# Moment of Inertia: Thin Spherical Shell

collinsmark
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It is thin but it is not infinitesimally thin. It is a real object. It is thin enough that the integrand does not change much over the thickness making the integral in the third dimension trivial, but the third dimension must be included.

As Jake discovered

dI of solid sphere =r²dm
=ρr²dv

dv=(rdφ)(dr)(rsinφdθ)

dI=ρr²r²sinφdrdθdφ

Except the first r is the radial distance from the axis and should be r sin(phi)

that is the correct reasoning and the correct integrand with the correct infinitesimal piece of volume. Now answer me this: does the correct moment of inertia for a thin sphere have “dr” in the expression, or does it have thickness? It has thickness. Why? Because that is what dr INTEGRATES to! Yes, you absolutely DO have to integrate in all three dimensions. The fact that the sphere is thin means that the radial integral is trivial, but it is there

\int_r1^r2 dr\ = r2-r1 = thickness
If you are deriving the formula for a spherical shell -- not a thin spherical shell, but a spherical shell that has an inner radius $a$ and an outer radius $b$ -- then yes you would integrate over $r$ from $a$ to $b$. The moment of intertia would not be a function of a $r$ but instead would be a function of $a$ and $b$. In that case a triple integral would be involved.

But the problem statement specifically stated "thin" spherical shell, implying that the thickenss between $a$ and $b$ is negligible. In such a case, the moment of inertia is a function of the radius $r$. Only a double integral is necessary. No $dr$ is involved.

jake010 and Dr.D
@Cutter Ketch wants to insist that this problem be treated as a thick shell. He says that this leads to a trivial integration on the radius, but here he is mistaken.

If the thick shell problem is properly formulated, the radial distance from the reference axis to the mass element will vary with the radial position of the mass element. Thus the radial integration will not be the simple integration of dr to get the thickness, but something more complicated. The inclusion of the word "thin" in the problem statement was intended to be significant simplifcation; it is ignored at the expense of considerably more work for no gain.

collinsmark
This isn't true for a problem such as this. (I.e., infinitesimally "thin," hollow shell.)

All that's necessary is to define surface density.

$\rho_s = \frac{M}{A}$

You can think of it as mass per unit area. (If it helps to think of it in familiar units, perhaps think of it as $\left[ \frac{\mathrm{kg}}{\mathrm{m^2}} \right]$.)

For this problem, the density of an infinitesimally thin, spherical shell is $\rho_s = \frac{M}{\mathrm{Surface \ area \ of \ a \ sphere}}$.

This shouldn't be too unfamiliar to you. You've probably done this before for linear density. For example, if you were finding the mass per unit length of a thin rod, you would start with $\lambda = \frac{\mathrm{mass \ of \ the \ rod}}{\mathrm{length \ of \ the \ rod}}$. It's the same idea here with $\rho_s$, except two dimensions are involved instead of just one.

The differential mass, $dm$ is calculated by

$dm = \rho_s (\mathrm{differential \ length})(\mathrm{differential \ width})$

A picture might help here. The following is a picture of a small, differential patch of surface on a sphere of radius $r$, where $\theta$ represents longitude and $\varphi$ represents latitude (well, lattidue with 0 deg and 180 deg defining the poles).
View attachment 252482

This differential mass can be used in calculating the moment of inertia, keeping care to properly specify the distance of $dm$ to the axis of rotation. The integral can then be carried out integrating over $\theta$ and $\varphi$ . Only two integrals are necessary.

No integration over the radius of the sphere $r$ is necessary. The radius of the sphere treated as a constant. (Don't confuse the radius of the sphere $r$ with the distance from $dm$ to the axis of rotation. They are different entities.)

Got it. I understand it now. Sorry I haven't been responding, I was studying on my own. Thank you all so so much for the help! I finally get it. I really appreciate you guys taking the time to help me!

collinsmark