Conserve angular momentum when radius changes

Predict a spin without adding a turning effort

Imagine two equal masses rotating on opposite sides of a vertical axis. A mechanism brings both closer to that axis without exerting a net external torque about it. Predict whether they slow down, maintain their angular velocity or speed up. Do not assume that a conserved quantity must be angular velocity.

In this investigation each point mass is 1 kg. Initially both lie 1 m from the axis. You choose an initial angular velocity and a final radius. The mechanism moves the masses symmetrically, has negligible inertia and supplies no net torque about the axis. Radial motion is stopped at both endpoints. The display compares those endpoints; it does not calculate the intervening path, duration or final angular orientation.

Name the conserved quantity

For this planar rotation, angular momentum about the fixed axis is L = Iω. L has units kg·m²/s; I is rotational inertia in kg·m² and ω is angular velocity in rad/s. Counterclockwise viewed from above is positive. A negative L means clockwise angular momentum, not a negative amount of energy.

Net external torque changes angular momentum. If it is zero throughout the process, initial and final angular momentum are equal: I_initial ω_initial = I_final ω_final. These subscripts identify the endpoints. Choose the same axis and system for both sides. The conservation claim depends on the external-torque condition, not merely on the absence of an obvious motor.

For two 1 kg point masses at the same radius r, I = 2r². At the initial radius of 1 m, I is 2 kg·m². Starting at +2 rad/s gives L = 2 × 2 = +4 kg·m²/s. OpenStax angular-momentum conservation explains the zero-external-torque condition and the resulting changes when inertia varies.

Pull inward and check the product

Set final radius to 0.5 m while keeping initial angular velocity +2 rad/s. Final inertia is 2 × 0.5² = 0.5 kg·m². Dividing conserved angular momentum by final inertia gives final angular velocity 4/0.5 = 8 rad/s. Halving radius reduces inertia to one quarter, so angular velocity increases fourfold.

Check both ledger products: 2 × 2 = 4 initially and 0.5 × 8 = 4 finally. The angular-momentum bars match on the same signed scale even though the angular velocities differ. Each mass’s speed changes from 1 × 2 = 2 m/s to 0.5 × 8 = 4 m/s. Linear speed doubles while angular velocity quadruples; confusing these quantities produces a different prediction.

The endpoint diagrams use a common distance scale. They deliberately align the two configurations so you can compare radius. They do not predict that the masses finish on the original line. Angular orientation requires information about how long and how the transition occurs.

Move outward, reverse direction and test rest

Capture radius 2 m with the same initial +2 rad/s. Final inertia is 8 kg·m², angular velocity is 0.5 rad/s and each mass’s speed is 1 m/s. The product 8 × 0.5 still gives L = 4 kg·m²/s. Greater inertia gives smaller angular velocity under this conservation condition.

Reverse initial angular velocity to −2 rad/s and repeat the inward case. Angular momentum becomes −4 kg·m²/s and final angular velocity −8 rad/s. Both endpoint kinetic energies remain nonnegative. With positive inertias, changing radius alone cannot reverse the spin direction while preserving nonzero L.

Finally start at zero angular velocity. L is zero and the final angular velocity remains zero for every selectable positive radius. The radial mechanism cannot create net angular momentum under the stated conditions. Radial motion and energy could occur during the transition, but neither is modeled; both endpoints have zero radial velocity.

Why zero torque does not always mean zero angular acceleration

The earlier equation τ = Iα applies in its simple form when I is constant. With changing inertia, use torque as the rate of change of Iω. During contraction, ω changes even though the product Iω stays constant. Applying τ = Iα while ignoring the changing I would incorrectly predict no angular acceleration.

The conservation equation is an endpoint constraint, not a complete dynamical simulation. To calculate acceleration and forces during the transition, we would need the radius as a function of time and a mechanical model. The current controls do not supply those details. Frictional torque, an inertia-bearing mechanism or angular momentum carried away by detached material would require different accounting.

Optional derivative connection

Differentiating L = Iω gives τ = Iα + ω dI/dt. At zero torque, a decreasing positive I requires increasing ω for a positive initial spin. This expression explains the changing-inertia term; it does not determine α numerically without a rate of radius change.

Transfer to a new inertia

A planar rotor changes from I = 3 kg·m² at −4 rad/s to I = 2 kg·m² with zero net external torque about its fixed axis. Predict final angular velocity. Then decide whether the same calculation remains justified if a bearing applies an unknown frictional torque during the change.

Check the angular momentum transfer

Initial L = 3 × (−4) = −12 kg·m²/s. Final angular velocity is −12/2 = −6 rad/s. The direction remains clockwise. An unknown external bearing torque can change angular momentum, so its angular impulse must be known or bounded before the same final prediction can be justified.