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

• Molar
In summary, the problem involves a uniform chain of length L hanging from a horizontal table, with a portion of its length l hanging off the table. The system is in equilibrium and the task is to find the Lagrangian of the system. The Lagrangian is equal to the kinetic energy minus the potential energy. In this case, the potential energy is equal to -mgl, where m is the mass of the chain and g is the acceleration due to gravity. The kinetic energy is equal to zero since the system is in equilibrium and there is no motion. However, it is suggested to introduce a friction force μ to the system, which would affect the calculations. The position of the center of mass of the chain is -l/
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 ?

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

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 ?

## What is the Lagrangian of a system where the kinetic energy is 0 and the potential energy is -mgl?

The Lagrangian of a system is a function that represents the total energy of the system, taking into account both kinetic and potential energy. In this specific case, where the kinetic energy is 0 and the potential energy is -mgl, the Lagrangian would simply be -mgl.

## How is the Lagrangian of a system related to the equations of motion?

The equations of motion for a system can be derived from the Lagrangian using the Euler-Lagrange equations. These equations describe the evolution of the system over time and can be used to predict its future behavior.

## What does it mean for the kinetic energy to be 0 in a system?

When the kinetic energy of a system is 0, it means that there is no motion or movement within the system. This could be due to the system being at rest, or all of its particles having the same velocity and canceling each other out.

## Why is the potential energy in this system equal to -mgl?

In this particular system, the potential energy is equal to -mgl because it is a simple pendulum in a gravitational field. The potential energy in a pendulum system is dependent on its mass, gravitational acceleration, and height above the ground, which in this case is given by the length of the pendulum.

## How does the Lagrangian of a system change if the kinetic and potential energy are not constant?

If the kinetic and potential energy of a system are not constant, then the Lagrangian will also change. This can occur if external forces are acting on the system, or if there are changes in the system's parameters, such as mass or position. In such cases, the Lagrangian will be a function of time and the system's variables, and the equations of motion will also be time-dependent.

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