid body systems, and no friction systems. This is the only restric-tion on the nature of the constraints: workless in a virtual displacement. Thisis again D’Alembert’s principle for the motion of a system, and what is goodabout it is that the forces of constraint are not there. This is great news, but itis not yet in a form that is useful for deriving equations of motion. Transformthis equation into an expression involving virtual displacements of the gener-alized coordinates. The generalized coordinates are independent of each otherfor holonomic constraints. Once we have the expression in terms of generalizedcoordinates the coeﬃcients of the δqi can be set separately equal to zero. Theresult is: d ∂T ∂T {[ ( )− ] − Qj }δqj = 0 dt ∂ qj ˙ ∂qj 5
6. 6. Lagrange’s Equations come from this principle. If you remember the indi-vidual coeﬃcients vanish, and allow the forces derivable from a scaler potentialfunction, and forgive me for skipping some steps, the result is: d ∂L ∂L ( )− =0 dt ∂ qj ˙ ∂qj1.5 Velocity-Dependent Potentials and The Dissipation FunctionThe velocity dependent potential is important for the electromagnetic forces onmoving charges, the electromagnetic ﬁeld. L=T −U where U is the generalized potential or velocity-dependent potential. For a charge mvoing in an electric and magnetic ﬁeld, the Lorentz forcedictates: F = q[E + (v × B)]. The equation of motion can be dervied for the x-dirction, and notice theyare identical component wise: m¨ = q[Ex + (v × B)x ]. x If frictional forces are present(not all the forces acting on the system arederivable from a potential), Lagrange’s equations can always be written: d ∂L ∂L ( )− = Qj . dt ∂ qj ˙ ∂qj where Qj represents the forces not arising from a potential, and L containsthe potential of the conservative forces as before. Friction is commonly, Ff x = −kx vx . Rayleigh’s dissipation function: 1 2 2 2 Fdis = (kx vix + ky viy + kz viz ). 2 i The total frictional force is: Ff = − v Fdis Work done by system against friction: dWf = −2Fdis dt 6
7. 7. The rate of energy dissipation due to friction is 2Fdis and the component ofthe generalized force resulting from the force of friction is: ∂Fdis Qj = − . ∂ qj ˙ In use, both L and Fdis must be speciﬁed to obtain the equations of motion: d ∂L ∂L ∂Fdis ( )− =− . dt ∂ qj ˙ ∂qj ∂ qj ˙1.6 Applications of the Lagrangian FormulationThe Lagrangian method allows us to eliminate the forces of constraint from theequations of motion. Scalar functions T and V are much easier to deal withinstead of vector forces and accelerations. Procedure: 1. Write T and V in generalized coordinates. 2. Form L from them. 3. Put L into Lagrange’s Equations 4. Solve for the equations of motion. Simple examples are: 1. a single particle is space(Cartesian coordinates, Plane polar coordinates) 2. atwood’s machine 3. a bead sliding on a rotating wire(time-dependent constraint). Forces of contstraint, do not appear in the Lagrangian formulation. Theyalso cannot be directly derived. 7
8. 8. Goldstein Chapter 1 Derivations Michael Good June 27, 20041 Derivations1. Show that for a single particle with constant mass the equation of motionimplies the follwing diﬀerential equation for the kinetic energy: dT =F·v dt while if the mass varies with time the corresponding equation is d(mT ) = F · p. dt Answer: dT d( 1 mv 2 ) = 2 = mv · v = ma · v = F · v ˙ dt dt with time variable mass, d(mT ) d p2 = ( ) = p · p = F · p. ˙ dt dt 22. Prove that the magnitude R of the position vector for the center of mass froman arbitrary origin is given by the equation: 1 M 2 R2 = M 2 mi ri − 2 mi mj rij . i 2 i,j Answer: MR = mi ri 1
9. 9. M 2 R2 = mi mj ri · rj i,j Solving for ri · rj realize that rij = ri − rj . Square ri − rj and you get 2 2 2 rij = ri − 2ri · rj + rj Plug in for ri · rj 1 2 2 2 ri · rj = (r + rj − rij ) 2 i 1 1 1 M 2 R2 = 2 mi mj ri + 2 mi mj rj − 2 mi mj rij 2 i,j 2 i,j 2 i,j 1 1 1 M 2 R2 = M 2 mi r i + M 2 mj rj − 2 mi mj rij 2 i 2 j 2 i,j 1 M 2 R2 = M 2 mi ri − 2 mi mj rij i 2 i,j3. Suppose a system of two particles is known to obey the equations of mo-tions, d2 R (e) M 2 = Fi ≡ F(e) dt i dL = N(e) dtFrom the equations of the motion of the individual particles show that the in-ternal forces between particles satisfy both the weak and the strong laws of ac-tion and reaction. The argument may be generalized to a system with arbitrarynumber of particles, thus proving the converse of the arguments leading to theequations above. Answer: First, if the particles satisfy the strong law of action and reaction then theywill automatically satisfy the weak law. The weak law demands that only theforces be equal and opposite. The strong law demands they be equal and oppo-site and lie along the line joining the particles. The ﬁrst equation of motion tellsus that internal forces have no eﬀect. The equations governing the individualparticles are (e) ˙ p1 = F1 + F21 (e) ˙ p2 = F2 + F12 2
