Moment of Inertia of Spherical Shell

Tags:
1. Oct 21, 2015

minimario

1. The problem statement, all variables and given/known data
Find the moment of inertia of a spherical shell (hollow) with mass M and radius R.

2. Relevant equations
$I = \int r^2 dm$

3. The attempt at a solution
This is method I use to find Moment of Inertia of solid sphere:

We use circular cross sections.

At some radius r, $\frac{dm}{M} = \frac{\pi * r^2 dz}{4/3 \pi R^3} \Rightarrow dm = \frac{3Mr^2}{4R^3} dz$.

Therefore, the total moment of inertia is $\int \frac{1}{2} (dm) r^2 = \frac{3M}{8R^3} \int r^4 dz$. We use $r = R^2 - z^2$, and integral is $\int (R^2-z^2)^2 dz$, and evaluated from -R to R, it is $\frac{16}{15} R^5$. So the total integral is $\frac{2}{5} MR^2$.

But with spherical shell, something is different. $\frac{dm}{M} = \frac{2 \pi * r dz}{4 \pi R^2} \Rightarrow dm = \frac{Mr}{2R^2} dz$. (Again, we use circles).

But when taking this integral, the answer is incorrect. So is there something wrong with my expression for dm?

2. Oct 21, 2015

rude man

Well, I sympathize because I also proceeded as you did & got the wrong answer. If I figure it out I'll post again later.

Meanwhile, you can take your result for I of a solid sphere, subtract I for a sphere slightly smaller than the first sphere, then take limits as the radius difference approaches zero, with the volume density taken into account of course.

3. Oct 21, 2015

rude man

Why the "1/2" in your expression for $\int \frac{1}{2} (dm) r^2$? Are you sure you did this integration right?

For the shell my take is that it's incorrect to say dm = 2πr(M/4πR2)dz. The reason is the strip of shell is curved. If you look at other derivations they all use dm = 2πr (M/4πR2) R dθ
where θ runs along the shell's surface from z = +R to z = -R. R dθ is the true width of the strip, not dz. 2πr R dθ is a differential arc segment of area, unlike 2πr dz.

4. Oct 21, 2015

minimario

Then, why, when integrating the sphere, it is ok to ignore the curved area? And the 1/2, it is right because we are summing the moments of each individual disk, which is 1/2 MR^2

5. Oct 21, 2015

rude man

I have to admit, I don't have an answer for that right now. Maybe figure it out later. Hopefully, others mighht pitch in.

6. Oct 21, 2015

rude man

If you want to pursue this further I would need to see what you got for I for the shell. I came up with an integral with an arc tangent coefficient which when evaluated came up with arc tan ∞ = π/2 but I'm always wary of dealing with infinities. Could be the math just won't accommodate dz = R dθ.

7. Oct 21, 2015

Nathanael

It's not hard to show (and good to know) that the differential surface area of a sphere around a differential radius increment dr is 2πRdr (where R is the entire sphere's radius) regardless of where along the diameter dr is located. This result comes from approximating the differential surface area by truncated cones (for which the surface area is the length of the slanted side times 2πravg).

Just to emphasize that it is wrong to approximate the surface of a sphere by the surface of differential cylinders (as you did); you can integrate 2π√(R2-z2)dz from z=-R to z=R and see that the result will not be 4πR2.

As for why it is okay to approximate a solid sphere by cylinders instead of truncated cones; well intuitively the difference between the volumes of a differential cylinder and a differential truncated cone is negligible (whereas the difference between their differential surface areas is not negligible).
I'm sure you can show this mathematically by showing that the difference in their volumes is dependent only on higher powers of dr (and so dV/dr is the same for both, as the higher order terms vanish in the limit of dr→0) but I will just leave you with this hand-wavy explanation instead.

8. Oct 21, 2015

rude man

Thank you for pitching in Nathanael.