Connect torque, rotational work and power

Ask where the motion energy comes from

The rotor starts at rest and gains kinetic energy. The actuator supplies that energy through work. A fixed axis does not imply zero work: each mass moves while the tangential forces act. Their net force can be zero while their torques and work add.

Predict the work and power for +2 N·m acting on the default two-mass rotor. Torque has units N·m, but a torque of 2 N·m is not a statement that 2 J has been transferred or that the actuator supplies 2 W.

Account for the same energy two ways

At radius 1 m, the two 1 kg masses have I = 2 kg·m². After two seconds under +2 N·m, angular velocity is 2 rad/s and each mass speed is 2 m/s. Each particle has kinetic energy ½ × 1 × 2² = 2 J. The pair has 4 J.

The compact rotational expression is K = ½Iω², where K is kinetic energy in joules, I is inertia about the fixed axis in kg·m² and ω is angular velocity in rad/s. Substitution gives ½ × 2 × 2² = 4 J. This is the same energy as the sum of particle kinetic energies. Adding 4 J of particle energy to another 4 J of rotational energy would count it twice. The axis is fixed; this example does not add center-of-mass translation.

For constant net torque, rotational work is W = τΔθ. Here Δθ is the signed angular displacement in radians. Starting from rest, the default trial turns through 2 rad, so W = 2 × 2 = 4 J, equal to the kinetic-energy increase. The connector’s inward forces do no work during ideal fixed-radius motion because they are perpendicular to instantaneous displacement. OpenStax rotational work and power develops the work–energy relation.

Constant torque does not mean constant power

Power P is the rate of energy transfer in watts, or joules per second. Rotational power is P = τω. A given torque supplies more power when the rotor turns faster.

At one second in the default trial, ω = 1 rad/s and θ = 0.5 rad. Work and kinetic energy are 1 J; instantaneous power is 2 W. At two seconds, ω = 2 rad/s and θ = 2 rad. Work and kinetic energy are 4 J; instantaneous power is 4 W.

Average power over the whole two-second interval is total work divided by duration: 4/2 = 2 W. That average differs from the final instantaneous power. At the starting instant, ω = 0 and power is zero even though torque and angular acceleration are nonzero. Immediately afterward the rotor moves and power becomes positive.

Compare the ledger at 0, 0.5, 1, 1.5 and 2 seconds. Explain why angle and work grow quadratically while power grows linearly in this particular constant-torque, rest-start experiment. Do not infer the same time dependence for a motor with constant power or a speed-dependent torque.

Reverse direction without making energy negative

Set torque to −2 N·m at radius 1 m. After two seconds, angle is −2 rad and angular velocity is −2 rad/s. Work is (−2) × (−2) = +4 J, and power is also +4 W at that instant. Kinetic energy remains 4 J because the velocity is squared.

A negative torque is not automatically a braking torque. Here it starts clockwise motion and then acts in the direction of rotation. Negative work would require torque opposing the displacement; that can remove kinetic energy from an already moving rotor. This widget always starts at rest and does not simulate such a braking trial.

Capture the positive and negative cases and identify the signed quantities. Direction is meaningful for angular velocity; a negative energy value would instead signal an accounting or model error in this example.

State what stays fixed in a comparison

Keep +2 N·m for two seconds, but change each radius to 2 m in a separately prepared trial. I becomes 8 kg·m², ω becomes 0.5 rad/s and angle becomes 0.5 rad. Work is now 1 J and final power 1 W. The larger inertia gives less acceleration, less angle and less work over the specified time.

This does not contradict the increase of K with I at fixed angular velocity. These two trials do not end at equal angular velocity. If both systems were instead brought to 2 rad/s, their energies would be 4 J and 16 J. If the same constant torque acted through the same angular displacement, its work would be the same regardless of inertia, but the elapsed times and final angular velocities would differ.

The actuator is an external energy source. The model does not estimate electrical consumption, motor efficiency, heating or connector strength. Input energy to a real motor can exceed the mechanical work delivered to the rotor. Keep those system boundaries explicit when discussing efficiency.

Transfer to a different rotor and duration

A fixed-axis rotor has I = 3 kg·m². A constant −6 N·m net torque acts from rest for one second. Predict angle, angular velocity, work, kinetic energy, final power and average power. Explain why the two power values differ.

Check the rotational work transfer

Angular acceleration is −2 rad/s², angular velocity is −2 rad/s and angle is −1 rad. Work is (−6) × (−1) = 6 J. Kinetic energy is ½ × 3 × (−2)² = 6 J. Final power is (−6) × (−2) = 12 W, while average power is 6 J/1 s = 6 W. Power rises from zero as the rotor gains speed; the final value is not its average.

Optional calculus connection

For constant inertia, angular acceleration is dω/dt and angular velocity is dθ/dt. Multiplying τ = I dω/dt by ω gives τω = d(½Iω²)/dt. Integrating over time connects actuator work to kinetic-energy change. If torque varies with angle, use the signed area under the torque–angle curve rather than multiplying a single torque by the full angle.