11.3 Angular Momentum of Particles and Rigid Bodies
The Skater's Secret
A figure skater spins slowly with arms extended, tracing a wide circle on the ice. Then she pulls her arms in close to her body. Instantly, she spins dramatically faster --- two, three, even four times the original rate. No one pushed her. No motor kicked in. No external torque acted on her.
Her angular velocity changed. Something else must have stayed the same.
This section is about finding that something --- the rotational quantity that plays the same role momentum plays in translational physics. You met linear momentum $\vec{p} = m\vec{v}$ in Chapter 8, and you saw that it is conserved when no external force acts. Now you will meet its rotational counterpart, and you will see the same conservation architecture appear in a new setting.
Before you read on: A skater spins with moment of inertia $I$ and angular velocity $\omega$. She pulls her arms in, reducing her moment of inertia to $I/2$. No external torque acts.
(a) What happens to her angular velocity?
(b) What happens to her rotational kinetic energy?
Commit to specific answers --- not just "increases" or "decreases," but by what factor.
The Guiding Question
What rotational quantity plays the role momentum played in translational interactions?
In Chapter 8, you discovered that $\vec{p} = m\vec{v}$ is the quantity that stays constant in collisions and interactions when no external force acts. Mass times velocity turned out to be the right combination --- not mass alone, not velocity alone, not kinetic energy.
Now ask the same question for rotation. A spinning object has a moment of inertia $I$ (its rotational analog of mass) and an angular velocity $\omega$ (its rotational analog of velocity). When no external torque acts, what quantity is conserved? Is it $\omega$? Is it $I$? Is it some combination? The skater's spin tells you it is not $\omega$ --- her angular velocity changed. It is not $I$ --- she changed that deliberately. It must be something else.
Exploration: What Stays Constant?
[Interactive: Spinning Figure (Top View). A figure is shown from directly above, spinning about a vertical axis. The figure has adjustable "arms" --- two masses on rods that extend outward from the center. Students control a slider labeled "Arm Extension" that moves the masses closer to or farther from the axis of rotation. As the arm extension changes, three quantities update in real time on the side panel:
- $I$ (moment of inertia), displayed with a bar graph
- $\omega$ (angular velocity), displayed with a bar graph and the visible rotation speed of the figure
- $L = I\omega$ (angular momentum), displayed with a bar graph and a bold numerical readout
The figure's rotation visually speeds up as arms are pulled in and slows down as arms extend outward. A dashed circle shows the path traced by the outermost mass.
Guided prompts appear in sequence:
- "Start with arms fully extended. Note the values of $I$, $\omega$, and $L$."
- "Slowly pull the arms in. Watch all three quantities. Which one increases? Which decreases? Which stays the same?"
- "Push the arms back out. Does the quantity that stayed constant before still stay constant?"
- "Try different starting configurations. Is the pattern always the same?"
The interactive should make it visually unmistakable that $L = I\omega$ remains constant while $I$ and $\omega$ trade off against each other.]
Spend some time with this interactive. Try pulling the arms in quickly and slowly. Try partial adjustments. The conclusion should be clear:
- $\omega$ changes. When the arms come in, $\omega$ increases.
- $I$ changes. When the arms come in, $I$ decreases.
- The product $L = I\omega$ does not change. It is the same before and after, regardless of how you redistribute the mass.
The quantity $I\omega$ is special. It is what is conserved when no external torque acts. It has a name: angular momentum.
The Concept: Angular Momentum
For rigid bodies about a fixed axis
The angular momentum of a rigid body rotating about a fixed axis is:
$$L = I\omega$$
where $I$ is the moment of inertia about that axis and $\omega$ is the angular velocity.
This is the rotational analog of linear momentum $p = mv$. The structural parallel is exact:
| Translational | Rotational |
|---|---|
| Mass $m$ | Moment of inertia $I$ |
| Velocity $v$ | Angular velocity $\omega$ |
| Momentum $p = mv$ | Angular momentum $L = I\omega$ |
Just as linear momentum tells you "how much translational motion" an object has, angular momentum tells you "how much rotational motion" it has. And just as a more massive object moving at the same speed has more linear momentum, an object with a larger moment of inertia spinning at the same rate has more angular momentum.
