Angular Momentum of a Rigid Body

Treat the body as particles of mass mᵢ locked at fixed separations, and take angular momentum about an origin on the axle, which runs along z. Particle i sits at perpendicular distance Rᵢ from the axle and moves on a circle at speed vᵢ = ωRᵢ, tangent to that circle. Split its position vector into a part along the axle and a part of length Rᵢ across it. Only the part across the axle, crossed with mᵢvᵢ, points along z, and it contributes mᵢRᵢvᵢ = mᵢRᵢ²ω. Height up the axle never enters, which is why sliding the origin along the axle leaves L_z unchanged. Rigidity gives every particle the same ω, so the sum factorises: L_z = (Σ mᵢRᵢ²)ω = Iω. On a light turntable at 2.0 rad s⁻¹, a 3.0 kg block at 0.40 m contributes 0.96 kg m² s⁻¹ and a 1.0 kg block at 0.80 m contributes 1.28 kg m² s⁻¹. The lighter block carries more, because it sits twice as far out.

A real body is continuous, so Σ mᵢRᵢ² becomes I = ∫R² dm. The skill is in the slicing: choose pieces on which all the mass sits at one distance from the axis, then write dm with a density. For a thin rod use dm = λ dx; for a flat disc use thin rings, dm = σ × 2πr dr; for a solid use shells or slabs with dm = ρ dV. A uniform disc of mass M and radius R has σ = M/(πR²), so dm = (2Mr/R²) dr, and integrating r² × (2Mr/R²) dr from 0 to R gives (2M/R²)(R⁴/4) = ½MR². A thin hoop of the same mass and radius has all of its mass at r = R, so I = MR². Spin each at 12 rad s⁻¹ with M = 2.0 kg and R = 0.25 m: the disc carries L = 0.75 kg m² s⁻¹ and the hoop 1.5 kg m² s⁻¹. Same mass, same size, same ω, and twice the angular momentum, because the disc keeps much of its mass near the axle where R² is small.

Any moment of inertia can be written I = Mk², where the radius of gyration k is the radius of a thin hoop with the same mass and the same I. Then L_z = Mk²ω, and comparing bodies becomes comparing lengths. A uniform disc on its axle has k = R/√2 = 0.707R; a solid sphere about a diameter, √(2/5) R = 0.632R; a thin rod of length ℓ about its centre, ℓ/√12 = 0.289ℓ; the same rod about one end, ℓ/√3 = 0.577ℓ. The same length tells a designer where to put material. A flywheel with most of its mass in a heavy rim might reach k = 0.90R, and at the same mass, radius and ω it carries (0.90/0.707)² = 1.62 times the angular momentum of a uniform disc. That is why flywheels are heavy rims on light spokes rather than solid plates.

A moment of inertia belongs to a body and an axis together. For any body, L about a fixed origin splits exactly into two parts: the centre of mass moving as a particle, r_cm × Mv_cm, plus the angular momentum about the centre of mass. Take a uniform rod of mass M and length ℓ swinging about a fixed axle through one end. Its centre moves on a circle of radius ℓ/2 at speed ωℓ/2, which gives an orbital part M(ℓ/2)²ω = Mℓ²ω/4. The rod also turns at ω about its own centre, which gives a spin part (Mℓ²/12)ω. Together L = (Mℓ²/12 + Mℓ²/4)ω = (Mℓ²/3)ω: the parallel-axis theorem I = I_cm + Md², written for angular momentum. Treating the rod as a point at its centre keeps only the orbital part and loses a quarter of L; using I_cm for an axle at the end keeps only the spin part and loses three quarters.

L_z = Iω is one component, so look at the whole vector. A particle at (x, y, z) turning about the z axis has velocity ω × r = ω(−y, x, 0), and m r × v = mω(−xz, −yz, x² + y²). Summing over the body returns L_z = Iω, and also Lₓ = −ω Σ mxz and L_y = −ω Σ myz. These cross-axis sums vanish whenever the mass at (x, y, z) is matched by equal mass at (−x, −y, z), as it is for any body symmetric about the axle: a wheel, disc, cylinder or sphere on its own axis, or a rod turning about a perpendicular line through its centre. Whenever both sums vanish, and only then, L is parallel to ω and L = Iω holds as a vector equation. Symmetry is the usual way to secure that, but not the only one: a flat body on an axle perpendicular to it has z = 0 for every particle, which is why the rod on an end axle above still carried L along its axle. Its direction follows the right-hand rule: curl the fingers with the rotation and the thumb points along both ω and L.

Now break the symmetry. Fix a light rod of half-length a, with mass m at each end, to a vertical axle through its midpoint so that the rod makes angle θ with the axle. At the instant the rod lies in the xz plane, the masses sit at ±(a sin θ, 0, a cos θ), so Σ mxz = 2ma² sin θ cos θ and L = 2ma²ω sin θ (−cos θ, 0, sin θ). That vector is perpendicular to the rod and makes 90° − θ with the axle. With m = 0.50 kg, a = 0.20 m, ω = 10 rad s⁻¹ and θ = 60°: I = 2m(a sin θ)² = 0.030 kg m², L_z = Iω = 0.30 kg m² s⁻¹ and L_⊥ = 0.17 kg m² s⁻¹, so L leans 30° off the axle. L_⊥ is locked to the rod and sweeps round the axle once per revolution, so L keeps changing direction although ω is constant. Something must drive that change; the next topic identifies it as a torque from the bearings, which is why wheels are balanced with weights on both faces.