Moment of Inertia of a Hollow Sphere

Result

$I = \frac{2}{3}MR^2 \quad \text{(about any diameter)}$

Derivation

Divide the shell into thin horizontal rings. At angle $\theta$ from the axis (polar angle):

• Ring radius: $r = R\sin\theta$
• Ring width along surface: $R \, d\theta$
• Ring mass: $dm = \frac{M}{4\pi R^2} \cdot 2\pi R\sin\theta \cdot R \, d\theta = \frac{M}{2}\sin\theta \, d\theta$

MI of this ring about the $z$-axis: $dI = r^2 \, dm = R^2\sin^2\theta \cdot \frac{M}{2}\sin\theta \, d\theta$

$I = \frac{MR^2}{2}\int_0^\pi \sin^3\theta \, d\theta$

$\int_0^\pi \sin^3\theta \, d\theta = \int_0^\pi (1-\cos^2\theta)\sin\theta \, d\theta = \left[-\cos\theta + \frac{\cos^3\theta}{3}\right]_0^\pi = \frac{4}{3}$

$I = \frac{MR^2}{2} \cdot \frac{4}{3} = \frac{2MR^2}{3}$

$\boxed{I = \frac{2}{3}MR^2}$

About a tangent

$I_{tangent} = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2$

Comparison table — all standard spherical shapes

The hollow sphere has $I = \frac{2}{3}MR^2$ — larger than the solid sphere. All mass at the surface means maximum average $r^2$.