What is the Lagrangian and equation of motion for a sliding rope on a table?

[stunner5000pt]

But *must* be done using algrangian mechanics
(because its so much fun taht way)
A uniform rope of length l and mass m lies on a horizontal table with one side just barely over the edge. The rope is given a nudge, taht is an initial speed Vo and begins to slide off
Assuming that friction is negligible, determine the Lagrangina of this system and the resulting equation of motion

$$T = \frac{1}{2} m \dot{r}^2$$
the force of gravity acts on the mass hanging off the edge only so
$$F = \frac{my}{L}g = - \nabla V(y)$$
$$V(y) = \frac{mgy^2}{2L}$$
can r be found in terms of y, though?

any help is greatly appreciated!

 

[StatusX]

You haven't made clear what r is. If it's the distance traveled by some point on the string, as it must be (up to a constant) for your formula to be correct, then the relation to y should be obvious.

 

[stunner5000pt]

r would be the direction vector for any point on the string

so fr example the right end of the rope at t=0 is y=0
then r'(0) = v0
after time t the end opf hte rope has traveled v0 + gt

so r(t) = v0 t + gt^2 / 2?
r'(t) = v0 + gt
is that correct?

 

[tomkeus]

Divide lagrangian in two parts such that L=Lh+Lt where Lh is lagrangian of piece of rope that is hanging over the edge and Lt is part of rope lying on the table. It it quite simple. If you have any trouble with that, just yell.