10. 10. Assuming the equation of motion to be true, then (e) (e) ˙ ˙ p1 + p2 = F1 + F21 + F2 + F12 must give F12 + F21 = 0 Thus F12 = −F21 and they are equal and opposite and satisfy the weak lawof action and reaction. If the particles obey dL = N(e) dt then the time rate of change of the total angular momentum is only equal tothe total external torque; that is, the internal torque contribution is null. Fortwo particles, the internal torque contribution isr1 × F21 + r2 × F12 = r1 × F21 + r2 × (−F21 ) = (r1 − r2 ) × F21 = r12 × F21 = 0 Now the only way for r12 × F21 to equal zero is for both r12 and F21 to lieon the line joining the two particles, so that the angle between them is zero, iethe magnitude of their cross product is zero. A × B = ABsinθ4. The equations of constraint for the rolling disk, dx − a sin θdψ = 0 dy + a cos θdψ = 0are special cases of general linear diﬀerential equations of constraint of the form n gi (x1 , . . . , xn )dxi = 0. i=1A constraint condition of this type is holonomic only if an integrating functionf (x1 , . . . , xn ) can be found that turns it into an exact diﬀerential. Clearly thefunction must be such that ∂(f gi ) ∂(f gj ) = ∂xj ∂xifor all i = j. Show that no such integrating factor can be found for either of theequations of constraint for the rolling disk. Answer: 3
11. 11. First attempt to ﬁnd the integrating factor for the ﬁrst equation. Note it isin the form: P dx + Qdφ + W dθ = 0 where P is 1, Q is −a sin θ and W is 0. The equations that are equivalent to ∂(f gi ) ∂(f gj ) = ∂xj ∂xi are ∂(f P ) ∂(f Q) = ∂φ ∂x ∂(f P ) ∂(f W ) = ∂θ ∂x ∂(f Q) ∂(f W ) = ∂θ ∂φ These are explicitly: ∂(f ) ∂(−f a sin θ) = ∂φ ∂x ∂(f ) =0 ∂θ ∂(−f a sin θ) =0 ∂θ Simplfying the last two equations yields: f cos θ = 0 Since y is not even in this ﬁrst equation, the integrating factor does notdepend on y and because of ∂f = 0 it does not depend on θ either. Thus ∂θ f = f (x, φ) The only way for f to satisfy this equation is if f is constant and thus appar-ently there is no integrating function to make these equations exact. Performingthe same procedure on the second equation you can ﬁnd ∂(f a cos θ) ∂f = ∂y ∂φ ∂f ∂f a cos θ = ∂y ∂φ and f sin θ = 0 4
12. 12. ∂f =0 ∂θ leading to f = f (y, φ)and making it impossible for f to satsify the equations unless as a constant. Ifthis question was confusing to you, it was confusing to me too. Mary Boas saysit is ‘not usually worth while to spend much time searching for an integratingfactor’ anyways. That makes me feel better.5. Two wheels of radius a are mounted on the ends of a common axle of lengthb such that the wheels rotate independently. The whole combination rolls with-out slipping on a palne. Show that there are two nonholonomic equations ofconstraint, cos θdx + sin θdy = 0 1 sin θdx − cos θdy = a(dφ + dφ ) 2 (where θ,φ, and φ have meanings similar to those in the problem of a singlevertical disk, and (x,y) are the corrdinates of a point on the axle midway betweenthe two wheels) and one holonomic equation of constraint, a θ = C − (φ − φ ) b where C is a constant. Answer: The trick to this problem is carefully looking at the angles and getting thesigns right. I think the fastest way to solve this is to follow the same procedurethat was used for the single disk in the book, that is, ﬁnd the speed of thedisk, ﬁnd the point of contact, and take the derivative of the x component,and y component of position, and solve for the equations of motion. Here thesteps are taken a bit further because a holonomic relationship can be found thatrelates θ, φ and φ . Once you have the equations of motion, from there its justslightly tricky algebra. Here goes: We have two speeds, one for each disk ˙ v = aφ ˙ v = aφ and two contact points, b b (x ± cos θ, y ± sin θ) 2 2 5
13. 13. The contact points come from the length of the axis being b as well as x andy being the center of the axis. The components of the distance are cos and sinfor x and y repectively. So now that we’ve found the speeds, and the points of contact, we want totake the derivatives of the x and y parts of their contact positions. This willgive us the components of the velocity. Make sure you get the angles right, theywere tricky for me. d b (x + cos θ) = vx dt 2 b ˙ x− ˙ sin θθ = v cos(180 − θ − 90) = v cos(90 − θ) = v cos(−90 + θ) = v sin θ 2 b ˙ ˙ x − sin θθ = aφ sin θ ˙ 2 Do this for the next one, and get: b ˙ ˙ x+ ˙ sin θθ = aφ sin θ 2 The plus sign is there because of the derivative of cos multiplied with thenegative for the primed wheel distance from the center of the axis. For the yparts: d b (y + sin θ) = vy dt 2 b ˙ ˙ y+ ˙ cos θθ = −v cos θ = −aφ cos θ 2 It is negative because I decided to have axis in the ﬁrst quadrent headingsouth-east. I also have the primed wheel south-west of the non-primed wheel.A picture would help, but I can’t do that on latex yet. So just think about it. Do it for the next one and get: b ˙ ˙ y− ˙ cos θθ = −aφ cos θ 2 All of the derivatives together so you aren’t confused what I just did: b ˙ ˙ x− ˙ sin θθ = aφ sin θ 2 b ˙ ˙ x+ ˙ sin θθ = aφ sin θ 2 b ˙ ˙ y + cos θθ = −aφ cos θ ˙ 2 b ˙ ˙ y − cos θθ = −aφ cos θ ˙ 2Now simplify them by cancelling the dt s and leaving the x and y’s on one side: 6
