Lagrange’s Equation
Lagrange’s equation is given by:
d
T
T
V
+
= Qn , i = 1,.., N
dt qi
qi
qi
where
T = Kinetic Energy, V = Potential Energy
qi = generalized coordinate
Qni = nonconservative generalized force
N = DOF
i
Generalized coordinates
A system having N degrees of freedom must have N independent
coordinates to describe its motion. Any set of N independent
coordinates is called generalized coordinates and is designated
by q1, q2, q3, …, qN.
In the triple pendulum shown, the angular
displacements i (i = 1,2,3) are used to
specify the locations of the masses at any
time. Since these angles are independent,
they form a set of generalized coordinates
and are denoted by qi = i (i = 1,2,3).
1
2
3
1
Other examples of generalized coordinates are shown below.
2
When external forces act on a system, the configuration of the
system changes. The new configuration can be obtained by
changing the generalized coordinates qi by qi. Any vector r can be
written as a function of the generalized coordinates q1, q2, …, qN as
follows:
r = r (q1 , q2 ,
, qN )
It follows that
r
r
q1 +
q2 + .....
q1
q2
r=
=
N
k =1
r
qk
qk
r is called a virtual displacement and is an infinitesimal
displacement that does not violate the constraints.
Generalized forces
The forces represented by Qni are the nonconservative generalized
force corresponding to the generalized coordinate qi. If Ui
denotes the work done (or virtual work) in changing the
generalized coordinate qi by an amount qi, the corresponding
generalized force Qi is defined as:
Ui
Qi =
qi
where Qi will be a force (moment) when qi is a linear (angular)
displacement. The virtual work, U, done by a physical force F
due to a displacement r is given by:
3
U = Fi r = Fi
= (F i
r
r
q1 +
q2 + .....
q1
q2
r
r
) q1 + ( F i
) q2 + .....
q1
q2
= ( Q1 q1 + Q2 q2 + .....) =
N
k =1
Qk qk
where
r
qk
is the generalized force corresponding to the generalized
coordinate qk.
Example: Obtain the equations of motion for the system shown.
x
Qk = F i
k1
k2
M
rC
C
l
F
Solution:
Here the end displacement is given by:
xend = x + l sin
in other words,
4
xend = r ( x, )
it follows that:
xend
x
x + end
x
= x + l cos
xend =
hence the work done is:
U = F i xend = F i( x + l cos
= ( F x + Fl cos
= ( Qx x + Q
)
)
)
from which the nonconservative generalized forces are:
Qx = F , Q = Fl cos
so now we have
q1 = x, q2= , Q1 = F and Q2 = F l cos
And the equations of motion will be determined from
d T
T
V
+
= Q1
dt x
x
x
and
d T
T
V
+
= Q2
dt
yielding
l 2
l
( M + m) x + (k1 + k2 ) x m
sin + m
cos = F
2
2
and
l
l
2
1
+ m cos x + mg sin = Fl cos
3 ml
2
2
5
Example (see also example on page 473)
x
k1
F
M
c1
k2
c2
m
Solution: here the nonconservative forces consist of the external
force F and the damping forces provided by the dashpots. The
work done is:
U = F i x ( c1 x )i x ( c2 x )i x
= ( F c1 x c2 x ) x
= ( Qx ) x
from which the nonconservative generalized forces are:
Q1 = F c1 x c2 x
Q2 = 0
where
q1 = x , q2 =
6