Angular Momentum and Angular Impulse

Lesson ~11 min read 8 MCQs

In simple terms: In simple terms, angular momentum is an object's "quantity of rotation," and angular impulse is the rotational "push" over time that changes this quantity of rotation.

Why this matters

Imagine you're at a playground with your younger cousin, pushing them on a merry-go-round. At first, it's tough to get moving. You give it a solid, sustained push. That push is a force, but because you're making it spin, it's really a torque. The fact that you push for a few seconds, not just for an instant, is also critical.

You've just applied an angular impulse: a torque applied over a time interval. The result? The merry-go-round, which had zero rotation, now has angular momentum. It's spinning! If you want it to spin faster, you give it another push in the same direction. You've used another angular impulse to add even more angular momentum. This lesson is all about that fundamental connection: how rotational pushes change rotational motion.

Concept overview

flowchart TD
    A[Torque τ] --> B{Applied over time Δt}
    B --> C[Creates Angular Impulse: τΔt]
    C --> D[Equals the Change in Angular Momentum: ΔL]
    E[Initial Angular Momentum: L₀] --> F(ΔL = Lf - L₀)
    D --> F
    F --> G[Final Angular Momentum: Lf]
    H[L = Iω or L = rmv sin(θ)]
    E --- H
    G --- H

Core explanation

Hey everyone, it's Saavi. Today we're diving into one of the most important concepts in rotational motion. It's the rotational sibling of the momentum and impulse ideas you've already mastered for linear motion. If you understood that a force applied over time changes an object's momentum, you're already halfway here.

What is Angular Momentum?

Just like an object moving in a straight line has linear momentum (p = mv), an object that is rotating has angular momentum, which we represent with the letter L. It's a measure of how much "rotational motion" an object has. It depends not just on how fast it's spinning, but also on how its mass is distributed.

We have two main ways to calculate it, depending on the situation.

1. For a Rigid, Rotating System

Think of a spinning bicycle wheel, a DVD, or a planet rotating on its axis. For any solid object rotating around a fixed axis, the angular momentum is:

L = Iω

Where:

L is the angular momentum (in kg·m²/s).
I is the rotational inertia (or moment of inertia). Remember, this is the rotational equivalent of mass. It tells you how hard it is to change the object's rotation.
ω (omega) is the angular velocity (in rad/s).

This equation should feel familiar. It looks just like p = mv, but with all the rotational versions of the variables!

2. For a Point Mass

Now, here's where most students get a little turned around. An object doesn't have to be moving in a circle to have angular momentum. A single object moving in any path has angular momentum relative to a chosen point (or axis).

Imagine a meteor flying in a straight line past the planet Mars. From the perspective of Mars, that meteor has angular momentum.

For a single particle or an object that can be treated as a point mass, the equation is:

L = rmv sin(θ)

Let's break this down:

r is the magnitude of the position vector from your reference point (our "pivot") to the object.
m is the mass of the object.
v is the speed of the object.
θ (theta) is the angle between the position vector r and the velocity vector v.

What is Angular Impulse?

Remember linear impulse? It's a force applied over a time interval (Impulse = FΔt) that causes a change in linear momentum.

Angular impulse is the exact same idea, but for rotation. It's a torque applied over a time interval.

Angular Impulse = τΔt

τ (tau) is the torque you apply.
Δt is the time interval you apply it for.

When you pushed that merry-go-round, you applied a torque for a few seconds. That was an angular impulse. The direction of the angular impulse is the same as the direction of the torque.

Just like with linear impulse, if you have a graph of torque vs. time, the area under the curve is the angular impulse. This is a very common type of question on the AP exam.

The Angular Impulse-Momentum Theorem

Now, let's put it all together. This is the main event.

The angular impulse you deliver to an object is exactly equal to the change in that object's angular momentum.

ΔL = τ_net Δt

This is the angular impulse-momentum theorem. It's one of the most powerful tools in this unit.

Let's see where it comes from. It's a direct result of Newton's second law for rotation:

Start with τ_net = Iα.
We know that angular acceleration α is the change in angular velocity over time: α = Δω / Δt.
Substitute that in: τ_net = I (Δω / Δt).
Now, multiply both sides by Δt: τ_net Δt = I Δω.
Since I is constant for a rigid body, I Δω is the same as the change in Iω. And what is Iω? It's angular momentum, L!
So, I Δω = Δ(Iω) = ΔL.

Which brings us right back to our theorem: τ_net Δt = ΔL.

This equation tells us so much.

