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Finding time taken to fall off table

  1. Apr 30, 2009 #1
    1. The problem statement, all variables and given/known data
    A uniform chain of length L lies on a smooth table with part of its length b initially hanging over the edge. Show that the time required for it to fall off the table when released from rest is given by

    t = √(L/g) arccosh (L/b)


    2. Relevant equations
    http://img10.imageshack.us/img10/2203/integral.png [Broken]


    3. The attempt at a solution
    Not sure how to begin, since the acceleration of the chain keeps increasing as b increases. b is in turn related to the acceleration of the chain.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 1, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Arctic! Welcome to PF! :smile:

    (have a square-root: √ and try using the X2 tag just above the Reply box :wink:)
    Either use good ol' Newton's second law, or (slightly quicker, since it gives you a first-order equation :wink:) simply use conservation of energy. :smile:

    (But if you use conservation of energy, don't forget you have to integrate to find the PE)
     
  4. May 2, 2009 #3
    Using conservation of energy I get:

    eq.latex?Initial%20PE%20=%20mg[(\frac{L}{2})(\frac{L-b}{L})+(\frac{L-b}{2})(\frac{b}{L})].gif
    This is by breaking the chain into 2 parts: the part of length L-b lying on the table and the part b over the edge. I also take PE = 0 at height L/2 below the table to simplify calculations later

    When the chain has fallen off the table, PE = 0 (since its center of gravity is at L/2 below the table) and
    eq.latex?PE%20=%200.5mv^{2}=%20mg[(\frac{L}{2})(\frac{L-b}{L})+(\frac{L-b}{2})(\frac{b}{L})].gif

    eq.latex?\Leftrightarrow%20v^{2}=%20g[(L-b)+(\frac{b(L-b)}{L})]=g(L-b)(1+\frac{b}{L}).gif
    Not sure how to continue from here, assuming this is right that is.

    As for Newton's 2nd law..
    F=ma
    mgb/L = ma
    a = gb/L
    dv/dt = gb/L
    d2b/dt2 = gb/L
    dt2 = d2b L/gb

    Again, I am stuck here
     
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