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Chapter 4. Lagrangian Dynamics
(Most of the material presented in this chapter is taken from Thornton and Marion, Chap.
4.1 Important Notes on Notation
In this chapter, unless otherwise stated, the following notation conventions will be used:
- Einstein’s summation convention. Whenever an index appears twice (an only
twice), then a summation over this index is implied. For example,
x
i
x
i
! x
i
x
i
i
"
= x
i
2
i
"
- The index i is reserved for Cartesian coordinates. For example, x
i
, for i = 1 , 2 , 3 ,
represents either x , y , or z depending on the value of i. Similarly, p
i
can represent
p
x
, p
y
, or p
z
. This does not mean that any other indices cannot be used for
Cartesian coordinates, but that the index i will only be used for Cartesian
coordinates.
- When dealing with systems containing multiple particles, the index! will be used
to identify quantities associated with a given particle when using Cartesian
coordinates. For example, if we are in the presence of n particles, the position
vector for particle! is given by r
!
, and its kinetic energy T
!
by
T
!
m
!
x
! , i
x
! , i
,! = 1 , 2 , ... , n and i = 1 , 2 , 3. ( 4. 2 )
Take note that, according to convention 1 above, there is an implied summation on
the Cartesian velocity components (the index i is used), but not on the masses
since the index! appears more than twice. Correspondingly, the total kinetic
energies is written as
T =
m
!
x!
! , i
! = 1
n
"
x!
! , i
m
!
x!
2
2
+! z
2
( )
! = 1
n
"
4.2 Introduction
Although Newton’s equation F = p! correctly describes the motion of a particle (or a
system of particles), it is often the case that a problem will be too complicated to solve
using this formalism. For example, a particle may be restricted in its motion such that it
follows the contours of a given surface, or that the forces that keep the particle on the
surface (i.e., the forces of constraints ), are not easily expressible in Cartesian
coordinates. It may not even be possible at times to find expressions for some forces of
constraints. Such occurrences would render it impossible to treat the problem with the
Newtonian formalism since this requires the knowledge of all forces acting on the
particles.
In this section we will study a different approach for solving complicated problems in a
general manner. The formalism that will be introduced is based on the so-called
Hamilton’s Principle , from which the equations of motion will be derived. These
equations are called Lagrange’s equations. Although the method based on Hamilton’s
Principle does not constitute in itself a new physical theory, it is probably justified to say
that it is more fundamental that Newton’s equations. This is because Hamilton’s Principle
can be applied to a much wider range of physical phenomena than Newton’s theory (e.g.,
quantum mechanics, quantum field theory, electromagnetism, relativity). However, as
will be shown in the following sections, the Lagrange’s equation derived from this new
formalism are equivalent to Newton’s equations when restricted to problems of
mechanics.
4.3 Hamilton’s Principle
Hamilton’s Principle is concerned with the minimization of a quantity (i.e., the action) in
a manner that is identical to extremum problems solved using the calculus of variations.
Hamilton’s Principle can be stated as follows:
The motion of a system from time t
1
to t
2
is such that the line integral
(called the action or the action integral ),
I = L dt
t 1
t 2
!
where L = T! U (with T , and U the kinetic and potential energies,
respectively), has a stationary value for the actual path of the motion.
Note that a “stationary value” for equation ( 4. 4 ) implies an extremum for the action, not
necessarily a minimum. But in almost all important applications in dynamics a minimum
occurs.
Because of the dependency of the kinetic and potential energies on the coordinates x
i
the velocities x!
i
, and possibly the time t , it is found that
L = L x
i
, x!
i
, t ( )
Hamilton’s Principle can now be expressed mathematically by
! L x
i
x
i
( , t ) dt = 0
t
1
t
2
"
L = T! U =
mR
2
2
mR
2
sin
2
2
+ F
"
R ". ( 4. 12 )
Upon inspection of the Lagrangian, we can see that there are two degrees of freedom for
this problem, i.e., !, and ". We now need to calculate the different derivatives that
compose the Lagrange equations
! L
= mR
2
2
sin( " )cos( " ) + F
"
R
! L
d
dt
! L
d
dt
mR
2
( )
= mR
2
d
dt
! L
d
dt
mR
2
sin
2
( )
= mR
2
sin "
cos "
sin
2
( )
applying equation ( 4. 7 ) for !, and "we find the equations of motion to be
F
!
= mR
2
sin(! )cos(! )
( )
0 = mR
2
sin!
sin!
cos!
( )
Incidentally, this problem was analyzed at the end of Chapter 1 on Newtonian Mechanics
(problem 2-2, with F
!