Units. Angular momentum has units of $\text{kg} \cdot \text{m}^2/\text{s}$. You can verify: $[I][\omega] = (\text{kg} \cdot \text{m}^2)(\text{rad/s}) = \text{kg} \cdot \text{m}^2/\text{s}$. There is no special named unit for angular momentum.
Direction. For rotation about a fixed axis, angular momentum is a vector along the axis of rotation. Its direction is given by the right-hand rule: curl the fingers of your right hand in the direction of rotation, and your thumb points in the direction of $\vec{L}$. As a vector equation:
$$\vec{L} = I\vec{\omega}$$
For a particle
A particle does not "spin," but it can still have angular momentum about a chosen point. If a particle of mass $m$ is at position $\vec{r}$ relative to a point $O$ and has velocity $\vec{v}$, its angular momentum about $O$ is:
$$\vec{L} = \vec{r} \times m\vec{v} = \vec{r} \times \vec{p}$$
This is a cross product, so $\vec{L}$ is a vector perpendicular to the plane containing $\vec{r}$ and $\vec{v}$.
The magnitude is:
$$L = rmv\sin\phi$$
where $\phi$ is the angle between $\vec{r}$ and $\vec{v}$. The factor $r\sin\phi$ is the perpendicular distance from $O$ to the line of motion --- exactly the lever arm you met when computing torque.
Pause and think: A ball moves in a straight line at constant velocity, passing near (but not through) a point $O$. Does the ball have angular momentum about $O$? If so, does that angular momentum change as the ball moves?
Check your answer
Yes, the ball has angular momentum about $O$ whenever its line of motion does not pass through $O$. The angular momentum is $L = mvd$, where $d$ is the perpendicular distance from $O$ to the line of motion. Remarkably, this angular momentum is constant even though the ball moves in a straight line. As the ball moves, $r$ changes and $\phi$ changes, but the product $r\sin\phi = d$ stays the same --- it is always the perpendicular distance from $O$ to the line of motion. No external torque acts on the ball (about $O$), so its angular momentum about $O$ is conserved. A particle moving in a straight line can have constant angular momentum about a point. Angular momentum is not just about spinning.How the two formulas connect
For a rigid body, the formula $L = I\omega$ is actually derived from the particle formula $\vec{L} = \vec{r} \times m\vec{v}$. Think of the rigid body as a collection of particles. Each particle at distance $r_i$ from the axis moves with speed $v_i = r_i\omega$. Its angular momentum about the axis is $L_i = m_i r_i v_i = m_i r_i^2 \omega$. Summing over all particles:
$$L = \sum m_i r_i^2 \omega = \left(\sum m_i r_i^2\right)\omega = I\omega$$
The moment of inertia $I = \sum m_i r_i^2$ collects all the geometric information about how mass is distributed. The formula $L = I\omega$ is the rigid-body summary of what is really a sum of particle angular momenta.
Returning to the Prediction
The skater starts with moment of inertia $I$ and angular velocity $\omega$. Her initial angular momentum is:
$$L_i = I\omega$$
She pulls her arms in, reducing her moment of inertia to $I/2$. No external torque acts, so angular momentum is conserved:
$$L_f = L_i$$
$$\frac{I}{2}\,\omega_f = I\omega$$
$$\omega_f = 2\omega$$
(a) Her angular velocity doubles. If you predicted this, good --- it follows directly from $L = I\omega$ being constant.
(b) Now check the kinetic energy:
$$KE_i = \frac{1}{2}I\omega^2$$
$$KE_f = \frac{1}{2}\left(\frac{I}{2}\right)(2\omega)^2 = \frac{1}{2}\left(\frac{I}{2}\right)(4\omega^2) = I\omega^2 = 2\left(\frac{1}{2}I\omega^2\right) = 2\,KE_i$$
Her kinetic energy doubles. She spins faster and has more energy --- even though no one pushed her and no external torque acted.
