Separate translational and rotational kinetic-energy ledgers

1. Frame the question

A spinning body can have kinetic energy even when its center of mass is at rest. Each small mass element moves at speed r|ω| relative to a fixed axis. Adding its energy gives Σ½m(rω)²=½(Σmr²)ω²=½Iω². This derivation uses the rigid body’s common angular velocity and the same axis for all particle distances.

2. Build the mechanism

Kinetic energy is scalar and nonnegative: reversing angular velocity leaves ½Iω² unchanged. Angular momentum, introduced next, retains a planar sign and is not interchangeable with energy. Doubling angular speed quadruples rotational kinetic energy at fixed inertia; doubling inertia only doubles it if angular speed is held fixed.

For a freely translating and rotating rigid body, total kinetic energy is ½MvCM²+½ICMω², with rotational inertia specifically about the center of mass. The decomposition follows from adding particle velocities relative to the center: the cross terms cancel because the mass-weighted relative velocities sum to zero. A supplied center-axis inertia must not be replaced by an unrelated pivot inertia.

A fixed-pivot body may instead be described by the complete expression ½Ipivotω², which already includes every particle’s motion about that pivot. Adding another center-of-mass translation term to that expression would double-count energy. Choose one consistent decomposition, show the reference axis, and check the physical constraint before combining terms.

3. Work an example

A disk with mass 2 kg and center-axis inertia 1 kg m² moves with center speed 3 m/s and spins at 4 rad/s. Translation contributes ½×2×9=9 J and rotation ½×1×16=8 J, for 17 J total. If the center is held at rest while that spin remains, the rotational 8 J does not disappear.

4. Predict and investigate

Predict first: Can two systems have the same total kinetic energy but different divisions between translation and spin?

Compare (vCM,ω)=(3,4) with (1,4), then with (3,0). Keep separate subtotals, predict a factor change before moving a control, and explain why no rolling relation is imposed on this free-body snapshot. Retain settings and raw evidence in the lesson notebook.

5. Transfer and explain

Try a new practice case

A body has M=4 kg, vCM=2 m/s, ICM=2 kg m² and ω=3 rad/s. Find total kinetic energy.

Your numerical result (J) Check number

Reveal worked check (practice, not mastery)

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

How would you avoid double-counting for a rod pivoted at one end? State which axis belongs in each alternative energy expression.

Retain your first response, system and axis choices, separate scalar energy and signed angular-momentum ledgers, assumptions and a changed-case prediction. Public worked checks are practice, not independent mastery evidence. Synthetic captures are not a performed physical experiment.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 6 topic mapping: 6.1; 6.5. Checked 30 October 2026. Planar signed analysis without required three-dimensional angular-momentum direction. Slipping evolution is qualitative, in accordance with the framework boundary; no rolling-resistance model. Framework cross-reference, not AP authorization.