Falling chain center of mass dynamics

  • #1
Not sure if this is in the right subforum but:

A chain of uniform mass density, length b, and mass M hangs in a loop from the ceiling (both its ends are adjacent to each other) At time t = 0, one end, end B is released. Find the tension in the chain at point A after end B has fallen a distance x by a) assuming free fall b) by using energy conservation


This is an example in Marion/Thornton's Classical dynamics of systems and particles.
They start of by stating that the time derivative of the center of mass momentum is equal to the weight minus the tension. I do not fundamentally understand this. Why is the tension not just equal to the instantaneous weight of the side of the chain that is immobile???

In other wods, I tried to use this:
T = (M+[tex]\rho[/tex]dx)g



Sorry, I'm not sure if this is in the right subforum... ?

but the book starts off with

[tex]\dot{P}[/tex] = Mg - T


What is the physical significance of that equation? I think I understand it sort of... but doesn't this mean that the downwards force could be greater than the tension... and then that the chain would fall off the ceiling? I don't really understand the physical significance of this
 
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Answers and Replies

  • #2
gabbagabbahey
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What is the physical significance of that equation? I think I understand it sort of... but doesn't this mean that the downwards force could be greater than the tension... and then that the chain would fall off the ceiling? I don't really understand the physical significance of this

The chain is falling...sort of. At [itex]t=0[/itex] one end is released and the chain begins to uncoil as that end falls downwards. During that free-fall period, the center of mass of the chain will be moving downwards, so the tension is not equal to the weight.
 

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