3. Übungsblatt zur Theoretischen Physik I im SS16: Mechanik & Spezielle Relativitätstheorie. Lagrange-Formalismus I

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1 3. Übungsblatt ur Theoretischen Physik I im SS16: Mechanik & Speielle Relativitätstheorie Lagrange-Formalismus I Aufgabe 7 Atwoodsche Fallmaschine Betrachten Sie das System aus wei Punktmassen und m 2 im homogenen Schwerefeld der Erde. Die beiden Massen seien über ein masseloses Seil der Länge L, welches über eine masselose Rolle mit Radius R geführt wird, miteinander verbunden. Reibungseffekte seien u vernachlässigen. R y m 2 x a Wie in der Vorlesung werde ein System mit N-Punktmassen, deren Ortsvektoren durch r 1,... r 2 gegeben sind durch die 3N-Koordinaten x n, n = 1,..., 3N der N Punktmassen beschrieben. Es gebe R holonome Zwangsbedingungen g α x 1,...x 3N, t, α = 1,..., R. Es sei nun mindestens eine der R-Zwangsbedingungen, hier o.b.d.a. als g 1 beeichnet, von der Form g 1 x 1,.., x 3N, t = x n a, wobei a eine Konstante sei. Zeigen Sie gan allgemein, dass bei der Beschränkung auf eine Ebene die Bewegung des Systems senkrecht u dieser Ebene von vornherein außer Acht gelassen werden kann. b Bestimmen Sie die Anahl der Freiheitsgrade für wei freie Punktmassen. Bestimmen Sie die Anahl der Zwangsbedingungen. Wie lauten die Zwangsbedingungen? Durch wie viele unabhängige Freiheitsgrade ist die Bewegung des Systems bestimmt? c Wie lauten die Lagrange-Gleichungen erster Art? Nuten Sie das Ergebnis aus Aufgabenteil a. um die Dynamik des Systems u beschreiben. Wie lautet der u bestimmende Lagrange-Parameter λ, ausgedrückt durch die das System charakterisierenden Konstanten? d Lösen sie die Bewegungsgleichungen durch Integration. Gilt Energieerhaltung? Begründen Sie Ihre Antwort. e Bestimmen Sie die Zwangskraft. Vergleichen Sie die Bewegung der Lösung x 1 t mit der des freien Falls. Was passiert für die Grenfälle = m 2 und m 2? Was passiert für = m 2 + δm mit δm 1? 1

2 Proposal for solution a We write for the constraint g 1 x 1,...x 3N, t = x n a = 0. 1 and numerate it, without loss of generality, as the first of the R constraints. This constraint restricts the motion of the systems orthogonal to the plane defined by x n = a. For the equations of motion of the n-th component, we then have the Lagrange equation of the first kind m n ẍ n = F n + λ 1 g 1 + R α=2 λ α g α. 2 As usual, the forces F n, n = 1,..., 3N are assumed to be known. The simple form of the constraint g 1 results in g1 = 1. We therefore find for the equations of motion of the nth component m n ẍ n = F n + λ 1 + R α=2 λ α g α. 3 Differentiating g 1 twice w.r.t the time t, we obtain the trivial result ẍ n = 0. Inserting the equations of motion and solving for λ 1 leads to λ 1 = F n Inserting this back in the equations of motion results in the trivial equation This equation has the solution linear in time R α=2 λ α g α. 4 m n ẍ n = 0. 5 x n = A + B t. 6 The constants A and B have to be chosen such that they are compatible with the constraints for arbitrary times t. The compatibility with the constraint g 1 therefore enforces the choices Thus, the solution for all times t is simply given by the constant A = a and B = 0. 7 x n t = a. 8 Since λ 1 does not appear in the equations of motion for the other components x m, m n as x m g 1 = x m x n a = 0, we can shorten the general procedure for finding the solution by simply omitting the coordinate x n from the very beginning, i.e. for constraints of the form x n = const., it is useful and possible to simply consider the 3 N 1 equations of motions for the remaining components x m, m n and the remaining R 1 constraint equations. The same is of course true if one has several constraints of the form x n = const. b In general, there are 6 coordinates x 1, y 1, 1, x 2, y 2, 2 for the system of two free point masses. In the problem, there are in total 5 constraints: 2 constraint equations for each point mass which constraint the motion to the one-dimensional x-axes and 1 constraint that connects and m 2 via the rope of constant length. Thus the number of independent degrees of freedom is 6 5 = 1. The length of the rope is L = l + π R, where l is the sum of the displacements x 1 and x 2, i.e. the relevant constraint is g 1 x 1, x 2 = x 1 + x 2 l =

