IB Physics HLRigid Body MechanicsHL onlyPaper 1 & 2~10 min read
Angular Impulse
Give a merry-go-round a long, steady push and it spins up gradually; give it a sharp, hard shove and it jumps to the same speed instantly. Either way, what matters is the torque and how long it acts. That combination is angular impulse — the rotational partner of the impulse you met in linear momentum. It’s the bridge between torque and a change in angular momentum, and it turns a torque–time graph into a tool you can read areas straight off.
📘 What you need to know
Resultant torque is the rate of change of angular momentum: τ = ΔL / Δt
Angular impulse is the change in angular momentum it produces: ΔL = τΔt = Δ(Iω)
It’s measured in kg m2 s−1, or equivalently N m s
This equation uses a constant torque; if it varies, use the average
A small torque over a long time has the same effect as a large torque over a short time
On a torque–time graph, the area under the graph equals the angular impulse (the change in angular momentum)
From torque to angular impulse
In linear motion, the resultant force is the rate of change of momentum, F = Δp/Δt, which rearranges into linear impulse Δp = FΔt. Rotation follows the exact same logic. The resultant torque is the rate of change of angular momentum:
Torque and angular momentumτ = ΔL / Δt
Rearranging gives the definition of angular impulse — an average resultant torque τ acting for a time Δt produces a change in angular momentum ΔL:
Angular impulse
ΔL = τΔt = Δ(Iω)
where τ is the resultant torque in N m, Δt is the time interval in seconds, and ΔL is the change in angular momentum in kg m2 s−1 (the same as N m s). Because it’s the product of torque and time, a small torque acting for a long time delivers the same angular impulse as a large torque acting briefly.
Angular impulse mirrors linear impulse: force becomes torque, and a change in linear momentum becomes a change in angular momentum, ΔL = τΔt.
WE 1
A constant torque of 6.0 N m acts on a wheel for 4.0 s. Calculate the angular impulse delivered.
Step 1 — angular impulse = τΔt
ΔL = τ × Δt
Step 2 — substituteΔL = 6.0 × 4.0ΔL = 24 kg m² s⁻¹ (N m s)This is also the change in the wheel’s angular momentum.
WE 2
A wheel with a moment of inertia of 2.0 kg m2 is already spinning at 3.0 rad s−1. A constant torque of 8.0 N m is applied in the same direction for 0.50 s. Find its new angular velocity.
Step 1 — find the angular impulse (change in angular momentum)ΔL = τΔt = 8.0 × 0.50 = 4.0 kg m² s⁻¹Step 2 — ΔL = IΔω, so Δω = ΔL ÷ IΔω = 4.0 ÷ 2.0 = 2.0 rad s⁻¹Step 3 — add to the starting angular velocityωf = 3.0 + 2.0ωf = 5.0 rad s⁻¹The impulse adds angular momentum, which raises the angular velocity.
Angular impulse on a torque–time graph
Because angular impulse is torque × time, the area under a torque–time graph gives the angular impulse — and therefore the change in angular momentum. This works even when the torque varies: just find the area of the shape under the curve.
When the torque varies with time, the area under the torque–time graph gives the angular impulse — equal to the change in angular momentum.
WE 3
An object has a moment of inertia of 5.0 kg m2 and at t = 0 is rotating at 2.0 rad s−1 clockwise. A torque–time graph shows a positive triangle (peak 10 N m, from 0 to 3 s) then a negative rectangle (−5 N m, from 3 to 5 s). Taking anticlockwise as positive, find the angular velocity at t = 5 s.
Step 1 — area of the positive triangle½ × 10 × 3 = 15 N m sStep 2 — area of the negative rectangle−5 × 2 = −10 N m sStep 3 — net angular impulse = change in angular momentumΔL = 15 − 10 = 5 N m sStep 4 — ΔL = I(ωf − ωi), with ωi = −2.0 (clockwise)5 = 5(ωf − (−2.0)) → ωf + 2 = 1ωf = −1.0 rad s⁻¹ (clockwise)The sign tells the direction: still clockwise, but slower than it started.
Sign convention is everything in these graph problems. Pick a positive direction at the start (the question usually tells you), and stick with it. Areas above the time axis are positive angular impulse, areas below are negative — add them with their signs to get the net change in angular momentum. A negative final angular velocity simply means the body is rotating in the direction you called negative.
Torque × time τΔt
= area under graph =
Angular impulse ΔL
= IΔω →
new ω
🛠️ Solving an angular impulse problem
Constant torque? Use ΔL = τΔt directly.
Varying torque? Find the area under the torque–time graph.
Set a positive direction and give areas above/below the axis the right sign.
Equate to the change in angular momentum: ΔL = I(ωf − ωi).
Solve for the unknown, and read the sign as the direction of rotation.
💡 Top tips
Units link up. Angular impulse is in kg m2 s−1, exactly the same as angular momentum.
Area = impulse. On any torque–time graph, the enclosed area is the change in angular momentum.
Mind the signs. Below-axis areas are negative — they reduce the angular momentum.
Same trade-off as linear. A small torque over a long time equals a big torque over a short time.
Quick recap: Resultant torque is the rate of change of angular momentum, τ = ΔL/Δt, which gives angular impulse ΔL = τΔt = Δ(Iω), in kg m2 s−1. On a torque–time graph the area under the line is the angular impulse. Keep a consistent sign convention so a negative result reads as the opposite direction of rotation.
⚠ Common mistakes
Forgetting to use the average torque when the torque isn’t constant
Ignoring the sign of areas below the time axis on a torque–time graph
Mixing up angular impulse (ΔL) with torque or with angular momentum itself
Not converting the initial angular velocity to the chosen sign convention (e.g. clockwise negative)
Reporting only a magnitude when the question asks for the direction too
You’ve now linked torque, time and angular momentum through angular impulse. There’s one last piece of the rotational toolkit to add: the energy a spinning body carries. Just as a moving mass has kinetic energy ½mv2, a rotating body has rotational kinetic energy — and that’s exactly where we finish this topic.
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