6.3 Angular Momentum and Angular Impulse

Angular momentum measures how much rotational motion an object or system has, found with $L = I\omega$ for a rigid system or $L = rmv\sin\theta$ for an object about a point. Angular impulse is torque applied over time ($\tau\Delta t$), and it equals the change in angular momentum, just like linear impulse changes linear momentum. These ideas help you connect torque, rotation, and momentum changes in spinning systems.

Why This Matters for the AP Physics 1 Exam

This topic gives you the rotational versions of two ideas you already know from linear momentum: momentum itself and the impulse-momentum theorem. On both multiple-choice and free-response questions, you will be asked to describe how changing one quantity (like moving mass closer to the axis or applying a torque for longer) affects another quantity. That kind of "if this changes, what happens to that?" reasoning shows up across Unit 6.

Free-response answers need more than naming an equation. If you write that angular momentum changes "because of the impulse-momentum theorem," that is not enough credit. You have to walk through the steps that connect the principle to your claim, especially on the first free-response question that focuses on reasoning. Practice explaining each link in your logic.

Key Takeaways

• Angular momentum of a rigid system about an axis is $L = I\omega$; for an object about a point it is $L = rmv\sin\theta$.
• The axis or reference point you pick changes the value of angular momentum, so always state your axis.
• Angular impulse is $\tau\Delta t$, points the same way as the torque, and equals the change in angular momentum: $\Delta L = \tau\Delta t$.
• The slope of an angular momentum vs. time graph is the net torque; the area under a net torque vs. time graph is the angular impulse.
• The rotational impulse-momentum theorem comes from Newton's second law in rotational form: $\tau_{\text{net}} = \frac{\Delta L}{\Delta t} = I\alpha$ (when rotational inertia is constant).
• Units of angular momentum are $\text{kg}\cdot\text{m}^2/\text{s}$, which is the same as $\text{N}\cdot\text{m}\cdot\text{s}$ for angular impulse.

Angular Momentum of Objects

Angular momentum depends on how mass is spread out, how fast something rotates, and the axis you measure it about.

Angular Momentum Equation

For a rigid system rotating about a fixed axis:

• $L = I\omega$ where $L$ is angular momentum, $I$ is rotational inertia (moment of inertia), and $\omega$ is angular velocity.
• A larger rotational inertia or a faster spin both mean more angular momentum.
• Angular momentum stays constant when the net external torque on the system is zero.

Angular Momentum About a Point

You can also find the angular momentum of an object relative to any reference point, even if the object is moving in a straight line.

• $L = rmv\sin\theta$ gives the magnitude of angular momentum about a given point.
• The point or axis you choose changes the value, so the same motion can have different angular momentum depending on your reference point.
• For an object traveling in a straight line, the angular momentum depends on:
  • the distance $r$ between the reference point and the object
  • the mass $m$ of the object
  • the speed $v$ of the object
  • the angle $\theta$ between the radial distance and the velocity

Angular Impulse From Torque

Angular impulse is the total effect of a torque acting over a time interval.

Angular Impulse

• $\text{Angular impulse} = \tau\Delta t$ where $\tau$ is torque and $\Delta t$ is the time interval.
• Angular impulse has the same direction or sign as the torque. AP Physics 1 handles this with one-dimensional sign conventions (for example, counterclockwise as positive), not full 3D vectors.
• Angular impulse equals the change in angular momentum: $\Delta L = \tau\Delta t$.
• Its units are $\text{N}\cdot\text{m}\cdot\text{s}$, the same as $\text{kg}\cdot\text{m}^2/\text{s}$.

Reading Torque vs. Time Graphs

You can pull angular impulse straight off a graph.

• The area under a net torque vs. time graph equals the angular impulse delivered.
• For constant net torque, that area is a rectangle: height $\tau_{\text{net}}$, width $\Delta t$.
• For changing torque, the angular impulse is still the area under the curve.

Change in Angular Momentum

Knowing how and why angular momentum changes is the heart of rotational dynamics.

Magnitude of the Change

• $\Delta L = L - L_0$ where $L$ is the final and $L_0$ is the initial angular momentum.
• This works for both increases and decreases.
• The size of the change depends on how strong the torque is and how long it acts.

Impulse-Momentum Theorem for Rotation

• $\tau\Delta t = \Delta L$ means angular impulse equals the change in angular momentum.
• This comes from Newton's second law in rotational form: $\tau_{\text{net}} = \frac{\Delta L}{\Delta t} = I\alpha$.
• When rotational inertia is constant, this lets you connect torque, time, and the change in angular velocity directly.

Torque and Angular Momentum Graphs

• The slope of an angular momentum vs. time graph equals the net torque.
• A constant slope means constant net torque.
• A steeper slope means a larger torque and a faster change in angular momentum.