3 The other constraints are given by g 2 = y 1 + R = 0, g 3 = y 2 R = 0, g 4 = 1 = 0, g 5 = 2 = The constraints 10 are all of the form discussed in part a. of the problem and constrain the motion to the x-direction. They can be omitted from the very beginning by focusing only on the relevant x 1, x 2 motion. c The Lagrange equations of the first kind for the x 1 and x 2 components are given by ẍ 1 = g + λ, m 2 ẍ 2 = m 2 g + λ. 11 Here we made as usual the ansat Z = λ g for the constraint forces Z and used g1 x 1 = g1 x 2 = 1. The Lagrange multiplier λ is of course the same in both equations of motion as it was derived from one and the same constraint g 1. We have however two constraint forces, pointing in the same direction but each acting on the corresponding point mass only. Differentiating g 1 twice w.r.t to the time t, we obtain Inserting the equations of motion for x 1 and x 2, we can solve for λ ẍ 1 + ẍ 2 = λ = 2 g m 2 + m d Inserting the solution for λ in the equations of motion we find by trivial integration The constants c 1, c 2, are to be determined by the initial data. x 1 t = m 2 + m 2 g t c 1 t + c Energy is conserved, because 1. Gravity is the only fundamental force in the system and the gravitational force is conservative and 2. Because the constraints do not depend on time holonomic, scleronomic. e The constraint force Z is the same on both point masses Z 1 =λ 1 g 1 x 1, x 2 = λ e x, 15 Z 2 =λ 2 g 1 x 1, x 2 = λ e x. 16 The solution represents a slowed down free fall. In case the masses are equal = m 2, the system is in equilibrium and the constraint forces Z 1 + Z 2 = 2 m g just compensate the gravitational force. 2 In the limit m 2, we have m1 m2 +m 2 = 1 2 m2 m + O 2 1 which recovers again the free fall for. The point of Atwood s construction is that the heavy mass, that determines the weight and thereby determines the accelerating force, is given by the difference m 2 while the inertial mass that has to be accelerated is given by the total mass, i.e. the sum of the two masses + m 2. As a result the acceleration is given by the ratio m1 m2 +m 2 g and is only a fraction of the gravitational acceleration of the earth g. Thus arranging for m := m 2 by fine tuning the difference between and m 2 to a very small amount m 2 =: δm with δm/m 1, one can slow down the free fall to very small accelerations a δm 2 m g g. 3