= 0 ), where Newton’s equation was used to solve the problem.
One can see how simpler the present treatment is. There was no need to calculate
relatively complex equations like
e
r
. Furthermore, it is important to realize that the
spherical coordinates! and " are treated as Cartesian coordinates when using the
Lagrangian formalism.
4.4 Degrees of Freedom and Generalized Coordinates
If a system is made up of n particles, we can specify the positions of all particles with
3 n coordinates. On the other hand, if there are m equations of constraints (for example, if
some particles were connected to form rigid bodies), then the 3 n coordinates are not all
independent. There will be only 3 n! m independent coordinates, and the system is said
to possess 3 n! mdegrees of freedom.
Furthermore, the coordinates and degrees of freedom do not have to be all given in
Cartesian coordinates, or any other systems. In fact, we can choose to have different
types of coordinate systems for different coordinates. Also, the degrees of freedom do not
even need to share the same unit dimensions. For example, a problem with a mixture of
Cartesian and spherical coordinates will have “lengths” and “angles” as units. Because of
the latitude available in selecting the different degrees in freedom, the name of
generalized coordinates is given to any set of quantities that completely specifies the
state of the system. The generalized coordinates are usually written as q
j
It is important to realize that there is not one unique way of setting up the generalized
coordinates, there are indeed many different ways of doing this. Unfortunately, there are
no clear rules for selecting the “best” set of generalized coordinates. The ultimate test is
whether or not the choice made leads to a simple solution for the problem at hand.
In addition to the generalized coordinates q
j
, a corresponding set of generalized
velocities q!
j
is defined. In general, the relationships linking the Cartesian and
generalized coordinates and velocities can be expressed as
x
! , i
= x
! , i
q
j
, t
( )
x!
! , i
= x!
! , i
q
j
, q!
j
, t
( )
,! = 1 , ... , n , i = 1 , 2 , 3 , j = 1 , ... , 3 n " m ,
or alternatively
q
j
= q
j
x
! , i
, t
( )
q!
j
= q!
j
x
! , i
, x!
! , i
, t ( )
We must also include the equations of constraints
f
k
x
! , i
, t ( )
= f
k
q
j
, t ( )
It follows naturally that Hamilton’s Principle can now be expressed in term of the
generalized coordinates and velocities as
! L q
j
, q!
j
, t ( )
dt = 0
t 1
t
2
"
with Lagrange’s equations given by
! L
! q
j
d
dt
! L
q
j
= 0 , j = 1 , 2 , ... , 3 n " m. ( 4. 19 )
Examples
- The simple pendulum. Let’s solve the problem of the simple pendulum (of mass m and
length !) by first using the Cartesian coordinates to express the Lagrangian, and then
transform into a system of cylindrical coordinates.
! L
= # mg !sin( " )
d
dt
! L
d
dt
m!
2
"
( )
= m!
2
""
and finally
g
sin!
- The double pendulum. Consider the case of two particles of mass m
1
and m
2
each
attached at the end of a mass less rod of length l
1
and l
2
, respectively. Moreover, the
second rod is also attached to the first particle (see Figure 4 - 2 ). Derive the equations
of motion for the two particles.
Solution. It is desirable to use cylindrical coordinates for this problem. We have two
degrees of freedom, and we will choose!
1
and!
2
as the independent variables. Starting
with Cartesian coordinates, we write an expression for the kinetic and potential energies
for the system
T =
m
1
x!
1
2
1
2
( )
2
x!
2
2
2
2
( )
U = m
1
gy
1
2
gy
2
But Figure 4 - 2 – The double pendulum.
x
1
= l
1
sin!
1
( )
y
1
= " l
1
cos!
1
( )
x
2
= l
1
sin!
1
( )
2
sin!
2
( )
y
2
= " l
1
cos!
1
( )
" l
2
cos!
2
( )
and
x!
1
= l
1
1
cos!
1
( )
y!
1
= l
1
1
sin!
1
( )
x!
2
= l
1
1
cos!
1
( )
2
2
cos!
2
( )
y!
2
= l
1
1
sin!
1
( )
2
2
sin!
2
( )
Inserting equations ( 4. 26 ) and ( 4. 27 ) in ( 4. 25 ), we get
T =
m
1
l
1
2
!
1
2
2
l
1
2
1
2
2
2
2
2
1
l
2
1
2
cos!
1
( )cos!
2
( ) + sin!
1
( )sin!
2
( ) ( { })
m
1
l
1
2
!
1
2
2
l
1
2
!
1
2
2
2
!
