Rankwise

Moment of inertia of a uniform disc about its central axis

I = MR²/2, radius of gyration k = R/√2

Axis: through the centre, perpendicular to its plane.

kg
m

Derivation

  1. Split the disc into thin rings of radius r and width dr. Mass per unit area σ = M/πR², so dm = σ × 2πr dr = (2M/R²) r dr.
  2. Each ring contributes r² dm, so I = ∫₀ᴿ (2M/R²) r³ dr.
  3. = (2M/R²) × R⁴/4 = MR²/2.

Moment of inertia of standard bodies

Questions students ask

What is the moment of inertia of a uniform disc through the centre, perpendicular to its plane?

I = MR²/2, so the radius of gyration is k = R/√2.

How is the moment of inertia of a uniform disc through the centre, perpendicular to its plane derived?

Split the disc into thin rings of radius r and width dr. Mass per unit area σ = M/πR², so dm = σ × 2πr dr = (2M/R²) r dr. Each ring contributes r² dm, so I = ∫₀ᴿ (2M/R²) r³ dr. = (2M/R²) × R⁴/4 = MR²/2.

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