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.

moment of inertia of sphere

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:

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


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

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,

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

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


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:

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


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:

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

  1. 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?

    • 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


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