14. 14. b dx = sin θ[ dθ + adφ] (1) 2 b dx = sin θ[− dθ + adφ ] (2) 2 b dy = − cos θ[ dθ + adφ] (3) 2 b dy = − cos θ[− dθ + adφ ] (4) 2 Now we are done with the physics. The rest is manipulation of these equa-tions of motion to come up with the constraints. For the holonomic equationuse (1)-(2). (1) − (2) = 0 = bdθ + a(dφ − dφ ) a dθ = − (dφ − dφ ) b a θ = − (φ − φ ) + C b For the other two equations, I started with b b (1) cos θ + (3) sin θ = cos θ sin θ[ dθ + adφ] − sin θ cos θ[ dθ + adφ] 2 2 cos θdx + sin θdy = 0 and (1) + (2) = 2dx = sin θa[dφ + dφ ] (3) + (4) = 2dy = − cos θa[dφ + dφ ] multiply dy by − cos θ and multiply dx by sin θ to yield yourself a − cos θdy = cos2 θ [dφ + dφ ] 2 2 a sin θdx = sin θ [dφ + dφ ] 2 Add them together and presto! a sin θdx − cos θdy = [dφ + dφ ] 26. A particle moves in the xy plane under the constraint that its velocity vectoris always directed towards a point on the x axis whose abscissa is some givenfunction of time f (t). Show that for f (t) diﬀerentiable, but otherwise arbitrary, 7
15. 15. the constraint is nonholonomic. Answer: The abscissa is the x-axis distance from the origin to the point on the x-axisthat the velocity vector is aimed at. It has the distance f (t). I claim that the ratio of the velocity vector components must be equal tothe ratio of the vector components of the vector that connects the particle tothe point on the x-axis. The directions are the same. The velocity vectorcomponents are: dy vy = dt dx vx = dt The vector components of the vector that connects the particle to the pointon the x-axis are: Vy = y(t) Vx = x(t) − f (t) For these to be the same, then vy Vy = vx Vx dy y(t) = dx x(t) − f (t) dy dx = y(t) x(t) − f (t) This cannot be integrated with f (t) being arbituary. Thus the constraint isnonholonomic. It’s nice to write the constraint in this way because it’s frequentlythe type of setup Goldstein has: ydx + (f (t) − x)dy = 0 There can be no integrating factor for this equation.7. The Lagrangian equations can be written in the form of the Nielsen’s equa-tions. ∂T˙ ∂T −2 =Q ∂q˙ ∂qShow this. 8
16. 16. Answer: I’m going to set the two forms equal and see if they match. That will showthat they can be written as displayed above. Lagrangian Form = Nielsen’s Form d ∂T ∂T ∂T˙ ∂T ( )− = −2 dt ∂ q ˙ ∂q ∂q˙ ∂q d ∂T ∂T ∂T˙ ( )+ = (5) dt ∂ q ˙ ∂q ∂q˙ ∂T˙ ˙ What is ∂q˙ you may ask? Well, lets solve for T ﬁrst. ˙ d T ≡ T (q, q, t) ˙ dt d Because dt is a full derivative, you must not forget the chain rule. ˙ d ∂T ∂T ∂T T ≡ T (q, q, t) = ˙ + q+ ˙ q ¨ dt ∂t ∂q ∂q˙ ∂T˙ Now lets solve for ∂q˙ , not forgetting the product rule ∂T˙ ∂ ∂T ∂T ∂T = [ + q+ ˙ q] ¨ ∂q˙ ∂ q ∂t ˙ ∂q ∂q˙ ∂T˙ ∂ ∂T ∂ ∂T ∂T ∂ q ˙ ∂ ∂T = + q+ ˙ + q ¨ ∂q˙ ∂ q ∂t ˙ ∂ q ∂q ˙ ∂q ∂ q ∂ q ∂ q ˙ ˙ ˙ ∂T˙ ∂ ∂T ∂ ∂T ∂T ∂ ∂T = + q+ ˙ + ( )¨ q ∂q˙ ∂t ∂ q ˙ ∂q ∂ q ˙ ∂q ∂q ∂q ˙ ˙ ∂T˙ Now we have ∂q˙ , so lets plug this into equation (5). d ∂T ∂T ∂ ∂T ∂ ∂T ∂T ∂ ∂T ( )+ = + q+ ˙ + ( )¨ q dt ∂ q ˙ ∂q ∂t ∂ q ˙ ∂q ∂ q ˙ ∂q ∂q ∂q ˙ ˙ d ∂T ∂ ∂T ∂ ∂T ∂ ∂T ( )= + q+ ˙ ( )¨ q dt ∂ q ˙ ∂t ∂ q ˙ ∂q ∂ q ˙ ∂q ∂q ˙ ˙ Notice that this is indeed true. d ∂T ∂ ∂T ∂ ∂T ∂ ∂T ( )= ( )+ ( )q + ˙ ( )¨ q dt ∂ q ˙ ∂t ∂ q ˙ ∂q ∂ q ˙ ∂q ∂q ˙ ˙ because T = T (q, q, t). ˙ 9
17. 17. If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange’sequations, show by direct substitution that dF (q1 , ..., qn , t) L =L+ dtalso satisﬁes Lagrange’s equations where F is any arbitrary, but diﬀerentiable,function of its arguments. Answer: Let’s directly substitute L into Lagrange’s equations. d ∂L ∂L − =0 dt ∂ q ˙ ∂q d ∂ dF ∂ dF (L + )− (L + )=0 dt ∂ q ˙ dt ∂q dt d ∂L ∂ dF ∂L ∂ dF [ + ]− − =0 dt ∂ q ˙ ∂ q dt ˙ ∂q ∂q dt d ∂L ∂L d ∂ dF ∂ dF − + − =0 dt ∂ q ˙ ∂q dt ∂ q dt ˙ ∂q dt On the left we recognized Lagrange’s equations, which we know equal zero.Now to show the terms with F vanish. d ∂ dF ∂ dF − =0 dt ∂ q dt ˙ ∂q dt d ∂F ˙ ∂F˙ = dt ∂ q ˙ ∂q This is shown to be true because ∂F˙ ∂F = ∂q˙ ∂q We have d ∂F ˙ d ∂F = dt ∂ q ˙ dt ∂q ∂ ∂F ∂ ∂F = + q ˙ ∂t ∂q ∂q ∂q ∂ ∂F ∂F ∂F˙ = [ + q] = ˙ ∂q ∂t ∂q ∂q 10
18. 18. Thus as Goldstein reminded us, L = T − V is a suitable Lagrangian, but itis not the only Lagrangian for a given system.9. The electromagnetic ﬁeld is invariant under a gauge transformation of thescalar and vector potential given by A→A+ ψ(r, t) 1 ∂ψ φ→φ− c ∂t where ψ is arbitrary (but diﬀerentiable). What eﬀect does this gauge trans-formation have on the Lagrangian of a particle moving in the electromagneticﬁeld? Is the motion aﬀected? Answer: 1 q L= mv 2 − qφ + A · v 2 c Upon the gauge transformation: 1 1 ∂ψ q L = mv 2 − q[φ − ] + [A + ψ(r, t)] · v 2 c ∂t c 1 q q ∂ψ q L = mv 2 − qφ + A · v + + ψ(r, t) · v 2 c c ∂t c q ∂ψ L =L+ [ + ψ(r, t) · v] c ∂t q ˙ L = L + [ψ] c In the previous problem it was shown that: d ∂ψ ˙ ˙ ∂ψ = dt ∂ q ˙ ∂qFor ψ diﬀerentiable but arbitrary. This is all that you need to show that theLagrangian is changed but the motion is not. This problem is now in the sameform as before: dF (q1 , ..., qn , t) L =L+ dt And if you understood the previous problem, you’ll know why there is noeﬀect on the motion of the particle( i.e. there are many Lagrangians that maydescribe the motion of a system, there is no unique Lagrangian).10. Let q1 , ..., qn be a set of independent generalized coordinates for a system 11
