Moment of inertia of a solid sphere derivation.

1. Dec 12, 2011

physicsod

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$\rho$$\pi$r2 dr

And then solved for the moment of inertia:

I = $\int$r2dm

= $\int$r2(4$\rho$$\pi$r2 dr)

= 4$\rho$$\pi$$\int$r4 dr

= (4/5)$\rho$$\pi$r5

And solving for $\rho$ we get the following:

$\rho$ = M/((4/3)$\pi$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.

2. Dec 12, 2011

6.28318531

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
3. Dec 12, 2011

Dick

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.

4. Dec 12, 2011

physicsod

@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?

5. Dec 12, 2011

Dick

Something like that, yes. One r is the radius of the shell. The other r depends on an angle variable.

6. Dec 12, 2011

6.28318531

yeah haha I think the disk idea is better..I cant type very fast atm, you dont need any angles

7. Dec 12, 2011

physicsod

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?

8. Dec 12, 2011

gnulinger

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?

9. Dec 12, 2011

physicsod

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. :)

10. Dec 12, 2011

Dick

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.