Where did the extra energy come from? From her muscles. Pulling her arms inward against the centrifugal tendency requires her to do work. That muscular work is the source of the extra kinetic energy. Angular momentum is conserved; kinetic energy is not. The skater adds energy to the system through internal forces, just as a person jumping off a boat adds kinetic energy to the system while conserving total momentum.
The Axis Matters
Here is something that surprises many students: angular momentum depends on your choice of axis (or reference point). The same object, with the same motion, can have different angular momentum values about different points.
Consider a ball of mass $m$ moving in a straight line at velocity $v$. Its angular momentum about a point $O$ is $L = mvd$, where $d$ is the perpendicular distance from $O$ to the line of motion. Choose a different point $O'$, at a different perpendicular distance $d'$, and the angular momentum is $L' = mvd'$. Different point, different value.
This is fundamentally different from linear momentum. The linear momentum $\vec{p} = m\vec{v}$ does not depend on your choice of origin. It is the same regardless of where you place your coordinate system. Why the asymmetry?
The answer lies in the definitions. Linear momentum $\vec{p} = m\vec{v}$ depends only on velocity, which is origin-independent (shifting your origin does not change how fast something moves). Angular momentum $\vec{L} = \vec{r} \times m\vec{v}$ involves the position vector $\vec{r}$, which depends explicitly on your choice of reference point. Change the reference point, and $\vec{r}$ changes, and so does $\vec{L}$.
This is not a flaw --- it is a feature. Angular momentum is fundamentally about rotation about a specific axis or point. You must always specify what axis or point you are computing angular momentum about. Forgetting this is one of the most common errors in rotational dynamics.
Pause and think: A particle moves along the $x$-axis. What is its angular momentum about the origin? What is its angular momentum about a point on the $y$-axis?
Check your answer
If the particle moves along the $x$-axis, its position is $\vec{r} = (x, 0, 0)$ and its velocity is $\vec{v} = (v_x, 0, 0)$. **About the origin:** $\vec{L} = \vec{r} \times m\vec{v} = m(x, 0, 0) \times (v_x, 0, 0) = \vec{0}$. The angular momentum is zero because $\vec{r}$ and $\vec{v}$ are parallel --- the particle moves directly toward or away from the origin. There is no "rotation" about the origin. **About a point on the $y$-axis**, say $O' = (0, d, 0)$: The position relative to $O'$ is $\vec{r}' = (x, -d, 0)$. Now $\vec{r}'$ and $\vec{v}$ are *not* parallel (they have different directions because $\vec{r}'$ has a $y$-component). The cross product gives $\vec{L}' = m\vec{r}' \times \vec{v} = m(x, -d, 0) \times (v_x, 0, 0) = m(0, 0, dv_x) = mdv_x\,\hat{z}$. The angular momentum is $L' = mdv_x$, directed along the $z$-axis. The same particle, the same motion, but a nonzero angular momentum about $O'$. The perpendicular distance $d$ from $O'$ to the line of motion is what creates the angular momentum.Connection: From $\vec{p} = m\vec{v}$ to $\vec{L} = I\vec{\omega}$
The structural parallel between linear and angular momentum is worth stating explicitly, because it will guide your thinking throughout the rest of this chapter.
| Linear (Chapter 8) | Angular (this section) |
|---|---|
| Momentum: $\vec{p} = m\vec{v}$ | Angular momentum: $\vec{L} = I\vec{\omega}$ |
| Inertia: mass $m$ | Rotational inertia: moment of inertia $I$ |
| Motion: velocity $\vec{v}$ | Rotational motion: angular velocity $\vec{\omega}$ |
| For a particle: $\vec{p} = m\vec{v}$ | For a particle: $\vec{L} = \vec{r} \times m\vec{v}$ |
| Independent of origin | Depends on choice of axis |
The same conservation story is coming. In Section 8.3, you saw that if the net external force on a system is zero, total linear momentum is conserved. In Section 11.5, you will see that if the net external torque on a system is zero, total angular momentum is conserved. The skater is a preview of that result. The architecture repeats.
