Rankwise

Moment of inertia of an annular disc (radii R₁ and R₂) about its central axis

I = M(R₁² + R₂²)/2, radius of gyration k = √((R₁² + R₂²)/2)

Axis: through the centre, perpendicular to its plane.

kg
m
m

Derivation

  1. dm = σ 2πr dr with σ = M/π(R₂² − R₁²).
  2. I = ∫ from R₁ to R₂ r² dm = σ × 2π × (R₂⁴ − R₁⁴)/4.
  3. = M(R₂⁴ − R₁⁴)/2(R₂² − R₁²) = M(R₁² + R₂²)/2.

Moment of inertia of standard bodies

Questions students ask

What is the moment of inertia of an annular disc (radii R₁ and R₂) through the centre, perpendicular to its plane?

I = M(R₁² + R₂²)/2, so the radius of gyration is k = √((R₁² + R₂²)/2).

How is the moment of inertia of an annular disc (radii R₁ and R₂) through the centre, perpendicular to its plane derived?

dm = σ 2πr dr with σ = M/π(R₂² − R₁²). I = ∫ from R₁ to R₂ r² dm = σ × 2π × (R₂⁴ − R₁⁴)/4. = M(R₂⁴ − R₁⁴)/2(R₂² − R₁²) = M(R₁² + R₂²)/2.

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