Rotational Kinetic Energy

Start from what you already trust: a particle of mass mᵢ moving at speed vᵢ has kinetic energy ½mᵢvᵢ², and the energy of a collection is the sum. What a rigid body adds is a constraint. Turning about a fixed axis, every particle moves on a circle around that axis and they all sweep out angle at the same rate ω, so vᵢ = ωrᵢ with rᵢ the perpendicular distance to the axis. Then K = Σ½mᵢ(ωrᵢ)² = ½ω²Σmᵢrᵢ² = ½Iω². Test it on a light rod carrying 0.20 kg at 0.10 m and 0.30 kg at 0.40 m from the axis, turning at 5.0 rad s⁻¹. The speeds are 0.50 and 2.0 m s⁻¹, the energies 0.025 J and 0.60 J, total 0.625 J. The short way: I = 0.20 × 0.10² + 0.30 × 0.40² = 0.050 kg m², and ½ × 0.050 × 5.0² = 0.625 J. The outer bead holds 96% of it.

I is not a property of the body alone; it belongs to the body and an axis, so K does too. A uniform rod of mass 1.2 kg and length 0.80 m turning at 6.0 rad s⁻¹ about its centre has I = ML²/12 = 0.064 kg m² and K = ½ × 0.064 × 6.0² = 1.15 J. Turn it at the same rate about one end and I = ML²/3 = 0.256 kg m², so K = 4.61 J: four times the energy at the same ω, because the far half now moves much faster. The other habit to fix here is units. The step vᵢ = ωrᵢ is true only for ω in rad s⁻¹; the radian is a ratio of two lengths, so kg m² s⁻² is a joule with nothing left over. A motor rated 3000 rev min⁻¹ runs at 3000 × 2π/60 = 314 rad s⁻¹. Feed 3000 into ½Iω² and the answer is about 91 times too big.

A thrown spanner turns about no fixed axis, but the particle sum still works. Write each velocity as vᵢ = v_cm + vᵢ′, where vᵢ′ is measured by an observer moving with the centre of mass. Then vᵢ² = v_cm² + 2v_cm ⋅ vᵢ′ + vᵢ′², and summing with ½mᵢ gives three pieces. The first is ½Mv_cm². The last is the kinetic energy seen from the centre of mass, where the body does nothing but turn, so it is ½I_cmω². The middle piece is v_cm ⋅ Σmᵢvᵢ′, and Σmᵢvᵢ′ is the time derivative of Σmᵢrᵢ′, which is zero because mass-weighted positions measured from the centre of mass sum to zero at every instant. So K = ½Mv_cm² + ½I_cmω², exactly. The centre of mass is the one moving reference point for which the cross term vanishes whatever the motion; split about any other moving point and that term generally survives.

Take the rod from before, swinging about one end at 6.0 rad s⁻¹. The pivot description gave K = ½I_endω² = 4.61 J, and the centre-of-mass description must agree. The centre moves on a circle of radius L/2 = 0.40 m, so v_cm = 6.0 × 0.40 = 2.4 m s⁻¹ and ½Mv_cm² = ½ × 1.2 × 2.4² = 3.46 J. The rod also turns about its own centre at the same ω, since a rigid body has one angular velocity whichever point you watch it from, and that adds ½ × 0.064 × 6.0² = 1.15 J. The total is 4.61 J again. In symbols this is the parallel-axis theorem multiplied by ½ω²: ½(I_cm + Md²)ω² = ½I_cmω² + ½M(ωd)². So ½I_Pω² already contains the motion of the centre of mass. Adding ½Mv_cm² to it gives 8.06 J and counts three quarters of the energy twice.

With K in hand, the energy methods you already know reach rotating bodies. Hold the same 0.80 m rod horizontal on a frictionless pivot at one end and let go. The pivot force acts at a point that does not move, so it does no work, and gravity's potential energy depends only on the height of the centre of mass, which falls L/2 by the time the rod is vertical. Then Mg(L/2) = ½(ML²/3)ω², so ω = √(3g/L) = √(3 × 9.81/0.80) = 6.07 rad s⁻¹, with no mass in the answer. The tip moves at ωL = 4.85 m s⁻¹, faster than the 3.96 m s⁻¹ of a stone dropped through the same 0.80 m, while the centre moves at 2.43 m s⁻¹, slower than the 2.80 m s⁻¹ of a stone dropped 0.40 m. The rigid constraint shares the energy along the rod; no particle of it is in free fall.

K = ½Iω² rewards speed and radius, not mass as such. A 12 kg solid disc of radius 0.25 m has I = ½MR² = 0.375 kg m²; at 3000 rev min⁻¹, ω = 314 rad s⁻¹ and K = ½ × 0.375 × 314² = 18.5 kJ. Move the same 12 kg out into a thin rim and I doubles to 0.75 kg m², so K = 37.0 kJ. Double the spin rate instead and either figure quadruples. Writing I = βMR² shows the pattern: K = ½βM(ωR)² = ½βMvᵣᵢₘ², so for a given shape factor β the energy per kilogram is fixed by rim speed alone, here 78.5 m s⁻¹. That is also where the model gives out. Real rims burst when the hoop stress, which grows as ρvᵣᵢₘ², exceeds what the material can carry, and well before that a wheel that flexes or a tank of sloshing liquid stops having one ω for all its parts, and ½Iω² with a fixed I stops describing it.