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Moment of Inertia using density of Earth

  • Thread starter musiliu
  • Start date
  • #1
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Homework Statement



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


Homework 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


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:

Answers and Replies

  • #2
44
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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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