Angular Momentum and Angular Impulse: Study Guide (Physics 1 6.3)

This guide is for the algebra-based Physics 1 course. It uses algebra and graphs only. Topic 6.2 showed how a torque does work on a rotating body. Here you meet the rotational partner of linear momentum, and the rotational partner of impulse.

The idea: a rotational version of momentum

In Unit 4 you met linear momentum, p = mv, and the impulse–momentum theorem, FΔt = Δp. A force acting for a time changes an object’s momentum.

Rotation has the same structure. Swap each linear quantity for its rotational partner:

Linear	Rotational
mass m	rotational inertia I
velocity v	angular velocity ω
momentum p = mv	angular momentum L = Iω
force F	torque τ
impulse FΔt	angular impulse τΔt
FΔt = Δp	τΔt = ΔL

If you understand the left column, you already understand most of this topic. The new idea is that angular momentum is always measured about an axis or a point, and the value depends on that choice.

Angular momentum of a rigid body

A rigid body spinning about a fixed axis has angular momentum

L = Iω

where I is its rotational inertia about that axis (kg·m²) and ω is its angular velocity (rad/s). The unit of L is kg·m²/s.

Example: a flywheel with I = 0.050 kg·m² spins at 40 rad/s. Its angular momentum is L = 0.050 × 40 = 2.0 kg·m²/s.

Two things increase L: spinning faster, or having more mass far from the axis. A wheel and a ring of the same mass and speed do not have the same L if the ring’s mass sits further out.

Signs, not directions. Pick one sense of rotation as positive and say so, for example “counterclockwise as seen from above is +”. Then a wheel turning clockwise has negative ω and negative L. This course only asks you to work with signs about one axis. Finding the three-dimensional direction of L (with the right-hand rule) is beyond its scope.

Angular momentum of an object moving in a straight line

An object does not need to go round in a circle to have angular momentum. Take any fixed reference point. An object of mass m moving with speed v has angular momentum about that point of

L = rmv sin θ

Here r is the distance from the reference point to the object, and θ is the angle between the line from the point to the object (the radial line) and the object’s velocity.

There is a simpler way to see it. The quantity r sin θ is the perpendicular distance d from the reference point to the object’s line of motion. So

L = mv × d

This is just like torque, where you multiply a force by its lever arm. Here you multiply the momentum by its “lever arm”.

Work through Figure 1 with the formula:

Position	r (m)	θ	sin θ	L = rmv sin θ (kg·m²/s)
P1 (approaching)	1.0	150°	0.50	1.0 × 0.20 × 3.0 × 0.50 = 0.30
P2 (closest)	0.50	90°	1.0	0.50 × 0.20 × 3.0 × 1.0 = 0.30
P3 (moving away)	1.0	30°	0.50	1.0 × 0.20 × 3.0 × 0.50 = 0.30

Three lessons come from this table.

1. The reference point matters. About A the angular momentum is 0.30 kg·m²/s. About B, on the line of motion, it is zero. About a point C that is 1.5 m from the line, it would be 0.20 × 3.0 × 1.5 = 0.90 kg·m²/s. Always say which point or axis you are using.
2. What L depends on. The distance from the point, the mass, the speed and the angle between r and v. If the velocity points straight towards or away from the point, sin θ = 0 and L = 0.
3. Constant velocity gives constant L. A ball moving at steady speed in a straight line keeps the same d, so its angular momentum about any fixed point stays the same. This makes sense: no net force acts, so no net torque acts about any point.

Angular impulse

A torque that acts for a time interval delivers an angular impulse:

angular impulse = τΔt

The unit is N·m·s, which is the same as kg·m²/s. The angular impulse has the same sign as the torque that delivers it. A large torque for a short time can give the same angular impulse as a small torque for a long time.

If the torque changes with time, use a graph. The angular impulse is the area under a torque–time graph, in the same way that linear impulse is the area under a force–time graph. Area below the time axis counts as negative.

The rotational impulse–momentum theorem

Start from Newton’s second law in rotational form, for a body whose rotational inertia I stays constant:

τ_net = Iα = I(Δω / Δt)

Multiply both sides by Δt:

τ_net Δt = IΔω = Iω − Iω₀ = L − L₀

So

τ_net Δt = ΔL

In words: the net angular impulse on an object or rigid system equals its change in angular momentum. The change in angular momentum is found by comparing the final and initial values, ΔL = L − L₀, with signs.

Two graph readings follow from this:

• Slope of an L–t graph = net torque. Rearranging, τ_net = ΔL / Δt. A straight L–t line means constant net torque. A curved one means the torque is changing.
• Area under a τ_net–t graph = ΔL. This is the graph form of the theorem.

Reading an angular momentum–time graph

A desk fan is switched off. Take its spin direction as +. Measurements give:

t (s)	0	2.0	4.0	6.0	8.0
L (kg·m²/s)	3.0	2.4	1.8	1.2	0.60

Plotted against time, these points lie on a straight line. Its slope is (0.60 − 3.0) ÷ 8.0 = −0.30 N·m. That is the net torque from friction in the bearings and air drag. It is constant and opposite to the spin. If it stays the same, the fan stops 3.0 ÷ 0.30 = 10 s after switch-off.

