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 or comment below.

Recall: Moment of inertia for a hoop: I = r2 dm


$$dI = r^{2} dm$$

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 sin3 θ, you’re done with this post. For those who needs a little bit more help, read on.

Now, we split the sin3 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 (UY1)

Derivation Of Moment Of Inertia Of Common Shapes:

What Others Are Saying:

  1. dan scribbled

    Hi I think instead of “Recall: Moment of inertia for a hoop: I = r2 dm”
    you mean just “m” and not “dm”

  2. esraa scribbled

    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

  3. Omar scribbled

    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?

    • scribbled

      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.

  4. Mark Nicholson scribbled

    for the dm=msin(theta)d(theta)/2, where did the 2 come from and how does it get into the denominator?

  5. scribbled

    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 ?

    • scribbled

      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.

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