# UY1: Calculation of moment of inertia of a thin spherical shell

## 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,

$$$$dI = r^{2} \, dm$$ \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:

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

Back To Mechanics (UY1)

Derivation Of Moment Of Inertia Of Common Shapes:

### 37 thoughts on “UY1: Calculation of moment of inertia of a thin spherical shell”

1. Please can you explain that simple integration in equation 7 for me.
I don’t know it
🥺🥺🥺

2. 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.

3. explain the integration part again

4. 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.

5. 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

6. Sir moment of inertia Ke sabhi part Hindi me derivation Kate

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

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

8. 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.

9. Thankyou for such a crisp derivation.

• 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

11. 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.

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

13. 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.

14. 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?

15. 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.

16. 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

17. 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.

18. 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! 🙂

19. 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.

20. 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 ?