But notice one key difference: angular momentum depends on the axis, while linear momentum does not depend on the origin. This means that when you use angular momentum conservation, you must specify your axis --- and different choices of axis can make a problem easier or harder. Choosing the right axis is a skill, and you will practice it.
Practice
Layer 1: Concrete
Problem 1. A solid disk of mass 4 kg and radius 0.3 m spins at 10 rad/s about its central axis. Its moment of inertia is $I = \frac{1}{2}mR^2$. What is the angular momentum of the disk?
Check your answer
First, compute the moment of inertia: $$I = \frac{1}{2}mR^2 = \frac{1}{2}(4)(0.3)^2 = \frac{1}{2}(4)(0.09) = 0.18 \text{ kg}\cdot\text{m}^2$$ Then the angular momentum: $$L = I\omega = (0.18)(10) = 1.8 \text{ kg}\cdot\text{m}^2/\text{s}$$ The direction of $\vec{L}$ is along the axis of rotation, determined by the right-hand rule.Problem 2. A 0.5 kg ball moves at 6 m/s along a straight line that passes 2 m from a point $O$. What is the magnitude of the ball's angular momentum about $O$?
Check your answer
For a particle, $L = mvd$, where $d$ is the perpendicular distance from $O$ to the line of motion: $$L = (0.5)(6)(2) = 6 \text{ kg}\cdot\text{m}^2/\text{s}$$ Note that this angular momentum is constant as the ball moves, even though the ball travels in a straight line. The perpendicular distance $d$ does not change, and no torque acts about $O$.Problem 3. A thin hoop of mass 2 kg and radius 0.5 m rolls without slipping at a translational speed of 4 m/s. Compute its angular momentum about its center. (For a hoop, $I = mR^2$.)
Check your answer
First, find the angular velocity. For rolling without slipping, $v = R\omega$, so: $$\omega = \frac{v}{R} = \frac{4}{0.5} = 8 \text{ rad/s}$$ The moment of inertia of a hoop about its center is $I = mR^2 = (2)(0.5)^2 = 0.5 \text{ kg}\cdot\text{m}^2$. The angular momentum about the center: $$L = I\omega = (0.5)(8) = 4 \text{ kg}\cdot\text{m}^2/\text{s}$$Layer 2: Pattern
Problem 4. A 3 kg particle moves at 5 m/s in a straight line. Compute its angular momentum about three different points:
(a) A point on the line of motion.
(b) A point 1 m from the line of motion.
(c) A point 4 m from the line of motion.
What pattern do you see?
Check your answer
Using $L = mvd$ where $d$ is the perpendicular distance from the point to the line of motion: **(a)** $d = 0$, so $L = (3)(5)(0) = 0$. The particle moves directly toward or away from the point. There is no rotational motion about it. **(b)** $d = 1$ m, so $L = (3)(5)(1) = 15 \text{ kg}\cdot\text{m}^2/\text{s}$. **(c)** $d = 4$ m, so $L = (3)(5)(4) = 60 \text{ kg}\cdot\text{m}^2/\text{s}$. **Pattern:** Angular momentum scales linearly with the perpendicular distance from the reference point to the line of motion. The same particle, the same velocity, but a different angular momentum about each point. The farther the point is from the line of motion, the larger the angular momentum. A point *on* the line of motion gives zero angular momentum. This is the geometric nature of angular momentum. It measures not just "how much motion" but "how much of that motion acts as rotation about this particular point."Problem 5. Two objects rotate about the same axis. Object A is a solid sphere ($I = \frac{2}{5}mR^2$) with mass 5 kg, radius 0.2 m, spinning at 20 rad/s. Object B is a thin ring ($I = mR^2$) with mass 1 kg, radius 0.4 m, spinning at 15 rad/s.
Which has more angular momentum? Which has more rotational kinetic energy?
