📂Classical Mechanics

Moment of Inertia of a Spherical Shell

Formulas

The moment of inertia of a spherical shell with a radius of $a$ and a mass of $m$ is as follows.

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

Derivation

Consider a uniform spherical shell with a radius of $a$ and a mass of $m$. The same idea is used as when calculating the moment of inertia of a sphere. However, there is a bit of a difference. Think of the spherical shell as a sum of many cylindrical shells, just as when calculating the moment of inertia of a sphere.

But if we calculate it the same way as in the case of a sphere, there is a problem. In the case of a sphere, integrating the volume of small discs results in the volume of the sphere.

$$\begin{align*} \int \nolimits _{-r} ^{r} \pi \left( r^{2} -z^{2} \right) dz &= \int \nolimits _{-r} ^{r} \pi r^{2} dz - \int \nolimits _{-r} ^{r} \pi z^{2}dz \\ &= \pi r^{2} \left[ z \right]_{-r}^{r} - \pi \left[ \frac{1}{3} z^{3} \right] _{-r} ^{r} \\ &= 2 \pi r^{3} - \dfrac{2}{3}\pi r^{3} \\ &= \dfrac{3}{4} \pi r^{3} \end{align*}$$

However, the limit of the sum of the surface areas of the small discs is not equal to the surface area of the sphere. The problem arises right here. It can actually be confirmed through the calculations below.

$$\int _{-r} ^{r} 2\pi x dz = \int _{-r} ^{r} 2\pi \sqrt{ r^{2}-z^{2}} dz$$

When substituting with $z \equiv r \sin \theta$,

$$\int _{-r} ^{r} \implies \int _{-\frac{\pi}{2}} ^{\frac{\pi}{2}} \quad \And \quad dz=r \sin \theta d\theta$$

$$\begin{align*} \int_{-\frac{\pi}{2}} ^{\frac{\pi}{2}} 2 \pi \sqrt{r^{2}-r^{2}\sin^{2} \theta} r \cos\theta d\theta &= 2\pi r \int_{-\frac{\pi}{2}} ^{\frac{\pi}{2}} \sqrt{r^{2}} \sqrt{1-\sin^{2}\theta} \cos\theta d\theta \\ &= 2 \pi r^{2} \int \nolimits _{-\frac{\pi}{2}} ^{\frac{\pi}{2}} \cos^{2} d\theta \\ &= \pi r^{2} \int \nolimits _{-\frac{\pi}{2}} ^{\frac{\pi}{2}} (1+\cos 2\theta) d\theta \\ &= \pi r^{2} \left[ \theta+\frac{1}{2}\sin 2\theta \right]_{\frac{\pi}{2}} ^{\frac{\pi}{2}} \\ &= \pi r^{2} (\pi )=\pi^{2} r^{2} < 4 \pi r^{2} \end{align*}$$

So it falls short of the surface area of the sphere. Therefore, the surface area of the sphere cannot be approximated as the sum of the surface areas of the discs. Now, let’s approximate the surface area of the small cylindrical shell a bit differently. Assume the height of the cylinder is not the vertical distance between the top and bottom surfaces, but the distance along the side surface and solve it. Integrating in this way yields the surface area of the cylinder. Try it yourself if you’re curious.

Since the mass is uniformly $m$ and the radius is $r$, the moment of inertia of the cylindrical shell is $I=mr^{2}$, so the moment of inertia of the spherical shell is as follows.

$$I_{\text{sphere shell}}=\int dl=\int x^{2}dm$$

Substituting the value of $dm$ obtained from the above figure gives the following.

$$\int x^{2} \rho 2 \pi \color{blue}{ x a d \theta}$$

At this point, using the relationship below to change the blue part from an integration over $\theta$ to an integration over $z$ yields the following.

$$z=a\sin\theta \implies dz=a\cos\theta d\theta,\quad x=a\cos\theta$$

The reason for specifically integrating over $z$ is because it is easier than integrating over $\theta$.

$$\begin{align*} \int ( a^{2}-z^{2})\rho 2 \pi \color{blue}{a \cos\theta ad\theta} &= \rho 2 \pi a \int \nolimits _ {-a} ^{a} ( a^{2}-z^{2})dz \\ &= \rho 2 \pi a \left[ a^{2}z-\frac{1}{3}z^{3} \right]_{-a}^{a} \\ &= \rho 2 \pi a (2a^{3}-\frac{2}{3}a^{3}) \\ &= \rho \dfrac{8}{3} \pi a^4 \end{align*}$$

And since the mass of the spherical shell is $m=\rho 4 \pi a^{2}$ and $\rho = \dfrac{m}{4\pi a^{2}}$, the following is obtained.

$$I_{z}=\rho \dfrac{8}{3} \pi a^4=\dfrac{m}{4\pi a^{2}} \dfrac{8}{3} \pi a^4 = \dfrac{2}{3}ma^{2}$$

Just like a sphere, the spherical shell is also symmetrical in all directions, so the following is true.

$$I_{x}=I_{y}=I_{z}=\dfrac{2}{3}ma^{2}$$

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