Angular momentum and its conservation — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Angular momentum for particles and rigid bodies, the torque-angular momentum relation, angular impulse, conservation of angular momentum, cross product calculation, and applications to collisions and changing moment of inertia problems.

You should already know: Torque for rotating rigid bodies, rotational moment of inertia, linear momentum and its conservation.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Angular momentum and its conservation?

Angular momentum is the rotational analog of linear momentum, quantifying the tendency of a rotating or moving system to maintain its rotational motion about a given pivot point. In the AP Physics C: Mechanics CED, Unit 5 (Rotation) accounts for 14-20% of total exam score, and this subtopic makes up roughly 4-6% of the total exam, appearing in both multiple choice (MCQ) and free response (FRQ) sections, often combined with collisions, energy, and torque concepts. Conservation of angular momentum states that the total angular momentum of an isolated system (with zero net external torque about a given pivot) remains constant over time, regardless of internal changes to the system like rearrangements of mass or inelastic collisions. This law is one of the three fundamental conservation laws (along with energy and linear momentum) that you will use to solve complex multi-step problems on the exam. Unlike linear momentum, angular momentum is always defined relative to a specific pivot point, so its value changes depending on which point you measure it from, a key distinction from linear momentum.

2. Calculating Angular Momentum for Particles and Rigid Bodies

Angular momentum for a point particle moving with velocity $\vec{v}$ at position $\vec{r}$ relative to a reference pivot point is defined via the vector cross product: $$\vec{L}=\vec{r}\times m\vec{v}$$ The magnitude of $\vec{L}$ is $L=rmv\sin \theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{v}$. This can be rewritten as $L=r_{\perp }mv=rm{v}_{\perp }$, where $r_{\perp }$ is the perpendicular distance from the pivot to the particle’s line of motion, and $v_{\perp }$ is the component of velocity perpendicular to $\vec{r}$. Direction follows the right-hand rule for cross products.

For a rigid body rotating about a fixed axis, the total angular momentum simplifies to: $$L=I\omega$$ where $I$ is the total moment of inertia of the body about the fixed rotation axis, and $\omega$ is the angular speed about that axis. If the rotation axis is not through the body’s center of mass, use the parallel axis theorem to find the correct $I$ about the axis: $I=I_{CM}+M{d}^2$, where $d$ is the distance from CM to the axis.

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. Recall that 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, $r_{\perp }=4.0$ m.
2. Use the magnitude formula for particle angular momentum: $L=r_{\perp }mv$.
3. Substitute values: $L=(4.0\text{m})(2.0\text{kg})(3.0\text{m/s})=24{\text{kgcdotpm}}^2/\text{s}$.
4. Confirm that the cross product rule gives the same result: $L=rmv\sin \theta =rmv(y/r)=ymv$, matching 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 $\vec{r}$ between the pivot and the particle’s current position.

3. Torque-Angular Momentum Relation and Angular Impulse

Starting from the definition of angular momentum for a system, take the time derivative: $$\frac{d\vec{L}}{dt}=\frac{d\vec{r}}{dt}\times m\vec{v}+\vec{r}\times \frac{d(m\vec{v})}{dt}$$ The first term simplifies to $\vec{v}\times m\vec{v}=0$, since the cross product of a vector with itself is zero. The second term is $\vec{r}\times {\vec{F}}_{net}={\vec{\tau }}_{net}$, giving the general form of Newton’s second law for rotation: $${\vec{\tau }}_{net,ext}=\frac{d\vec{L}}{dt}$$ This relation is more general than $\tau =I\alpha$, which only holds when the moment of inertia $I$ is constant. $\tau =dL/dt$ works for systems with changing $I$, such as a skater pulling in their arms.

Integrating both sides over time gives the angular impulse-momentum theorem: $$\Delta L={\int }_{t_1}^{t_2}{\tau }_{net,ext}dt$$ The integral on the right is called angular impulse, equal to the total change in angular momentum of the system, just as linear impulse equals change in linear momentum.

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, the change in angular momentum equals the total angular impulse. Initial angular momentum is 0, so final $L={\int }_0^4\tau (t)dt$.
2. Integrate term-by-term: $\int (12t-3t^2)dt=6t^2-t^3+C$.
3. Evaluate the definite integral from 0 to 4: $L=\left[6(4{)}^2-(4{)}^3\right]-0=96-64=32{\text{kgcdotpm}}^2/\text{s}$.
4. Confirm units: N·m·s = (kg·m/s²)·m·s = kg·m²/s, which matches angular momentum units.

Exam tip: Always use $\tau =dL/dt$ instead of $\tau =I\alpha$ when a problem involves changing moment of inertia; the latter will give an incorrect result because it ignores the $\omega dI/dt$ term from the product rule of differentiation.

4. Conservation of Angular Momentum

From the general relation ${\tau }_{net,ext}=dL/dt$, if the net external torque on a system about a given pivot is zero, then $dL/dt=0$, meaning the total angular momentum of the system is constant: $$L_{i,total}=L_{f,total}$$ This is the law of conservation of angular momentum. Key points for AP exam problems:

1. Conservation only applies for a given pivot point where net external torque is zero. It does not guarantee conservation about other points.
2. 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 the torque from this force about the pivot is zero (since $r=0$ for the force), so angular momentum is conserved.
3. Angular momentum conservation does not imply kinetic energy conservation. Internal forces can do work on the system, changing the 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: $L_i=mvr=(0.5\text{kg})(10\text{m/s})(0.2\text{m})=1{\text{kgcdotpm}}^2/\text{s}$.
3. Calculate final total moment of inertia: $I_{total}=I_{disk}+I_{clay}=\frac{1}{2}M{R}^2+m{R}^2=0.5(4.5)(0.2{)}^2+0.5(0.2{)}^2=0.11{\text{kgcdotpm}}^2$.
4. Use conservation: $L_i=I_{total}\omega$, so $\omega =\frac{1}{0.11}\approx 9.1\text{rad/s}$.

