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A rod, a ball, garvitational Potential Energy (U), and the power series expns.

  1. Feb 8, 2008 #1


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    1. The problem statement, all variables and given/known data

    Mass of rod: M
    mass of ball: m
    Length of Rod: L
    distance between rod and ball: x
    GPE is zero at infinty

    The questiopn asks to take the GPE of the rod/ball system, using the Power Series Expansion for ln(1+x) .

    2. Relevant equations

    U = -GMm/r

    3. The attempt at a solution

    I'm not quite sure where to start - the Power Series expansion has confused me slightly, as otherwise I wouold hqave just put the variables in the above equation?

    TFM :confused:
  2. jcsd
  3. Feb 8, 2008 #2
    It just means you can approximate ln(1+x) with x, for x << 1.
  4. Feb 8, 2008 #3


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    Where does the Ln(x+1)come from?

  5. Feb 8, 2008 #4
    It'll probably appear as you solve the problem. They say to use the power series to make it easier to solve.
  6. Feb 8, 2008 #5


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    Do you still use the U=-GMm/r, with U = 0, r = infinty, giving:


  7. Feb 8, 2008 #6
    I guess you want to integrate over the length of the rod and end up with the integral of
    (1/l+1) dl from 0 to L... that should give you ln(x+1)

    U= -GmM/L [integral 0 to L (dl /sqrt. of x^2+l^2)] and I guess you can say that the square root of x^2+l^2 is x+C where C is some constant, though this makes no sense I can't think of any other way

    btw, the way I got that is by saying dU= -Gm(dM)/r.. then setting dM=dl(M/L) and r=sqrt. (x^2+l^2)
    Last edited: Feb 8, 2008
  8. Feb 8, 2008 #7
    take the intergral of 1/r, which is ln(r), setting your limits from infinity to x. Do you know the expansion to ln(r)?
  9. Feb 8, 2008 #8
    from infinity to x? how in the world do you integrate from infinity to x?
  10. Feb 9, 2008 #9


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    Why do I need to take the integral of 1/r?

  11. Feb 9, 2008 #10


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    I'm Still rather cionfused about what I should be doing:frown:

    Any Help?

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