Download Lagrangian Mechanics

Alain J. Brizard
Saint Michael's College
Lagrangian Mechanics
1
Principle of Least Action
The con¯guration of a mechanical system evolving in an n-dimensional space, with coordinates x = (x1 ; x2 ; :::; xn), may be described in terms of generalized coordinates q =
(q1 ; q2; :::; q k ) in a k-dimensional con¯guration space, with k < n.
The Principle of Least Action (also known as Hamilton's principle) is expressed in terms
of a function L(q; q;
_ t) known as the Lagrangian, which appears in the action integral
A[q] =
Z tf
_ t) dt;
L(q; q;
(1)
ti
where the action integral is a functional of the vector function q(t), which provides a path
from the initial point q i = q(ti) to the ¯nal point qf = q(tf ). The variational principle
¯
"
Ã
!#
Z tf
¯
d
@L
d @L
¯
j
0 = ±A[q] =
A[q + ² ±q]¯
=
±q
¡
dt;
¯
d²
ti
@ qj
dt @ q_j
²=0
where the variation ±q is assumed to vanish at the integration boundaries (±q i = 0 = ±qf ),
yields the Euler-Lagrange equation for the generalized coordinate qj (j = 1; :::; k)
d
dt
Ã
@L
@ q_ j
!
=
@L
;
@ qj
(2)
The Lagrangian also satis¯es the second Euler equation
Ã
d
@L
L ¡ q_j
dt
@ q_ j
!
=
@L
;
@t
(3)
and thus for time-independent Lagrangian systems (@L=@t = 0) we ¯nd that L¡ q_ j @L=@ q_ j
is a conserved quantity whose interpretation will be discussed shortly.
The form of the Lagrangian function L(r; r_ ; t) is dictated by our requirement that
Newton's Second Law m Är = ¡ rU(r; t) describing the motion of a particle of mass m in a
nonuniform (possibly time-dependent) potential U(r; t) be written in the Euler-Lagrange
form (2). One easily obtains the form
L(r; r_ ; t) =
m 2
j_rj ¡ U (r; t);
2
1
(4)
which is simply the kinetic energy of the particle minus its potential energy. For a timeindependent Lagrangian (@L=@t = 0), we also ¯nd that the energy function
r_ ¢
m 2
@L
¡ L =
j_rj + U (r) = E;
@ r_
2
is a constant of the motion. Hence, for a simple mechanical system, the Lagrangian function
is obtained by computing the kinetic energy of the system and its potential energy and then
construct Eq. (4).
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Examples
The construction of a Lagrangian function for a system of N particles proceeds in three
steps as follows.
Step I. De¯ne k generalized coordinates fq1 (t); :::; q k(t)g that represent the instantaneous con¯guration of the system of N particles.
Step II. Construct the position vector ra (q; t) and its associated velocity
va (q; q;
_ t) =
k
X
@r a
@ra
+
q_j
@t
@q j
j=1
for each particle (a = 1; :::; N).
Step III. Construct the kinetic energy
_ t) =
K(q; q;
1 X
_ t)j 2
ma jv a (q; q;
2 a
and the potential energy U (q; t) for the system and combine them to obtain the Lagrangian
L(q; q;
_ t) = K (q; q;
_ t) ¡ U(q; t);
from which the Euler-Lagrange equations (2) are derived.
2.1
Example I: Pendulum
Consider a pendulum composed of an object of mass m and a massless string of constant
length ` in a constant gravitational ¯eld with acceleration g.
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Although the motion of the pendulum is two-dimensional, a single generalized coordinate is needed to describe the con¯guration of the pendulum: the angle µ measured from
the negative y-axis (see Figure above). Here, the position of the object is given as
x(µ) = ` sin µ and y(µ) = ¡ ` cos µ;
with associated velocity components
_ = ` µ_ cos µ and y(µ;
_ = ` µ_ sin µ:
x(µ;
_ µ)
_ µ)
Hence, the kinetic energy of the pendulum is
K =
m 2 _2
` µ ;
2
and choosing the zero potential energy point when µ = 0 (see Figure above), the gravitational potential energy is
U = mg` (1 ¡ cos µ):
The Lagrangian L = K ¡ U is, therefore, written as
_ = m `2µ_ 2 ¡ mg` (1 ¡ cos µ);
L(µ; µ)
2
and the Euler-Lagrange equation for µ is
@L
d
= m`2 µ_ !