How to Use This on the AP Physics 1 Exam

Problem Solving

• Decide which angular momentum equation fits: use $L = I\omega$ for a rotating rigid body and $L = rmv\sin\theta$ for an object about a point.
• For impulse problems, set $\tau\Delta t = \Delta L$. If the object starts at rest, $\Delta L$ is just the final angular momentum.
• Check that your axis or reference point stays the same throughout the problem, since changing it changes $L$.

Graph Reading

• For an angular momentum vs. time graph, read net torque as the slope.
• For a net torque vs. time graph, read angular impulse as the area under the curve.
• Watch units: angular momentum in $\text{kg}\cdot\text{m}^2/\text{s}$, angular impulse in $\text{N}\cdot\text{m}\cdot\text{s}$.

Free Response

• When you justify a claim, do not stop at naming the theorem. Spell out each step from the principle to your conclusion.
• If you say angular speed increases, explain why: for example, no net external torque means $L$ is constant, so reducing $I$ forces $\omega$ to rise.
• Tie your equations back to the physical situation and keep your units consistent.

Common Misconceptions

• Angular momentum is not a fixed property of an object. Its value depends on the axis or reference point you choose, so the same motion can give different numbers.
• An object moving in a straight line can still have angular momentum about a point. It only equals zero when $\theta$ is 0 or when $r$ is zero.
• Angular impulse and angular momentum are not the same thing. Angular impulse ($\tau\Delta t$) is the cause; the change in angular momentum ($\Delta L$) is the result.
• A larger torque does not automatically mean a larger change in angular momentum. The change depends on both the torque and how long it acts.
• The relation $\tau_{\text{net}} = I\alpha$ assumes rotational inertia stays constant. If $I$ changes (like a skater pulling in their arms), you reason with angular momentum instead.

Practice Problem 1: Angular Momentum Calculation

To solve this problem, we need to:

1. Find the moment of inertia of the disk
2. Use the angular momentum equation

The moment of inertia for a uniform disk about its center is $I = \frac{1}{2}mr^2$

$I = \frac{1}{2} \times 2.0 \text{ kg} \times (0.25 \text{ m})^2 = 0.0625 \text{ kg}\cdot\text{m}^2$

Now we can calculate the angular momentum: $L = I\omega = 0.0625 \text{ kg}\cdot\text{m}^2 \times 12 \text{ rad/s} = 0.75 \text{ kg}\cdot\text{m}^2/\text{s}$

Therefore, the angular momentum of the disk is 0.75 kg·m²/s.

Practice Problem 2: Angular Impulse

To solve this problem:

1. Calculate the angular impulse
2. Use the impulse-momentum theorem to find the change in angular momentum
3. Determine the final angular velocity

Angular impulse = $\tau \times \Delta t = 15 \text{ N}\cdot\text{m} \times 3.0 \text{ s} = 45 \text{ N}\cdot\text{m}\cdot\text{s}$

Since the wheel starts from rest, its initial angular momentum is zero. Using the impulse-momentum theorem: $\tau\Delta t = \Delta L = L_{final} - 0 = L_{final}$

This means $L_{final} = 45 \text{ kg·m}^2/\text{s}$

With $L = I\omega$, we can find the angular velocity: $\omega = \frac{L}{I} = \frac{45 \text{ kg·m}^2/\text{s}}{2.5 \text{ kg·m}^2} = 18 \text{ rad/s}$

Therefore, the wheel's final angular velocity is 18 rad/s.

Practice Problem 3: Change in Angular Momentum

To solve this problem:

1. Recognize that angular momentum is conserved
2. Calculate the initial moment of inertia
3. Use conservation of angular momentum to find the final angular velocity

First, let's find the initial moment of inertia: $L = I\omega$ $I_{initial} = \frac{L}{\omega_{initial}} = \frac{12 \text{ kg·m}^2/\text{s}}{4.0 \text{ rad/s}} = 3.0 \text{ kg·m}^2$

Since angular momentum is conserved (no net external torque): $L_{initial} = L_{final}$ $I_{initial} \times \omega_{initial} = I_{final} \times \omega_{final}$

Solving for final angular velocity: $\omega_{final} = \frac{I_{initial} \times \omega_{initial}}{I_{final}} = \frac{3.0 \text{ kg·m}^2 \times 4.0 \text{ rad/s}}{1.5 \text{ kg·m}^2} = 8.0 \text{ rad/s}$

Therefore, the skater's angular velocity increases to 8.0 rad/s when she pulls her arms in.

Related AP Physics 1 Guides

• 6.4 Conservation of Angular Momentum
• 6.5 Rolling
• 6.6 Motion of Orbiting Satellites
• 6.1 Rotational Kinetic Energy
• 6.2 Torque and Work