Assignment 1. g i (x 1,..., x n ) dx i = 0. i=1

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1 Assignment 1 Golstein 1.4 The equations of motion for the rolling isk are special cases of general linear ifferential equations of constraint of the form g i (x 1,..., x n x i = 0. i=1 A constraint conition of this type is holonomic only if an integrating function, f(x 1,..., x 2 can be foun that turns it into an exact ifferential. Clearly the function must be such that (fg i x j = (fg j x i for all i j. Show that no such integrating factor can be foun for either of Eqs. (1.39. The ifferential constraint above can be written out as 0 = g x x + g y y + This woul be integrable if there were some function, f(x, y,, which, which multiplying this turne it into a perfect ifferential, i.e 0 = F = F F x + x y y + = f g x x + f g y y + In such a case, we woul actually have a holonomic constraint, F = constant, i.e. an algebraic relation between the coorinates. To be able to have an integrating factor, f, it must be the case that the mixe partial erivatives must be equal, i.e (fg x = (fg y y x an so on for all the mixe partial erivatives. In Golstein s notation, this is (fg i x j = (fg j x i for i j. Applying this to 0 = x a sin θ φ we note g x = 1, g y = 0, g θ = 0 an g φ = a sin θ. We have a bunch of conitions: (fg x y (fg x θ (fg x φ (fg y θ (fg y φ (fg θ φ = (fg y x = (fg θ x = (fg φ x = (fg θ y = (fg φ y = (fg φ θ 1

2 These reuce to Note that the final equation can be written y = 0 θ = 0 = a sin θ φ x (f sin θ = 0 θ f cos θ + sin θ θ = 0 Thus, unless θ = π/2, the only solution to the above equations is f = 0 an no integrating factor exists. For the case that θ = π/2, the isk is rolling parallel to the x-axis an x = aφ which can be integrate to x = a (φ φ 0 where φ 0 is a constant. Applying this to 0 = y + a cos θ φ we get g x = 0, g y = 1, g θ = 0, an g φ = a cos θ. This time the conitions become x = 0 θ = 0 = a cos θ φ y sin θ f + cos θ θ = 0 Again, unless θ = 0, the only solution to these equations is f = 0. So no integrating factor exists in general to rener this a holonomic constraint. 2

3 Golstein 1.5 Two wheels of raius a are mounte on the ens of a common axle of length b such that the wheels rotate inepenently. The whole combination rolls without slipping on a plane. Show that there are two nonholonomic equations of constraint, cos θ x + sin θ y = 0 sin θ x cos θ y = 1 2 a ( φ + φ (where θ, φ, an φ have meanings similar to those in the problem of a single vertical isk, an (x, y are the coorinates of a point on the axle miway between the two wheels an one holonomic equation of constraint, where C is a constant. θ = C a b ( φ φ, As in the previous problem with one wheel, each wheel here of the pair satisfies the same constraints as a single rolling isk. We have two wheels so we have two sets of equations in obvious notation: x 1 a sin θ φ 1 = 0 y 1 + a cos θ φ 1 = 0 x 2 a sin θ φ 2 = 0 y 2 + a cos θ φ 2 = 0 The center of the axle can be consiere the center of mass so that so that we can reuce our four equations to two: x = 1 ( x1 + x 2 2 y = 1 ( y1 + y 2 2 x a 2 sin θ ( φ 1 + φ 2 = 0 y + a 2 cos θ ( φ 1 + φ 2 = 0 Through two combinations involving sin θ an cos θ, we can write these as cos θ x + sin θ y = 0 sin θ x cos θ y = a ( φ1 + φ 2 2 It may look like we are one, but what we have constraine is that the wheels roll in unison. But our approach so far, leaves open the possibility that they coul, in fact, change their istance between them. To fix the constraint that the istance between the wheels remains at a constant value, b, such that x 2 x 1 = b cos θ an y 2 y 1 = b sin θ, we impose ẋ 2 ẋ 1 = b θ sin θ Using the relations between the x s an the φ s, we can write ẋ 2 ẋ 1 = a sin θ ( φ2 φ 1 Equating these, ropping the sin θ, an integrating we have θ = C a b 3 ( φ2 φ 1

