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Homework Help: Moment of inertia of a solid sphere derivation.

  1. Dec 12, 2011 #1
    Hello! I'm trying to derive the formula for the moment of inertia of a solid sphere, and I keep running into a strange solution.

    I set up the infinitesimally mass of an infinitesimally thin "shell" of the sphere:

    dm = 4[itex]\rho[/itex][itex]\pi[/itex]r2 dr

    And then solved for the moment of inertia:

    I = [itex]\int[/itex]r2dm

    = [itex]\int[/itex]r2(4[itex]\rho[/itex][itex]\pi[/itex]r2 dr)

    = 4[itex]\rho[/itex][itex]\pi[/itex][itex]\int[/itex]r4 dr

    = (4/5)[itex]\rho[/itex][itex]\pi[/itex]r5

    And solving for [itex]\rho[/itex] we get the following:

    [itex]\rho[/itex] = M/((4/3)[itex]\pi[/itex]r3).

    Substituting that into the previously solved equation for I, I get the following:

    I = (3/5)Mr3.

    What am I doing wrong? I know the formula involves a coefficient of 2/5, not 3/5, but I can't find my problem.

    Thank you in advance!
  2. jcsd
  3. Dec 12, 2011 #2
    oops bad idea, think about adding lots of thin disks together

    so dI=(1/2)(r^2)dm=(1/2)(r^2)(rho)dV=1/2(r^4)(rho)(pi)dh

    then integrate from -R to R, and sub (r^4) as (R^2-h^2)^2
    Last edited: Dec 12, 2011
  4. Dec 12, 2011 #3


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    One problem is that you have two different r's in there. The r in dm is the spherical distance of the shell from the origin. The r in the r^2 of the I calculation is supposed to be the distance from the axis of rotation. Revisit finding I for the shell. You'll need to split the shell up into rings.
  5. Dec 12, 2011 #4
    @6.283...: Well if you integrate both sides of the original equation, you get m = (4/3)*(pi)*(r^3)*(rho), so the volume part is correct I think....

    @Dick: Oh, so I treat the first r (let's call the second one R) as a constant in the first line for evaluating the integral? So I just bring out "r" and integrate R^2 dR?
  6. Dec 12, 2011 #5


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    Something like that, yes. One r is the radius of the shell. The other r depends on an angle variable.
  7. Dec 12, 2011 #6
    yeah haha I think the disk idea is better..I cant type very fast atm, you dont need any angles
  8. Dec 12, 2011 #7
    Still not sure exactly how to relate R to r. Is it like, r is the distance from the origin of the shell, and R is the radius of the infinitesimally small disk? Can anybody tell me where to go from my mistake above?
  9. Dec 12, 2011 #8
    The r that you want to calculate should be the distance from the axis of rotation.

    Let's say that the axis of ration is the z-axis. Using spherical coordinates, what is the distance from the z-axis in terms of r?
  10. Dec 12, 2011 #9
    Wait okay, I think I got it. So, instead of involving spherical coordinates, I just let the radius of the sphere be R, and the distance to the infinitesimally thin disk be z, and then the radius of the disk, y, would be sqrt(R^2 - z^2), and so you integrate [(1/2)*(rho)*(pi)*(y^4) dz] since you're just using the inertia of a disk, right?

    Substitute for y, remove the constants and you get I = (1/2)*(rho)*(pi) * (integral<-R, R> of [R^2 - z^2]^2 dz). If you integrate that, you get I = (8/15)*(rho)*(pi)*(R^5). I think that works, because then you substitute in (rho), and you get I = (2/5)MR^2.

    Did I get lucky, or is that one way to do it? I don't particularly like spherical coordinates, so I try and avoid them whenever possible. :)
  11. Dec 12, 2011 #10


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    I think that's ok. If you want to it completely in spherical coordinates, the distance to the z axis where the polar angular coordinate, phi, goes from 0 at one pole to pi at the other pole is r*sin(phi). So you want to integrate (r*sin(phi))^2 times the spherical volume element r^2*sin(phi)*d(phi)*d(theta)*dr. If you integrate r from 0 to R. phi from 0 to pi and theta from 0 to 2*pi, you get (8/15)*pi*R^5. You can not like spherical coordinates, but they do work all the same.
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