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Homework Help: Moment of Inertia using density of Earth

  1. Apr 18, 2009 #1
    1. The problem statement, all variables and given/known data

    The density of the Earth, at any distance r from its center, is approximately

    p = [14.2 - 11.6 r/R] x 10^3 kg/m^3

    where R is the radius of the Earth. Show that this density leads to a moment of inertia I = 0.330MR^2 about an axis through the center, where M is the mass of the Earth

    2. Relevant equations

    My teacher gave hints to the class and told us to:

    --Find the mass M = integral(dm) where dm = pdV
    --Break Earth into pieces with same r ---> spherical shells
    so dI = (2/3)r^2 dm and I = (2/3) integral( (r^2) dm) where dm = pdV
    -- dV = area x height = 4(pi r^2)dr

    3. The attempt at a solution

    calculated the integral for I and got
    I = (2/3) [ (56800pi / 5)r^5 - (46400pi / 6R)r^6

    I calculated the intral for M and got
    M = (56800pi / 3)r^3 - (46400pi / 4R)r^4

    now i am stuck and do not know how to show that I = 0.330MR^2
    also i think my calculations are wrong..

    please can someone help me figure out the problem..thanks
    Last edited: Apr 18, 2009
  2. jcsd
  3. Apr 18, 2009 #2
    ok, I don't need help on this problem anymore. A tutor helped me and i figured out that there are limits of integration from 0 to R..

    thanks anyway.
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