4 Golstein 1.8 If L is a Lagrangian for a system of n egrees of freeom satisfying Lagrange s equations, show by irect substitution that L = L + F (q 1,..., q n, t t also satisfies Lagrange s equations where F is any arbitrary, but ifferentiable, function of its arguments. We know L satisfies Lagrange s equations an we nee only show the same for L. To that en, we note that we can write F t = F n t + F q i so that we can also say i=1 ( F = ( F + q j t t q j = ( F t q j Further, becuase F is inepenent of the q i, we can conclue F q i = F i=1 Constructing Lagrange s equations in terms of L, we get ( F q j q i ( L L = ( L L + [ ( F ] ( F t q i t q i t q i t t The first two terms on the right vanish because L satisfies Lagrange s equations an the thir an fourth terms can be written ( L L = [ F ] ( F t q i t t which, of course, vanishes. Thus L as a shift of L by a total erivative also satisfies Lagrange s equations. 4

5 Golstein 1.10 Let q 1,..., q n be a set of inepenent generalize coorinates for a system of n egrees of freeom, with a Lagrangian L(q, q, t. Suppose we transform to another set of inepenent coorinates s 1,...s n by means of transformation equations q i = q i (s 1,..., s n, t, i = 1,..., n. (Such a transformation is calle a point transformation. Show that if the Lagrangian function is expresse as a function of s j, ṡ j, an t through the equations of tranformation, then L satisfies Lagrange s equations with respect to the s coorinates: ( L L = 0. t ṡ j s j In other wors, the form of Lagrange s equations is invariant uner a point transformation. Again, we have to be careful when we talk about what various quantities epen on. In particular we are assuming that there are n coorinates, q i, that epen on new coorinates s i an t. We also assume that the inverse relation exists, namely s i = s i (q j, t Note, importantly, that a point transformation oes not epen on the erivatives of the new coorinates, namely ṡ i. However, we o assume that the Lagrangian can epen on the erivatives of the new coorinates. Explicitly, we can write L = L ( q i (s j, t, q i (s j, ṡ j, t, t If we now write the Lagrange equations, we have 0 = L t = j=1 ( L q i L s j s j + L ṡ j ṡ j t [ L s j + L ṡ ] j s j q i ṡ j q i where we have use the chain rule twice because the erivatives of L epen on both s j an ṡ j. From a couple of erivations ientical to those in problem 1.8, we can establish that an ṡ j q i = s j ( sj = ṡ j t Using these, we can simplify our Lagrange equations as 0 = j=1 L s j s j + L ṡ j ( sj [ L s j + L s ] j t t s j q i ṡ j s i Note that the first term in square brackets vanishes becuase s j oes not epen on the velocities so that s j q i = 0 Expaning the final term in a prouct rule an rearranging a bit, we have 0 = j=1 L s j ( L sj + L s j t ṡ j ṡ j 5 ( sj L t ṡ j ( sj t

6 The final two terms cancel an we have 0 = { L ( L } sj s j t ṡ j j=1 Provie the new coorinates are inepenent, their erivatives with respect to the ol coorinates will be invertible an the only way this can vanish is if the quantity in curly braces vanishes for each j. This of course is just Lagrange s equations in the new coorinates. Thus Lagrange s equations are invariant in form uner a point transformation. Golstein 1.12 The escape velocity of a particle on Earth is the minimum velocity require at Earth s surface in orer that the particle can escape from Earth s gravitational fiel. Neglecting the resistance of the atmosphere, the system is conservative. From the conservation theorem for potential plus kinetic energy show that the escape velocity for Earth, ignoring the presence of the Moon, is 11.2 km/s. This is a straigtforwar recall of a problem from introuctory physics. For a particle of mass m, conservation is just E = V + T = 1 2 mv2 GM em r where everything has the usual meaning, i.e. E is the constant total energy, G is Newton s constant, M e is the mass, r is the raial istance of the particle from the center of the earth, an v is the velocity of the particle. The notion of the escape velocity, v esc, or a minimum velocity suggests that this velocity is exactly that which woul allow the particle to have no kinetic energy an infinite istance from the source. In this case, the value of E woul be zero an we have Solving this for v esc an putting in numbers, we have 0 = 1 2 mv2 esc GM em R e v esc = 2GMe R e 11.2 km/s 6