19. 19. of n degrees of freedom, with a Lagrangian L(q, q, t). Suppose we transform ˙to another set of independent coordinates s1 , ..., sn by means of transformationequations qi = qi (s1 , ..., sn , t), i = 1, ..., n. (Such a transformatin is called a point transformation.) Show that if theLagrangian function is expressed as a function of sj , sj and t through the equa- ˙tion of transformation, then L satisﬁes Lagrange’s equations with respect to thes coordinates d ∂L ∂L − =0 dt ∂ sj ˙ ∂sj In other words, the form of Lagrange’s equations is invariant under a pointtransformation. Answer: We know: d ∂L ∂L − =0 dt ∂ qi ˙ ∂qiand we want to prove: d ∂L ∂L − =0 dt ∂ sj ˙ ∂sj ∂L ∂L If we put ∂ sj and ∂sj in terms of the q coordinates, then they can be ˙substitued back in and shown to still satisfy Lagrange’s equations. ∂L ∂L ∂qi = ∂sj i ∂qi ∂sj ∂L ∂L ∂ qi ˙ = ∂ sj ˙ i ∂ qi ∂ sj ˙ ˙ We know: ∂qi ∂ qi ˙ = ∂sj ∂ sj ˙ Thus, ∂L ∂L ∂qi = ∂ sj ˙ i ∂ qi ∂sj ˙ ∂L ∂L Plug ∂ sj ˙ and ∂sj into the Lagrangian equation and see if they satisfy it: d ∂L ∂qi ∂L ∂qi [ ]−[ ]=0 dt i ∂ qi ∂sj ˙ i ∂qi ∂sj 12
20. 20. ∂qi Pulling out the summation to the right and ∂sj to the left, we are left with: d ∂L ∂L ∂qi [ − ] =0 i dt ∂ qi ˙ ∂qi ∂sj This shows that Lagrangian’s equations are invariant under a point trans-formation. 13
21. 21. Goldstein Chapter 1 Exercises Michael Good July 17, 20041 Exercises11. Consider a uniform thin disk that rolls without slipping on a horizontalplane. A horizontal force is applied to the center of the disk and in a directionparallel to the plane of the disk. • Derive Lagrange’s equations and ﬁnd the generalized force. • Discuss the motion if the force is not applied parallel to the plane of the disk.Answer: To ﬁnd Lagrangian’s equations, we need to ﬁrst ﬁnd the Lagrangian. L=T −V 1 1 T = mv 2 = m(rω)2 V =0 2 2 Therefore 1 L= m(rω)2 2 Plug into the Lagrange equations: d ∂L ∂L − =Q dt ∂ x ˙ ∂x d ∂ 1 mr2 ω 2 2 ∂ 1 mr2 ω 2 − 2 =Q dt ∂(rω) ∂x d m(rω) = Q dt m(rω ) = Q ¨ 1
22. 22. If the motion is not applied parallel to the plane of the disk, then theremight be some slipping, or another generalized coordinate would have to beintroduced, such as θ to describe the y-axis motion. The velocity of the diskwould not just be in the x-direction as it is here.12. The escape velocity of a particle on Earth is the minimum velocity re-quired at Earth’s surface in order that that particle can escape from Earth’sgravitational ﬁeld. Neglecting the resistance of the atmosphere, the system isconservative. From the conservation theorme for potential plus kinetic energyshow that the escape veolcity for Earth, ingnoring the presence of the Moon, is11.2 km/s.Answer: GM m 1 = mv 2 r 2 GM 1 = v2 r 2 Lets plug in the numbers to this simple problem: (6.67 × 10−11 ) · (6 × 1024 ) 1 = v2 (6 × 106 ) 2 This gives v = 1.118 × 104 m/s which is 11.2 km/s.13. Rockets are propelled by the momentum reaction of the exhaust gasesexpelled from the tail. Since these gases arise from the raction of the fuels carriedin the rocket, the mass of the rocket is not constant, but decreases as the fuelis expended. Show that the equation of motion for a rocket projected verticallyupward in a uniform gravitational ﬁeld, neglecting atmospheric friction, is: dv dm m = −v − mg dt dtwhere m is the mass of the rocket and v’ is the velocity of the escaping gasesrelative to the rocket. Integrate this equation to obtain v as a function of m,assuming a constant time rate of loss of mass. Show, for a rocket starting initallyfrom rest, with v’ equal to 2.1 km/s and a mass loss per second equal to 1/60thof the intial mass, that in order to reach the escape velocity the ratio of thewight of the fuel to the weight of the empty rocket must be almost 300!Answer: This problem can be tricky if you’re not very careful with the notation. Buthere is the best way to do it. Deﬁning me equal to the empty rocket mass,mf is the total fuel mass, m0 is the intitial rocket mass, that is, me + mf , anddm m0 dt = − 60 as the loss rate of mass, and ﬁnally the goal is to ﬁnd the ratio of 2
23. 23. mf /me to be about 300. The total force is just ma, as in Newton’s second law. The total force onthe rocket will be equal to the force due to the gas escaping minus the weightof the rocket: d ma = [−mv ] − mg dt dv dm m = −v − mg dt dt The rate of lost mass is negative. The velocity is in the negative direction,so, with the two negative signs the term becomes positive. Use this: dv dm dv = dm dt dt Solve: dv dm dm m = −v − mg dm dt dt dv dm v dm =− −g dm dt m dt dv v 60g =− + dm m m0 Notice that the two negative signs cancelled out to give us a positive farright term. v 60g dv = − dm + dm m m0 Integrating, me me dm 60g dv = −v + dm m0 m m0 m0 me 60g v = −v ln + (me − m0 ) m0 m0 me me − me − mf v = −v ln + 60g me + mf me + mf me + mf mf v = v ln − 60g me me + mf Now watch this, I’m going to use my magic wand of approximation. This iswhen I say that because I know that the ratio is so big, I can ignore the empty 3
