More Theoretical Mechanics - Holonomic Constraints and Lagrangian Multipliers

More complex aspects of a kind of general "circular" motion can now be examined along  with computing their Lagrangians. (See e.g. the previous two posts on theoretical mechanics and circular motion. )  Consider the diagram below for a body rolling without slipping:

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:

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.