# Angular Momentum and Its Conservation

> AP Physics C: Mechanics · Unit 5 Rotation
> Source: https://www.owlsprep.com/study/ap-physics-c-mech-u5-angular-momentum-and-its-conservation/

This module covers angular momentum for particles and rigid bodies, the torque-angular momentum relation, angular impulse, and conservation of angular momentum, including applications to collisions and changing moment of inertia problems for AP Physics C: Mechanics.

**Prerequisites:** Torque for rotating rigid bodies; Rotational moment of inertia; Linear momentum and its conservation

## Learning objectives

- Calculate angular momentum for point particles and rigid rotating bodies about a given pivot
- Relate net external torque to the rate of change of total angular momentum
- Apply the angular impulse-momentum theorem to solve problems
- Use conservation of angular momentum to solve collision and changing moment of inertia problems
- Distinguish when angular momentum, linear momentum, and kinetic energy are conserved

## Definition and Calculation of Angular Momentum

Angular momentum is the rotational analog of linear momentum, quantifying a system’s tendency to maintain its rotational motion relative to a specific pivot point. Unlike linear momentum, its value depends on the pivot point you choose.

**Angular Momentum (Point Particle)** — For a point particle with mass $m$, velocity $\vec{v}$, at position $\vec{r}$ relative to a pivot, angular momentum is the cross product $\vec{L} = \vec{r} \times m\vec{v}$. Magnitude is $L = rmv\sin\theta = r_\perp mv$, where $r_\perp$ is the perpendicular distance from the pivot to the particle's path.

*Notation:* \vec{L}

**Angular Momentum (Rigid Body)** — For a rigid body rotating about a fixed axis, total angular momentum simplifies to $L = I\omega$, where $I$ is the moment of inertia about the rotation axis and $\omega$ is angular speed. Use the parallel axis theorem $I = I_{CM} + Md^2$ for off-center axes.

*Notation:* L

**Worked example:** A 2.0 kg particle moves at 3.0 m/s along the horizontal line $y = 4.0$ m in the positive $x$-direction. What is the magnitude of the particle’s angular momentum about the origin?

1. For straight-line motion, the perpendicular distance from the origin to the particle’s path is equal to the constant $y$-coordinate of the path, so $r_\perp = 4.0$ m.
2. Use the magnitude formula for particle angular momentum:
3. $$L = r_\perp mv$$
4. Substitute values:
5. $$L = (4.0\ \text{m})(2.0\ \text{kg})(3.0\ \text{m/s}) = 24\ \text{kg·m}^2/\text{s}$$
6. Confirming with the cross product rule gives the same result: $L = rmv\sin\theta = r mv (y/r) = y mv$, which matches our calculation.

> **Exam tip:** When calculating angular momentum for a straight-moving particle, always use the perpendicular distance from the pivot to the path, not the full magnitude of the position vector $\vec{r}$.

## Torque-Angular Momentum Relation and Angular Impulse

The general form of Newton's second law for rotation connects net external torque to the rate of change of total angular momentum. This relation is more general than $\tau = I\alpha$, which only applies when moment of inertia is constant.

**Derivation:** Derive the general torque-angular momentum relation

*Starting from:* Definition of angular momentum $\vec{L} = \vec{r} \times m\vec{v}$

1. Take the time derivative of $\vec{L}$:
2. $$\frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt} \times m\vec{v} + \vec{r} \times \frac{d(m\vec{v})}{dt}$$
3. The first term simplifies to zero because the cross product of a vector with itself is zero: $\frac{d\vec{r}}{dt} = \vec{v}$, so $\vec{v} \times m\vec{v} = 0$.
4. The second term simplifies using Newton's second law $d(m\vec{v})/dt = \vec{F}_{net}$, so:
5. $$\frac{d\vec{L}}{dt} = \vec{r} \times \vec{F}_{net} = \vec{\tau}_{net,ext}$$

*Conclusion:* The relation $\vec{\tau}_{net,ext} = d\vec{L}/dt$ holds for all systems, including those with changing moment of inertia.

Integrating both sides over time gives the angular impulse-momentum theorem, which states that the total change in angular momentum equals the total angular impulse (the integral of net torque over time):

$$\Delta L = \int_{t_1}^{t_2} \tau_{net,ext} dt$$

**Worked example:** A net torque $\tau(t) = 12t - 3t^2$ (units: N·m for torque, s for time) acts on a stationary rotating platform from $t=0$ to $t=4$ s. What is the final angular momentum of the platform?

