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

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Homework Help Overview

The problem involves a uniform chain of length L positioned on a horizontal table, with a portion of its length (l) hanging off the edge. The original poster seeks to find the Lagrangian of the system, given that it is in equilibrium, and has noted the potential energy as V = -mgl. The inquiry centers around the kinetic energy of the system under these conditions.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster questions the kinetic energy of the system, reasoning that since it is in equilibrium, the kinetic energy should be zero. Other participants challenge this assumption, suggesting that friction or external forces may be necessary for equilibrium.

Discussion Status

The discussion is active, with participants exploring different interpretations of equilibrium and the forces at play. Some have suggested introducing friction to maintain the system's state, while others are clarifying the potential energy calculations based on the center of mass of the chain.

Contextual Notes

Participants are considering the implications of static friction and the position of the center of mass in their calculations. There is an ongoing examination of the assumptions regarding the system's equilibrium and the role of external forces.

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

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