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Chain hanging from a table

  1. Sep 27, 2014 #1
    1. The problem statement, all variables and given/known data
    There is a chain of uniform density on a table with negligible friction. The length of the entire chain is 1 m. Initially, one-third of the chain is hanging over the edge of the table. How long will it take the chain (in seconds) to slide off the table?

    2. Relevant equations


    3. The attempt at a solution
    I tried to use conservation of energy.
    Initially position of CM of hanging part of chain wrt to table is 1/3.
    And mass of part of chain hanging down the table is 2m/3.
    So ##{ E }_{ pi }=\frac { 2mg }{ 9 } ##
    If the chain falls by distance '##x##'
    then position of CM of hanging part of chain wrt to table will be ##(3x+2)/6##
    And mass of part of chain hanging down the table will be ##(3x+2)m/3##
    So ##{ E }_{ px }=\frac { { (2+3x) }^{ 2 } }{ 18 } mg##
    The potential energy is converted into kinetic energy.

    So ##\frac { { (2+3x) }^{ 2 } }{ 18 } mg-\frac { 2mg }{ 9 } =\frac { 1 }{ 2 } \times \frac { (2+3x)m }{ 3 } { v }^{ 2 }##

    on simplifying I got
    ##\frac { { (2+3x) }^{ 2 } }{ 18 } mg-\frac { 2mg }{ 9 } =\frac { 1 }{ 2 } \times \frac { (2+3x)m }{ 3 } { v }^{ 2 }##

    or ##\frac { 1 }{ { v }^{ 2 } } =\frac { (3x+2) }{ x(3x+4)g } ##

    or ##\int { dt } =\int { \sqrt { \frac { (3x+2) }{ x(3x+4)g } } dx } ##
    Is this correct till here?
     
  2. jcsd
  3. Sep 27, 2014 #2

    Orodruin

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    This would bean the CoM of the hanging part would initially be at 1/3 m, i.e., at the end of the hanging part.

    The same thing here, initially the chain would have 2/3 of its mass hanging down the side.
     
  4. Sep 27, 2014 #3
    I understood that.Is my last equation in #post1 wrong(I think so),if yes then could you please explain why is it wrong.
    How should I proceed to solve this question.
     
  5. Sep 27, 2014 #4
    EDIT: (I don't think so):(
     
  6. Sep 27, 2014 #5
    I would do this differently. Work out the problem generically, then plug in the details. For example. Let ##x## denote the vertical coordinate of the chain's free end. The potential energy of the chain is ##U(x)##. The mass of the moving part of the chain is ##m(x)##. Conservation of energy then gives $$U(x_0) - U(x) = {m(x) v^2 \over 2}.$$ Which yields $$ t = \int\limits_{x_0}^x dx \sqrt{m(x) \over 2\left[U(x_0) - U(x)\right]}.$$ Now express ##U(x)## and ##m(x)## correctly and you have the problem solved.
     
  7. Sep 28, 2014 #6

    ehild

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    I agree, but I imagine that the whole chain moves, so m does not depend on x. KE=1/2 mv2. The potential energy depends on the hanging mass, which is mx/L. Initially L/3 is the hanging length.

    ehild hangingchain.JPG
     
    Last edited: Sep 28, 2014
  8. Sep 28, 2014 #7
    As the chains free end is at distance ##x## so the CoM of chain will be at a distance ##\frac { x }{ 2 } ##.
    As the mass is distributed uniformly so mass of chain is ##\frac { M }{ L } x##

    So ##M(x)=\frac { M }{ L } x##

    and ## U(x)=\frac { M{ x }^{ 2 } g}{ 2L } ##
     
  9. Sep 28, 2014 #8
    Unfortunately, it is not clear from the initial description whether the entire chain moves, or a part of it. Satvik must clarify this.
     
  10. Sep 28, 2014 #9
    How come the potential energy increases as ##x## increases?

    Please see my previous message, too.
     
  11. Sep 28, 2014 #10

    ehild

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    Anyway, the length of the chain is constant and all its parts must reach the edge somehow. It can be said that the whole thing is piled up at the edge of the table and has zero volume but it is far from being physical. You are right, it must have been specified by the problem writer.
    I used to give the same problem to my pupils, but the chain was a snake, and I added the picture.

    ehild
     
  12. Sep 28, 2014 #11

    ehild

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    The potential energy is negative. Add a minus sign.
     
  13. Sep 28, 2014 #12
    Oh! Sorry as chain comes down its potential decreases so there should be minus sign.

    As ##\triangle U=\frac { 1 }{ 2 } m{ v }^{ 2 }##

    So ##\frac { 2mg }{ 9 } -\frac { m{ x }^{ 2 }g }{ 2 } =\frac { m{ v }^{ 2 } }{ 2 } ##

    or ##\frac { 2g }{ 9 } -\frac { { x }^{ 2 }g }{ 2 } =\frac { { v }^{ 2 } }{ 2 } ##
     
  14. Sep 28, 2014 #13
    Sorry.:s
     
  15. Sep 28, 2014 #14
    Snake instead of chain!:eek:.
    Nice substitution.:)
     
  16. Sep 28, 2014 #15

    haruspex

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    If the chain is heaped up at the edge of the table initially, or if it is laid out straight, but you are going to ignore the horizontal velocity of the descending part, then you cannot use conservation of energy. (This problem is come up so many times on this forum, in different guises.). Use momentum.
    If you want to take it as initially straight and incorporate the horizontal movement in descent, the problem becomes very difficult. You need to consider the shape of the descending arc in order to get the PE right. See 'chain fountain'.
     
  17. Sep 28, 2014 #16
    Where does 2/9 come from?

    This is hardly a clarification.
     
  18. Sep 28, 2014 #17
    Is this what you intended to say? Frankly, I do not see why (I assume the volume of the chain is negligible).
     
  19. Sep 28, 2014 #18

    Initially ##2/3## of chain is hanging from the table sa ##-\frac { m{ x }^{ 2 }g }{ 2 } ## becomes ##2mg/9##
    also L(length of the chain is 1m).
     
  20. Sep 28, 2014 #19
    Really? Then why does the problem say "one-third of the chain is hanging over the edge of the table"?
     
  21. Sep 28, 2014 #20
    I saw this question online on a website.I think this is poorly stated problem.However ehild's explanation in #post6 sound good to me.
     
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