1. By the angular impulse-momentum theorem, change in angular momentum equals total angular impulse. Initial angular momentum is zero, so final $L$ equals the integral of $\tau(t)$ from 0 to 4.
2. Integrate term-by-term:
3. $$\int (12t - 3t^2) dt = 6t^2 - t^3 + C$$
4. Evaluate the definite integral from 0 to 4:
5. $$L = \left[6(4)^2 - (4)^3\right] - 0 = 96 - 64 = 32\ \text{kg·m}^2/\text{s}$$
6. Confirm units: N·m·s = (kg·m/s²)·m·s = kg·m²/s, which matches the units of angular momentum.

> **Exam tip:** Always use $\tau = dL/dt$ instead of $\tau = I\alpha$ when a problem involves changing moment of inertia; $\tau = I\alpha$ ignores the $\omega dI/dt$ term from the product rule, leading to incorrect results.

## Conservation of Angular Momentum

From the relation $\tau_{net,ext} = dL/dt$, if net external torque about a given pivot is zero, $dL/dt = 0$, so total angular momentum of the system remains constant. This is the law of conservation of angular momentum, one of the three fundamental conservation laws in classical mechanics.

- Conservation only applies for a specific pivot point where net external torque is zero.
- Angular momentum can be conserved even when linear momentum is not: for collisions with a fixed pivot, the pivot exerts an external force (so linear momentum is not conserved), but its torque about the pivot is zero, so angular momentum is conserved.
- Angular momentum conservation does not imply kinetic energy conservation: internal forces can do work on the system, changing total kinetic energy even when angular momentum is constant.

**Worked example:** A 0.5 kg blob of clay is thrown horizontally at 10 m/s and sticks to the outer edge of a stationary solid disk free to rotate about its center. The disk has mass 4.5 kg and radius 0.2 m. Find the angular speed of the disk after the collision.

1. The pivot at the disk’s center exerts an external force during collision, so linear momentum is not conserved. Net torque about the pivot is zero, so angular momentum about the pivot is conserved.
2. Initial angular momentum comes only from the clay:
3. $$L_i = m v r = (0.5\ \text{kg})(10\ \text{m/s})(0.2\ \text{m}) = 1\ \text{kg·m}^2/\text{s}$$
4. Calculate final total moment of inertia, sum of disk and clay:
5. $$I_{total} = I_{disk} + I_{clay} = \frac{1}{2}MR^2 + mR^2 = 0.5(4.5)(0.2)^2 + 0.5(0.2)^2 = 0.11\ \text{kg·m}^2$$
6. Use conservation of angular momentum $L_i = I_{total}\omega$, solve for $\omega$:
7. $$\omega = \frac{L_i}{I_{total}} = \frac{1}{0.11} \approx 9.1\ \text{rad/s}$$

> **Exam tip:** When solving collision problems with rotation on FRQs, always explicitly state that net external torque about the pivot is zero to earn full points for your application of conservation of angular momentum.

## Additional AP-Style Worked Examples

**Worked example:** A figure skater spins at 5 rad/s with arms outstretched, with a total moment of inertia of 4 kg·m². She pulls her arms inward, decreasing her total moment of inertia to 1 kg·m². By what factor does her rotational kinetic energy change?

1. Net torque about the skater’s rotation axis is zero, so angular momentum is conserved. Calculate initial angular momentum:
2. $$L_i = I_1\omega_1 = (4)(5) = 20\ \text{kg·m}^2/\text{s}$$
3. Conservation of angular momentum gives $L_f = L_i$, so solve for final angular speed:
4. $$\omega_2 = \frac{L_i}{I_2} = \frac{20}{1} = 20\ \text{rad/s}$$
5. Calculate initial and final rotational kinetic energy:
6. $$KE_i = \frac{1}{2}I_1\omega_1^2 = 0.5(4)(25) = 50\ \text{J}, \quad KE_f = 0.5(1)(20^2) = 200\ \text{J}$$
7. The ratio $KE_f/KE_i = 4$, so kinetic energy increases by a factor of 4.

**Worked example:** A uniform rod of mass $M$ and length $L$ is pivoted at one end, held stationary horizontally, then released from rest. (a) Find angular speed when the rod reaches vertical. (b) Find angular momentum about the pivot when vertical. (c) A 1/4 length segment breaks off from the pivot while the rod is vertical; no external torque acts during the break. Find the new angular speed of the remaining 3/4 rod.

