📂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.

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.

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

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.

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

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