We may write: dS/ dt
- R (dq /dt)
= 0
Or: S’
- R q’ = 0
Which implies: å c a q’ a
= 0
Then:
ò [dS/dt -
R (dq /dt ) ]
= const.
Or: S
- R q =
const. = 0
This condition is
what defines a holonomic constraint.
If such integration is not possible the constraint is non-holonomic. By looking at such constraints the aim is always to reduce the number of coordinates needed to describe the system.
Next, consider a
disk rolling down an inclined plane as shown in the diagram below:
We want to
calculate the constraints assuming no slipping, where the moment of inertia of
the disk is given by:
I =
½ m R 2
And we have the
differential equation:
x’
- R q’ = 0
Which when
integrated yields:
x - R q =
const. = 0
The kinetic energy
of the system can then be written:
T
= ½ m x’ 2 + ½ I q’ 2
Or: T =
½ m x’ 2 +
¼ m R 2 q’ 2
The potential
energy will be expressed:
V = mg
(L – x) sin f
So the Lagrangian
of the system can be written:
L = T –
V =
½ m x’ 2 + ¼ m R 2 q’ 2 - mg
(L – x) sin f
Using the angular
–linear relation seen earlier, e.g.
x’ = R q’
We can simplify
further:
L = ¾ m x’ 2 - mg (L – x) sin f
Then it can be
shown:
d/dt (¶ L/¶ x’ )
- ¶ L/¶ x = 0
Working:
3/2 m x”
- m g sin f = 0
And: q” = 2/3 g sin f /R
Lagrangian Multipliers:
The Lagrange
equations can be rewritten in terms of multipliers, i.e.
d/dt (¶ L/¶ q’ k) - (¶ L/¶ q k) =
l 1 (¶f1/¶ q
k ) - l 2 (¶f2 /¶
q k) + …..
Where l 1 and
l 2 denote
Lagrange multipliers, with one
multiplier per each equation of constraint.
Example: Find the Lagrangian multiplier for the system discussed above:
We have for the
relevant partial differential equations:
i)
d/dt (¶ L/¶ x )
- ¶ L/¶ x - l ((¶f/¶x ) = 0
ii) d/dt (¶ L/¶ q’ ) - ¶ L/¶ q
- l ((¶f/¶q ) = 0
From which we
obtain:
i)
mx”
- m g sin f - l (1)
= 0
ii)
½ m R 2 q”
+ l
R = 0
We find from the
preceding equations:
l
= - ½ m R 2 q” = -
½ mx”
But:
x” =
2/3 g sin f
And: q”
= 2/3 g sin f /R
Hence:
l
= - ½ m(2/3 g
sin f )
Or: l =
- m g sin f / 3
Problems:
1). Find
the Lagrangian for the double pendulum shown below for the least number of coordinates possible.
2) Consider a cone and particle situated on its inside surface:
with a force : F = - mg k exerted:
Let the potential
energy V = mg z
And: z = ar
The equation for
the total energy is given by:
T
= ½ m( r” 2 + r
q’ 2 + z ' 2 )
Find:
Find:
a)The
Lagrangian after applying constraints and
eliminating one coordinate (z).
b)The new Lagrange
equations, viz.
d/dt (¶ L/¶ r’ )
- ¶ L/¶ r = 0
d/dt (¶ L/¶ q )
- ¶ L/¶ q = 0
c) The force acting along the radial direction.
Show how this is
obtained from the preceding equations and an undetermined multiplier l.
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