Rankwise

Moment of inertia of a thin hollow sphere (spherical shell) about a diameter

I = 2MR²/3, radius of gyration k = R√(2/3)

Axis: about a diameter.

kg
m

Derivation

  1. By symmetry the three diameters x, y and z have equal I, and Ix + Iy + Iz = ∫ 2(x² + y² + z²) dm.
  2. Every point is at distance R from the centre, so this sum is 2MR².
  3. So each I = 2MR²/3.

Moment of inertia of standard bodies

Questions students ask

What is the moment of inertia of a thin hollow sphere (spherical shell) about a diameter?

I = 2MR²/3, so the radius of gyration is k = R√(2/3).

How is the moment of inertia of a thin hollow sphere (spherical shell) about a diameter derived?

By symmetry the three diameters x, y and z have equal I, and Ix + Iy + Iz = ∫ 2(x² + y² + z²) dm. Every point is at distance R from the centre, so this sum is 2MR². So each I = 2MR²/3.

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