Lösungsvorschlag 3

3. Übungsblatt zur Theoretischen Physik I im SS16:
Mechanik & Spezielle Relativitätstheorie
Lagrange-Formalismus I
Aufgabe 7
Atwoodsche Fallmaschine
Betrachten Sie das System aus zwei Punktmassen m1 und m2 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 zu vernachlässigen.
R
y
m2
m1
x
a) Wie in der Vorlesung werde ein System mit N -Punktmassen, deren Ortsvektoren durch ~r1 , ...~r2
gegeben sind durch die 3N -Koordinaten xn , n = 1, ..., 3N der N Punktmassen beschrieben. Es
gebe R holonome Zwangsbedingungen gα (x1 , ...x3N , t), α = 1, ..., R. Es sei nun mindestens eine der
R-Zwangsbedingungen, hier o.B.d.A. als g1 bezeichnet, von der Form g1 (x1 , .., x3N , t) = xn − a,
wobei a eine Konstante sei. Zeigen Sie ganz allgemein, dass bei der Beschränkung auf eine Ebene
die Bewegung des Systems senkrecht zu dieser Ebene von vornherein außer Acht gelassen werden
kann.
b) Bestimmen Sie die Anzahl der Freiheitsgrade für zwei freie Punktmassen. Bestimmen Sie die Anzahl 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? Nutzen Sie das Ergebnis aus Aufgabenteil a.) um
die Dynamik des Systems zu beschreiben. Wie lautet der zu 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 x1 (t) mit der des freien
Falls. Was passiert für die Grenzfälle m1 = m2 und m1 m2 ? Was passiert für m1 = m2 + δm mit
|δm| 1?
1
Proposal for solution
a) We write for the constraint
g1 (x1 , ...x3N , t) = xn − 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 xn = a. For the equations of motion
of the n-th component, we then have the Lagrange equation of the first kind
R
mn ẍn = Fn + λ1
∂ g1 X
∂gα
+
.
λα
∂xn α=2
∂xn
(2)
As usual, the forces Fn , n = 1, ..., 3N are assumed to be known. The simple form of the constraint
∂ g1
= 1. We therefore find for the equations of motion of the nth component
g1 results in ∂x
n
mn ẍn = Fn + λ1 +
R
X
λα
α=2
∂gα
.
∂xn
(3)
Differentiating g1 twice w.r.t the time t, we obtain the trivial result ẍn = 0. Inserting the equations
of motion and solving for λ1 leads to
λ1 = −Fn −
R
X
λα
α=2
∂gα
.
∂xn
(4)
Inserting this back in the equations of motion results in the trivial equation
mn ẍn = 0.
(5)
This equation has the solution linear in time
xn = 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 g1 therefore enforces the choices
A=a
and B = 0 .
(7)
Thus, the solution (for all times t) is simply given by the constant
xn (t) = a .
(8)
Since λ1 does not appear in the equations of motion for the other components xm , m 6= n (as
∂
∂
∂xm g1 = ∂xm (xn − a) = 0), we can shorten the general procedure for finding the solution by simply
omitting the coordinate xn from the very beginning, i.e. for constraints of the form xn = const., it is
useful and possible to simply consider the 3 N −1 equations of motions for the remaining components
xm , m 6= n and the remaining R − 1 constraint equations. The same is of course true if one has
several constraints of the form xn = const.
b) In general, there are 6 coordinates (x1 , y1 , z1 , x2 , y2 , z2 ) 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 m1 and m2 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 x1 and x2 , i.e. the relevant
constraint is
g1 (x1 , x2 ) = x1 + x2 − l = 0 .
2
(9)
The other constraints are given by
g2 = y1 + R = 0,
g3 = y2 − R = 0,
g4 = z1 = 0,
g5 = z2 = 0 .
(10)
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
x1 , x2 motion.
c) The Lagrange equations of the first kind for the x1 and x2 components are given by
m1 ẍ1 = m1 g + λ,
m2 ẍ2 = m2 g + λ .
(11)
~ = λ ∇g
~ for the constraint forces Z
~ and used ∂g1 = ∂g1 = 1. The
Here we made as usual the ansatz Z
∂x1
∂x2
Lagrange multiplier λ is of course the same in both equations of motion as it was derived from one
and the same constraint g1 . We have however two constraint forces, pointing in the same direction
but each acting on the corresponding point mass only. Differentiating g1 twice w.r.t to the time t,
we obtain
ẍ1 + ẍ2 = 0 .
(12)
Inserting the equations of motion for x1 and x2 , we can solve for λ
λ = −2 g
m1 m2
.
m1 + m2
(13)
d) Inserting the solution for λ in the equations of motion we find by trivial integration
x1 (t) =
m1 − m2 g t 2
+ c1 t + c2 .
m1 + m2 2
(14)
The constants c1 , c2 , are to be determined by the initial data.
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).