7 Golstein 1.13 Rockets are propelle by the momentum reaction of the exhaust gases expelle from the tail. Since these gases arise from the reaction of the fuels carrie in the rocket, the mass of the rocket is not constant, but ecreases as the fuel is expene. Show that the equation of motion for a rocket projecte vertically upwar in a unifrorm gravitational fiel, neglecting atmosperic friction, is m v t = v m t mg, where m is the mass of the rocket an v is the velocity of the escaping gases relative 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 initially from rest, with v equal to 2.1 km/s an a mass loss per secon eequal to 1/60th of the initial mass, that in orer to reach the escape velocity the ratio of the weight of the fuel to the weight of the empty rocket must be almost 300! Consier the rocket (an its unburne fuel at some time t moving with respect to the earth with a vertical velocity of v an momentum mv. At a later time t + t there are two parts to the system, namely the rocket (an some still unburne fuel of mass m m moving with v + v an the expelle fuel of mass m which moves with velocity v f with respect to the earth. The total momentum of the system at the later time is p + p = (v + v(m m + v f m = p + mv + (v f vm + where we ignore the higher orer terms. Thus we have mg = p t = m v m + v t t where v = v f v is the spee of the expelle fuel with respect to the rocket. If we assume v is a constant, we can integrate this equation in t to get v(t = gt v ln m(t + C where C = v 0 + v ln m 0 where v 0 an m 0 are the initial spee an mass of the rocket, respectively. However, we want the velocity as a function of m. We can o this by writing v m = ġ m v m assuming v an ṁ are constants an integrating to get v(m = v(m 0 + ġ m ( m0 m v ln( m m 0 Assuming the initial spee is zero, we can write this in terms of the values given, namely v = 2.1 km/s, ṁ = m 0 /60s an the escape velocity: = ( 1 x 2100 ln(x where x = m/m 0. Using Mathematica or Maple to solve this equation, we fin x Thus the initial mass of the rocket on the launch pa must be about 275 times the mass of the rocket (presumably empty at escape velocity. 7

8 Golstein 1.20 A particle of mass m moves in one imension such that it has the Lagrangian L = 1 12 m2 ẋ 4 + m ẋ 2 V (x V 2 (x where V is some ifferentiable function of x. Fin the equation of motion for x(t an escribe the physical nature of the system on the basis of the equation. The Lagrange equations become 0 = L x ( L t ẋ = m ẋ 2 V (x 2V (xv (x t( 1 3 m2 ẋ 3 + 2mẋ V (x 1 1 = 2V (x( 2 mẋ2 V (x 2m( 2 m ẋ2 ẍ + ẍ V (x + ẋ 2 V (x 1 ( = 2( 2 mẋ2 + V (x mẍ + V (x The right han sie must vanish. We thus have two possibilities. Interpreting x as the generalize one imensional coorinate an V (x as a one imensional potential, either the sum of the kinetic an potential energies must vanish or the corresponing force is conservative an a particle of mass m moves along the x irection uner the influence of the conservative force given by V (x. 8

