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Classical Dynamics -- Falling chain and energy conservation

  1. Nov 26, 2015 #1
    1. The problem statement, all variables and given/known data

    The statement of the question is:


    A chain of uniform linear mass density ##\rho##, length ##b## and mass ##M## hands as shown in the figure below. At time t=0, the ends A and B are adjacent, but end B is released. Find the tension in the chain at point A after end B has fallen a distance x by using energy conservation.



    Refer to the figure:

    http://i.stack.imgur.com/RGmud.jpg


    I can solve this example (from Marion and Thornton Classical Dynamics 5e) easily from assuming a free fall and summing force. I however do not follow the example's solution using conservation of energy.


    The example states:


    We treat the chain as one-dimensional motion ignoring the small horizontal motion. Let the potential energy ##U## be measured relative to the fixed end of the chain so that the initial potential energy ##U(t=0)= U_0 = -\rho g b^2 /4##. A careful geometric construction shows that the potential energy after the chain has dropped a distance ##x## is


    ##U=-\frac{1}{4} \rho g (b^2 +2bx - x^2)##


    I can reproduce the initial potential energy via integration, that is to say, with ##U(x=0)=0##


    ##U=2mgh=2g\rho \int_{\frac{b}{2}}^{0} x dx =- \rho g b^2 /4##


    but am still unable to reproduce the falling chain potential. Can anyone help me derive the expression for the potential energy as the chain falls,

    ##U=-\frac{1}{4} \rho g (b^2 +2bx - x^2)##


    Thank you.


    2. Relevant equations


    ##U=mgh##


    3. The attempt at a solution



    Let ##U(x=0) = 0##. The initial potential is derived above. When the chain falls by an amount x on the B side, the same length ##x## is added to the A side.


    On the ##A## side:


    ##U=mgh = g \rho \int_{\frac{b}{2} + x}^{0} x dx = -\frac{g\rho (x+b/2)^2}{2}##


    Similarly, on the ##B## side:


    ##U=mgh = g \rho \int_{\frac{b}{2} - x}^{x} x dx =\frac{g \rho [ x^2 - (b/2 -x)^2 ]}{2}##


    We next sum these two expressions yielding:


    ##U = \frac{-g\rho (x+b/2)^2 + g \rho [ x^2 - (b/2 -x)^2 ]}{2}##


    which, after expanding we see the ##bx## terms cancel, and therefore does not reproduce the solution
     
    Last edited: Nov 26, 2015
  2. jcsd
  3. Nov 26, 2015 #2

    andrewkirk

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    The length of the A side is ##\frac{b+x}{2}##, not ##\frac{b}{2}+x##. The length of the B side adjusts accordingly.

    Make these changes and it will work.
     
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