4 Aufgabe 8 Pleuelstange An einer drehbaren masselosen Scheibe mit Radius r ist eine masselose Pleuelstange mit konstanter Länge l durch ein Gelenk befestigt. Am Gelenk sei eine Masse befestigt. Am anderen Ende der Pleuelstange sei die Masse m 2 befestigt die auf einer Führungschiene, welche mit der x-achse des Koordinatensystems usammenfällt, reibungsfrei gleite. Die Scheibe sei mit ihrem Mittelpunkt im Ursprung des Koordinatensystems befestigt und ihre Rotationsachse falle mit der y-achse des Koordinatensystems usammen. Es wirke keine äussere Kraft außer der Gravitationskraft. m 2 l r x a Wieviele unabhängige Freiheitsgrade hat das System? Wie lauten die Zwangsbedingungen? b Welche Grösse bietet sich als generalisierte Koordinate an? Stellen Sie mithilfe dieser generalisierten Koordinate die Lagrangefunktion für das System auf. Proposal for solution a The six degrees of freedom x 1, y 1, 1 and x 2, y 2, 2 of the two point masses and m 2 are constraint by five constraint equations. The first three are trivial and constrain the motion of the point particle 1 to the x-plane and particle 2 to the x-axis respectively. g 1 y 1 = y 1 = 0, g 2 y 2 = y 2 = 0, g 3 2 = 2 = From the first problem, we know that we can simply omit these degrees of freedom from the very beginning and focus only on the remaining three x 1, 1 and 2. The remaining two constraints are easily found by trigonometric considerations. The motion of the point mass is constraint to lie on a circle with radius r in the x-plane g 4 x 1, 1 = x r 2 = Finally, using the trigonometric relations we can express the x-coordinate of the second point mass m 2 in terms of x 1, 1 and l This can be written in constraint form as l 2 = x 2 x g 5 x 1, 1, x 2 = x 2 x 1 2 l 2 = Counting independent degrees of freedom, we started with six degrees of freedom describing the free motion and five constraint equations relating the six coordinates. Thus, we are left with one independent degree of freedom. It is clear from the figure that the system is fully characterised by the angle ϕ that determines the amount of rotation of the wheel. Thus, the natural choice for the generalied coordinate is q = ϕ. 4

5 b First we express the Cartesian coordinates x 1, 1 in terms of the generalied coordinate ϕ by choosing polar coordinates with a fixed radius r. x 1 = r cos ϕ, 1 = r sin ϕ, r = const. 21 The coordinate x 2 expressed in terms of the generalied coordinate then becomes x 2 = x 1 l 2 21 = r cos ϕ l 2 r 2 sin 2 ϕ. 22 In order to calculate the kinetic energy, we calculate the time derivative of x 1, 1 and x 2, ẋ 1 = ϕ r sin ϕ, ż 1 = ϕ r cos ϕ, ẋ 2 = ϕ r sin ϕ + ϕ r2 sin ϕ cos ϕ l2 r 2 sin 2 ϕ. 23 The kinetic energy is then given by T = 2 The potential energy is ẋ2 1 + ż m 2 2 ẋ2 2 = 2 r2 ϕ 2 + m 2 2 ϕ2 r 2 sin 2 ϕ Finally, the Lagrange function is then found to be L = T V = 2 r2 ϕ 2 + m 2 2 ϕ2 r 2 sin 2 ϕ 1 r cos ϕ l2 r 2 sin 2 ϕ V = g 1 = g r sin ϕ r cos ϕ l2 r 2 sin 2 ϕ 2 g r sin ϕ. 26 Aufgabe 9 Zwangskraft auf Skifahrer Ein Skifahrer mit Masse m gleite im homogenen Schwerefeld der Erde reibungsfrei eine Piste hinab, deren Form durch einen Viertelkreis mit konstantem Radius R in der x -Ebene beschrieben sei siehe Abbildung. Er starte am höchsten Punkt in Ruhe und gleite dann reibungsfrei herab, bis u dem Punkt an dem er schließlich abhebt. ϑ R x a Wie lautet die Zwangsbedingung hr, ϑ = 0 bis um Abhebepunkt in Polarkoordinaten? 5

