Moment of inertia of a solid sphere about a diameter
I = 2MR²/5, radius of gyration k = R√(2/5)
Axis: about a diameter.
Derivation
- Slice the sphere into discs of thickness dz at height z. Each has radius √(R² − z²) and mass dm = ρπ(R² − z²) dz, with ρ = 3M/4πR³.
- Each disc gives dI = ½ dm (R² − z²) = ½ρπ(R² − z²)² dz.
- I = ½ρπ ∫ from −R to R (R² − z²)² dz = ½ρπ × 16R⁵/15 = 2MR²/5.
Moment of inertia of standard bodies
| Body and axis | I |
|---|
| Thin ring, through the centre, perpendicular to its plane | MR² |
| Thin ring, about a diameter | MR²/2 |
| Thin ring, about a tangent in its plane | 3MR²/2 |
| Thin ring, about a tangent perpendicular to its plane | 2MR² |
| Uniform disc, through the centre, perpendicular to its plane | MR²/2 |
| Uniform disc, about a diameter | MR²/4 |
| Uniform disc, about a tangent in its plane | 5MR²/4 |
| Uniform disc, about a tangent perpendicular to its plane | 3MR²/2 |
| Thin rod, through the centre, perpendicular to its length | ML²/12 |
| Thin rod, through one end, perpendicular to its length | ML²/3 |
| Solid sphere, about a diameter | 2MR²/5 |
| Thin hollow sphere (spherical shell), about a diameter | 2MR²/3 |
| Solid sphere, about a tangent | 7MR²/5 |
| Thin hollow sphere, about a tangent | 5MR²/3 |
| Solid cylinder, about its own axis | MR²/2 |
| Thin hollow cylinder, about its own axis | MR² |
| Solid cylinder, through the centre, perpendicular to its axis | M(R²/4 + L²/12) |
| Rectangular plate (sides a and b), through the centre, perpendicular to the plate | M(a² + b²)/12 |
| Annular disc (radii R₁ and R₂), through the centre, perpendicular to its plane | M(R₁² + R₂²)/2 |
| Solid cone, about its own axis | 3MR²/10 |
Questions students ask
What is the moment of inertia of a solid sphere about a diameter?
I = 2MR²/5, so the radius of gyration is k = R√(2/5).
How is the moment of inertia of a solid sphere about a diameter derived?
Slice the sphere into discs of thickness dz at height z. Each has radius √(R² − z²) and mass dm = ρπ(R² − z²) dz, with ρ = 3M/4πR³. Each disc gives dI = ½ dm (R² − z²) = ½ρπ(R² − z²)² dz. I = ½ρπ ∫ from −R to R (R² − z²)² dz = ½ρπ × 16R⁵/15 = 2MR²/5.
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