Integrate rotational relaxation and its work budget

Integrate rotational relaxation and its work budget

For a rigid body rotating about a fixed axis with constant inertia, torque balance is I dω/dt = τnet. A constant applied torque does not imply constant net torque if resistance changes with angular speed. The supplied body has I = 2 kg m², drive torque 4 N m and resisting torque −bω, initially at rest.

Solving the first-order equation gives terminal speed 4/b and relaxation time 2/b. Angular speed is (4/b)[1 − exp(−t/(2/b))]. Integrating it gives angular displacement (4/b)[t − (2/b)(1 − exp(−t/(2/b)))]. The initial angle can be chosen zero without changing work or speed.

At b = 1 N m s and t = 2 s, speed is 4(1 − exp(−1)) ≈ 2.52848 rad/s and angle is 8exp(−1) ≈ 2.94304 rad. Drive work is four times the angle, about 11.7721 J. Rotational kinetic energy is Iω²/2 ≈ 6.39322 J, leaving about 5.37892 J dissipated.

Drive power at the endpoint is 4ω, a rate, not the accumulated work. Multiplying that endpoint power by the whole duration would overstate the work because the body started at rest. The fixed-axis model includes one viscous energy conversion; it does not identify a physical bearing, temperature history or safety rating.

Changed-case check

Predict before reading the comparison. At the same b = 1 but t = 4 s, speed is 4(1 − exp(−2)) ≈ 3.45866 rad/s. It approaches 4 rad/s without reaching it at any finite time in the ideal model.

Retained investigation

Derive speed and angle from the torque equation, then close drive work as kinetic gain plus viscous dissipation on two time intervals. Why must endpoint power be integrated rather than multiplied by an arbitrary elapsed time?

Source and boundary

Primary conceptual reference checked 30 October 2026: OpenStax University Physics Volume 1. All models, packets and teaching prose are original. These are bounded mathematical investigations, not measured apparatus, physical procedures, practical certification or institutional credit.