## 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.

**Note**: If you are lost at any point, please visit the beginner’s lesson (Calculation of moment of inertia of uniform rigid rod) or comment below.

### Initial ingredients

Notice that the thin spherical shell is made up of nothing more than lots of thin circular hoops.

Recall that from Calculation of moment of inertia of cylinder:

$$\text{Moment of inertia for a thin circular hoop}: I \, = Mr^{2}$$

Hence,

$$ \begin{equation} dI = r^{2} \, dm \end{equation} \tag{1}$$

In order to continue, we will need to find an expression for $dm$ in Equation 1.

$$\begin{aligned} dm &= \frac{M}{A} \, dA \end{aligned} \tag{2} $$

,where $A$ is the total surface area of the shell – $4 \pi R^{2}$

### Finding $dA$

If $A$ is the total surface area of the shell, $dA$ is the area of one of the many thin circular hoops. With reference to the picture, each thin circular hoops can be thought to be a thin rectangular strip. The area for each hoop, $dA$, is the product of the “length” (circumference of the hoop) and the “breadth” ($dx$ in the picture or known as the arc length). [*Recall: The equation for normal arc length is $R \theta$.*]

$dA$ can be expressed with:

$$\begin{aligned} dA &= \text{length} \times \text{breadth} \\ &= \text{circumference} \times \text{arc length} \\ &= 2 \pi r \times R \: d \theta \end{aligned} \tag{3}$$

Now, in Equation 3, notice that you will have different $r$ for different hoops. Hence, we have to find a way to relate $r$ with $\theta$.

### Relating $r$ with $\theta$

Consider the above picture, notice that there is a right-angle triangle with angle $\theta$ at the centre of the circle. Hence,

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

$$r = R \, \text{sin} \: \theta \tag{4}$$

### Substitutions

Hence, using Equation 4 in Equation 3, $dA$ can be expressed by:

$$dA = 2 \pi R^{2} \text{sin} \: \theta \: d \theta \tag{5}$$

Substituting the Equation 5 into the Equation 2, we have:

$$dm = \frac{M \text{sin} \: \theta}{2} d \theta \tag{6}$$

Substituting Equation 6 and the Equation 4 into Equation 1, we have:

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

### Integration

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

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

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

Now, we split the $\sin^{3}{\theta}$ 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: $u = \cos{\theta}$. Hence,

$$I = \frac{MR^{2}}{2} \int\limits_{1}^{-1} u^{2} – 1 \: du \tag{7}$$

## Final Result

I’m pretty sure you can handle the simple integration in Equation 7 by yourself. Hence, we have:

$$I = \frac{2}{3} MR^{2} \tag{8}$$

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

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

Why integrating between 0 to π..why not 0 to 2π??

because what used here is a full size ring, not half. So 0 to pi actually includes the full spherical shell.

Don’t you need to multiply the final integral by 2 because you’ve only integrated between pi and 0 (not 2 pi). Because sin^3(x) integrated between pi and 0 is 2/3 right?

This is when using

Integral of sin^3(x) = 1/3 cos^3(x) – cos(x)

I don’t know why it gives 2/3 instead of 4/3

Sir moment of inertia Ke sabhi part Hindi me derivation Kate

Thank you sir/ ma’am for such a guide to us

Why not derive it like the moment inertia of a cylinder?

You can’t just multiply Rdø by 2pi*r can you? They are not perpendicular.

rd theta is height. 2πr is area. V =Ah. Rdtheta is so small that we can consider it perpendicular.

Thankyou for such a crisp derivation.

Why are u taking area 2πr not πr square please help

The circumference of the hoop multiplied by the “thickness” of the hoop gives the area of the hoop.

pi r squared is the area of a circle isnt it? this is just a hoop and a hoop is just the edge circle with a slight thickness. so the area of this hoop is just the circumference of a circle x thickness