Physics › Engineering physics › Torque, angular momentum and rotational power
Torque, angular momentum and rotational power
The rotational twins of force, momentum and power, and the bookkeeping that ties them together. A skater's spin, a lorry's clutch and a neutron star all obey the same line: with no external torque, Iω does not change.
Builds on Rotational motion and moment of inertia and Momentum and impulse.
IN THIS TOPIC
- Calculate torque as Fr and link it to angular acceleration through T = Iα.
- Use angular momentum Iω, conserve it when no external torque acts, and explain the classic cases.
- Apply angular impulse TΔt = Δ(Iω) when a torque acts for a time.
- Find work and power in rotation from W = Tθ and P = Tω.
WHAT YOU PROBABLY THINK
A skater pulling in her arms spins faster because her muscles give her an extra twist.
Torque, the turner
A force accelerates a body along a line. To accelerate a body around an axis you need a torque, a force applied at a distance from the axis. For a force F applied tangentially at radius r,
in newton metres. The same force turns harder from further out, the entire working principle of spanners, door handles and gear wheels.
Torque plays force's role in the second law. Where F = ma, rotation has
a net torque T giving angular acceleration α = T/I. Large moment of inertia, sluggish response, exactly as large mass resists a force. When several torques act, the net torque is what counts, and a body spinning at constant ω has zero net torque on it, however fast it turns.
WORKED EXAMPLE
Spinning up a flywheel
A flywheel of moment of inertia 12 kg m2 starts from rest under a constant net torque of 30 N m. Find its angular velocity after 20 s, and show the work done on it equals its kinetic energy.
α = T/I = 30/12 = 2.5 rad s−2, so ω = αt = 2.5 × 20 = 50 rad s−1.
Angle turned: θ = αt2/2 = 2.5 × 202/2 = 500 rad. Work done W = Tθ = 30 × 500 = 1.5 × 104 J.
Ek = ½Iω2 = ½ × 12 × 502 = 1.5 × 104 J. The two agree, as they must. Torque times angle is how energy enters a rotating machine.
Angular momentum, and what conserves it
Momentum's rotational twin is angular momentum,
in kg m2 s−1 (equivalently N m s). Its power comes from its conservation law. With no external torque, the total angular momentum of a system does not change. Internal forces, muscles, clutches, gravity acting at the axis, cannot alter it.
Now the skater. Spinning with arms out, she pulls them in. Her mass moves towards the axis, so I falls, and since Iω must stay fixed, ω rises. No twist was added anywhere. Her spin rate can triple without any torque at all, and the lie at the top of the page is dead.
One subtlety upgrades the answer to full marks. Her kinetic energy ½Iω2 increases, because with Iω fixed the energy can be written as half Iω times ω, and ω rose. The extra energy is the work her muscles did hauling her arms inward against the spin. Momentum is conserved; energy was bought. The same physics spins up a collapsing star. When a giant star's core shrinks to a neutron star a hundred thousand times smaller, I collapses, and rotation once a month becomes many times a second.
YOUR TURN
The skater in numbers
A skater spins at 2.0 rev s−1 with arms out, moment of inertia 4.8 kg m2. She pulls her arms in, reducing it to 1.6 kg m2. Find her new spin rate, and the factor her kinetic energy rises by.
Show the working
Iω conserved: 4.8 × 2.0 = 1.6 × f, so f = 6.0 rev s−1. Revolutions per second serve fine here, because only the ratio matters.
Ek = ½Iω2: the ratio is (1.6 × 6.02)/(4.8 × 2.02) = 3.0. Three times the energy, supplied by her own muscles doing work as they pull her arms in.
When a torque does act, it changes angular momentum at a rate you can bank on. Acting for a time Δt, it delivers an angular impulse:
the rotational twin of FΔt = Δ(mv). A small torque for a long time or a large torque briefly moves the same angular momentum, the reason a heavy flywheel is stopped gently over seconds rather than snatched to rest.
