Derivation of moment of inertia of a thin spherical shell

 

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

Imagine that it is a hoop and not a disk.

Imagine that it is a hoop and not a disk.

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 = r2 dm

Hence,

$$dI = r^{2} dm$$

Finding dm,

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

Where A is the total surface area of the shell: $4 \pi R^{2}$

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,

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

$$r = R \, \text{sin} \: \theta$$

Hence, dA becomes:

$$dA = 2 \pi R^{2} \text{sin} \: \theta \: d \theta$$

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

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

 

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

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

 

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

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

 

For those who knows how to integrate sin3 θ, you’re done with this post. For those who needs a little bit more help, read on.

Now, we split the sin3 θ into two,

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

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

 

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

 

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

 

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: