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Sphere's Moment of Inertia 📂Classical Mechanics

Sphere's Moment of Inertia

Formula

The moment of inertia of a sphere with radius $a$ and mass $m$ is as follows.

$$ I=\frac{2}{5}ma^{2} $$

Proof

The idea of finding the moment of inertia of a sphere is slightly different from other rigid bodies. The key idea is to think of the sphere as a sum of infinitely many discs, similar to the method of integration.

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Adding up the moment of inertia of these infinitely many discs yields the moment of inertia of the sphere. The moment of inertia of a disc perpendicular to the axis of rotation is $I = \dfrac{1}{2}mr^{2}$ ($r=$ radius, $m=$ mass), so the moment of inertia of the sphere can be determined as follows.

$$ I_{\text{sphere}} = \int dI = \int \frac{1}{2}r^{2} dm $$

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Now, just calculate.

$$ \begin{align*} I_{z} &= \int \frac{1}{2}r^{2}dm=\int_{-a}^{a} \frac{1}{2}x^{2} \rho\pi x^{2} dz \\ &= \int_{-a}^{a} \frac{1}{2}\rho\pi x^{4} dz \\ &= \frac{1}{2} \rho \pi\int_{-a}^{a} (a^{2}-z^{2})^{2}dz \\ &= \frac{1}{2} \rho\pi \int_{-a}^{a} (a^{4}-2a^{2}z^{2}+z^{4})dz \\ &= \frac{1}{2} \rho \pi \left[ a^{4}z-\frac{2}{3}a^{2}z^{3}+\frac{1}{5}z^{5} \right]_{-a}^{a} \\ &= \frac{1}{2} \rho \pi \left(2a^{5}-\frac{4}{3}a^{5}+\frac{2}{5}a^{5} \right) \\ &= \rho \pi \left(a^{5}-\frac{2}{3}a^{5}+\frac{1}{5}a^{5} \right) \\ &= \rho \pi \frac{8}{15}a^{5} \end{align*} $$

And since the mass of the sphere is $m = \rho \pi \dfrac{4}{3} a^{3}$, substituting $I_{z}$ gives the following.

$$ I_{z} = \rho \pi \frac{8}{15} a^{5} = \left( \rho \pi \frac{4}{3}a^{3} \right) \left( \dfrac{2}{5}a^{2} \right) = \frac{2}{5}ma^{2} $$

Furthermore, since the sphere is symmetrical in all directions, we obtain the following result.

$$ I_{x} = I_{y} = I_{z} $$