24. 24. rocket mass as compared to the fuel mass. me << mf . Let me remind you, weare looking for this ratio as well. The ratio of the fuel mass to empty rocket,mf /me . me + mf mf v = v ln − 60g me me + mf mf mf v = v ln − 60g me mf v + 60g mf = ln v me v + 60g mf exp[ ]= v me Plug in 11,200 m/s for v, 9.8 for g, and 2100 m/s for v . mf = 274 me And, by the way, if Goldstein hadn’t just converted 6800 ft/s from his secondedition to 2.1 km/s in his third edition without checking his answer, he wouldhave noticed that 2.07 km/s which is a more accurate approximation, yields aratio of 296. This is more like the number 300 he was looking for.14. Two points of mass m are joined by a rigid weightless rod of length l, thecenter of which is constrained to move on a circle of radius a. Express the kineticenergy in generalized coordinates.Answer: T1 + T2 = T Where T1 equals the kinetic energy of the center of mass, and T2 is the ki-netic energy about the center of mass. Keep these two parts seperate! Solve for T1 ﬁrst, its the easiest: 1 1 ˙ ˙ T1 = M vcm = (2m)(aψ)2 = ma2 ψ 2 2 2 2 Solve for T2 , realizing that the rigid rod is not restricted to just the X-Yplane. Don’t forget the Z-axis! 1 T2 = M v 2 = mv 2 2 Solve for v 2 about the center of mass. The angle φ will be the angle in thex-y plane, while the angle θ will be the angle from the z-axis. 4
25. 25. If θ = 90o and φ = 0o then x = l/2 so: l sin θ cos φ x= 2 If θ = 90o and φ = 90o then y = l/2 so: l y= sin θ sin φ 2 If θ = 0o , then z = l/2 so: l z= cos θ 2 Find v 2 : x2 + y 2 + z 2 = v 2 ˙ ˙ ˙ l ˙ ˙ x= ˙ (cos φ cos θθ − sin θ sin φφ) 2 1 ˙ ˙ y = (sin φ cos θθ + sin θ cos φφ) ˙ 2 l ˙ z = − sin θθ ˙ 2 Carefully square each: l2 ˙ l ˙l ˙ l 2 ˙ x2 = ˙ cos2 φ cos2 θθ2 − 2 sin θ sin φφ cos φ cos θθ + sin2 θ sin2 φφ2 4 2 2 4 l2 ˙ l ˙l ˙ l 2 ˙ y2 = ˙ sin2 φ cos2 θθ2 + 2 sin θ cos φφ sin φ cos θθ + sin2 θ cos2 φφ2 4 2 2 4 l2 ˙ sin2 θθ2 z2 = ˙ 4 Now add, striking out the middle terms: l2 ˙ ˙ ˙ ˙ ˙x2 +y 2 +z 2 =˙ ˙ ˙ [cos2 φ cos2 θθ2 +sin2 θ sin2 φφ2 +sin2 φ cos2 θθ2 +sin2 θ cos2 φφ2 +sin2 θθ2 ] 4 Pull the ﬁrst and third terms inside the brackets together, and pull thesecond and fourth terms together as well: l2 ˙ ˙ ˙ v2 = [cos2 θθ2 (cos2 φ + sin2 φ) + sin2 θφ2 (sin2 φ + cos2 φ) + sin2 θθ2 ] 4 5
26. 26. l2 ˙ ˙ ˙ v2 = (cos2 θθ2 + sin2 θθ2 + sin2 θφ2 ) 4 l2 ˙ 2 ˙ v2 =(θ + sin2 θφ2 ) 4 Now that we ﬁnally have v 2 we can plug this into T2 2 ˙ l ˙ ˙ T = T1 + T2 = ma2 ψ 2 + m (θ2 + sin2 θφ2 ) 4 It was important to emphasize that T1 is the kinetic energy of the total massaround the center of the circle while T2 is the kinetic energy of the masses aboutthe center of mass. Hope that helped.15. A point particle moves in space under the inﬂuence of a force derivable froma generalized potential of the form U (r, v) = V (r) + σ · Lwhere r is the radius vector from a ﬁxed point, L is the angular momentumabout that point, and σ is a ﬁxed vector in space. 1. Find the components of the force on the particle in both Cartesian and spherical poloar coordinates, on the basis of Lagrangian’s equations with a generalized potential 2. Show that the components in the two coordinate systems are related to each other as in the equation shown below of generalized force 3. Obtain the equations of motion in spherical polar coordinates ∂ri Qj = Fi · i ∂qjAnswer: This one is a fairly tedious problem mathematically. First lets ﬁnd thecomponents of the force in Cartesian coordinates. Convert U (r, v) into Cartesianand then plug the expression into the Lagrange-Euler equation. d ∂ ∂Qj = [V ( x2 + y 2 + z 2 )+σ ·(r ×p)]− [V ( x2 + y 2 + z 2 )+σ ·(r ×p)] dt ∂ qj ˙ ∂qj d ∂ ∂Qj = i+y j+z ˆ i+p j+p ˆ [σ·[(xˆ ˆ k)×(pxˆ y ˆ z k)]− [V ( x2 + y 2 + z 2 )+σ·(r×p)] dt ∂ vj ˙ ∂xj 6
27. 27. d ∂ ˆ ∂Qj = [σ·[(ypz −zpy )ˆ i+(zpx −xpz )ˆ j+(xpy −px y)k]− [V ( x2 + y 2 + z 2 )+σ·(r×p)] dt ∂ vj ˙ ∂xj d ∂ ∂Qj = [mσx (yvz −zvy )+mσy (zvx −xvz )+mσz (xvy −vx y)]− [V ( x2 + y 2 + z 2 )+σ·(r×p)] dt ∂ vj ˙ ∂xj Where we know that mσx (yvz − zvy ) + mσy (zvx − xvz ) + mσz (xvy − vx y) = σ · (r × p) So lets solve for just one component ﬁrst and let the other ones follow byexample: d ∂Qx = (mσy z−mσz y)− [V ( x2 + y 2 + z 2 )+mσx (yvz −zvy )+mσy (zvx −xvz )+mσz (xvy −vx y)] dt ∂x 1Qx = m(σy vz − σz vy ) − [V ( x2 + y 2 + z 2 )(x2 + y 2 + z 2 )− 2 x − mσy vz + mσz vy ] x Qx = 2m(σy vz − σz vy ) − V r If you do the same for the y and z components, they are: y Qy = 2m(σz vx − σx vz ) − V r z Qz = 2m(σx vy − σy vx ) − V r Thus the generalized force is: r F = 2m(σ × v) − V r Now its time to play with spherical coordinates. The trick to this is settingup the coordinate system so that σ is along the z axis. Thus the dot productsimpliﬁes and L is only the z-component. U = V (r) + mσ(xy − y x) ˙ ˙ With spherical coordinate deﬁnitions: x = r sin θ cos φ y = r sin θ sin φ z = r cos θ Solving for (xy − y x) ˙ ˙ 7
