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Help setting up integral

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data
    This is actually just a minor part of a larger problem - I need to find the potential energy of a string of mass m and length L that is hanging over the edge of a table.


    2. Relevant equations



    3. The attempt at a solution
    If we define V = 0 at the level of the talbe, then the potential energy of a mass element dm below the able is given by V = -gydm where y is the height of dm below the table. But here I blank - how can I use this to find the total potential energy o hte cord?
     
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  3. Nov 15, 2009 #2

    LCKurtz

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    What about thinking in terms of a mass density [itex]\delta[/itex] so the potential of a string segment of length dy is -[itex]\delta[/itex]ydy and integrating with respect to y?
     
  4. Nov 16, 2009 #3
    I'm not sure I follow

    the liear mass density would be m/L assuming that it is uniform, but how can I turn that into the integral?
     
  5. Nov 16, 2009 #4

    LCKurtz

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    For each segment dy of cable that is hanging over the edge a distance y it's potential is [itex]-\delta y\,dy[/itex]. You have to add all these up, which you do by integrating with appropriate y limits.
     
  6. Nov 16, 2009 #5
    So we get:

    [tex]\int_0^y(\frac{-mg}{L}y)dy = \frac{-mg}{2L}y^2[/tex]

    Is that right, assuming that a length y is hanging over the edge?


    The only problem is that I am not explicitly given L in the question, so I am not sure if I can use it. Is there a way to get the potential of a string of mass m hanging a distance y over the edge of a table without using the length? I can't think of anything.
     
  7. Nov 16, 2009 #6

    LCKurtz

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    y is the variable. You don't want it in the upper limit. If h is the length of the cable hanging over the edge your integral would go from 0 to h.
     
  8. Nov 17, 2009 #7
    Alright - if you replace y by h in the above, is it correct? ^_^
     
  9. Nov 17, 2009 #8

    LCKurtz

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    It looks OK to me.
     
  10. Nov 17, 2009 #9
    Thanks :)

    Now that it's done, I see you can get the same thing without the integral by using the centre of mass of the part of the cord that is hanging over the edge.
     
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