Physics 325
Lecture 19
More Examples using Lagrangians
Example
: Moveable plane pendulum
m
1
moves along x without friction.
2
m
swings from a massless rod of length
that is
suspended from
1
m
and restricted to move in the
x

y
plane.
For generalized coordinates we choose
1
x
and
. With
(
0)
0
Uy
, we have
22
2
cos
U
m gy
mg
where
cos
and
sin
yx
The kinetic energy of
m
1
is
2
1
1 1
1
2
T
m x
For
m
2
we have a contribution from
, and a contribution from the motion of
m
1
(
1
x
).
This gives
21
1
ˆ
ˆ
ˆˆ
cos
sin
v
x x
x
x
y
so that
2
2
2 2
2
2
2
2
1
1
1
2
1
2
cos
2
T
m v
m x
x
x
y
m
1
m
2
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View Full DocumentThe total kinetic energy is
T
1
+T
2
, and the Lagrangian is
2
2
2
1
2
1
2
2
1
2
11
cos
cos
22
L
T U
m
m x
m
m
x
m g
The partial derivatives are as follows
2
1
2
1
1
2
1
2
1 1
2
1
1
2
21
0
sin
sin
cos
cos
cos
LL
m
x
m g
x
L
m
m
x
m
m x
m
x
x
L
mx
The Lagrange equation for
x
1
again gives conservation of linear momentum.
We have
1 1
2
1
1
2
cos
0
xx
dd
m x
m x
p
p
dt
dt
We can work out the time derivative to give
2
1
2
1
2
2
cos
sin
0
m
m x
m
m
(19.1)
The Lagrange equation for
provides
2
2
1
1
2
1
cos
sin
sin
0
m
x
dx
m
g
x
or
1
cos
sin
0
xg
(19.2)
Equations (19.1) and (19.2) are the differential equations of motion for the system. They
are second order (as always – think
F
mx
), and they are nonlinear.
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 Spring '08
 Staff
 mechanics, Friction, Kinetic Energy, Mass, Sin, mr 2

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