Rotational dynamics and angular momentum
Work, power and τ = Iα
Push a merry-go-round in a park. The harder you push, and the farther from the centre, the faster it speeds up.
A force F acts at the rim, at distance r from the axis. As the body turns through a small angle dθ, the point where the force acts moves a distance r dθ.
The work done is F times that distance, F r dθ, which is τ dθ. So in rotation, dW = τ dθ.
Divide by time and you get power: P = τω. It matches P = Fv for straight-line motion.
This work becomes rotational kinetic energy, ½Iω². Rate of work equals rate of gain of kinetic energy, so τω = Iωα.
Cancel ω, and we get τ = Iα: torque equals moment of inertia times angular acceleration.
Compare it with F = ma. Torque plays force, moment of inertia plays mass, and angular acceleration plays acceleration.
The same torque gives a big α to a body with small I, and a small α to a body with large I.
L = Iω and its conservation
For a body turning about a fixed axis, the angular momentum along the axis is L = Iω.
Then τ = dL/dt becomes τ = d(Iω)/dt. If I does not change, this is just τ = Iα again.
Now suppose there is no external torque. Then Iω stays constant, even if I changes.
Sit on a stool that turns freely, holding a dumbbell in each hand with your arms stretched out. A friend gives you a gentle spin.
Now pull your arms in. Your moment of inertia drops, so your angular speed shoots up, keeping Iω the same: I₁ω₁ = I₂ω₂.
Stretch your arms out again and you slow down. No one touched you; there was no external torque.
Skaters and dancers use the same trick. A diver curls into a ball in mid-air to spin faster, then straightens out to slow down before entering the water.
One surprise: L is conserved, but kinetic energy is not. When you pull your arms in, your muscles do work, and ½Iω² goes up.
Try it: torque and angular acceleration
- Same torque on both bodies. The one with smaller I gets the bigger α = τ/I.
- Double the torque and watch α double.
- Compare the work done τθ with the kinetic energy ½Iω². They always match.
Try it: the rotating stool
- Press “Give it a spin”, then pull the dumbbells in. Up/Down arrow keys also move them.
- Watch L = Iω: it does not change while ω and I do.
- Check the kinetic energy as you pull in. Where does the extra energy come from?
Work it out
A flywheel with moment of inertia 0.5 kg m² starts from rest. A constant torque of 2 N m acts on it. What is its angular speed after 5 s?
- 4 rad/s
- 5 rad/s
- 20 rad/s
- 50 rad/s
Answer
C. α = τ/I = 2/0.5 = 4 rad/s². Then ω = αt = 4 × 5 = 20 rad/s. Its kinetic energy is then ½ × 0.5 × 20² = 100 J, equal to the work done τθ = 2 × 50 = 100 J.
Quick check
A person on a freely rotating stool has I = 6 kg m² and spins at 2 rad/s with arms out. Pulling the arms in reduces I to 2 kg m². What is the new angular speed?
- 2 rad/s
- 4 rad/s
- 6 rad/s
- 12 rad/s
Answer
C. No external torque, so I₁ω₁ = I₂ω₂: 6 × 2 = 2 × ω₂, giving ω₂ = 6 rad/s. The kinetic energy rises from 12 J to 36 J; the extra comes from the work done by the person’s arms.
Summary
- Second law for rotation:
τ = Iα - Work and power:
W = τθ, P = τω - Angular momentum (fixed axis):
L = Iω - No external torque:
I₁ω₁ = I₂ω₂ - Rotational kinetic energy:
K = ½Iω²
- τ = Iα is the rotational form of F = ma: torque, moment of inertia and angular acceleration replace force, mass and acceleration.
- A torque turning a body through angle θ does work τθ, and delivers power τω.
- With no external torque, Iω stays constant: smaller I means faster spin (skaters, divers, the rotating stool).
- Angular momentum is conserved, but kinetic energy can change, because internal forces can do work.