# Lagrangian of a falling rod

**1. The problem statement, all variables and given/known data**

A rigid rod of length ##\ell## and mass ##m## has its lower end in contact with a frictionless horizontal floor. Initially, the rod is at an angle ##\alpha_o## to the upward vertical when it is released from rest. The subsequent motion takes place in a vertical plane.

a)Taking as generalized coordinates ##x##, the horizontal displacement of the centre of the rod and ##\alpha##, the angle between the rod and the upward vertical, show that the Lagrangian is $$L = \frac{1}{2}m\dot{x}^2 + \frac{1}{24}m\ell^2 \dot{\alpha}^2 (1+3\sin^2 \alpha) – \frac{1}{2}mg\ell \cos \alpha$$

b)Deduce the Euler-Lagrange equations, showing that ##x## remains constant in the motion.

**2. Relevant equations**

Lagrange’s Equations, kinetic and potential energies.

**3. The attempt at a solution**

a)I don’t see why ##x## and ##\alpha## are independent coordinates. If ##\ell/2## is the distance of the centre of the rod from the contact point, then it seems to me that the x component of the centre of the rod can be adequately described by simply ##\frac{\ell}{2}\sin \alpha##. Since they are related, I don’t see why they can both be generalized coordinates in the problem.

b)I solved for the EoMs and did obtain that ##x## is a constant. But I can’t seem to picture this. Initially, ##x_{\text{centre of rod}} = \frac{\ell}{2}\sin \alpha_o## As ##\alpha## increases, ##\sin \alpha## increases since it makes sense for ##0 \leq \alpha \leq \pi/2 ##. Therefore ##x## must increase.

Perhaps I am misinterpreting something in the question.

Many thanks

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