2
2
1
l
2
1
2
cos!
1
2
( ( ))
U = & m
1
2
( )
gl
1
cos!
1
( )
& m
2
gl
2
cos!
2
( )
and for the Lagrangian
L = T! U
m
1
l
1
2
!
1
2
2
l
1
2
!
1
2
2
2
!
2
2
1
l
2
1
2
cos "
1
2
( ( ))
1
2
( )
gl
1
cos "
1
( )
2
gl
2
cos "
2
( )
Inspection of equation ( 4. 29 ) tells us that there are two degrees of freedom for this
problem, and we choose!
1
, and!
2
as the corresponding generalized coordinates. We
now use this Lagrangian with equation ( 4. 19 )
Figure 4 - 3 – A simple pendulum attached to a rotating rim.
The kinetic and potential energies, and the Lagrangian are
T =
m a
2
2
2
2
" $sin( " )cos(! t ) # sin(! t )cos( " )
( )
m a
2
2
2
2
" sin( " #! t )
( )
U = mg a sin(! t ) # b cos( " )
( )
L = T # U
m a
2
2
2
!
2
" sin( " #! t )
( )
# mg a sin(! t ) # b cos( " )
( )
We now calculate the derivatives for the Lagrange equation using! as the sole
generalized coordinate
! L
= mab #
" cos( " $ # t ) $ mgb sin( " )
d
dt
! L
"
= mb
2
!!
" + mab #
( )
cos( " $ # t ).
Finally, the equation of motion is
a
b
2
cos(! " # t ) +
g
b
sin(! " # t ) = 0. ( 4. 37 )
Figure 4 - 4 – A slides along a smooth wire that rotates about the z - axis.
- The sliding bead. A bead slides along a smooth wire that has the shape of a parabola
z = cr
2
(see Figure 4 - 4 ). At equilibrium, the bead rotates in a circle of radius R when
the wire is rotating about its vertical symmetry axis with angular velocity !. Find the
value of c.
Solution. We choose the cylindrical coordinates r , !, and z as generalized coordinates.
The kinetic and potential energies are
T =
m r!
2
2
!
2
2
( )
U = mgz.
We have in this case some equations of constraints that we must take into account,
namely
z = cr
2
z! = 2 cr! r ,
and
! = " t
Inserting equations ( 4. 39 ) and ( 4. 40 ) in equation ( 4. 38 ), we can calculate the Lagrangian
for the problem
any further. However, if the constants A
i
, and B are such that equation ( 4. 46 ) can be
expressed as
! f
! x
i
! x
i
! t
! f
! t
or more simply
df
dt
then it can be integrated to give
f x
i
, t ( )
! cste = 0. ( 4. 49 )
Using generalized coordinates, and by slightly changing the form of equation ( 4. 48 ), we
conclude that, as was stated above, constraints that can be written in the form of a
differential
df =
! f
! q
j
dq
j
! f
! t
dt = 0 ( 4. 50 )
are similar to the constraints considered at the beginning of this section, that is
f q
j
, t ( )
! cste = 0. ( 4. 51 )
Problems involving constraints such as the holonomic kind discussed here can be handled
in exactly the same manner as was done in the chapter on the calculus of variations. This
is done by introducing the so-called Lagrange undetermined multipliers. When this is
done, we find that the following form for the Lagrange equations
! L
! q
j
d
dt
! L
! q!
j
k
( t )
! f
k
! q
j
where the index j = 1 , 2 , ... , 3 n! m , and k = 1 , 2 , ... , m.
Although the Lagrangian formalism does not require the insertion of the forces of
constraints involved in a given problem, these forces are closely related to the Lagrange
undetermined multipliers. The corresponding generalized forces of constraints can be
expressed as
Q
j
k
( t )
" f
k
" q
j
Examples
- The rolling disk on an inclined plane. We now solve the problem of a disk of mass m
and of radius R rolling down an inclined plane (see Figure 4 - 5 ).
Solution. Referring to Figure 4 - 1 , and separating the kinetic energies in a translational
rotational part, we can write
T =
my!
2
I
!
2
my!
2
mR
2
!
!
2
where I = mR
2
2 is the moment of inertia of the disk about its axis of rotation. The
potential energy and the Lagrangian are given by
U = mg ( l! y )sin( " )
L = T! U
my!
2
mR
2
!
2
! mg ( l! y )sin( " ),
where l , and! are the length and the angle of the inclined plane, respectively. The
equation of constraint given by
f = y! R " = 0. ( 4. 56 )
This problem presents itself with two generalized coordinates ( y and !) and one
equation of constraints, which leaves us with one degree of freedom. We now apply the
Lagrange equations as defined with equation ( 4. 52 )
Figure 4 - 5 - A disk rolling on an incline plane without slipping.
