Uniform chain of length L is kept of a horizontal table in such a way that l of its length keeps hanging from the table. If the whole system is in equilibrium, find the Lagrangian of the system.
Lagrangian of the system = Kinetic energy (T) - Potential energy (V)
T = (mv^2)/2
V = mgh
The Attempt at a Solution
In this case, V = 0 - mgl = -mgl
My question is what will be the kinetic energy of the system ?
I think as the system is in equilibrium, there is no motion in any portion of the chain. So, K.E. = 0
Am I wrong ?