Unit 6.2 | Angular Momentum and Angular Impulse

Angular Impulse

A net torque changes angular momentum, just as a net force changes momentum. The change equals the torque times the time it acts, the angular impulse:

[katex] \tau_{net}\Delta t = \Delta L = I\omega_f – I\omega_i [/katex]

If the torque changes with time, the angular impulse is the area under the torque vs. time graph. And since [katex] \tau_{net} = \frac{\Delta L}{\Delta t} [/katex], the slope of an angular momentum vs. time graph is the net torque. With no net torque, [katex] L [/katex] stays constant: that is the next lesson.

A force can change an object’s momentum without changing its angular momentum: a push straight through the center of mass moves the object but does not spin it. A push off-center does both.

Every angular momentum question on the AP exam is one of the four types below.

Problem type	The clue	The setup
1. Angular momentum of a moving point	an object moving in a line “relative to point P”	[katex] L = mvr_\perp [/katex], with the perpendicular distance to the path
2. Angular impulse	a torque acts “for [katex] 3.0\,\text{s} [/katex]”, a brake stops a wheel	[katex] \tau\Delta t = I\omega_f – I\omega_i [/katex]
3. Torque vs. time graphs	a graph of [katex] \tau [/katex] against [katex] t [/katex]	Area = [katex] \Delta L [/katex]
4. Angular momentum vs. time graphs	a graph of [katex] L [/katex] against [katex] t [/katex]	Slope = net torque

Problem Type 1: Angular momentum of a moving point

How to spot it: An object moves in a straight line (or passes by), and the question asks for its angular momentum about a point that is not on its path, or ranks it for several points.

Setup: Find [katex] r_\perp [/katex], the perpendicular distance from the point to the line of motion, not the straight-line distance to the object. Then [katex] L = mvr_\perp [/katex]. Give it a sign: + if the object goes counterclockwise around the point.

A [katex] 0.20\,\text{kg} [/katex] puck slides at [katex] 3.0\,\text{m/s} [/katex] in a straight line. Its path passes [katex] 0.50\,\text{m} [/katex] from a post at its closest. Find its angular momentum about the post when it is closest, and again when it is [katex] 2.0\,\text{m} [/katex] farther along.

[katex] L = (0.20)(3.0)(0.50) = 0.30 \text{ kg·m}^2\text{/s} [/katex] at the closest point. Farther along, the puck is farther from the post, but [katex] r_\perp [/katex] is still [katex] 0.50\,\text{m} [/katex], so [katex] L [/katex] is still [katex] 0.30\,\text{kg·m}^2\text{/s} [/katex]. No torque acts on the puck about the post, so its angular momentum cannot change.

AP trap: Using the distance from the point to the object instead of [katex] r_\perp [/katex]. Only the part of the position perpendicular to the velocity counts, exactly like the lever arm of a force.

Problem Type 2: Angular impulse

How to spot it: A torque acts for a known time (“for [katex] 4.0\,\text{s} [/katex]”), a brake or friction stops a spinning object, or a question asks how long a spin-up takes.

Setup: Find the net torque (a brake or friction torque is negative). Then [katex] \tau_{net}\Delta t = I\omega_f – I\omega_i [/katex]. Use it whenever time is given or asked for; use energy (Unit 6.1) when an angle is.

A grinding wheel with [katex] I = 0.40 \text{ kg·m}^2 [/katex] spins at [katex] 30\,\text{rad/s} [/katex]. When the motor is switched off, a friction torque of [katex] 2.0\,\text{N·m} [/katex] acts at the axle. How long does the wheel take to stop?

[katex] \Delta L = 0 – (0.40)(30) = -12 \text{ kg·m}^2\text{/s} [/katex]. [katex] -(2.0)\Delta t = -12 [/katex], so [katex] \Delta t = 6.0 \text{ s} [/katex]. Same answer as [katex] \alpha = \frac{\tau}{I} [/katex] plus kinematics, in one line.

AP trap: Mixing up linear and angular impulse. A force through the center of mass gives a linear impulse but no angular impulse about the center. The same force off-center gives the same linear impulse and an angular impulse as well.

Problem Type 3: Torque vs. time graphs

How to spot it: A graph of net torque against time, followed by “angular momentum at t = …”, “angular speed at …” or “when is L greatest”.

Setup: The area under the [katex] \tau [/katex] vs. [katex] t [/katex] graph is [katex] \Delta L [/katex]. Area below the axis counts as negative. [katex] L [/katex] keeps growing as long as the torque is positive, even while the torque shrinks.

Net torque on a wheel

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A wheel with [katex] I = 2.0 \text{ kg·m}^2 [/katex] starts from rest. The net torque on it is shown above. How fast is it spinning at [katex] 5\,\text{s} [/katex]?

Area: [katex] \tfrac{1}{2}(2)(4) + (3)(4) = 16 \text{ kg·m}^2\text{/s} [/katex]. Starting from rest, [katex] L = 16 [/katex], so [katex] \omega = \frac{L}{I} = \frac{16}{2.0} = 8.0 \text{ rad/s} [/katex].

AP trap: Picking the time of the largest torque as the time of the largest angular momentum. [katex] L [/katex] peaks where the torque crosses zero, the moment it stops adding and starts taking away.

Problem Type 4: Angular momentum vs. time graphs

How to spot it: A graph of [katex] L [/katex] (or [katex] \omega [/katex]) against time, and the question asks about the torque: when it is largest, when it is zero, or how two torques compare.

Setup: The slope is the net torque: [katex] \tau_{net} = \frac{\Delta L}{\Delta t} [/katex]. Steep means a large torque, flat means zero torque, and a straight line means a constant torque.

A flywheel’s angular momentum drops in a straight line from [katex] 24\,\text{kg·m}^2\text{/s} [/katex] to zero in [katex] 8.0\,\text{s} [/katex]. What does the graph tell you about the brake?

Slope: [katex] \frac{0 – 24}{8.0} = -3.0 \text{ N·m} [/katex]. The brake applies a constant torque of [katex] 3.0\,\text{N·m} [/katex] against the rotation. A straight line means the torque never changed.

AP trap: Reading the height of the graph instead of its slope. Where [katex] L [/katex] is largest and flat, the net torque is zero. Where [katex] L [/katex] crosses zero, the torque can be large.

Lesson 6.3 Preview

In the next lesson the net torque is zero, so angular momentum is conserved. That one idea explains spinning skaters, collapsing stars and clay that sticks to a spinning disk. ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-6-2-angular-momentum-and-angular-impulse/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-6-2-angular-momentum-and-angular-impulse/