Uniform ladder sliding on smooth surface

[jncarter]

Homework Statement

A uniform ladder of length L and mass M has one end on a smooth horizontal floor and the other end against a smooth vertical wall. The ladder is initially at rest in a vertical plane perpendicular to the wall and makes an angle $\theta$₀ with the horizontal.

(a) Write down the Lagrangian and derive the equations of motion.

(b) Prove the ladder leaves the wall when its upper end (call it y') has fallen to a height of $\frac{2}{3}Lsin\theta$₀.

Homework Equations

L = $\frac{1}{2}$mV²+$\frac{1}{2}$I$\omega^2$ - U
U = -mgh, where h is the height from the horizontal.
$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}$

The Attempt at a Solution

I think I can just look at U on the center of mass, located at L/2. Right now I'm using y'/2, where y' depends on $\theta$, which is a function of time. So h = $\frac{1}{2}Lsin/theta$.
V is the velocity of the center of mass. CM = $\frac{L}{2} (sin\theta + cos\theta)$
I = $\frac{mL^2}{12}$ and I think $\omega$ is just $\dot{\theta}$
I haven't gotten to part (b) yet, I figure once I have the first part, the second will be easy.
I'm just looking to see if I'm going in the right direction here and if there is anything I am missing. I just seem to have a hard time with rigid bodies. Thanks for any help!

 

[Matterwave]

You need to include the constraints.

 

[jncarter]

$\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \Sigma \lambda \frac{\partial f}{\partial q}$

Or are you looking for the actual constraint? See, I've been trying to decide how best to approach that part of the problem. There is the most obvious one $L^2 = y'^2 + x'^2$, which can be written in the CM frame. Are there any other constraints? Also I haven't quite figured out how to use Lagrange multipliers (as it seems you are suggesting) to determine the condition for the ladder's leaving the vertical wall.