_
dt
@µ
or
Ã
!
@L
= m`2 µÄ
_
@µ
@L
= ¡ mg` sin µ
@µ
g
µÄ +
sin µ = 0
`
3
2.2
Example II: Bead on a Rotating Hoop
Consider a bead of mass m sliding freely on a hoop of radius R rotating with angular
velocity !0 in a constant gravitational ¯eld with acceleration g.
Here, since the bead of the rotating hoop moves on the surface of a sphere of radius R,
we use the generalized coordinates given by the two angles µ (measured from the negative
z-axis) and ' (measured from the positive x-axis), where '_ = !0 . The position of the bead
is given as
x(µ; t) = R sin µ cos('0 + !0t);
y(µ; t) = R sin µ sin(' 0 + !0t);
z(µ; t) = ¡ R cos µ;
where '(t) = '0 + !0 t and its associated velocity components are
³
´
³
´
_ t) = R µ_ cos µ cos ' ¡ !0 sin µ sin ' ;
x(µ;
_ µ;
_ t) = R µ_ cos µ sin ' + !0 sin µ cos ' ;
y(µ;
_ µ;
_ t) = R µ_ sin µ;
z(µ;
_ µ;
so that the kinetic energy of the bead is
2³
´
_ = m jvj 2 = m R µ_ 2 + ! 2 sin2 µ :
K(µ; µ)
0
2
2
The gravitational potential energy is
U(µ) = mgR (1 ¡ cos µ);
where we chose the zero potential energy point at µ = 0 (see Figure above). The Lagrangian
L = K ¡ U is, therefore, written as
2³
´
_ = m R µ_ 2 + !2 sin2 µ ¡ mgR (1 ¡ cos µ);
L(µ; µ)
0
2
4
and the Euler-Lagrange equation for µ is
@L
d
2 _
=
mR
µ
!
dt
@ µ_
or
Ã
!
@L
= mR2 µÄ
_
@µ
@L
= ¡ mgR sin µ
@µ
+ mR2 !02 cos µ sin µ
µ
¶
g
2
Ä
µ + sin µ
¡ !0 cos µ = 0
R
2.3
Example III: Rotating Pendulum
Consider a pendulum of mass m and length b attached to the edge of a disk of radius a
rotating at angular velocity ! in a constant gravitational ¯eld with acceleration g.
Placing the origin at the center of the disk, the coordinates of the pendulum mass are
x = ¡ a sin !t + b cos µ
y = a cos !t + b sin µ
so that the velocity components are
x_ = ¡ a! cos !t ¡ b µ_ sin µ
y_ = ¡ a! sin !t + b µ_ cos µ
and the squared velocity is
v2 = a2!2 + b2 µ_2 + 2 ab ! µ_ sin(µ ¡ !t):
Setting the zero potential energy at x = 0, the gravitational potential energy is
U = ¡ mg x = mga sin !t ¡ mgb cos µ:
5
The Lagrangian L = K ¡ U is, therefore, written as
h
i
_ t) = m a 2!2 + b2µ_ 2 + 2 ab ! µ_ sin(µ ¡ !t)
L(µ; µ;
2
¡ mga sin !t + mgb cos µ;
(5)
and the Euler-Lagrange equation for µ is
@L
= mb2 µ_ + m ab ! sin(µ ¡ !t) !
@ µ_ Ã !
d @L
= mb2 µÄ + m ab ! (µ_ ¡ !) cos(µ ¡ !t)
dt @ µ_
and
@L
= m ab ! µ_ cos(µ ¡ !t) ¡ mg b sin µ
@µ
or
g
a
µÄ +
sin µ ¡ !2 cos(µ ¡ !t) = 0
b
b
We recover the standard equation of motion for the pendulum when a or ! vanish.
