### 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 or comment below.

Recall: Moment of inertia for a hoop: I = r^{2} 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 sin^{3} θ, you’re done with this post. For those who needs a little bit more help, read on.

Now, we split the sin^{3} 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:**

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

esraaJune 3, 2014 at 11:16 PMwhy 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

Mini PhysicsJune 4, 2014 at 1:29 PMNote 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: http://www.miniphysics.com/uy1-calculation-of-moment-of-inertia-of.html

OmarDecember 31, 2013 at 4:02 AMUhhhhh, 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?

Mini PhysicsJanuary 29, 2014 at 5:03 PMWhen 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.

Harald EskerudSeptember 26, 2013 at 7:42 PMwhy do you integrate from 0 to Pi, and not 0 to 2pi?

Mini PhysicsSeptember 26, 2013 at 8:03 PMIf 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.

Harald EskerudSeptember 26, 2013 at 8:05 PMah, thanks!

Mark NicholsonJanuary 30, 2013 at 1:12 AMfor the dm=msin(theta)d(theta)/2, where did the 2 come from and how does it get into the denominator?

Mini PhysicsJanuary 30, 2013 at 9:25 AMYou 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.

خالد عصامDecember 31, 2012 at 4:32 PMI 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 ?

Mini PhysicsDecember 31, 2012 at 11:50 PMI 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.