A thin uniform spherical shell has a radius of R and mass M. Calculate its moment of inertia about any axis through its centre.

Note: If you are lost at any point, please visit the beginner’s lesson or comment below.

Recall: Moment of inertia for a hoop: I = r^{2} dm

Hence,

$dI = r^{2} dm$

Finding dm,

$dm = \frac{M}{A} dA$

Now, dA is the area of the ring.

$dA = R \: d \theta \times 2 \pi r$

**Note:** 2πr is the circumference of the hoop while R dθ is the “thickness” of the hoop (its dx in the above picture). The R dθ comes from the equation for arc length: S = Rθ.

Now, we have to find a way to relate r with θ. Consider the above picture, notice that there is a right-angle triangle with angle θ at the centre of the circle. Hence,

$ sin \: \theta = \frac{r}{R}$

$r = R sin \: \theta$

Substituting the equation for dA and the equation above into the equation for dm, we have:

$dm = \frac{M sin \: \theta}{2} d \theta$

Substituting the equation above and the equation for r into the equation for dI, we have:

$dI = \frac{MR^{2}}{2} sin^{3} \: \theta \: d \theta$

Integrating with the proper limits, (from one end to the other)

$I = \frac{M R^{2}}{2} \int_{0}^{\pi} sin^{3} \theta \: d \theta$

For those who knows how to integrate sin^{3} θ, you’re done with this post. For those who needs a little bit more help, read on.

Now, we split the sin^{3} θ into two,

$I = \frac{M R^{2}}{2} \int_{0}^{\pi} sin^{2} \theta \: sin \: \theta \: d \theta$

$I = \frac{M R^{2}}{2} \int_{0}^{\pi} (1 – cos^{2} \theta) \: sin \: \theta \: d \theta$

Now, at this point, we will use the substitution: u = cos θ. Hence,

$I = \frac{MR^{2}}{2} \int_{1}^{-1} u^{2} – 1 \: du$

I’m pretty sure you can handle this simple integration by yourself. Hence, we have:

$I = \frac{2}{3} MR^{2}$

Back To Rotation And Moment Of Intertia

**Derivation Of Moment Of Inertia Of Common Shapes:**

- Uniform Rigid Rod (“Beginners’ Lesson”)
- Hollow/solid Cylinder
- Uniform Solid Sphere