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.
The system includes two masses, m1 and m2, at two angles to the vertical, f1 and f2, respectively.
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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