Integrate rotational energy without double counting

Start with particles, not a memorized shape

For particles sharing a fixed-axis angular speed, each speed is v_i=r_i omega. Summing K_i=m_i v_i²/2 gives K=omega² Σm_i r_i²/2=I omega²/2. Energy is scalar and nonnegative; reversing every velocity reverses angular momentum but leaves kinetic energy unchanged. In the continuum, dK=(omega x)² dm/2 for a rod rotating about its perpendicular endpoint axis.

Hold mass fixed while changing shape

The model rod has M=3 kg, L=2 m and lambda(x)=lambda0(1+beta x/L). Normalize first: M=lambda0 L(1+beta/2), so lambda0=M/[L(1+beta/2)]. Then Iend=integral x² lambda dx=ML²(1/3+beta/4)/(1+beta/2). Comparing beta values without renormalizing would mix redistribution with adding material.

Separate center motion only once

The center is xCM=L(1/2+beta/3)/(1+beta/2). With Icm=Iend−M xCM² and vCM=omega xCM, K=Iend omega²/2=M vCM²/2+Icm omega²/2. These are two equivalent accounts, not three terms to add. The endpoint-axis expression already includes center translation. For general rigid planar motion, center translation plus center-axis rotation is the appropriate split.

Make a graph prediction

The plotted density dK/dx is proportional to x² lambda(x), not just lambda(x). Explain why the endpoint near the axis contributes little even when its mass density is nonzero. Predict what happens to the graph and area when omega doubles, and what happens when omega changes sign. Doubling speed multiplies every energy density by four; sign reversal leaves the graph unchanged.

Test a stationary center

A uniform disk with Icm=2 kg m² spins at 3 rad/s while its center stays fixed. Predict its energy before using the formula: K=9 J, despite zero center translational energy. The rim particles still move. Changing the spin to −3 rad/s leaves 9 J unchanged; kinetic energy has no clockwise or counterclockwise sign.

Worked original case

At beta=0, Iend=4 kg m² and xCM=1 m. With omega=2 rad/s, K=8 J. The center contribution is 6 J and center-axis contribution is 2 J. Adding another 6 J to the endpoint result would count the same motion twice.

Investigation

Compare beta=0,1,2 at omega=2. Integrate the normalized density independently, calculate the two energy decompositions and verify their sums. Reverse omega and retain the changed angular-momentum sign alongside unchanged energy. Explain why holding lambda0 fixed would answer a different question.

Public numerical transfer practice

Try a new practice case

A fixed-axis body has I=3 kg m² and omega=−4 rad/s. What is its rotational kinetic energy in joules?

Numerical result in the requested units Check this practice result

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

Sources and evidence boundary

Original derivations and assigned numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax angular momentum. These are supplied or generated records, not physical measurements or performed laboratory time. Do not construct rotating equipment, release loads or conduct launch experiments. Public feedback and retained self-reports do not certify independent mastery.