6 b Drücken Sie die Lagrange-Gleichungen 1. Art m r = F g + Z 27 in Polarkoordinaten aus. Hier beeichnet F g die Gravitationskraft und Z die Zwangskraft. Hinweis: Die Einheitsvektoren e r und e θ sowie der Gradient in Polarkoordinaten sind gegeben durch sin ϑ cos ϑ e r =, e cos ϑ ϑ =, = e sin ϑ r r + e 1 ϑ. r ϑ Indem Sie 27, ausgedrückt in Polarkoordinaten, jeweils mit den orthonormierten Einheitsvektoren e r bw. e ϑ multipliieren, erhalten Sie die r bw. ϑ Komponenten der Lagrange-Gleichung 1. Art. c Bestimmen Sie die auf den Skifahrer wirkende Zwangskraft Z. Hinweis: Nuten Sie die Energieerhaltung um ϑ 2 u eliminieren. d Bei welchem Winkel ϑ = ϑ j hebt der Skifahrer ab? Proposal for solution a The motion takes place in the x-plane. In terms of polar coordinates the constraint equation in polar coordinates is obviously of the form b We can write the position r in polar coordinates x r = = The first and second derivatives are then given by + r ϑ and The gravitational force reads F g = V = r = ṙ sin ϑ cos ϑ x =r sin ϑ, 28 =r cos ϑ, 29 hr, ϑ = hr = r R = r sin ϑ r cos ϑ cos ϑ sin ϑ r = r e r + ṙ e r + ṙ ϑ e ϑ + r ϑ e ϑ + r ϑ e ϑ. 31 = ṙ e r + r ϑ e ϑ, 32 = r e r + ṙ ϑ e ϑ + ṙ ϑ e ϑ + r ϑ e ϑ r ϑ 2 e r = r r ϑ 2 e r + r ϑ + 2 ṙ ϑ e ϑ. 33 e x x + e m = e r r + e 1 ϑ r m g r cosϑ ϑ = m g cosϑ e r + m g sinϑ e ϑ. 34 6

7 With the ansat for the constraint force Z = λ h, we have Z = λ hr = λ e r r + e 1 ϑ hr = λ e r. 35 r ϑ The Lagrange equation of the first kind in vectorial notation is given by m r = F g + Z. 36 Inserting the corresponding expressions, we can collect terms with e r and e ϑ and write the Lagrange equations of the first kind for r and ϑ componentwise projecting with e r and e ϑ m r =m r ϑ 2 g cos ϑ + λ, 37 m r ϑ = m 2 ṙ ϑ g sin ϑ. 38 c Differentiating the constraint equation twice leads to ḣ = ṙ = 0, ḧ = r = Inserting this into the equation of motion for r and solving for λ yields λ = m g cos ϑ m r ϑ 2 = m g cos ϑ R ϑ Here, we have used the constraint equation in the last equality in order to express r by R. In order to eliminate ϑ 2 from the expression for λ, we use conservation of energy. On the one hand, in general, the energy is given by E = T + V = 1 2 m ṙ 2 + r 2 ϑ2 + m g r cos ϑ = 1 2 m R2 ϑ2 + m g R cos ϑ. 41 Here, we have again used the constraint equation in the form h = 0 and ḣ = 0 in order to express r by R and in order to eliminate the ṙ 2 term. On the other hand, during the start, the skier is at rest which corresponds to ϑ 0 = 0 and ϑ 0 = 0 in polar coordinates. Thus the energy at the starting point is given by Conservation of energy now implies E E 0 = 0 Solving this equation for ϑ 2 yields or E 0 = R m g m R2 ϑ2 + m g R cos ϑ 1 = ϑ 2 = 2 g R cos ϑ Of course, the same equation could have been obtained by first using de/dt = 0 and afterwards using the equations of motion to eliminate ϑ as well as setting r = R, ṙ = r = 0. The result 44 can be inserted in the equation for λ The constraint force Z can then finally be calculated as λ = m g 3 cos ϑ Z = λ h = λ e r = m g 3 cos ϑ 2 e r. 46 d During the start of the ride ϑ 0 we have Z m g e r, i.e. initially the constraint force has to compensate the total weight of the ski-driver. With growing ϑ, the weight component on the surface of the ski-piste reduces and so that the centrifugal force reduces the tracking force. When the ski drives reaches the angle ϑ j, where the constraint force vanishes completely, the skier takes off. Z = 0 cos ϑ j =