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Homework Help: Change in potential energy problem

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data
    A uniform cord of length 25cm and mass 15g is initially stuck to a ceiling. Later, it hangs vertically from the cieling with only one end still stuck. What is the change in the gravitational potential energy of the cord with this change in orientation? (Hint: Consider a differential slice of the cord and then use integral calculus)


    2. Relevant equations
    Ug=mgh


    3. The attempt at a solution
    Well in calculus we have only recently started integrals so my understanding of integrals is shaky at best, so i am not really sure how to apply them to this problem... any push in the right direction would be appreciated
     
  2. jcsd
  3. Feb 14, 2010 #2
    A better idea, use centre of mass instead.
     
  4. Feb 14, 2010 #3

    tiny-tim

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    Hi Cherrybawls! :smile:
    Yes (you could lower the whole cord half-way, and then rotate it!), but you'd better do it the way they've hinted, since it's to give you practice (which you need! :wink:) at integration.

    The general method is to slice the length (or area or volume) into bits of thickness d(something), treat each bit separately, and then "add" all the bits.

    So choose a slice between x and x + dx, decide how much PE it gets, and integrate. :smile:
     
  5. Feb 14, 2010 #4
    so would I be right in writing:

    Ug= ∫0.25mgx dx? If I set the lowest point of the rope equal to 0 and x cooresponds to the height of the rope?
     
  6. Feb 14, 2010 #5

    tiny-tim

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    Yes, except the mass of the slice from x to x + dx isn't m, is it? :wink:
     
  7. Feb 15, 2010 #6
    I'm not quite sure, is it another variable or is it something like m/x?
     
  8. Feb 15, 2010 #7

    tiny-tim

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    Well, it's not m/x (where x is the distance from the slice to one end of the cord), but it's something similar.

    The mass of the whole cord is 15g, the cord has length 0.25m, so what is the mass of a bit of the cord of length dx ?
     
  9. Feb 15, 2010 #8
    m/dx? I am just gessing really...
     
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