Moment of inertia of particle systems

Every particle of a rigid body shares one angular velocity ω and one angular acceleration α, but each moves on a circle of its own radius r, so its speed is v = rω and its tangential acceleration is a = rα. Follow one particle through the chain. A 0.20 kg bead 0.30 m from the axis, given α = 5.0 rad s⁻², needs a = rα = 1.5 m s⁻²; that takes a tangential force F = ma = 0.30 N; and that force acting at radius r is a torque rF = 0.090 N m. One factor of r turned α into a linear acceleration and the other turned a force into a torque, so the torque is (mr²)α with mr² = 0.018 kg m². The kinetic energy ½mv² = ½(mr²)ω² collects the same combination. Because ω is shared, it factors out of the sum over particles, K = ½(Σmᵢrᵢ²)ω², and what remains depends on the mass layout alone.

An axis is a line, and rᵢ is the shortest distance from particle i to that line, measured at right angles to it. Sliding a particle parallel to the axis changes nothing. For the z-axis this means rᵢ² = xᵢ² + yᵢ², with zᵢ absent. A bead at (0.30, 0.40, 1.20) m is 0.50 m from the z-axis, so r² = 0.25 m²; its distance from the origin is 1.30 m, and using that would inflate its contribution 6.76 times. About the x-axis the same bead has r² = 0.40² + 1.20² = 1.60 m². For an axis through the origin along any direction û, subtract the part of the displacement that runs along the axis: r² = |d|² − (d ⋅ û)², or equivalently r = |d × û|. Both forms need û to be a unit vector; forget to normalise it and the subtraction removes the wrong amount.

Four 0.50 kg sensors sit at the corners of a light 0.60 m × 0.20 m rectangular frame. About the line through the centre parallel to the long sides, each sensor is 0.10 m away: I = 4 × 0.50 × 0.10² = 0.020 kg m². About the central line parallel to the short sides, each is 0.30 m away and I = 0.18 kg m². About the axis through the centre perpendicular to the frame, r² = 0.30² + 0.10² = 0.10 m², giving 0.20 kg m². About one long edge, two sensors lie on the axis and add nothing while the other two are 0.20 m out: I = 0.040 kg m². Same frame, same mass, a factor of ten between axes. That is why “the moment of inertia of the frame” means nothing until the axis is stated, and why the frame is far easier to start and stop about its long central line than about the perpendicular one.

Put particles on a line at positions xᵢ and let the rotation axis cross that line at right angles at x = s. Then I(s) = Σmᵢ(xᵢ − s)², a quadratic in s. Its derivative dI/ds = −2Σmᵢ(xᵢ − s) = −2M(x_cm − s) vanishes only at s = x_cm, and d²I/ds² = 2M is positive, so the centre of mass gives the smallest I of all these parallel axes. Expanding about that point gives I(s) = I_cm + M(s − x_cm)²: moving the axis a distance D from the centre of mass always adds MD². Take 2.0 kg at x = 0 and 1.0 kg at x = 0.90 m. Then x_cm = 0.30 m and I_cm = 2.0(0.30)² + 1.0(0.60)² = 0.54 kg m², which is μd² with μ = 2/3 kg. About the 2.0 kg bead, I = 0.81 kg m² = 0.54 + 3.0(0.30)²; about the 1.0 kg bead, I = 1.62 kg m² = 0.54 + 3.0(0.60)².

Divide I by the total mass and take the root: k = √(I/M). Because I/M = Σmᵢrᵢ²/Σmᵢ is the mass-weighted mean of r², k is a root-mean-square distance, and like any rms value it is at least as large as the ordinary mean, equal only when every particle sits at the same r. For the pair above, about its centre of mass, k = √(0.54/3.0) = 0.424 m, while the mass-weighted mean distance is (2.0 × 0.30 + 1.0 × 0.60)/3.0 = 0.400 m. About the 2.0 kg end, k = √(0.81/3.0) = 0.520 m, although the centre of mass is only 0.300 m from that axis. The payoff is comparison across designs: a 1.2 kg wheel with k = 0.30 m has I = Mk² = 0.108 kg m², and a lighter wheel can match that only by pushing its k further out.

The sum treats each object as a point, so test that before trusting it. Take an object of mass m whose centre of mass is a distance d from the axis and whose parts all lie within a/2 of that centre. The worst case puts half the mass at d − a/2 and half at d + a/2, and the contribution becomes ½m(d − a/2)² + ½m(d + a/2)² = md² + m(a/2)², because the cross terms cancel. So the point model can be low by at most the fraction (a/2d)². A 10 cm weight centred 1.00 m from the axis is off by at most (0.05)² = 0.25%; centred 0.20 m out it can be off by 6.25%, and the point model has stopped being honest. Two more assumptions hide in the sum: the rods and frame are light compared with the beads, and the distances stay fixed. If a bead slides along its spoke while the body turns, I changes with time and must be evaluated at each instant.