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Chain falling on weighing scale-find total reading

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data
    A chain of length l, mass M falls on a weighing scale vertically down. We need to find the reading of the scale when a length 'x' has fallen on the scale.


    2. Relevant equations
    F= dP/dt

    W=Mg
    dm= (M/l)x
    dm/dt=m(v^2)/l


    3. The attempt at a solution
    So initially, I used this approach:
    F=dP/dt.
    P=Mv, where M is instantaneous mass of rope fallen on the scale.

    So, dP/dt=Mdv/dt+vdM/dt.
    dv/dt=g, as the rope falls under the influence of the rope.

    dM/dt=mv^2/l
    Since the centre of mass of the rope has fallen by an amount x/2, the velocity of that portion of rope will be (gx)^(1/2) using energy conservation.

    So, F=2Mgx/l.
    But this is only impulsive force. To this, we need to add the weight of the body.
    So, F=3Mgx/l.

    A few doubts I had with this approach:
    1) Can we indeed conserve mechanical energy, since normal component of weight is acting upon the rope which hits the ground?
    2)If we can, can we conserve the mechanical energy by saying that the change in potential energy of centre of mass of rope (Mgx/2) = (1/2)Mv^2? (This is because, the velocity that you get by this = root (gx). If we just say since the body falls under influence of gravity, it would be root (2gx).
    3) And this is the most important bit. When we say that F = M dv/dt + vdm/dt, does this include the weight of the object, since dv/dt=g?

    (i.e. is the impulsive force of fall ONLY due to the changing mass or also due to changing velocity?)
    Suppose we have an object which has constant mass. We keep it on the table, without any velocity. IT exerts a force=Mg on the table.
    Now suppose we take it to some height and leave it. It hits the table with some impulsive force greater than Mg. But, dm/dt is 0 and dv/dt is g, since it is free fall. So where does this impulsive force come from? Is it separate from the weight of the object that it would normally exert on the table??
    Another approach: v=root (2gx) for falling body, dv/dt is not considered in the impulsive bit. Therefore, F=mv^2/l, and F=@Mgx/l. To this we add weight. So total reading = 3Mgx/l.

    Which of these approaches is correct, and why?
     
  2. jcsd
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