28. 28. ˙ ˙ x = r(− sin θ sin φφ + cos φ cos θθ) + r sin θ cos φ ˙ ˙ ˙ ˙ y = r(sin θ cos φφ + sin φ cos θθ) + r sin θ sin φ ˙ ˙ Thus xy − y x is ˙ ˙ ˙ ˙ = r sin θ cos φ[r(sin θ cos φφ + sin φ cos θθ) + r sin θ sin φ] ˙ ˙ ˙ −r sin θ sin φ[r(− sin θ sin φφ + cos φ cos θθ) + r sin θ cos φ]. ˙ ˙ Note that the r terms drop out as well as the θ terms. ˙ ˙ ˙ xy − y x = r2 sin2 θ cos2 φφ + r2 sin2 θ sin2 φφ ˙ ˙ ˙ ˙ ˙ xy − y x = r2 sin2 θφ Thus ˙ U = V (r) + mσr2 sin2 θφ Plugging this in to Lagrangian’s equations yields: For Qr : ∂U d ∂U Qr = − + ( ) ∂r dt ∂ r ˙ dV ˙ d Qr = − − 2mσr sin2 θφ + (0) dr dt dV ˙ Qr = − − 2mσr sin2 θφ dr For Qθ : ˙ Qθ = −2mσr2 sin θφ cos θ For Qφ : d Qφ = (mσr2 sin2 θ) dt ˙ Qφ = mσ(r2 2 sin θ cos θθ + sin2 θ2rr) ˙ ˙ Qφ = 2mσr2 sin θ cos θθ + 2mσrr sin2 θ ˙ For part b, we have to show the components of the two coordinate systemsare related to each other via ∂ri Qj = Fi · i ∂qj Lets take φ for an example, ∂r ∂x ∂y ∂z Qφ = F · = Fx + Fy + Fz ∂φ ∂φ ∂φ ∂φ 8
29. 29. Qφ = Qx (−r sin θ sin φ) + Qy (r sin θ cos φ) + Qz (0) x yQφ = [2m(σy vz −σz vy )−V ](−r sin θ sin φ)+[2m(σz vx −σx vz )−V ](r sin θ cos φ)+0 r r Because in both coordinate systems we will have σ pointing in only the zdirection, then the x and y σ’s disappear: x y Qφ = [2m(−σz vy ) − V ](−r sin θ sin φ) + [2m(σz vx ) − V ](r sin θ cos φ) r r Pull out the V terms, plug in x and y, see how V terms cancel Qφ = V (x sin θ sin φ − y sin θ cos φ) − 2mr sin θσ[vy sin φ + vx cos φ]Qφ = V (r sin2 θ cos φ sin φ − r sin2 θ sin φ cos φ) − 2mr sin θσ[vy sin φ + vx cos φ] Qφ = −2mr sin θσ[vy sin φ + vx cos φ] Plug in vy and vx : ˙ ˙ ˙ Qφ = −2mr sin θσ[sin φ(r sin θ cos φφ + r sin φ cos θθ + r sin θ sin φ) ˙ ˙ ˙ + cos φ(−r sin θ sin φφ + r cos φ cos θθ + r sin θ cos φ)]. ˙ ˙ Qφ = 2mσr sin θ[r sin2 φ cos θθ + r sin θ sin2 φ ˙ ˙ +r cos2 φ cos θθ + r sin θ cos2 φ]. Gather sin2 ’s and cos2 ’s: ˙ ˙ Qφ = 2mσr sin θ[r cos θθ + r sin θ] This checks with the derivation in part a for Qφ . This shows that indeedthe components in the two coordinate systems are related to each other as ∂ri Qj = Fi · i ∂qj Any of the other components could be equally compared in the same proce-dure. I chose Qφ because I felt it was easiest to write up. For part c, to obtain the equations of motion, we need to ﬁnd the general-ized kinetic energy. From this we’ll use Lagrange’s equations to solve for eachcomponent of the force. With both derivations, the components derived fromthe generalized potential, and the components derived from kinetic energy, theywill be set equal to each other. In spherical coordinates, v is: 9
30. 30. ˙ˆ ˙ˆ v = rˆ + rθθ + r sin θφφ ˙rThe kinetic energy in spherical polar coordinates is then: 1 1 ˙ ˙ T = mv 2 = m(r2 + r2 θ2 + r2 sin2 θφ2 ) ˙ 2 2For the r component: d ∂T ∂T ( )− = Qr dt ∂ r ˙ ∂r d ˙ ˙ (mr) − mrθ2 − mr sin2 θφ2 = Qr ˙ dt ˙ ˙ m¨ − mrθ2 − mr sin2 θφ2 = Qr rFrom part a, ˙ Qr = −V − 2mσr sin2 θφSet them equal: ˙ ˙ ˙ m¨ − mrθ2 − mr sin2 θφ2 = Qr = −V − 2mσr sin2 θφ r ˙ ˙ ˙ m¨ − mrθ2 − mr sin2 θφ2 + V + 2mσr sin2 θφ = 0 r ˙ ˙ ˙ m¨ − mrθ2 + mr sin2 θφ(2σ − φ) + V = 0 rFor the θ component: d ∂T ∂T ( )− = Qθ ˙ dt ∂ θ ∂θ d ˙ ˙ (mr2 θ) − mr2 sin θφ2 cos θ = Qθ dt ¨ ˙˙ ˙ mr2 θ + 2mrrθ − mr2 sin θφ2 cos θ = QθFrom part a, ˙ Qθ = −2mσr2 sin θ cos θφSet the two equal: ¨ ˙˙ ˙ ˙ mr2 θ + 2mrrθ − mr2 sin θφ2 cos θ + 2mσr2 sin θ cos θφ = 0 ¨ ˙˙ ˙ ˙ mr2 θ + 2mrrθ + mr2 sin θ cos θφ(2σ − φ) = 0 10
31. 31. For the last component, φ we have: d ∂T ∂T ( )− = Qφ ˙ dt ∂ φ ∂φ d ˙ (mr2 sin2 θφ) − 0 = Qφ dt d ˙ ˙ mr2 (sin2 θφ) + 2mrr sin2 θφ = Qφ ˙ dt ¨ ˙˙ ˙ mr2 sin2 θφ + 2mr2 sin θ cos θθφ + 2mrr sin2 θφ = Qφ ˙ From part a, ˙ Qφ = 2mσr2 sin θ cos θθ + 2mσrr sin2 θ ˙ Set the two equal: ¨ ˙˙ ˙ ˙ ˙mr2 sin2 θφ+2mr2 sin θ cos θθφ+2mrr sin2 θφ−2mσr2 sin θ cos θθ−2mσrr sin2 θ = 0 ˙ ¨ ˙ ˙ ˙ mr2 sin2 θφ + 2mr2 sin θ cos θθ(φ − σ) + 2mrr sin2 θ(φ − σ) = 0 ˙ That’s it, here are all of the equations of motion together in one place: ˙ ˙ ˙ m¨ − mrθ2 + mr sin2 θφ(2σ − φ) + V = 0 r ¨ ˙˙ ˙ ˙ mr2 θ + 2mrrθ + mr2 sin θ cos θφ(2σ − φ) = 0 ¨ ˙ ˙ ˙ mr2 sin2 θφ + 2mr2 sin θ cos θθ(φ − σ) + 2mrr sin2 θ(φ − σ) = 0 ˙16. A particle moves in a plane under the inﬂuence of a force, acting toward acenter of force, whose magnitude is 1 r2 − 2¨r ˙ r F = (1 − ) r2 c2where r is the distance of the particle to the center of force. Find the generalizedpotential that will result in such a force, and from that the Lagrangian for themotion in a plane. The expression for F represents the force between two chargesin Weber’s electrodynamics. Answer: This one takes some guess work and careful handling of signs. To get fromforce to potential we will have to take a derivative of a likely potential. Notethat if you expand the force it looks like this: 11
