Why Does a Sliding Plank Lose Contact with the Wall at Two-Thirds Its Height?

[Zeebo]

Homework Statement

A Planck of length L2 leans against a wall. It starts to slip downwards without friction. Show that the top of the plank loses contact with the wall at 2/3 of it's original height.

Homework Equations

F*R = torque
Moment of Inertia of Rod (I) = 1/3 M4L^{2}
Mgh = \frac{1}{2}MV^{2} + \frac{1}{2}I\omega^{2}

The Attempt at a Solution

I've been trying this problem for over two hours now and can't seem to get it to work. My initial idea was to work with torques, but I didn't seem to be getting anywhere so I tried the Mgh formula above, setting h as the height of the center of mass (1/2h) and trying to solve for h knowing that at the point the rod leaves the wall the horizontal velocity should be equal to the horizontal component of the radial velocity. So far I can't seem to get the equations to work out, any help would be very much appreciated.

The problem also has a hint:
Only a single variable is needed to describe the system. Note the motion of the center of mass.

 

[kuruman]

The hint is very valuable. Can you find what type of curve the center of mass describes while the plank is in contact with the wall? If not, assume that the plank never loses contact and draw a sequence of planks at different angles and connect their midpoints.