If you want to change an object's spin (change its L), you need to apply a net torque over some amount of time.
If you have a graph of angular momentum vs. time, the slope of that graph (ΔL / Δt) is the net torque.

This relationship between torque, time, and angular momentum is fundamental to understanding how spinning things work, from a simple top to the complex orbits of planets.

Worked examples

Let's walk through a couple of problems to see how these concepts work in practice.

Example 1: Spinning Up a Carousel

A playground carousel with a rotational inertia of 150 kg·m² is initially at rest. Priya comes along and pushes tangentially on the edge for 4.0 seconds, applying a constant torque of 50 N·m. What is the carousel's final angular velocity?

1. Identify the Goal and the Principle The question asks for the final angular velocity (ω_f). We're given a torque applied over time, which changes the carousel from rest to spinning. This is a perfect job for the angular impulse-momentum theorem: τΔt = ΔL.

2. Break Down the Theorem

We know τ = 50 N·m and Δt = 4.0 s.
The change in angular momentum is ΔL = L_f - L_i.
The final angular momentum is L_f = Iω_f.
The initial angular momentum is L_i = Iω_i. Since the carousel starts from rest, ω_i = 0, which means L_i = 0.

3. Set Up the Equation Let's substitute everything into the theorem: τΔt = L_f - L_i τΔt = Iω_f - 0 τΔt = Iω_f

4. Solve for the Unknown Now we just need to rearrange and plug in our numbers to solve for ω_f.

ω_f = (τΔt) / I ω_f = (50 N·m * 4.0 s) / 150 kg·m² ω_f = 200 N·m·s / 150 kg·m² ω_f ≈ 1.33 rad/s

Example 2: Torque vs. Time Graph

A flywheel is initially spinning with an angular momentum of 20 kg·m²/s. The graph below shows the net torque applied to the flywheel over a 6-second interval. What is the flywheel's final angular momentum at t = 6 s?

(Imagine a graph where torque is plotted on the y-axis and time on the x-axis. From t=0 to t=4s, the torque is a constant +10 N·m. From t=4s to t=6s, the torque is a constant -5 N·m.)

1. Identify the Goal and the Principle We need the final angular momentum, L_f. We're given a graph of torque vs. time. The key principle is that the area under a τ-t graph gives the angular impulse, and that impulse equals the change in angular momentum (ΔL).

2. Calculate the Angular Impulse (Area) We need to find the total area under the curve from t=0 to t=6s. Let's break it into two parts:

Area 1 (0 to 4s)
This is a rectangle. Area₁ = base × height = (4 s) × (10 N·m) = +40 N·m·s
Area 2 (4 to 6s)
This is another rectangle, but it's below the axis. Area₂ = base × height = (6 s - 4 s) × (-5 N·m) = (2 s) × (-5 N·m) = -10 N·m·s

The total angular impulse is the sum of these areas: Angular Impulse = Area₁ + Area₂ = 40 - 10 = +30 N·m·s

3. Apply the Theorem The angular impulse is equal to the change in angular momentum: Angular Impulse = ΔL = L_f - L_i +30 N·m·s = L_f - L_i

4. Solve for the Final Angular Momentum We were given the initial angular momentum, L_i = 20 kg·m²/s.

30 = L_f - 20 L_f = 30 + 20 L_f = 50 kg·m²/s

Try it yourself

Problem 1: A large potter's wheel has a rotational inertia of 2.5 kg·m² and is spinning freely at 5.0 rad/s. The potter then uses her foot on a pedal to apply a frictional torque of -3.0 N·m to slow it down. For how long must she apply the torque to bring the wheel to a complete stop?

Problem 2: A 0.2 kg hockey puck is sliding on frictionless ice at a constant 15 m/s. It passes by the center of the rink, with its straight-line path missing the center by 0.8 m. What is the magnitude of the puck's angular momentum relative to the center of the rink?

TL;DR

In simple terms, angular momentum is an object's "quantity of rotation," and angular impulse is the rotational "push" over time that changes this quantity of rotation.

Syntax pattern

L = Iω

Key terms

Angular momentum Angular impulse Ap physics 1 Rotational motion Impulse-momentum theorem Torque Rotational dynamics L=Iω Change in angular momentum Ap physics study guide ===END===

You can now…

6.3.A: Describe the angular momentum of an object or rigid system.
6.3.B: Describe the angular impulse delivered to an object or rigid system by a torque.
6.3.C: Relate the change in angular momentum of an object or rigid system to the angular impulse given to that object or rigid system.

