Revisiting Lagrangian Dynamics (2)

Continuation... 

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

 Using the polar coordinate unit vectors (n,, l) this can be rewritten as:

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)

 Where the last derivative can be evaluated from:

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

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.

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

Working:

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

 

Problems:

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