Note that the terms [(m=2) a 2!2] and [¡ mga sin !t] in the Lagrangian (5) play no role
in determining the dynamics of the system. In fact, as can easily be shown, a Lagrangian L
is always de¯ned up to an exact time derivative, i.e., the Lagrangians L and L0 = L+ df=dt,
where f (q; t) is an arbitrary function, lead to the same Euler-Lagrange equations. In the
present case,
f (t) = [(m=2) a2! 2] t + (mga=!) cos !t
and thus this term can be omitted from the Lagrangian (5) without changing the equations
of motion.
2.4
Example IV: Compound Atwood Machine
Consider a compound Atwood machine composed three masses (labeled m1 , m2, and m3 )
attached by two massless ropes through two massless pulleys in a constant gravitational
¯eld with acceleration g. The two generalized coordinates for this system (see Figure) are
the distance x of mass m1 from the top of the ¯rst pulley and the distance y of mass m2
from the top of the second pulley; here, the lengths `a and `b are constants.
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The coordinates and velocities of the three masses m1 , m2, and m3 are
x1 = x ! v1 = x;
_
x2 = `a ¡ x + y ! v2 = y_ ¡ x;
_
x3 = `a ¡ x + `b ¡ y ! v3 = ¡ x_ ¡ y;
_
respectively, so that the total kinetic energy is
m1 2
m2
m3
K =
x_ +
(y_ ¡ x)
_ 2 +
(x_ + y)
_ 2:
2
2
2
Placing the zero potential energy at the top of the ¯rst pulley, the total gravitational
potential energy, on the other hand, can be written as
U = ¡ g x (m1 ¡ m2 ¡ m3 ) ¡ g y (m2 ¡ m3) ;
where constant terms were omitted. The Lagrangian L = K ¡ U is, therefore, written as
m1 2
m2
m3
L(x; x;
_ y; y)
_ =
x_ +
( x_ ¡ y)
_ 2 +
(x_ + y)
_ 2
2
2
2
+ g x (m1 ¡ m2 ¡ m3 ) + g y (m2 ¡ m3) :
The Euler-Lagrange equation for x is
@L
= (m1 + m2 + m3 ) x_ + (m3 ¡ m2) y_ !
@ x_ Ã !
d @L
= (m1 + m2 + m3) xÄ + (m3 ¡ m2 ) yÄ
dt @ x_
@L
= g (m1 ¡ m2 ¡ m3 )
@x
7
while the Euler-Lagrange equation for y is
@L
= (m3 ¡ m2) x_ + (m2 + m3) y_ !
@ y_
Ã
!
d @L
= (m3 ¡ m2 ) xÄ + (m2 + m3 ) yÄ
dt @ y_
@L
= g (m2 ¡ m3)
@y
or
(m1 + m2 + m3 ) xÄ + (m3 ¡ m2) yÄ = g (m1 ¡ m2 ¡ m3 )
(m3 ¡ m2) xÄ + (m2 + m3) yÄ = g (m2 ¡ m3)
3
Symmetries and Conservation Laws
The Noether theorem states that for each symmetry of the Lagrangian there corresponds
a conservation law (and vice versa). When the Lagrangian L is invariant under a time
translation, a space translation, or a spatial rotation, the conservation law involves energy,
linear momentum, or angular momentum, respectively.
When the Lagrangian is invariant under time translations, t ! t + ±t, the Noether
theorem states that energy is conserved, dE=dt = 0, where
E =
dq @L
¢
¡ L
dt @ q_
de¯nes the energy invariant. When the Lagrangian is invariant under spatial translations
or rotations, ® ! ® + ±®, the Noether theorem states that the component
p® =
@L
@q @L
=
¢
@ ®_
@® @ q_
of the canonical momentum p = @L=@ q_ is conserved, dp® =dt = 0. Note that if ® is a
linear spatial coordinate, p® denotes a component of the linear momentum while if ® is an
angular spatial coordinate, p® = L³ denotes a component of the angular momentum (with
³ denoting the axis of symmetry about which ® is measured).
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