~ is the same on both point masses
e) The constraint force Z
~ 1 g1 (x1 , x2 ) = λ ~ex ,
Z~1 =λ ∇
~ 2 g1 (x1 , x2 ) = λ ~ex .
Z~2 =λ ∇
(15)
(16)
The solution represents a “slowed down” free fall. In case the masses are equal m1 = m2 , the system
~1 + Z
~ 2 = −2 m ~g just compensate the gravitational force.
is in equilibrium and the constraint forces Z
2
m2
m2
1 −m2
In the limit m1 m2 , we have m
=
1
−
2
+
O
≈ 1 which recovers again the free
m1 +m2
m1
m1
fall for m1 . 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 m1 − m2 while the inertial
mass that has to be accelerated is given by the total mass, i.e. the sum of the two masses m1 + m2 .
1 −m2
As a result the acceleration is given by the ratio m
m1 +m2 g and is only a fraction of the gravitational
acceleration of the earth g. Thus arranging for m := m1 ≈ m2 by fine tuning the difference between
m1 and m2 to a very small amount (m1 − m2 ) =: δm with δm/m 1, one can slow down the free
fall to very small accelerations a ≈ 2δm
m g g.
3
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 m1 befestigt. Am anderen Ende der Pleuelstange
sei die Masse m2 befestigt die auf einer Führungschiene, welche mit der x-Achse des Koordinatensystems
zusammenfällt, reibungsfrei gleite. Die Scheibe sei mit ihrem Mittelpunkt im Ursprung des Koordinatensystems befestigt und ihre Rotationsachse falle mit der y-Achse des Koordinatensystems zusammen. Es
wirke keine äussere Kraft außer der Gravitationskraft.
z
m1
l
r
x
m2
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 x1 , y1 , z1 and x2 , y2 , z2 of the two point masses m1 and m2 are constraint
by five constraint equations. The first three are trivial and constrain the motion of the point particle
1 to the xz-plane and particle 2 to the x-axis respectively.
g1 (y1 ) = y1 = 0,
g2 (y2 ) = y2 = 0,
g3 (z2 ) = z2 = 0 .
(17)
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 x1 , z1 and z2 . The remaining two constraints are
easily found by trigonometric considerations. The motion of the point mass m1 is constraint to lie
on a circle with radius r in the xz-plane
g4 (x1 , z1 ) = x21 + z12 − r2 = 0 .
(18)
Finally, using the trigonometric relations we can express the x-coordinate of the second point mass
m2 in terms of x1 , z1 and l
l2 = z12 + (x2 − x1 )2 .
(19)
This can be written in constraint form as
g5 (x1 , z1 , x2 ) = z12 + (x2 − x1 )2 − l2 = 0 .
(20)
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
generalized coordinate is q = ϕ.
4
b) First we express the Cartesian coordinates x1 , z1 in terms of the generalized coordinate ϕ by choosing
polar coordinates with a fixed radius r.
x1 = r cos ϕ,
z1 = r sin ϕ,
r = const.
(21)
The coordinate x2 expressed in terms of the generalized coordinate then becomes
q
q
x2 = x1 − l2 − z12 = r cos ϕ − l2 − r2 sin2 ϕ.
(22)
In order to calculate the kinetic energy, we calculate the time derivative of x1 , z1 and x2 ,
ẋ1 = −ϕ̇ r sin ϕ,
ż1 = ϕ̇ r cos ϕ,
r2 sin ϕ cos ϕ
ẋ2 = −ϕ̇ r sin ϕ + ϕ̇ p
.
l2 − r2 sin2 ϕ
(23)
The kinetic energy is then given by
m2 2
m1 2
m1 2 2 m2 2 2 2
T =
ẋ1 + ż12 +
ẋ =
r ϕ̇ +
ϕ̇ r sin ϕ
2
2 2
2
2
1− p
r cos ϕ
l2 − r2 sin2 ϕ
!2
.
(24)
The potential energy is
V = m1 g z1 = m1 g r sin ϕ .
(25)
Finally, the Lagrange function is then found to be
m1 2 2 m2 2 2 2
L=T −V =
r ϕ̇ +
ϕ̇ r sin ϕ
2
2
Aufgabe 9
1− p
r cos ϕ
l2 − r2 sin2 ϕ
!2
− m1 g r sin ϕ .
(26)
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 − z-Ebene beschrieben sei (siehe
Abbildung). Er starte am höchsten Punkt in Ruhe und gleite dann reibungsfrei herab, bis zu dem Punkt
an dem er schließlich abhebt.
z
R
ϑ
x
a) Wie lautet die Zwangsbedingung h(r, ϑ) = 0 bis zum Abhebepunkt in Polarkoordinaten?
5
b) Drücken Sie die Lagrange-Gleichungen 1. Art
~
m ~r¨ = F~g + Z
(27)
~ die Zwangskraft.
in Polarkoordinaten aus. Hier bezeichnet F~g die Gravitationskraft und Z
Hinweis: Die Einheitsvektoren ~er und ~eθ sowie der Gradient in Polarkoordinaten sind gegeben durch
1 ∂
∂
sin ϑ
cos ϑ
~
+ ~eϑ
.