Worked example 1: angular impulse from a torque–time graph

Question. Take counterclockwise (seen from above) as +. A potter’s wheel has rotational inertia I = 0.60 kg·m² and is turning at ω₀ = +2.0 rad/s. A motor then applies the torque shown in Figure 2, and friction is negligible. Find the wheel’s angular speed (a) at t = 2.0 s and (b) at t = 4.0 s.

1. Initial angular momentum: L₀ = Iω₀ = 0.60 × 2.0 = +1.2 kg·m²/s.
2. Angular impulse from 0 to 2.0 s (triangle): ½ × 2.0 s × 3.0 N·m = +3.0 N·m·s.
3. (a) L at 2.0 s = 1.2 + 3.0 = 4.2 kg·m²/s, so ω = 4.2 ÷ 0.60 = 7.0 rad/s.
4. Angular impulse from 2.0 s to 4.0 s (rectangle): 2.0 s × 3.0 N·m = +6.0 N·m·s. Total angular impulse = 9.0 N·m·s.
5. (b) L at 4.0 s = 1.2 + 9.0 = 10.2 kg·m²/s, so ω = 10.2 ÷ 0.60 = 17 rad/s.

Check. The average torque over the 4.0 s is 9.0 ÷ 4.0 = 2.25 N·m, which sits sensibly between 0 and 3.0 N·m. A common slip is to forget the starting angular momentum and give 9.0 ÷ 0.60 = 15 rad/s. The theorem gives the change in L, so you must add L₀.

Worked example 2: deriving a stopping time

Question. A wheel with rotational inertia I spins at ω₀ on a frictionless axle. A brake pad presses on the rim at radius R with a constant friction force of size f, tangent to the rim. (a) Derive an expression for the time t the wheel takes to stop. (b) Evaluate it for I = 0.12 kg·m², ω₀ = 25 rad/s, f = 6.0 N and R = 0.25 m. (c) Predict what happens to t if the pad is moved to act at radius 2R with the same force.

Take the wheel’s initial spin as +.

1. The friction force is tangent to the rim, so its lever arm is R. The torque it exerts opposes the spin: τ_net = −fR.
2. Initial angular momentum: L₀ = Iω₀. Final: L = 0. So ΔL = 0 − Iω₀ = −Iω₀.
3. Theorem: τ_net Δt = ΔL gives (−fR)t = −Iω₀.
4. (a) Solve: t = Iω₀ / (fR).
5. (b) Torque size: fR = 6.0 × 0.25 = 1.5 N·m. L₀ = 0.12 × 25 = 3.0 kg·m²/s. t = 3.0 ÷ 1.5 = 2.0 s.
6. (c) t is inversely proportional to R. Doubling R halves the stopping time: 1.0 s.

Check the expression. Units: (kg·m²)(1/s) ÷ (N·m) = kg·m²/s ÷ (kg·m²/s²) = s. Correct. Limiting cases: a heavier wheel (bigger I) or faster start (bigger ω₀) takes longer to stop; a harder press (bigger f) stops it sooner. All are sensible. You can also check with Unit 5 methods: α = fR / I = 12.5 rad/s², and t = ω₀ / α = 2.0 s again.

Factors of change

The theorem is ideal for “what happens if…” questions. Write the relationship, then hold everything else fixed.

• Same torque, twice the time: angular impulse doubles, so ΔL doubles.
• Same angular impulse given to a wheel with four times the rotational inertia: ΔL is the same, so Δω is one quarter as big.
• A ball moving in a straight line passes twice as far from a point at the same speed: L about that point doubles, because d doubles.
• A rigid body at the same ω with mass moved further out: I rises, so L rises in the same ratio.

Common misconceptions

• “Only spinning things have angular momentum.” An object moving in a straight line has angular momentum about any point off its path (Figure 1).
• “Angular momentum has one true value.” It depends on the axis or point chosen. About a point on the line of motion, it is zero.
• “L = rmv, whatever the angle.” Only when the velocity is perpendicular to r. In general use rmv sin θ, or mv times the perpendicular distance.
• “A bigger torque always gives a bigger change in L.” The time matters too. Compare τΔt, not τ alone.
• “The theorem gives the final angular momentum.” It gives the change. Add L₀ to find L (Worked example 1).
• Forgetting signs. If a torque opposes the spin, its angular impulse is negative. Mixing up signs is the most common way to lose marks here.
• Using one torque instead of the net torque. The theorem uses the net external torque on the object. Include friction if it is not negligible.
• Confusing angular impulse with linear impulse. Force × time has unit N·s. Angular impulse needs the lever arm: torque × time, N·m·s.

Where this leads

Topic 6.4 (Conservation of Angular Momentum) asks what happens when the net external torque on a system is zero, so that ΔL = 0. That one idea explains a turntable that slows when a ring lands on it and a spinner that speeds up when it pulls its mass in. Read the Topic 6.4 study guide next. First, try the practice questions, then use the revision notes and the checklist to consolidate. You can also go back to Topic 6.2 or the course roadmap.