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

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

  2. 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

  3. 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

  4. 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?

  5. 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

  6. 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.

  7. 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


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