Moment of inertia of a thin ring about a perpendicular tangent
I = 2MR², radius of gyration k = R√2
Axis: about a tangent perpendicular to its plane.
Derivation
- Start from the central axis: I = MR².
- Parallel axis theorem with d = R: I = MR² + MR² = 2MR².
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 thin ring about a tangent perpendicular to its plane?
I = 2MR², so the radius of gyration is k = R√2.
How is the moment of inertia of a thin ring about a tangent perpendicular to its plane derived?
Start from the central axis: I = MR². Parallel axis theorem with d = R: I = MR² + MR² = 2MR².
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