Lagrangian of System: K.E.=0, V=-mgl

[Molar]

Homework Statement

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.

Homework Equations

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 ?

 

[Dilemma]

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[Molar]

What was that @Dilemma Can you please fill in the blanks ?

 

[Dilemma]

I tried to solve your problem calculating the forces acting on the system but then I noticed that you were using the energy changes of the system. However, if I were you, I would introduce a friction force of μ to the system. There is no way this system can stay like that.

 

[hilbert2]

There's no way for that system to be in equilibrium unless there's static friction or if the chain is being pulled from the left side end. If you want to use the formula $V = mgh$ for the potential energy, you must interpret $h$ as the vertical (z) coordinate of the center of mass of the chain. If we set $z=0$ on the top of the table, what is the $z$ coordinate of the CM?

 

[Molar]

In that case, the position of the CM will be -l/2 k hat ,right ?

Yeah friction force is holding the chain above the table in its position, and that is F = μm₁g, where m₁ is the mass of the chain above the table
And the corresponding energy will be E = μm₁g(L-l), right ?