Integrate torque over an accessible angular path

Derive the angular work element

For a tangential displacement ds=r dtheta and tangential force F_t, dW=F_t ds=(r F_t)dtheta=tau dtheta. Only the torque component about the chosen fixed axis enters this planar expression. Integrating net torque over angle gives net work; integrating one selected torque gives only its work, not necessarily the total energy change.

Connect the integral to kinetic energy

For fixed I, tau=I dω/dt and dtheta=omega dt. Thus tau dtheta=I omega domega, and integral tau dtheta=I(omega_f²−omega_i²)/2. The model uses I=2 kg m², initial K=4 J and tau(theta)=A−B theta. Therefore K(theta)=4+A theta−B theta²/2. This is an energy equation, not a claim that every theta is physically reachable.

Stop the selected positive branch

The body initially rotates positively. For B>0, its first positive zero of K is theta_stop=[A+sqrt(A²+8B)]/B. The plotted branch ends at the smaller of this root and the target 2 rad. Beyond a zero, blindly taking a square root of negative K is invalid. A subsequent reversal must be treated with its own direction and dynamics; the energy expression alone does not specify elapsed time.

Read power at an instant

P=dW/dt=tau omega, so a negative torque can remove energy while positive rotation continues. Zero angular speed means zero instantaneous power even if torque is nonzero. The area under torque versus angle is work; the area under torque versus time is angular impulse. These axes cannot be exchanged without a motion relation.

Worked original case

With A=0 and B=4, K=4−2theta² reaches zero at sqrt(2) rad, before the target 2. Net work to that first stop is −4 J and final omega is zero. Extending the initial branch to 2 would wrongly demand K=−4 J.

Investigation

Compare A=0,B=4 with A=4,B=0. Derive work and accessible endpoint before using the workbench. Sketch signed torque-angle areas and mark where instantaneous power vanishes. Retain a rejected target and explain why clipping negative energy is not a valid continuation.

Public numerical transfer practice

Try a new practice case

A fixed I=2 kg m² body starts with K=4 J. Constant net torque 3 N m acts through positive angle 2 rad. What is final kinetic energy in joules?

Numerical result in the requested units Check this practice result

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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.