Essential knowledge (exam-tested)

6.3.A.1: The magnitude of the angular momentum of a rigid system about a specific axis can be described with the equation L = Iω.
6.3.A.2: The magnitude of the angular momentum of an object about a given point is L = rmv sin θ.
6.3.A.2.i: The selection of the axis about which an object is considered to rotate influences the determination of the angular momentum of that object.
6.3.A.2.ii: The measured angular momentum of an object traveling in a straight line depends on the distance between the reference point and the object, the mass of the object, the speed of the object, and the angle between the radial distance and the velocity of the object.
6.3.B.1: Angular impulse is defined as the product of the torque exerted on an object or rigid system and the time interval during which the torque is exerted. Relevant equation: angular impulse = τΔt
6.3.B.2: Angular impulse has the same direction as the torque exerted on the object or system.
6.3.B.3: The angular impulse delivered to an object or rigid system by a torque can be found from the area under the curve of a graph of the torque as a function of time.
6.3.C.1: The magnitude of the change in angular momentum can be described by comparing the magnitudes of the final and initial angular momenta of the object or rigid system: ΔL = L - L₀
6.3.C.2: A rotational form of the impulse-momentum theorem relates the angular impulse delivered to an object or rigid system and the change in angular momentum of that object or rigid system.
6.3.C.2.i: The angular impulse exerted on an object or rigid system is equal to the change in angular momentum of that object or rigid system. Relevant equation: ΔL = τΔt
6.3.C.2.ii: The rotational form of the impulse-momentum theorem is a direct result of the rotational form of Newton's second law of motion for cases in which rotational inertia is constant: τ_net = ΔL/Δt = I(Δω/Δt) = Iα
6.3.C.3: The net torque exerted on an object is equal to the slope of the graph of the angular momentum of an object as a function of time.
6.3.C.4: The angular impulse delivered to an object is equal to the area under the curve of a graph of the net external torque exerted on an object as a function of time.

Concept map

flowchart TD
    A[Torque τ] --> B{Applied over time Δt}
    B --> C[Creates Angular Impulse: τΔt]
    C --> D[Equals the Change in Angular Momentum: ΔL]
    E[Initial Angular Momentum: L₀] --> F(ΔL = Lf - L₀)
    D --> F
    F --> G[Final Angular Momentum: Lf]
    H[L = Iω or L = rmv sin(θ)]
    E --- H
    G --- H

Read what Saavi narrates

(Sound of a gentle, thoughtful intro music fades)

Hey everyone, it's Saavi from Shrutam.

Have you ever been to a playground and pushed a friend on a merry-go-round? Think about that moment. To get it started, you have to give it a good, solid push, and you have to keep pushing for a few seconds. That rotational push is a torque, and the fact that you apply it over time is what really matters today.

You've just performed an angular impulse. And the result is that the merry-go-round, which was still, is now spinning. It has angular momentum.

That's the core idea we're exploring. Just like a linear push, or impulse, changes an object's linear momentum... a rotational push, or angular impulse, changes an object's angular momentum. It’s a beautiful parallel.

Let's look at how this works with an example. Imagine a playground carousel. Let's say it has a rotational inertia of 150 kilogram meters squared, and it's starting from rest. Now, Priya comes along and pushes it for 4 seconds, applying a steady torque of 50 Newton-meters. We want to find its final angular velocity.

This is a perfect time to use the angular impulse-momentum theorem, which says that the angular impulse equals the change in angular momentum. In equation form, that's torque times the time interval... equals the change in L.

The angular impulse is the torque, 50, times the time, 4. That's 200 Newton-meter-seconds.

This impulse is equal to the change in angular momentum. Since it started from rest, the initial angular momentum was zero. So the change is just equal to the final angular momentum, L final.

So, L final is 200.

But we want the final angular *velocity*, omega. We know that angular momentum, L, is equal to the rotational inertia, I, times the angular velocity, omega.

So, 200 equals 150 times omega. To find omega, we just divide 200 by 150. That gives us about 1.33 radians per second. And there you have it. We connected the push to the final spin.

Now, one of the biggest mistakes I see students make is forgetting about initial momentum. In that last problem, the initial momentum was zero, but it won't always be. If the carousel was already spinning, we would have had to calculate the impulse, which is the *change*, and then add that change to the starting momentum to find the final value. Always remember: impulse equals the *change* in momentum, not the final value. It's a small detail that can make a big difference on the exam.

Keep practicing this relationship. It's a powerful tool, and once it clicks, you'll see it everywhere in the world of rotation. You've got this.

(Outro music fades in)

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