9 Golstein 1.21 Two mass points of mass m 1 an m 2 are connecte by a string passing through a hole in a smooth table so that m 1 rests on the table suface an m 2 hangs suspene. Assuming m 2 moves only in a vertical line, what are the generalize coorinates for the system? Write the Lagrange equations for the system, an if possible, iscuss the physical significance any of them might have. Reuce the problem to a single secon-orer ifferential equation an obtain a first integral of the equation. What is its physical significance? (Consier the motion only until m 1 reaches the hole. The system woul have three egrees of freeom if the problem were that m 1 moves freely on the table an m 2 only moves up an own. However, we will assume that the string between the two mass remains taut an that the raial position of m 1 is etermine by the vertical height of m 2 (an vice versa. Thus there are two egrees of freeom. Generalize coorinates can be thought of as the two coorinates in the plane for m 1, r an θ. The vertical istance from the hole for m 2 is given by x 0 r where x 0 is the length of the string. If we set the zero of potential to be when m 1 is on the hole, we have the Lagrangian L = 1 2 m 1 [ ] ṙ 2 + r 2 θ m 2 (ẋ0 ṙ 2 m2 g [ x 0 (x 0 r ] The equations of motion from this Lagrangian become 0 = ( m 1 r θ 2 m 2 g (m 1 + m 2 r ( 0 = 0 m 1 r 2 θ t The secon equation can be immeiately integrate an yiels conservation of angular momentum: l = m 1 r 2 θ Having integrate one equation, we can use this result to eliminate θ in the first equation. We have, (m 1 + m 2 r = l2 m 1 r 3 m 2g There is a further integral of the motion that we can obtain by multiplying this equation by ṙ an integrating in time: (m 1 + m 2 ṙ r = l2 ṙ m 1 r 3 m 2gṙ 1 2 (m 1 + m 2 ṙ 2 = l2 1 2m 1 r 2 m 2gr + C 0 On rearranging this a bit, one can see that this is a statement of conservation of energy. 9

10 Golstein 1.22 Obtain the Lagrangian an equations of motion for the ouble penulum illustrate in Fig. 1.4, where the lengths of the penula are l 1 an l 2 with corresponing masses m 1 an m 2. We can work with two pairs of coorinates for the two masses, (x 1, y 1 an (x 2, y 2. They are not inepenent, of course. We can write these Cartesian coorinates (with an assume origin at the top pivot point in terms of the (generalize angular coorinates given in the figure: The Lagrangian is given by x 1 = l 1 sin θ 1 y 1 = l 1 cos θ 1 x 2 = x 1 l 2 sin θ 2 y 2 = y 1 l 2 cos θ 2 L = 1 2 m 1(ẋ2 1 + ẏ m (ẋ ẏ2 2 m1 gy 1 m 2 gy 2 = 1 2 (m 1 + m 2 l 2 1 θ m 2 l 2 2 θ 2 2 m 2 l 1 l 2 θ1 θ2 cos(θ 1 + θ 2 + (m 1 + m 2 gl 1 cos θ 1 + m 2 gl 2 cos θ 2 From this Lagrangian, we get the equations of motion (m 1 + m 2 l 2 1 θ 1 m 2 l 1 l 2 θ2 cos(θ 1 + θ 2 + m 2 l 1 l 2 θ2 2 sin(θ 1 + θ 2 + (m 1 + m 2 gl 1 sin θ 1 = 0 m 2 l 2 2 θ 2 m 2 l 1 l 2 θ1 cos(θ 1 + θ 2 + m 2 l 1 l 2 θ2 1 sin(θ 1 + θ 2 + m 2 gl 2 sin θ 2 = 0 Note that some constants in both can be factore out. Golstein 1.23 Obtain the equation of motion for a particle falling vertically uner the influence of gravity when frictional forces obtainable from a issipation function kv 2 /2 are present. Integrate the equation to obtain the velocity as a function of time an show that the maximum possible velocity for a fall from rest is v = mg/k. The Lagrangian is Lagrange s equation is As an equation for v = ż, this is L = T V = 1 2 mż2 mgz 0 = L t ż L z + F ż m v + kv + mg = 0 = m z + mg + kż This can be solve with an integrating factor, e kt/m, so that ( me kt/m v = mg e kt/m t Integrating an rearranging, we get v(t = e kt/m[ mg ] k ekt/m + C The integration constant can be foun from the initial conition v(0 = v 0, so the solution becomes v(t = v 0 e kt/m + mg [ ] e kt/m 1 k The maximum possible velocity is obtaine as t. This leas to v mg/k. The minus sign enotes only that we are falling in the negative z irection. 10