Angular Momentum and Its Conservation
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Direct answer
An ice skater who pulls in her arms spins faster without any external push — the visible face of angular momentum conservation. For a rigid body L = Iω; for a particle L = mvr about the chosen axis, directed along the axis by the right-hand rule. Since τ = dL/dt, angular momentum stays constant whenever the net external torque about that axis is zero: conservation reads I1ω1 = I2ω2, so pulling mass toward the axis lowers I and raises ω, the kinetic energy changing only because the pulling muscles or gravity perform work. Gravity exerts no torque about the sun, which is why each planet conserves L and sweeps out equal areas in equal times.
What you must remember
- Two forms: rigid body L = Iω; particle L = mvr sin θ about the axis; the rotational analogue of p = mv.
- Rotational Newton's second law: τ = dL/dt; if τ_ext = 0 about an axis, L about that axis is conserved even while I and ω individually change.
- The skater effect: arms pulled in decrease I, so ω increases by the inverse factor; kinetic energy (1/2) Iω^2 rises, supplied by muscular work done against the centrifugal tendency.
- Planetary link: gravity is central, so L = mvr is constant; the areal velocity dA/dt = L/2m, which is Kepler's second law in one line.
- Kinetic energy bookkeeping: with L fixed, KE = L^2/2I, so halving I doubles the energy — the examiner checks whether you know where that energy came from.
- Bohr connection: in the hydrogen atom the electron's angular momentum is quantised as mvr = nh/2π — the same quantity in a different chapter.
- Units: kg m^2 s^-1 for L, N m for τ; both are axial vectors.
A platform problem from start to finish
A man stands on a freely rotating platform with his arms extended: I1 = 10 kg m^2 and ω1 = 6 rad s^-1. He pulls his arms in, dropping the moment of inertia to I2 = 4 kg m^2. No external torque acts about the vertical axis (the axle is frictionless in the idealisation), so L = I1ω1 = 60 kg m^2 s^-1 survives unchanged and ω2 = 60/4 = 15 rad s^-1 — two and a half times faster. The energy audit is the part NEET loves: kinetic energy goes from (1/2)(10)(36) = 180 J to (1/2)(4)(225) = 450 J, an increase of 270 J supplied by the work his muscles do pulling the masses inward. Nothing is violated; the same conservation that holds ω in check while I changes permits the energy to rise, because conservation applies to L, never to rotational kinetic energy on its own.
Where the questions twist
The recurring trap is conserving the wrong quantity. When a droplet or a child lands on a merry-go-round, or an insect crawls on a rotating disc, angular momentum about the axle is conserved but the event is fully inelastic — rotational kinetic energy drops, and a question asking for the "energy lost" expects you to compute both states from (1/2) Iω^2. The second twist is the direction of change: a person walking outward on the platform increases I and therefore slows ω — candidates memorise "pull in, spin faster" and then misapply it to the opposite case. Third, the conservation axis matters: gravity exerts zero torque about the centre of a planet's orbit but not about an arbitrary point, so L is conserved about the sun specifically. Finally, the parallel-axis trap: if the platform plus man is given as I about the centre but the man walks along a tangent, compute the new I from Σmr^2 afresh rather than patching the old value.
Frequently asked questions
Under what condition is angular momentum conserved?
When the net external torque about the chosen axis is zero; internal torques, like internal forces, cancel in pairs.
Why does a skater spin faster after pulling in the arms?
Pulling mass toward the axis lowers I, and constancy of L = Iω forces ω to rise in inverse proportion.
Where does the extra rotational kinetic energy of the skater come from?
From muscular work done against the outward centrifugal tendency while pulling the arms in — L is conserved, energy is not.
How does angular momentum conservation explain Kepler's second law?
The sun's gravitational force is central, so τ = 0 and mvr is constant; the areal velocity dA/dt = L/2m is therefore constant, giving equal areas in equal times.
What happens to the angular speed when a mass is dropped onto a rotating disc?
The moment of inertia increases by mr^2, so ω falls by the factor I_old/(I_old + mr^2) while angular momentum stays fixed and kinetic energy decreases.