Sphere Moment of Inertia

Moment of inertia of a sphere: solid (2/5)·M·R² vs hollow (2/3)·M·R²

For a solid sphere rotating about a diameter: I = (2/5)·M·R². For a thin hollow sphere (spherical shell): I = (2/3)·M·R². The hollow sphere has a larger I at the same M and R because its mass sits farther from the axis — and I scales with r². Example: a 5 kg, 0.2 m solid sphere has I = (2/5)·5·0.04 = 0.08 kg·m²; the same hollow sphere has 0.133 kg·m². Earth has I ≈ 0.33·M·R² (less than the uniform 0.4), which reveals a dense iron-nickel core. On a ramp, a hollow sphere arrives after a solid one because more of its energy goes into rotation; both lose to a solid cylinder (I = (1/2)·M·R²) since (2/5) < (1/2). For an axis offset by distance d from the center, use the parallel-axis theorem I = I_cm + M·d².

Applications

Planetary science (Earth's I/MR² ≈ 0.33 indicates internal mass distribution), rolling races on inclines, gyroscopes and inertial navigation, ball bearings, and simple atomic models. Used in physics olympiads, mechanical engineering, and aerospace stability analysis.

FAQ

Why does a hollow sphere lose the rolling race? A larger I means more rotational kinetic energy per unit of translational energy, so less of the released potential energy becomes linear speed.

Does the axis direction matter? No — a sphere is symmetric, so I is the same about any axis through the center.

How do I compute I about an axis outside the sphere? Apply the parallel-axis theorem: I = I_cm + M·d², where d is the perpendicular distance from the center of mass.