## 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

Please can you explain that simple integration in equation 7 for me.

I don’t know it

🥺🥺🥺

Hello, This is a great content and I love it.

But

I have a stupid question to ask if you don’t mind.

That derivation of sin^3 over there looks like a magic to me.

PLEASE HELP ME OUT!!!

explain the integration part again

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

Here x=Rcos(theta)

so, dx=-Rsin(theta)d(theta)

Why are we writing dx=Rd(theta)

$dx$ is not used in the derivation. $R d(\theta)$ is the arc length.

Please read the comments below to find out why $dx$ is not used in the derivation.

I don’t quite understand why are we taking the limits from 0 to pi.

Please help.

I do no understand why does the change in area, dA = R dθ * 2(pi)r, how is the circumference of the hoop(cross section) related to the dA? Thanks

Visualise dA as a long, rectangular strip. The length of the rectangular strip will be the circumference of the hoop, while the width is the arc length.

I did similar integration but instead of using angle I did the integration over x, and then it holds r=sqrt(R^2-x^2), and dA=2rpi*dx, but i got 3pi/16mR^2 instead of 2/3mR^2. I checked the calculation several times and it seems to be correct. Why is there a difference and when is it made? Is my whole idea wrong or is it something else?

Hello. Your situation is the same as his: https://www.miniphysics.com/uy1-calculation-of-moment-of-inertia-of-thin-spherical-shell.html/comment-page-1#comment-217

Hi I think instead of “Recall: Moment of inertia for a hoop: I = r2 dm”

you mean just “m” and not “dm”

In this case, I am treating the spherical shell as being made up of “infinite” hoops. Hence, dm is the mass of one of those “infinite” hoops.

why dm=m/A *dA i made it dm= (den)*2*PI*r*R*d0 because dm=(den)*dv , dv=2*PI*r*dL , dL=Rd0

and r=Rsin0 but the result with me wasnt 2/3 m*R^2

Note that the calculations in my post is done using $ dI = r^{2} \, dm $ (moment of inertia of a HOOP as the start). This is why I use $ dm = \frac{m}{A} dA $, and not dV.

Since you use dV in your calculations, you are actually finding moment of inertia of a sphere and not a spherical shell. You can see the calculations for that of a sphere by following this link: https://www.miniphysics.com/uy1-calculation-of-moment-of-inertia-of.html

Uhhhhh, this might sound silly but 2 questions:

Where did the sin go when we substituted with u?

Wouldn’t the last integration give us a 0?

When you use integration by substitution, u = cos theta becomes du = -sin theta d(theta). Hence, the sin theta d(theta) in the original integral is substituted by du.

The last integration gives 4/3. I do not see how you got 0. Maybe you could share the steps for the integration that you did so I could take a look.

why do you integrate from 0 to Pi, and not 0 to 2pi?

If you integrate from 0 to 2 pi, you’ll be double counting as I’ve used a complete hoop in the integration. Have a closer look at the figure.

ah, thanks! 🙂

for the dm=msin(theta)d(theta)/2, where did the 2 come from and how does it get into the denominator?

You have to include the total surface area of the shell, A which is equal to 4 pi R^2. Sub. A into the equation for dm.

I have understood this solution but why don’t you use the same method in finding the moment of inertia of the solid sphere

dm = (M / A ) r . dx

since r = sqrt ( R^2 – x^2 )

I used this method but the result was I = ( 3 pi / 16 ) MR^2 and i don’t know why ?

I believe you meant dm = surface area density * area of a hoop? In this case, your original equation for dm should be dm = (M/A) 2 pi r dx.

However, you cannot use this method to find the moment of inertia of the thin spherical shell. You are trying to find the area of a thin hoop by visualising it as a hollow cylinder with height dx and radius r. Consider moving the thin hoop closer and closer to the poles of the shell. (in this picture above, it will be towards the right) You will see that the dx is no longer the “height” of the hollow cylinder. This is because at the right pole of the sphere, the surface becomes horizontal. Hence, 2 pi r dx is not surface area of the hoop when x is ~ R.

The equation for dm breaks down at the poles and no longer describes the situation properly. Hence, you are unable to obtain the correct result.

dx canot be used bloc dx is a perpendicular to the axis while d theta is an arc length which actually vary with thé shell.dx cant be use blc it is not parallel to the axis