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Falling chain center of mass dynamics

  1. Mar 22, 2010 #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
    Last edited: Mar 22, 2010
  2. jcsd
  3. Mar 22, 2010 #2


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    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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