32. 32. 1 r2 ˙ 2¨ r F = 2 − 2 2+ 2 r c r c r We know that ∂U d ∂U − + =F ∂r dt ∂ r ˙ So lets focus on the time derivative for now. If we want a r we would have ¨to take the derivative of a r. Let pick something that looks close, say c2rr : ˙ 2 ˙ d 2r ˙ 2r˙ r˙ 2¨r 2r 2 ˙ 2¨r ( 2 ) = 2 (− 2 ) + 2 = − 2 2 + 2 dt c r c r c r c r c r Excellent! This has our third term we were looking for. Make this stay thesame when you take the partial with respect to r. ˙ ∂ r2 ˙ 2r˙ = 2 ∂ r c2 r ˙ c r 2 So we know that the potential we are guessing at, has the term cr2 r in it. Lets ˙add to it what would make the ﬁrst term of the force if you took the negativepartial with respect to r, see if it works out. That is, ∂ 1 1 − = 2 ∂r r r So 1 r2 ˙ U= + 2 r c r might work. Checking: ∂U d ∂U − + =F ∂r dt ∂ r ˙ We have ∂U 1 r2 ˙ =− 2 − 2 2 ∂r r c r and d ∂U d 2r ˙ 2r ˙ r ˙ 1 2¨r 2r2 ˙ 2¨ r = 2r = 2 (− 2 ) + ( 2 ) = − 2 2 + 2 dt ∂ r ˙ dt c c r r c c r c r thus ∂U d ∂U 1 r2 ˙ 2r 2 ˙ 2¨r − + = 2+ 2 2− 2 2+ 2 ∂r dt ∂ r ˙ r c r c r c r ∂U d ∂U 1 r2 ˙ 2¨ r − + = 2− 2 2+ 2 ∂r dt ∂ r ˙ r c r c r 12
33. 33. This is indeed the force unexpanded, 1 r2 − 2¨r ˙ r 1 r2 ˙ 2¨ r F = 2 (1 − 2 )= 2 − 2 2 + 2 r c r c r c r 2 Thus our potential, U = 1 + cr2 r works. To ﬁnd the Lagrangian use L = r ˙T − U . In a plane, with spherical coordinates, the kinetic energy is 1 ˙ T = m(r2 + r2 θ2 ) ˙ 2 Thus 1 1 r2 ˙ L= ˙ ˙ ˙ m(r2 + r2 θ2 ) − (1 + 2 ) 2 r c17. A nucleus, originally at rest, decays radioactively by emitting an electronof momentum 1.73 MeV/c, and at right angles to the direction of the electrona neutrino with momentum 1.00 MeV/c. The MeV, million electron volt, is aunit of energy used in modern physics equal to 1.60 × 10−13 J. Correspondingly,MeV/c is a unit of linear momentum equal to 5.34 × 10−22 kg·m/s. In whatdirection does the nucleus recoil? What is its momentum in MeV/c? If the massof the residual nucleus is 3.90 × 10−25 kg what is its kinetic energy, in electronvolts? Answer: If you draw a diagram you’ll see that the nucleus recoils in the oppositedirection of the vector made by the electron plus the neutrino emission. Placethe neutrino at the x-axis, the electron on the y axis and use pythagorean’stheorme to see the nucleus will recoil with a momentum of 2 Mev/c. Thenucleus goes in the opposite direction of the vector that makes an angle 1.73 θ = tan−1 = 60◦ 1from the x axis. This is 240◦ from the x-axis. To ﬁnd the kinetic energy, you can convert the momentum to kg · m/s, thenconvert the whole answer that is in joules to eV, p2 [2(5.34 × 10−22 )]2 1M eV 106 eV T = = · · = 9.13eV 2m 2 · 3.9 × 10−25 1.6 × 10−13 J 1M eV 13
34. 34. 18. A Lagrangian for a particular physical system can be written as m K L = (ax2 + 2bxy + cy 2 ) − (ax2 + 2bxy + cy 2 ). ˙ ˙˙ ˙ 2 2where a, b, and c are arbitrary constants but subject to the condition thatb2 − ac = 0. What are the equations of motion? Examine particularly the twocases a = 0 = c and b = 0, c = −a. What is the physical system describedby the above Lagrangian? Show that the usual Lagrangian for this system asdeﬁned by Eq. (1.57’): dF L (q, q, t) = L(q, q, t) + ˙ ˙ dtis related to L by a point transformation (cf. Derivation 10). What is thesigniﬁcance of the condition on the value of b2 − ac?Answer:To ﬁnd the equations of motion, use the Euler-Lagrange equations. ∂L d ∂L = ∂q dt ∂ q ˙For x ﬁrst: ∂L − = −(−Kax − Kby) = K(ax + by) ∂x ∂L = m(ax + by) ˙ ˙ ∂x ˙ d ∂L = m(a¨ + b¨) x y dt ∂ x ˙Thus −K(ax + by) = m(a¨ + b¨) x yNow for y: ∂L − = −(−Kby − Kcy) = K(bx + cy) ∂y ∂L = m(bx + cy) ˙ ˙ ∂x ˙ d ∂L = m(b¨ + c¨) x y dt ∂ x ˙Thus −K(bx + cy) = m(b¨ + c¨) x y 14
35. 35. Therefore our equations of motion are: −K(ax + by) = m(a¨ + b¨) x y −K(bx + cy) = m(b¨ + c¨) x y Examining the particular cases, we ﬁnd: If a = 0 = c then: −Kx = m¨ x − Ky = −m¨ y If b = 0, c = −a then: −Kx = m¨ x − Ky = −m¨ y The physical system is harmonic oscillation of a particle of mass m in twodimensions. If you make a substitution to go to a diﬀerent coordinate systemthis is easier to see. u = ax + by v = bx + cy Then −Ku = m¨ u −Kv = m¨ v The system can now be more easily seen as two independent but identicalsimple harmonic oscillators, after a point transformation was made. √ When the condition b2 − ac = 0 is violated, then we have b = ac, and Lsimpliﬁes to this: m √ √ K √ √ L = ( ax + cy)2 − ( ax + cy)2 ˙ ˙ 2 2 Note that this is now a one dimensional problem. So the condition keeps theLagrangian in two dimensions, or you can say that the transformation matrix a b b cis singluar because b2 − ac = 0 Note that u a b x = . v b c y So if this condition holds then we can reduce the Lagrangian by a pointtransformation.19. Obtain the Lagrange equations of motion for spherical pendulum, i.e., amass point suspended by a rigid weightless rod. 15