~er =
,
~eϑ =
,
∇ = ~er
cos ϑ
−sin ϑ
∂r
r ∂ϑ
Indem Sie (27), ausgedrückt in Polarkoordinaten, jeweils mit den orthonormierten Einheitsvektoren
~er bzw. ~eϑ multiplizieren, erhalten Sie die r bzw. ϑ Komponenten der Lagrange-Gleichung 1. Art.
~
c) Bestimmen Sie die auf den Skifahrer wirkende Zwangskraft Z.
Hinweis: Nutzen Sie die Energieerhaltung um ϑ̇2 zu eliminieren.
d) Bei welchem Winkel ϑ = ϑj hebt der Skifahrer ab?
Proposal for solution
a) The motion takes place in the xz-plane. In terms of polar coordinates
x =r sin ϑ,
(28)
z =r cos ϑ,
(29)
the constraint equation in polar coordinates is obviously of the form
h(r, ϑ) = h(r) = r − R = 0.
b) We can write the position ~r in polar coordinates
x
r sin ϑ
~r =
=
.
z
r cos ϑ
The first and second derivatives are then given by
sin ϑ
cos ϑ
~r˙ = ṙ
+ r ϑ̇
= ṙ ~er + r ϑ̇ ~eϑ ,
cos ϑ
−sin ϑ
(30)
(31)
(32)
and
~r¨ = r̈ ~er + ṙ ~e˙ r + ṙ ϑ̇ ~eϑ + r ϑ̈ ~eϑ + r ϑ̇ ~e˙ ϑ
= r̈ ~er + ṙ ϑ̇ ~eϑ + ṙ ϑ̇ ~eϑ + r ϑ̈ ~eϑ − r ϑ̇2 ~er
= r̈ − r ϑ̇2 ~er + r ϑ̈ + 2 ṙ ϑ̇ ~eϑ .
(33)
The gravitational force reads
∂
∂
1 ∂
∂
~
~
+ ~ez
m z = − ~er
+ ~eϑ
m g r cosϑ
Fg = −∇V = − ~ex
∂x
∂z
∂r
r ∂ϑ
= −m g cosϑ ~er + m g sinϑ ~eϑ .
(34)
6
~ = λ ∇h,
~ we have
With the ansatz for the constraint force Z
1 ∂
∂
~
~
+ ~eϑ
h(r) = λ ~er .
Z = λ ∇h(r) = λ ~er
∂r
r ∂ϑ
The Lagrange equation of the first kind in vectorial notation is given by
~.
m ~r¨ = F~g + Z
(35)
(36)
Inserting the corresponding expressions, we can collect terms with ~er and ~eϑ and write the Lagrange
equations of the first kind for r and ϑ componentwise (projecting with ~er and ~eϑ )
m r̈ =m r ϑ̇2 − g cos ϑ + λ,
(37)
m r ϑ̈ = − m 2 ṙ ϑ̇ − g sin ϑ .
(38)
c) Differentiating the constraint equation twice leads to
ḣ = ṙ = 0,
ḧ = r̈ = 0.
Inserting this into the equation of motion for r and solving for λ yields
λ = m g cos ϑ − m r ϑ̇2 = m g cos ϑ − R ϑ̇2 .
(39)
(40)
Here, we have used the constraint equation in the last equality in order to express r by R.
In order to eliminate ϑ̇2 from the expression for λ, we use conservation of energy. On the one hand,
in general, the energy is given by
1
1 E = T + V = m ṙ2 + r2 ϑ̇2 + m g r cos ϑ = m R2 ϑ̇2 + m g R cos ϑ .
(41)
2
2
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 ṙ2 term.
On the other hand, during the start, the skier is at rest which corresponds to ϑ0 = 0 and ϑ̇0 = 0 in
polar coordinates. Thus the energy at the starting point is given by
E0 = R m g .
(42)
1
m R2 ϑ̇2 + m g R (cos ϑ − 1) = 0.
2
(43)
Conservation of energy now implies
E − E0 = 0
or
Solving this equation for ϑ̇2 yields
g
(cos ϑ − 1) .
(44)
R
Of course, the same equation could have been obtained by first using dE/dt = 0 and afterwards
using the equations of motion to eliminate ϑ̈ as well as setting r = R, ṙ = r̈ = 0. The result (44)
can be inserted in the equation for λ
ϑ̇2 = −2
λ = m g (3 cos ϑ − 2) .
(45)
~ can then finally be calculated as
The constraint force Z
~ = λ∇h
~ = λ ~er = m g (3 cos ϑ − 2) ~er .
Z
(46)
~ ≈ m g ~er , i.e. initially the constraint force has to
d) During the start of the ride (ϑ ≈ 0) we have Z
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 ϑj , where the constraint force vanishes completely, the skier takes off.
~ =0
Z
⇒
7
cos ϑj =
2
.
3
(47)