Conservation of angular momentum

Differentiate L⃗ = Σ r⃗ᵢ × p⃗ᵢ about a fixed point O. The v⃗ᵢ × p⃗ᵢ terms vanish because the two vectors are parallel, leaving the torque of every force on each particle, and each force is either external or exerted by another particle of the system. Newton's third law pairs the internal ones: F⃗ on particle 1 from particle 2 and −F⃗ on 2 from 1, both along the line joining them. Their torques sum to r⃗₁ × F⃗ + r⃗₂ × (−F⃗) = (r⃗₁ − r⃗₂) × F⃗ = 0, because r⃗₁ − r⃗₂ is parallel to F⃗. So dL⃗/dt = Στ⃗ext, and the size of the internal forces appears nowhere. Clutch plates may grip with hundreds of newtons and a crumpling puck may peak at thousands; neither number is needed. What does need care is the boundary. Draw it round everything that pushes on everything else, and audit only the forces that cross it.

For each force crossing the boundary, ask whether its line of action passes through your point or runs parallel to the axis you care about. A pivot pin's force passes through the pivot; an axle bearing's force passes through the axle. Gravity on a free body acts at the centre of mass, so it has no torque about that point, and dL⃗/dt = Στ⃗ext still holds about the centre of mass while it accelerates. Because the law is a vector equation it can hold for one component alone: a vertical force has no torque about a vertical axis, so a turntable conserves L_z however gravity and its bearing push. What usually survives the audit is bearing friction, and it is judged by size, comparing the angular impulse τΔt with L. A 0.020 N m drag acting for 3.0 s removes 0.060 kg m² s⁻¹ from a platform carrying 12 kg m² s⁻¹, only 0.5%, so call L conserved. Over a 300 s spin-down the same drag removes 6.0 kg m² s⁻¹, half of it, and the law no longer describes the motion.

When a body changes shape about a fixed axis with no external torque about that axis, I₁ω₁ = I₂ω₂, and the path between the two shapes never enters. Build I for each shape from its parts, with parallel-axis terms for anything off the axis. A satellite spinning at 3.0 rad s⁻¹ with I = 40 kg m² unfolds two booms, each carrying a 4.0 kg instrument pod out to 3.5 m. The pods add 2 × 4.0 × 3.5² = 98 kg m² and the booms themselves 12 kg m², so I₂ = 150 kg m² and ω₂ = 3.0 × 40 ÷ 150 = 0.80 rad s⁻¹. Nothing about the deployment, how fast the hinges swung or whether they jerked, appears in the answer, because every force the hinges exert is internal. Nor can any internal trick restore the old spin with the pods left out: that takes hauling them back in, or thrusters whose exhaust carries angular momentum away.

At fixed L, K = L²/(2I), so halving I doubles K and something must pay the difference. Take a 0.25 kg puck circling an air table at r = 0.80 m and 2.0 rad s⁻¹ on a cord that runs down through a hole at the centre. The cord pulls radially, so it has no torque about the hole, and L = mr²ω = 0.32 kg m² s⁻¹ holds while the cord is drawn slowly in. Keeping the puck on its circle needs tension T = mω²r = L²/(mr³), which climbs as 1/r³: 0.80 N at r₁ = 0.80 m, 6.4 N at r₂ = 0.40 m. The work done is W = ∫ L²/(mr³) dr from r₂ to r₁ = L²/(2m) × (1/r₂² − 1/r₁²) = 0.2048 kg m⁴ s⁻² × (6.25 − 1.5625) m⁻² = 0.96 J, exactly the rise in K from 0.32 J to 1.28 J. The spin reached 8.0 rad s⁻¹ because of L; the energy to get there came from the hand on the cord.

During an impact lasting milliseconds, finite forces such as gravity deliver negligible angular impulse, so the audit shrinks to the impulsive external forces. A pinned bar has exactly one, at the pin. About the pin that force has no lever arm, so L is conserved there: m v d = (I_P + m d²) ω for a puck of mass m and speed v that sticks square-on at distance d. Linear momentum is not conserved, because the pin can deliver an impulse J. For a uniform bar of mass M and length ℓ, momentum bookkeeping gives J = m v M ℓ (d/2 − ℓ/3)/(I_P + m d²) along the puck's velocity: backward for a hit inside 2ℓ/3, forward outside it, and zero at d = 2ℓ/3 whatever the puck's mass. That point is the centre of percussion. About the bar's centre the pin's impulse has lever arm ℓ/2, so L about the centre changes by an amount you do not know. The same collision is a conservation law about one point and an unknown about the other.

Bring two rotors on one axis together by friction, as clutch plates or a disk dropped onto a turntable, and the friction torques are internal, so ω_f = (I₁ω₁ + I₂ω₂)/(I₁ + I₂). Kinetic energy is not shared so kindly. Subtracting before from after gives ΔK = −I₁I₂(ω₁ − ω₂)²/[2(I₁ + I₂)], negative for any mismatch in rates: the rotational form of a perfectly inelastic collision, with I₁I₂/(I₁ + I₂) standing in for the reduced mass. For two equal disks, one at rest, exactly half goes. A 0.050 kg m² disk at 90 rad s⁻¹ joining an identical disk ends at 45 rad s⁻¹, and K falls from 202.5 J to 101.25 J, the difference heating the sliding faces. So the order of work is fixed: take the final rate from L, take the heat from the energy difference, and never find the rate from energy.