36. 36. Answer: The kinetic energy is found the same way as in exercise 14, and the potentialenergy is found by using the origin to be at zero potential. 1 2 ˙2 ˙ T = ml (θ + sin2 θφ2 ) 2 If θ is the angle from the positive z-axis, then at θ = 90◦ the rod is alignedalong the x-y plane, with zero potential. Because cos(90) = 0 we should expecta cos in the potential. When the rod is aligned along the z-axis, its potentialwill be its height. V = mgl cos θ If θ = 0 then V = mgl. If θ = 180 then V = −mgl. So the Lagrangian is L = T − V . 1 2 ˙2 ˙ L= ml (θ + sin2 θφ2 ) − mgl cos θ 2 To ﬁnd the Lagrangian equations, they are the equations of motion: ∂L d ∂L = ∂θ ˙ dt ∂ θ ∂L d ∂L = ∂φ ˙ dt ∂ φ Solving these yields: ∂L ˙ = ml2 sin θφ2 cos θ + mgl sin θ ∂θ d ∂L ¨ = ml2 θ dt ∂ θ˙ Thus ˙ ¨ ml2 sin θφ2 cos θ + mgl sin θ − ml2 θ = 0 and ∂L =0 ∂φ d ∂L d ˙ ¨ ˙ = (ml2 sin2 θφ) = ml2 sin2 θφ + 2φml2 sin θ cos θ ˙ dt ∂ φ dt Thus ¨ ˙ ml2 sin2 θφ + 2φml2 sin θ cos θ = 0 16
37. 37. Therefore the equations of motion are: ˙ ¨ ml2 sin θφ2 cos θ + mgl sin θ − ml2 θ = 0 ¨ ˙ ml2 sin2 θφ + 2φml2 sin θ cos θ = 020. A particle of mass m moves in one dimension such that it has the Lagrangian m2 x4 ˙ L= + mx2 V (x) − V2 (x) ˙ 12where V is some diﬀerentiable function of x. Find the equation of motion forx(t) and describe the physical nature of the system on the basis of this system.Answer: I believe there are two errors in the 3rd edition version of this question.Namely, there should be a negative sign infront of mx2 V (x) and the V2 (x) ˙should be a V 2 (x). Assuming these are all the errors, the solution to thisproblem goes like this: m2 x4 ˙ L= − mx2 V (x) − V 2 (x) ˙ 12 Find the equations of motion from Euler-Lagrange formulation. ∂L = −mx2 V (x) − 2V (x)V (x) ˙ ∂x ∂L m2 x3 ˙ = + 2mxV (x) ˙ ∂x˙ 3 d ∂L = m2 x2 x + 2mV (x)¨ ˙ ¨ x dt ∂ x˙ Thus mx2 V + 2V V + m2 x2 x + 2mV x = 0 ˙ ˙ ¨ ¨is our equation of motion. But we want to interpret it. So lets make it looklike it has useful terms in it, like kinetic energy and force. This can be done by 1dividing by 2 and seperating out 2 mv 2 and ma’s. mx2 ˙ mx2 ˙ V +VV + m¨ + m¨V = 0 x x 2 2 Pull V terms together and m¨ terms together: x mx2 ˙ mx2 ˙ ( + V )V + m¨( x +V)=0 2 2 Therefore: mx2 ˙ ( + V )(m¨ + V ) = 0 x 2 17
38. 38. mx2 ˙ Now this looks like E · E = 0 because E = 2 + V (x). That would mean d 2 E = 2EE = 0 dt Which allows us to see that E 2 is a constant. If you look at t = 0 and thestarting energy of the particle, then you will notice that if E = 0 at t = 0 thenE = 0 for all other times. If E = 0 at t = 0 then E = 0 all other times whilem¨ + V = 0. x21. Two mass points of mass m1 and m2 are connected by a string passingthrough a hole in a smooth table so that m1 rests on the table surface andm2 hangs suspended. Assuming m2 moves only in a vertical line, what are thegeneralized coordinates for the system? Write the Lagrange equations for thesystem and, if possible, discuss the physical signiﬁcance any of them might have.Reduce the problem to a single second-order diﬀerential equation and obtain aﬁrst integral of the equation. What is its physical signiﬁcance? (Consider themotion only until m1 reaches the hole.)Answer: The generalized coordinates for the system are θ, the angle m1 moves roundon the table, and r the length of the string from the hole to m1 . The wholemotion of the system can be described by just these coordinates. To write theLagrangian, we will want the kinetic and potential energies. 1 1 ˙ T = m2 r2 + m1 (r2 + r2 θ2 ) ˙ ˙ 2 2 V = −m2 g(l − r) The kinetic energy is just the addition of both masses, while V is obtainedso that V = −mgl when r = 0 and so that V = 0 when r = l. 1 1 ˙ L=T −V = (m2 + m1 )r2 + m1 r2 θ2 + m2 g(l − r) ˙ 2 2 To ﬁnd the Lagrangian equations or equations of motion, solve for eachcomponent: ∂L =0 ∂θ ∂L ˙ = m1 r2 θ ˙ ∂θ d ∂L d ˙ = (m1 r2 θ) = 0 ˙ dt ∂ θ dt d ∂L ¨ ˙˙ = m1 r2 θ + 2m1 rrθ = 0 ˙ dt ∂ θ 18
39. 39. Thus d ˙ ¨ ˙˙ (m1 r2 θ) = m1 r(rθ + 2θr) = 0 dt and ∂L ˙ = −m2 g + m1 rθ2 ∂r ∂L = (m2 + m1 )r ˙ ∂r ˙ d ∂L = (m2 + m1 )¨r dt ∂ r ˙ Thus ˙ m2 g − m1 rθ2 + (m2 + m1 )¨ = 0 r Therefore our equations of motion are: d ˙ ¨ ˙˙ (m1 r2 θ) = m1 r(rθ + 2θr) = 0 dt ˙ m2 g − m1 rθ2 + (m2 + m1 )¨ = 0 r ˙ See that m1 r2 θ is constant. It is angular momentum. Now the Lagrangian ˙can be put in terms of angular momentum. We have θ = l/m1 r2 . 1 l2 L= (m1 + m2 )r2 + ˙ − m2 gr 2 2m1 r2 The equation of motion ˙ m2 g − m1 rθ2 + (m2 + m1 )¨ = 0 r Becomes l2 (m1 + m2 )¨ − r + m2 g = 0 m1 r3 The problem has been reduced to a single second-order diﬀerential equation.The next step is a nice one to notice. If you take the derivative of our newLagrangian you get our single second-order diﬀerential equation of motion. d 1 l2 l2 ( (m1 + m2 )r2 + ˙ − m2 gr) = (m1 + m2 )r¨ − ˙r r − m2 g r = 0 ˙ ˙ dt 2 2m1 r2 m1 r 3 l2 (m1 + m2 )¨ − r − m2 g = 0 m1 r3 Thus the ﬁrst integral of the equation is exactly the Lagrangian. As far as in-terpreting this, I will venture to say the the Lagrangian is constant, the system isclosed, the energy is conversed, the linear and angular momentum are conserved. 19