ME 5550 Intermediate Dynamics
Constraint Relaxation Method: The Meaning of Lagrange Multipliers
Previously, we noted that if a dynamic system is described using "n" generalized
coordinates qk (k  1, , n) , and if the system is subjected to "m" independent
configuration constraint equations of the form
n
a
k 1
jk
qk  a j 0  0
( j  1, , m)
(1.1)
then, we can find the equations of motion of the system by using one of the following two
forms of Lagrange's equations with Lagrange multipliers.
d  K

dt  qk
m
 K


F

 j a jk


qk

q
j 1
k

(k  1, , n)
(1.2)
or
d  L 
L
 Fqk

 
dt  qk 
qk
 
m
nc
   j a jk
(k  1, , n)
(1.3)
j 1
Here, K is the kinetic energy of the system, Fqk is the generalized force associated with the
generalized coordinate qk , L is the Lagrangian of the system, V is the potential energy
function for the conservative forces and torques,  Fqk  is the generalized force associated
nc
with qk for the non-conservative forces and torques, only,  j is the Lagrange multiplier
associated with the j th constraint equation, and a jk ( j  1, , m; k  1, , n) are the
coefficients from the constraint equations. Equations (1.1) and Equations (1.2) or (1.3)
form a set of n  m differential/algebraic equations for the n generalized coordinates and
the m Lagrange multipliers.
Alternatively, we can relax (or remove) some or all the constraints and replace them with
force and/or torque components that are required to maintain the constraints. Then, we
formulate the n Lagrange's equations in terms of the n generalized coordinates and the m
constraint force (or torque) components. Together with the constraint equations, this forms
a set of n  m differential/algebraic equations for the n generalized coordinates and the m
constraint force and/or torque components. If all the constraints are relaxed, then Equations
(1.2) and (1.3) can be written as
Kamman – ME 5550 – page: 1/3
d  K

dt  qk
 K
 Fqk  Fqk


q
k

 
constraints
(k  1, , n)
(1.4)
and
d  L 
L
 Fqk

 
dt  qk 
qk
   F 
nc
qk
constraints
(k  1, , n)
(1.5)
Comparing equation (1.2) with equation (1.4) or by comparing equation (1.3) with equation
(1.5), we find
 
Fqk
m
constraints
   j a jk
(k  1, , n)
(1.6)
j 1
So, the Lagrange multipliers are directly related to the forces and/or torques required to
maintain the constraints.
Kamman – ME 5550 – page: 2/3
Example: The Simple Pendulum
For the simple pendulum shown at the right, we will
use q1  x and q2  y as the generalized coordinates,
and we will relax the length constraint of the pendulum
in the formulation. In this case, Lagrange's equations
can be written in the form
d  L

dt  qk
 L
 Fqk


q
k

   F 
nc
qk
(1.7)
constraint
where L  12 m  x 2  y 2   mgy and
F 
qk
nc
 0 , and the contributions of the constraint
force to the right hand sides of the equations are
 Fx constraint  T   v
 Fy constraint  T   v

 

y   T    x L  i   y L  j     xi  yj 
x   T   x L  i   y L  j   xi  yj x  T  x L 
(1.8)
y  T  y L 
(1.9)
Substituting into Lagrange's equations (1.4) and supplementing with the twice differentiated
constrain equation give the following equations of motion
mx   Lx  T  0
my  mg 
 Ly T  0
(1.10)
xx  yy  x 2  y 2  0
Using Lagrange multipliers, we showed in previous notes that the equations for the
pendulum could be written as
mx   x  0
my  mg   y  0
(1.11)
xx  yy  x 2  y 2  0
Comparing Equations (1.7) and (1.8), we see that the Lagrange multiplier  is equal to
T L the force per unit length of the pendulum.
Kamman – ME 5550 – page: 3/3