Figure 4 - 6 – An arrangement of a spring, mass, and mass less pulleys.
- Consider the system of Figure 4 - 6. A string joining two mass less pulleys has a length
of l and makes an angle !
with the horizontal. This angle will vary has a function of the
vertical position of the mass. The two pulleys are restricted to a translational motion by
frictionless guiding walls. The restoring force of the spring is! kx and the arrangement is
such that when! = 0 , x = 0 , and when! = " 2 , x = l. Find the equations of motion for
the mass.
Solution. There are two equations of constraints for this problem
f = x! l 1! cos( " )
( )
g = y! l sin "
and we identify three generalized coordinates x , y , and !. We now calculate the energies
and the Lagrangian
T =
my!
2
U =
kx
2
! mgy
L = T! U =
m
y
2
kx
2
Using equations ( 4. 64 ) and ( 4. 65 ), we can write the Lagrange equations while using the
appropriate undetermined Lagrange multipliers
! L
! q
j
d
dt
! L
q
j
f
! f
! q
j
g
! g
! q
j
where q
j
can represent x , y , or !. Calculating the necessary derivatives
! L
! x
= " kx
! L
! y
= mg
! L
d
dt
! L
x
d
dt
! L
d
dt
! L
! y!
= m! y !,
and
! kx + "
1
mg! m! y! + "
2
1
l sin( # )! "
2
l cos( # ) = 0.
We, therefore, have
1
= kx = kl 1 " cos # ( ( ))
= kl 1 " 1 "
y
l
2
2
1
tan #
= " kl 1 " 1 "
y
l
2
y
l 1 "
y
l
2
= " ky 1 "
y
l
2
"
1
2
and with the insertion of these equations in the second of equations ( 4. 68 ) we finally get
! y!! g +
k
m
y 1!
y
l
2
!
1
2
4.7.1 The Kinetic Energy
In a Cartesian coordinates system the kinetic energy of a system of particles is expressed
as
T =
m
!
x!
! , i
x!
! , i
! = 1
n
"
where a summation over i is implied. In order to derive the corresponding relation using
generalized coordinates and velocities, we go back to the first of equations ( 4. 15 ), which
relates the two systems of coordinates
x
! , i
= x
! , i
q
j
, t
( )
, j = 1 , 2 , ... , 3 n " m. ( 4. 78 )
Taking the time derivative of this equation we have
x!
! , i
" x
! , i
" q
j
q!
j
" x
! , i
" t
and squaring it (and summing over
i )
x!
! , i
x!
! , i
" x
! , i
" q
j
" x
! , i
" q
k
q!
j
q!
k
" x
! , i
" q
j
" x
! , i
" t
q!
j
" x
! , i
" t
" x
! , i
" t
An important case occurs when a system is scleronomic , i.e., there is no explicit
dependency on time in the coordinate transformation, we then have
! x
" , i
! t
and the kinetic energy can be written in the form
T = a
jk
q
j
q
k
with
a
jk
m
!
! = 1
n
"
x
! , i
q
j
x
! , i
q
k
where a summation on i is still implied. Just as was the case for Cartesian coordinates,
we see that the kinetic energy is a quadratic function of the (generalized) velocities. If we
next differentiate equation ( 4. 81 ) with respect to
q
l
, and then multiply it by
q
l
(and
summing), we get
q
l
! T
! q!
l
= 2 a
jk
q
j
q
k
= 2 T ( 4. 83 )
since
a
jk
is not a function of the generalized velocities, and it is symmetric in the
exchange of the j and k indices.
4.7.2 Conservation of Energy
Consider a general Lagrangian, which will be a function of the generalized coordinates
and velocities and may also depend explicitly on time (this dependence may arise from
time variation of external potentials, or from time-dependent constraints). Then the total
time derivative of L is
dL
dt
! L
! q
j
q!
j
! L
! q!
j
! q!
j
! L
! t
But from Lagrange’s equations,
! L
! q
j
d
dt
! L
q
j
and equation ( 4. 84 ) can be written as
dL
dt
d
dt
! L
! q!
j
q
j
! L
! q!
j
q
j
! L
! t
d
dt
! L
! q!
j
q!
j
! L
! t
It therefore follows that
d
dt
! L
! q!
j
q!
j
" L
! L
! t
dH
dt
! L
! t
or
dH
dt
" L
" t
where we have introduced a new function
