Solving Double Pendulum Motion w/ Hit on Lower Rod

[JohanL]

If you have have double pendulum of two rods where the rods moves with the same frequency w at the equilibrium position and at the eq. position you hit the lower rod at a distance d from the point that connects the rods. The hit results in a motion where the upper rod moves with frequency w and the lower rod with frequency -w. If the rods lengths are l and their masses m what are the the distance d.

I have no problem to find the equations of motion for the ordinary double pendulum. With the hit i think it gets something like

Lagrange equations = F

Where F is different for theta1 and theta2.
If you find these equations you can solve for which d the resulting motion is w for the upper rod and -w for the lower rod, right?
But how do you write F?

Thank you

 

[Crosson]

You need to right down the lagrangian:

L = T - U

Where T is the kinetic energy and U is the potential energy.

In this case, the lagrangian will be a function of four variables:

$$L(\theta_1 , \theta_2 , \theta_1 ', \theta_2' )$$

Where primes denote differentiation with respect to time. So, after you have expressed the lagrangian in terms of those four variables, your equations of motion are the Euler-Lagrange equations (one for each position variable, so two total).

 

[JohanL]

Thank you.
But as i tried to say the problem for me is to find the generalized impulse F.
Because Lagrange equations leads to
F = Generalized impulse = change in generalized momentum
If I find F the problem is solved.
and how do i get the distance d into F?