Exam tip: When solving collision problems with rotation, always explicitly state why angular momentum is conserved (net external torque about the pivot is zero) to earn full points on FRQs.

5. Common Pitfalls (and how to avoid them)

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

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

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? A) Decreases by a factor of 4 B) Remains unchanged C) Increases by a factor of 2 D) Increases by a factor of 4

Worked Solution: Net torque about the skater’s rotation axis is zero, so angular momentum is conserved. Initial angular momentum $L_i=I_1{\omega }_1=(4)(5)=20$ kg·m²/s. Final angular momentum $L_f=I_2{\omega }_2=L_i$, so ${\omega }_2=20/1=20$ rad/s. Initial kinetic energy $K{E}_i=\frac{1}{2}{I}_1{\omega }_1^2=0.5(4)(25)=50$ J. Final kinetic energy $K{E}_f=0.5(1)(20^2)=200$ J. The ratio $K{E}_f/K{E}_i=4$, so kinetic energy increases by a factor of 4. Correct answer: D.

Question 2 (Free Response)

A uniform rod of mass $M$ and length $L$ is pivoted at one end, held stationary horizontally, then released from rest. (a) Use conservation of mechanical energy to find the angular speed of the rod when it reaches the vertical position. (b) What is the angular momentum of the rod about the pivot when it is vertical? (c) A 1/4 length segment of the rod breaks off, starting from the pivot, while the rod is vertical. Assuming no external torque acts during the break, what is the new angular speed of the remaining 3/4 length rod?

Worked Solution: (a) The center of mass of the rod falls by $L/2$ when rotating to vertical. Change in gravitational potential energy: $\Delta U=-Mg(L/2)$, so $\Delta KE=MgL/2=\frac{1}{2}I{\omega }^2$. Moment of inertia of a rod about one end is $I=\frac{1}{3}M{L}^2$. Substitute: $$MgL/2=\frac{1}{2}\left(\frac{1}{3}M{L}^2\right){\omega }^2⟹\omega =\sqrt{\frac{3g}{L}}$$

(b) Angular momentum is $L=I\omega =\left(\frac{1}{3}M{L}^2\right)\sqrt{\frac{3g}{L}}=ML\sqrt{\frac{gL}{3}}$.

(c) No external torque, so angular momentum is conserved. The remaining rod has mass $M^{\prime }=3M/4$ and length $L^{\prime }=3L/4$. Moment of inertia about the pivot is: $$I_{new}=\frac{1}{3}{M}^{\prime }{L}^{\prime 2}=\frac{1}{3}\left(\frac{3M}{4}\right){\left(\frac{3L}{4}\right)}^2=\frac{9M{L}^2}{64}$$ Set initial angular momentum equal to final angular momentum: $$\frac{1}{3}M{L}^2\sqrt{\frac{3g}{L}}=\frac{9M{L}^2}{64}{\omega }_{new}⟹{\omega }_{new}=\frac{64}{27}\sqrt{\frac{3g}{L}}\approx 2.37\sqrt{\frac{3g}{L}}$$

Question 3 (Application / Real-World Style)

A communication satellite in circular orbit around Earth has an initial moment of inertia about its spin axis of 1200 kg·m², and spins at 0.10 rev/s. Two 100 kg solar panels unfurl from the satellite’s center, extending 10 m out from the spin axis on opposite sides, acting as point masses. What is the new spin rate of the satellite, and what is the change in rotational kinetic energy of its spin?

Worked Solution: No net external torque acts on the satellite about its spin axis, so angular momentum is conserved. Initial angular momentum $L_i=I_i{\omega }_i=1200\times 0.10=12$ kg·m²·rev/s. The added moment of inertia from the two panels is $I_{panels}=2m{r}^2=2(100)(10^2)=20000$ kg·m², so total final $I_f=1200+20000=21200$ kg·m². Conservation gives ${\omega }_f=L_i/I_f=12/21200\approx 5.7\times 10^{-4}$ rev/s. Converting to rad/s, the change in kinetic energy is $\Delta KE=\frac{1}{2}{I}_f{\omega }_f^2-\frac{1}{2}{I}_i{\omega }_i^2\approx -2300$ J. In context, the satellite’s rotational kinetic energy decreases by ~2300 J, which is used to do work extending the solar panels against the centrifugal effect of rotation.

7. Quick Reference Cheatsheet

8. What's Next

This chapter completes the core conceptual framework for rotational motion, the central topic of Unit 5. Next, you will apply angular momentum and its conservation to rolling motion without slipping, rolling collisions, and rotational simple harmonic motion, the remaining topics in Unit 5. Without mastering conservation of angular momentum, you cannot solve the multi-step collision and rotation problems that are common full-point FRQ questions on the AP exam. 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 in all areas of physics. Follow-on topics: Rolling motion and rolling friction Gravitational orbital motion Rotational simple harmonic motion