TRY IT UNSEEN
Braking a flywheel
A flywheel of moment of inertia 0.90 kg m2 spins at 40 rad s−1. A brake pad applies a steady friction torque of 6.0 N m. Find the time it takes to stop.
Show the working
The momentum to remove is Δ(Iω) = 0.90 × 40 = 36 kg m2 s−1.
TΔt = Δ(Iω) gives Δt = 36/6.0 = 6.0 s. Half the torque would take twice as long, delivering the identical angular impulse either way.
Work, power and smoothing
A torque turning a shaft through angle θ does work
and delivered continuously that is a power
the rotational twin of P = Fv, and the single most used equation in engine questions. It is why a car's power and torque peak at different engine speeds, and why every quoted engine power implies a shaft speed alongside it.
It also explains the flywheel's second job, smoothing. A piston engine delivers torque in bangs, one per power stroke, with gaps between. A flywheel on the crankshaft barely changes speed under these fluctuations, because its large I turns torque spikes into tiny changes of ω. It absorbs angular momentum during each bang and pays it back through each gap, so the shaft the machinery sees turns nearly uniformly. The same trick in reverse protects machinery through a coupling: connect a spinning shaft to a stationary one and both settle at a shared speed set by conservation, with some kinetic energy lost as heat in the slip.
WORKED EXAMPLE
Reading an engine's badge
An engine delivers 90 kW at 3000 revolutions per minute. Find the torque at that speed.
ω = 3000 × 2π/60 = 314 rad s−1.
T = P/ω = 90 000/314 = 287 N m.
The division is the insight. The same 90 kW at half the shaft speed would need double the torque, so gearboxes exist to trade ω for T while P, minus losses, passes through unchanged.
THE EXAM BIT
- T = Iα wants the net torque. Where a drive torque fights a friction torque, subtract before dividing by I, and say so in your working.
- Conservation answers are sentences first. No external torque acts, so Iω stays constant; I falls, so ω rises. Then, if asked about energy, the kinetic energy rises because the skater does work pulling mass inward.
- Angular impulse questions are momentum questions. Find Δ(Iω) first, then divide by the torque for the time or by the time for the torque.
- P = Tω needs ω in rad s−1, so convert rev min−1 before dividing. Torques from powers at named rpm are this lesson's standard calculation.
- Coupling problems conserve angular momentum, never kinetic energy. Find the shared ω from I1ω1 = (I1 + I2)ω, then show energy was lost if the question asks where it went.
CHECK YOURSELF
A spinning disc of moment of inertia 0.60 kg m2 rotating at 90 rad s−1 is lowered onto an identical-axis stationary disc of moment of inertia 0.30 kg m2, and friction brings them to a common speed. Find that speed, and the kinetic energy lost.
Show a hint
No external torque acts on the pair, so conserve Iω; then compare the two kinetic energies.
Show the answer
Angular momentum before: 0.60 × 90 = 54 kg m2 s−1. Shared afterwards by I = 0.90 kg m2, so ω = 54/0.90 = 60 rad s−1.
Ek before = ½ × 0.60 × 902 = 2430 J. Ek after = ½ × 0.90 × 602 = 1620 J.
Lost: 2430 − 1620 = 810 J, a third of the store, heating the slipping surfaces while they grind to a shared speed. Momentum survived the coupling; energy did not.
Torque is Fr, and it does for rotation what force does for a line: T = Iα.
No external torque means Iω is fixed, whatever internal rearranging happens.
Work is Tθ and power is Tω, the equations every engine question turns on.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Calculate torque as Fr and link it to angular acceleration through T = Iα.
- Use angular momentum Iω, conserve it when no external torque acts, and explain the classic cases.
- Apply angular impulse TΔt = Δ(Iω) when a torque acts for a time.
- Find work and power in rotation from W = Tθ and P = Tω.
Open the full revision checklist to track your progress across the whole unit.