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 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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

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




Similar Discussions: Moment of inertia of a solid sphere derivation.
Loading...