Monday, May 10, 2021

Solutions To Lagrangian Dynamics Suggested Problem

 Suggested Problem:  Hamilton's principle of least action applies given the missile trajectory is a parabola  which motion is made in least time, i.e.  for which  the action is a minimum:

S    =   ò  t2   t1      [½ m  dx(t) 2/ dt 2    -  mg x(t)] dt

Or:  Letting y =  x =  x(t) to obtain the least time context:


S    =   ò  t2   t1      [½ m  dx 2/ dt 2    -  mgx] dt

Say between time t = t1 and t = t2.  The path taken is always such that we have a least difference between kinetic energy and potential energy  on average.  Hence,

ò  t2   t1  [T  -  V] dt  =  min

I.e. the action is a minimum for the (true, i.e. x) path that satisfies the differential equation:  

[- m  dx 2/ dt 2    -  V' x] =  0   

 We see at missile launch the KE is  high then becomes zero at apex, while PE is zero initially then becomes a maximum at apex. Throughout the trajectory the balance of PE (V) and KE (T) changes but the average remains a minimum to comport with least time, i.e. from launch point to end point.  

If y = A sin x defines the true (x) path then another expression representing a variation on it might be:  y =  3A sin  x  or in the least time coordinates: e.g.  x(t) = 3 A sin wtGiven a small variation h   then  the variation in the action would be:

 dS =  ò  t2   t1      [½ m (dx / dt) (d h/ dt)     - h V' x] dt   

where x defines the true or natural path.

The Lagrangian for the natural path (x)  is:

L   =  ½ m  dx 2/ dt 2    -  mg x

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