1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Moment of Inertia using density of Earth
Loading...