# Moment of inertia of a uniform solid sphere

1. Apr 7, 2006

### alcoholicsephiroth

Posted this question in the calculus section but I guess it's more of a basic physics question, so I've copied it here -

Taking a uniform solid sphere of radius R and mass M, with the centre of mass at the origin, I divided it into infinitesimal disks of thickness dx, and radius y. I need to find the moment of inertia about the x-axis, so taking an arbitrary disk at some horizontal distance x from the centre of mass, I obtain ;

y^2 + x^2 = R^2, (fairly obviously),

density, rho = dm/dV,

dV = (pi)(y^2)dx => dm = (rho)(pi)(y^2)dx

So using the standard definition for moment of inertia :

I = integral of (y^2)dm

I = integral of (y^2)(rho)(pi)(y^2)dx -with x limits R and -R

= (rho)(pi) integral of ((R^2 - x^2)^2) dx

which simplifies down to I = (16/15)(pi)(rho)R^5,

and using M = (4/3)(pi)R^3, I obtain I = (4/5)MR^2.

Of course my textbook is telling me it should be (2/5)MR^2, and as far as my understanding goes, this is a consequence of each infinitesimal disk having a moment of inertia of (1/2)dm(r^2).

Logically then, using dI = (1/2)dm(r^2), such that :

I = integral of (pi)(rho)((R^2 - x^2)^2)dx with x limits R and 0, the answer comes out correctly as (2/5)MR^2.

Unfortunately, I am not a particularly sophisticated mathematician and I am worried that my own method, using I = integral of (y^2)dm as described, is giving me an answer which is out by a factor of 2.

I fear I may have made a trivial mistake, but if not, I'd greatly appreciate some insight as to the cause of the discrepancy.

Many thanks!

Trev

2. Apr 7, 2006

### Hootenanny

Staff Emeritus
3. Apr 7, 2006

### alcoholicsephiroth

Thanks for the link, although I don't actually have any problems with that derivation. What I don't understand is the why my own approach [just integrating (y^2)(dm) to obtain I ] and the derivation shown in the link [ starting with dI = (1/2)(y^2)(dm) ] are any different.

When finding I for a solid cylinder for example, you would use thin concentric cylindrical shells as the infinitesimal mass elements, but it is not necessary to consider what the moment of inertia of each shell is. In this case you can start with I = integral of (y^2)(dm) to obtain I = (1/2)MR^2. But with a sphere, all the literature suggests that the moment of inertia of each mass element must be considered, since all the derivations start with dI = (1/2)(y^2)(dm).

4. Apr 7, 2006

### Staff: Mentor

In your derivation, you are adding up disks, not shells. The rotational intertia of each disk is not $y^2 dm$ as you assume, but $1/2 y^2 dm$.

5. Apr 7, 2006

### alcoholicsephiroth

But what of the general definition of moment of inertia ?

I = sum of mr^2 for each particle in the solid.

Does it not follow from this definition that the moment of inertia of any continuous solid can be found by turning the above sum into an integral, without the need to consider the rotational inertia of the infinitesimal element of choice, as I tried ?

6. Apr 7, 2006

### Staff: Mentor

That's true for particles, not for disks.

Only if your element is a point mass. You took a shortcut by taking disks as your mass element. If you started with a particle mass element, and did the full integration, you'd get dI = 1/2 dm r^2 for each disk.

7. Apr 7, 2006

### alcoholicsephiroth

Wow I see the light (I think) :P

I'll just take a quick guess here, and say that a small addition to your answer to my second question would be 'only if your element is a point mass, or the element has moment of inertia = dm(r^2) ' ?

I think this follows, looking at the working for a solid cylinder, my textbook uses shells, which each have moment of inertia dm(r^2), and that derivation just uses I = integral of r^2 dm, which is the same working for a point mass element as you point out.

Anyhow, I'll go and review this stuff. Thank you for your explanation!

8. Apr 7, 2006

### Staff: Mentor

Exactly right.