It is also useful to consider the unit vectors associated with polar coordinates in central force problems: n pointing in direction of increasing r, and l, in the direction of increasing q .
The velocity components can then be written:
v r = dr/ dt, v q = r dq /dt
v = dr/ dt n + r dq /dt l
For the change in
the radial coordinate alone
v = dr/ dt = d(r n)/ dt = ( dr/ dt n + r dn /dt)
dn /dt = dn / dq (dq /dt)
To evaluate dn / dq one makes use of the fact that by radian measure of angles:
D n = D q and in the limit as D q ® 0,
‖D n ‖ / ‖D q ‖ = 1
So: dn / dq = l
i.e. As D q ® 0, D n becomes perpendicular to n and assumes the direction of l .
In an analogous way we may obtain the relation:
dl / dq = -n
Finally, we have: dn /dt = (dq /dt) l
And: dl /dt = - (dq /dt ) n
The Nature of Constraints
Consider the diagram below:
We may write: dS/ dt - R (dq /dt) = 0
Or: S’ - R q’ = 0
Which implies: å c a q’ a = 0
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.
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
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
3/2 m x” - m g sin f = 0
And: q” = 2/3 g sin f /R
Use of 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 Problem: Find the Lagrangian multiplier for the system discussed above:
Solution: 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
1) Find the Lagrangian of the double pendulum shown in the sketch below which includes two masses, m1 and m2, at two angles to the vertical, f1 and f2.
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 )
a)Obtain the Lagrangian then apply constraints and eliminate one coordinate (z).
b)Write the new Lagrange equations, viz.
d/dt (¶ L/¶ r’ ) - ¶ L/¶ r = 0
d/dt (¶ L/¶ q ) - ¶ L/¶ q = 0
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