What is the Force Between Links in a Falling Chain?

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SUMMARY

The discussion focuses on the dynamics of a uniform chain of length l and mass M as it falls freely from a height, specifically analyzing the forces between the links and the time taken for the topmost link to reach the table. The time to reach the table is calculated using the formula sqrt(2s/a), derived from the equations of motion v=u+at and v^2 = u^2 + 2as. The tension in the chain is also examined, leading to the equation T=Mg-Ma, where T represents tension, M is the mass of the chain, and a is the acceleration. The complexities of falling chain problems are highlighted, indicating ongoing research in this area.

PREREQUISITES
  • Understanding of classical mechanics principles, particularly Newton's laws of motion.
  • Familiarity with kinematic equations, specifically v=u+at and v^2 = u^2 + 2as.
  • Knowledge of tension forces in a system of connected bodies.
  • Basic grasp of variable mass systems and their dynamics.
NEXT STEPS
  • Study the dynamics of variable mass systems in classical mechanics.
  • Research the concept of tension in falling objects and its implications on acceleration.
  • Explore advanced kinematic problems involving chains and linked bodies.
  • Read the paper "Falling chains as variable mass systems: theoretical model and experimental analysis" by C.A. de Sousa et al. for deeper insights.
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Students of physics, educators teaching mechanics, and researchers interested in the dynamics of falling objects and variable mass systems will benefit from this discussion.

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Homework Statement


A uniform chain of length l and mass M contains many links. It is held above a table so that one end is just touching the table top. The chain is released freely. What is the force between the links? What is the time for the topmost link to fall to the table?

Homework Equations



v=u+at, v^2 = u^2 + 2as

The Attempt at a Solution


I think that the time to reach the table is sqrt(2s/a), because v=at, so (at)^2=2as, so rearranging that would give me that answer. I'm trying to visualise the part about the force though; is it that the only force exerted on each chain the weight of the chain below it? Or is it tension, in which case Mg-T=Ma, so T=Mg-Ma?
 
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Falling chain problems are still somewhat of an open research area because the correct model is not entirely clear. Here is a paper on the topic:

"Falling chains as variable mass systems: theoretical model and experimental analysis," by C.A. de Sousa, P.M. Gordo, and P. Costa, Physics Education, 28 May, 2012.

The particular problem stated by the OP is not addressed in this paper, but some of the relevant modeling questions are.
 
PhysicsKid99 said:

Homework Statement


A uniform chain of length l and mass M contains many links. It is held above a table so that one end is just touching the table top. The chain is released freely. What is the force between the links? What is the time for the topmost link to fall to the table?

Homework Equations



v=u+at, v^2 = u^2 + 2as

The Attempt at a Solution


I think that the time to reach the table is sqrt(2s/a), because v=at, so (at)^2=2as, so rearranging that would give me that answer. I'm trying to visualise the part about the force though; is it that the only force exerted on each chain the weight of the chain below it? Or is it tension, in which case Mg-T=Ma, so T=Mg-Ma?
Suppose there is some tension in the chain as it falls. What does that tell you about the vertical acceleration of the top link? What does it tell you about the vertical acceleration of the bottom link that's not yet in contact with the table?
 

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