1. (a) The center of mass of the rod falls by $L/2$, so change in potential energy is $-MgL/2$. By conservation of mechanical energy: $\Delta KE = MgL/2 = \frac{1}{2}I\omega^2$. Moment of inertia of a rod about one end is $I = \frac{1}{3}ML^2$. Substitute:
2. $$Mg\frac{L}{2} = \frac{1}{2}\left(\frac{1}{3}ML^2\right)\omega^2 \implies \omega = \sqrt{\frac{3g}{L}}$$
3. (b) Calculate angular momentum using $L = I\omega$:
4. $$L = \left(\frac{1}{3}ML^2\right)\sqrt{\frac{3g}{L}} = ML\sqrt{\frac{gL}{3}}$$
5. (c) No external torque, so angular momentum is conserved. The remaining rod has mass $M' = 3M/4$ and length $L' = 3L/4$. Calculate new moment of inertia about the pivot:
6. $$I_{new} = \frac{1}{3}M'L'^2 = \frac{1}{3}\left(\frac{3M}{4}\right)\left(\frac{3L}{4}\right)^2 = \frac{9ML^2}{64}$$
7. Set initial angular momentum equal to final angular momentum and solve for $\omega_{new}$:
8. $$\frac{1}{3}ML^2\sqrt{\frac{3g}{L}} = \frac{9ML^2}{64}\omega_{new} \implies \omega_{new} = \frac{64}{27}\sqrt{\frac{3g}{L}} \approx 2.37\sqrt{\frac{3g}{L}}$$

## Common pitfalls

- **Wrong:** Calculating angular momentum of a straight-moving particle as $L = rmv$ instead of $L = r_\perp mv$
  - Why it fails: Students confuse the magnitude of the position vector with the perpendicular distance from the pivot to the particle’s path.
  - Correct: Always draw the line of motion, measure the perpendicular distance from the pivot to this line, and use that value for $r_\perp$.
- **Wrong:** Claiming linear momentum is conserved for a collision with a fixed pivot
  - Why it fails: Students assume all conservation laws apply automatically when angular momentum is conserved, forgetting the requirement of zero net external force for linear momentum.
  - Correct: Explicitly note that the pivot exerts an external force, so only angular momentum about the pivot is conserved for this scenario.
- **Wrong:** Using $\tau = I\alpha$ for problems with changing moment of inertia
  - Why it fails: Students learn $\tau = I\alpha$ first and forget it is only a special case of the general $\tau = dL/dt$ that requires constant $I$.
  - Correct: Always use $\tau = dL/dt$ or conservation of angular momentum when moment of inertia changes over time.
- **Wrong:** Assuming rotational kinetic energy is conserved when angular momentum is conserved
  - Why it fails: Students associate momentum conservation with energy conservation by habit, forgetting internal forces can do non-zero work.
  - Correct: Always recalculate final kinetic energy as $\frac{1}{2}I\omega^2$ after finding $\omega$ from angular momentum conservation; only assume KE is conserved for explicitly elastic collisions.
- **Wrong:** Calculating angular momentum of a rigid body rotating about an off-center axis using $L = I_{CM}\omega$
  - Why it fails: Students memorize $L = I\omega$ for rotation about CM and forget to adjust $I$ for a different rotation axis.
  - Correct: Use the parallel axis theorem to find $I$ about the actual fixed rotation axis before calculating $L$.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Particle angular momentum (about a point) | $\vec{L} = \vec{r} \times m\vec{v}, \quad L = r_\perp mv$ | $r_\perp$ = perpendicular distance from pivot to the particle's path |
| Rigid body angular momentum (fixed axis) | $L = I\omega$ | $I$ = moment of inertia about the *actual rotation axis*, not just CM |
| General Newton's 2nd Law for rotation | $\vec{\tau}_{net,ext} = \frac{d\vec{L}}{dt}$ | Holds for constant or changing moment of inertia |
| Angular Impulse-Momentum Theorem | $\Delta L = \int \tau_{net} dt$ | Relates total torque impulse to change in angular momentum |
| Conservation of Angular Momentum | $L_{i,total} = L_{f,total}$ | Applies if and only if net external torque about the pivot is zero |
| Parallel Axis Theorem for I | $I = I_{CM} + Md^2$ | Used to find I for rotation about an axis not through CM |
| Collisions with fixed pivots | $L$ is conserved about the pivot | Pivot exerts external force, so linear momentum is not conserved |

## What's next

This module completes the core conceptual framework for rotational motion, the central topic of Unit 5 for AP Physics C: Mechanics. Mastering angular momentum conservation is critical to solving the multi-step collision and rotation problems that commonly appear as full-point FRQ questions on the AP exam. Next, you will apply these concepts to rolling motion without slipping, rolling collisions, and rotational simple harmonic motion, the remaining topics in Unit 5. Beyond Unit 5, angular momentum is a foundational concept that reappears in gravitational orbital motion, where it explains Kepler’s second law of planetary motion, and it is a core conservation law across all areas of physics.

- [Oscillations Overview](https://www.owlsprep.com/study/ap-physics-c-mech-u6-overview/)
- [Simple Harmonic Motion](https://www.owlsprep.com/study/ap-physics-c-mech-u6-simple-harmonic-motion/)
- [Mass-spring systems and simple pendulum](https://www.owlsprep.com/study/ap-physics-c-mech-u6-mass-spring-systems-and-simple/)

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