Lagrangian for two identical rods connected by frictionless joint.

[center o bass]

Homework Statement

Hello. I have a problem with setting up the lagrangian for a system here.
The problem is stated at page 8 problem 2.3 with a diagram at the following

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2. The attempt at a solution
I used two generalized coordinates corresponding to the angle between the first rod
and the roof as well as the smalles angle between the two rods.
My probem here lies in finding an expression for the kinetic energy.
The second rods velocity is dependent on the motion of first rod so I don't see any simple
moment of inertia relations to use here.

My first attempt was to express the possition of a little mass element dm on the second rod in terms of a vector in cartesian coordinates and then differentiate and square it to get that elements kinetic energy followed by integrating along the length of the rod to sum up the kinetic energy for each mass element. I then ended up with the following expression for the kinetic energy of the second rod:

[tex] K_2 = \frac{1}{2}m \left( \frac{1}{12}l^2 (\dot{\phi} + \dot{\theta})^2 + l^2 \dot{\theta}^2\right).[/tex]

I'm not too confident here. And even if the expression is right, I suspect there is an easier way to do it. Any suggestions?

 

[Päällikkö]

You've identified the two angles as the two degrees of freedom. You should therefore not think in terms of movement of the rods, but rather in terms of these angles. Clearly the rotations about the angles are independent (the bottom rod should rotate at a constant speed, no matter what the top angle does).

Try using the parallel axis theorem to calculate the moment of inertia.