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Gravitational Potential Due to a Thin Rod of Varying Density.

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data

    A thin rod of length L lies along the +y-axis, with one end at the
    origin (see diagram).
    • The rod has length only- no thickness in other directions.
    • The density of the rod increases proportionally to the
    y-coordinate: λ = ky, where k is a known constant and λ is in
    • Gravitational potential is zero at infinity: φ (∞) = 0

    a) Find the gravitational potential φ ( x) at a point (x,0) by direct integration.
    b) Find the gravitational field g at a point (x,0) by direct integration.

    2. Relevant equations

    dφ = -(G dm)/r

    3. The attempt at a solution

    Still stuck on part a, so that's really the brunt of my question for now (though assistance with part b is more than welcome!).

    Using the given density function to solve for dm and substituting √(x^2+y^2) for r, I have an expression for dφ:

    dφ = -(Gk y dy)/√(x^2+y^2)

    ...but I have no idea how to manipulate this to get a soluble integral :( I've been messing around with partial derivatives and polar coordinates for hours, but nothing seems to work.
  2. jcsd
  3. Feb 27, 2012 #2
    Ah, I think I may have figured it out... x isn't actually changing here, so my attempts to relate it to y were completely unnecessary? The answer then (I think) would be:
    φ(x) = --Gk[√(x^2+L^2) - x] .

    Part b has me a bit stumped though.... help would be lovely.
  4. Feb 27, 2012 #3
    Alrighty, figured that bit out on my own as well... polar coordinates work out nicely.
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