Find the energy source during a change of radius

A conserved angular momentum can accompany increasing energy

The inward trial starts with two 1 kg masses at radius 1 m and angular velocity +2 rad/s. It ends at radius 0.5 m and +8 rad/s. Angular momentum stays +4 kg·m²/s. Predict the energy at each endpoint before examining the ledger. The presence of one conservation law does not automatically make every quantity constant.

Both endpoints have zero radial velocity, so their total particle kinetic energy is rotational kinetic energy K = ½Iω². During the transition the masses could also have radial kinetic energy; that intermediate motion is not calculated here.

Find the missing energy source

Initially I = 2 kg·m², so K_initial = ½ × 2 × 2² = 4 J. Finally I = 0.5 kg·m², so K_final = ½ × 0.5 × 8² = 16 J. The increase is 12 J. The radial mechanism must do net positive work of 12 J on the selected two-mass system under the ideal assumptions.

The accompanying mechanism energy change is −12 J. This is a change in its energy reservoir, not a claim that its absolute stored energy becomes negative. The masses gain what the ideal mechanism supplies. Real devices can also dissipate energy, so electrical or biological energy consumption need not equal this ideal mechanical work.

The OpenStax treatment of changing inertia distinguishes angular-momentum conservation from kinetic-energy change. Here the two-mass endpoint example makes the energy transfer explicit rather than attributing the gain to conservation itself.

How can a zero-torque force do work?

Torque about the axis depends on the perpendicular lever arm of a force. A radial force has zero torque about that axis. Work instead depends on force along displacement. While a mass moves inward, an inward force can act along part of its displacement and do positive work despite zero torque.

For a fixed-radius circle, a radial force is perpendicular to motion and does no work. For a changing radius, the path has a radial component. These are different geometries. The fixed-axis rigid-body formula W = τΔθ does not account for this radial work during deformation; it cannot be applied as though the radius remained fixed.

The selected system in the ledger is the two masses. The mechanism exchanges energy with that system. If we include the mechanism and its reservoir in a larger ideal system, the ledger’s equal-and-opposite energy changes balance. Choosing a larger system changes whether the transfer is called external work or internal conversion, but does not remove the need to account for it.

Let the masses move outward

Keep the initial +2 rad/s but set final radius to 2 m. Angular-momentum conservation gives final I = 8 kg·m² and ω = 0.5 rad/s. Final K = ½ × 8 × 0.5² = 1 J. The mechanism does −3 J of net work on the masses and gains 3 J in the ideal energy ledger.

Negative work means energy leaves the masses. It does not automatically mean that a real apparatus stores the energy usefully; it could dissipate it. This investigation specifies an ideal exchange and does not choose a motor, brake, spring or recovery efficiency. Nor does an outward move imply that the mechanism always pushes outward: force direction and detailed motion depend on the chosen protocol.

Compare inward and outward captures. Explain the signs of work using endpoint kinetic energies. Reverse the initial spin and repeat: signs of angular velocity and angular momentum reverse, but the same radius change gives the same kinetic energies and net work.

Use the conservation condition in the energy formula

With L fixed and I positive, replacing ω with L/I gives K = L²/(2I). Read this as: at a specified angular momentum, smaller inertia means greater kinetic energy. This differs from K = ½Iω² interpreted at specified angular velocity; in that comparison greater inertia means greater energy. Always say which variable is held constant.

At final radius 0.5 m, the energy ratio is 4 for every nonzero initial spin in the widget. At radius 2 m, it is 0.25. At zero initial spin both endpoint energies are zero, so dividing final energy by initial energy is undefined. A ratio of zero divided by zero must not be displayed as evidence of conservation or as a percentage gain.

The net work alone does not determine power. Power needs a rate of transfer, and the experiment supplies no transition duration or force history. Two mechanisms can accomplish the same ideal endpoint change with different power requirements and losses.

Transfer to an unfamiliar energy budget

A rotor with initial inertia 4 kg·m² spins at +3 rad/s. It changes to inertia 1 kg·m² with zero external torque, and no radial motion remains at either endpoint. Find final angular velocity, initial and final kinetic energies, and net work on the rotating masses. State one extra fact needed to estimate average power.

Check the redistribution energy transfer

L = 4 × 3 = 12 kg·m²/s. Final angular velocity is 12/1 = 12 rad/s. Initial kinetic energy is ½ × 4 × 3² = 18 J; final kinetic energy is ½ × 1 × 12² = 72 J. Net work is +54 J, with an ideal reservoir change of −54 J. The transition duration is needed for average mechanical power, 54 J divided by that duration. Real input-power estimates also need losses or efficiency.