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Build Any Moment of Inertia from Standard Shapes Plus Two Theorems

Moment of inertia I = Σ m r² depends on how mass is spread about the chosen axis. For almost every problem you build I from a short list of standard shapes, shift axes with the parallel-axis theorem, and use the perpendicular-axis theorem for flat bodies.

NSEJS · Rotational Motion · 12 linked questions · 100% coverage

Know this first

Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.

Distance · Length of the path travelledDisplacement · Straight-line change in positionResultant force · Net force including direction

Study deeper: NCERT Science · Rotational Motion

Standard values and radius of gyration

About the symmetry axis: ring M R², solid disc or solid cylinder M R²/2, solid sphere (2/5) M R², hollow sphere (2/3) M R². A thin rod is M L²/12 about its centre and M L²/3 about one end.

The radius of gyration k is defined by

I = M k²: a 10 kg disc with k = 0.20 m has I = 10 x 0.04 = 0.4 kg m²

A thin rod of length L spun about one end at n revolutions per second has

ω = 2πn and kinetic energy (1/2)(mL²/3)(4π²n²) = (2/3) m L² π² n²

Adding, shifting and flat bodies

Moments of inertia about the same axis add: a disc of mass M with four point masses m on its rim has

I = (1/2) M R² + 4 m R²

The parallel-axis theorem

I = I_{cm} + M d²

For a square frame of four rods of mass M and side ℓ about the central perpendicular axis, each rod has I = Mℓ²/12 + M(ℓ/2)² = Mℓ²/3, so the frame has (4/3) M ℓ². For three identical shells each with I_cm = (2/3) m r², add m d² for the ones whose centres are displaced from the axis: an axis touching two shells and passing through the diameter of the third gives 4 m r².

For a flat body in the x-z plane, the perpendicular-axis theorem says Iₓ +

I_z = I_y, so if Iₓ = 80 and I_y = 100 kg m², then I_z = 20 kg m²

For point masses, take each distance from the axis: masses at (0,0), (a,a), (2a,0) and (a,-a) about the y-axis have distances 0, a, 2a and a, giving I = 6 m a².

Comparing shapes and choosing axes

For equal moments of inertia about diameters, (2/5) M rₛ² = (2/3) M rₕ², so the ratio of the solid sphere radius to the hollow sphere radius is sqrt 5 : sqrt 3. A solid cylinder has the same I about its own axis and about a perpendicular axis through the centre when L²/12 + R²/4 = R²/2, which gives L = sqrt(3) R. A body has the least moment of inertia about the axis that keeps most of its mass closest to the axis, and two masses on a light rod need the least work to spin when the axis passes through their centre of mass, since I is then smallest.

OlympIQ exam lens

✓Ring M R², disc/cylinder M R²/2, solid sphere 2M R²/5, hollow sphere 2M R²/3, rod M L²/12 (centre) and M L²/3 (end)
✓I = M k²; moments about the same axis add
✓Parallel axis: I = I_cm + M d²; perpendicular axis for flat bodies: I_z = Iₓ + I_y
✓For point masses use I = Σ m r² with r measured from the axis, not from the origin
✓Least I and least work to spin: axis through the centre of mass

OlympIQ trap

Do not use the distance from the origin as r in I = Σ m r². The distance must be the perpendicular distance from the chosen axis, so a mass on the axis contributes zero, and the same masses give different I for different axes.

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