### 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 $\sin^{3} \theta$, 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 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:

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

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

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

Kaushlendra scribbled

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

YHJ scribbled

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

Alice scribbled

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?

Alice scribbled

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

SHREERAM VERMA scribbled

Sir moment of inertia Ke sabhi part Hindi me derivation Kate

Jack scribbled

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

Ashutosh Ukey scribbled

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

Mayank mittal scribbled

Thankyou for such a crisp derivation.

utkarsh scribbled

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

Mini Physics scribbled

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

yustineo scribbled

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