Check your answer
**Object A (solid sphere):** $$I_A = \frac{2}{5}(5)(0.2)^2 = \frac{2}{5}(5)(0.04) = 0.08 \text{ kg}\cdot\text{m}^2$$ $$L_A = I_A\omega_A = (0.08)(20) = 1.6 \text{ kg}\cdot\text{m}^2/\text{s}$$ $$KE_A = \frac{1}{2}I_A\omega_A^2 = \frac{1}{2}(0.08)(400) = 16 \text{ J}$$ **Object B (thin ring):** $$I_B = (1)(0.4)^2 = 0.16 \text{ kg}\cdot\text{m}^2$$ $$L_B = I_B\omega_B = (0.16)(15) = 2.4 \text{ kg}\cdot\text{m}^2/\text{s}$$ $$KE_B = \frac{1}{2}I_B\omega_B^2 = \frac{1}{2}(0.16)(225) = 18 \text{ J}$$ Object B has more angular momentum ($2.4 > 1.6$) *and* more kinetic energy ($18 > 16$), despite having less mass and a lower angular velocity. The larger moment of inertia --- due to mass being farther from the axis --- is what tips the balance. This reinforces: mass distribution matters as much as mass itself.Layer 3: Structure
Problem 6. Why does angular momentum $L$ depend on the choice of axis, while linear momentum $p$ does not depend on the choice of origin?
Check your answer
The answer is in the definitions. **Linear momentum** is $\vec{p} = m\vec{v}$. Velocity is the rate of change of position, but it does not depend on where you place your origin. If you shift your origin by a constant vector, every position changes, but the *difference* in positions (and hence the velocity) stays the same. So $\vec{p}$ is origin-independent. **Angular momentum** is $\vec{L} = \vec{r} \times m\vec{v}$. The position vector $\vec{r}$ is measured from a specific reference point. If you change that reference point, $\vec{r}$ changes, and so does the cross product. Angular momentum is built from position, not just velocity. Position is origin-dependent, so angular momentum is too. Physically, this makes sense. Angular momentum measures rotational motion *about a point or axis*. A car driving past you in a straight line has angular momentum about you, but zero angular momentum about a point on its own path. The "rotation" is relative to where you sit. Linear momentum, by contrast, measures translational motion --- "how much stuff is going how fast in what direction" --- which does not depend on where you observe from. This is not a flaw in angular momentum. It reflects the fact that rotation is inherently relational: you always rotate *about something*. Translation is not --- you just move.Layer 4: Debug
Problem 7. A uniform rod of mass $m$ and length $\ell$ is hinged at one end and swings freely. A student computes the rod's angular momentum by using $I = \frac{1}{12}m\ell^2$ (the moment of inertia about the center of mass). The student writes $L = \frac{1}{12}m\ell^2\omega$.
What is wrong?
Check your answer
The student used the wrong moment of inertia. The formula $I = \frac{1}{12}m\ell^2$ is the moment of inertia about the **center of mass** of the rod. But the rod is hinged at one **end** --- it rotates about its end, not about its center. The correct moment of inertia about the end of a uniform rod is: $$I_{\text{end}} = \frac{1}{3}m\ell^2$$ (This can be obtained from the parallel axis theorem: $I_{\text{end}} = I_{\text{cm}} + md^2 = \frac{1}{12}m\ell^2 + m(\ell/2)^2 = \frac{1}{12}m\ell^2 + \frac{1}{4}m\ell^2 = \frac{1}{3}m\ell^2$.) The correct angular momentum is: $$L = I_{\text{end}}\,\omega = \frac{1}{3}m\ell^2\omega$$ The student's answer is off by a factor of 4 (they got $\frac{1}{12}m\ell^2\omega$ instead of $\frac{1}{3}m\ell^2\omega = \frac{4}{12}m\ell^2\omega$). **The lesson:** The moment of inertia you use must match the axis about which you compute the angular momentum. The center-of-mass moment of inertia is a common default, but it is only correct when the object actually rotates about its center of mass. When the rotation axis is elsewhere, you must use the moment of inertia about *that* axis. Always ask: "What is the axis of rotation?"Reflection
What does angular momentum measure physically --- in your own words?
Here is one way to think about it. Linear momentum $\vec{p} = m\vec{v}$ measures how much "straight-line motion" an object has --- how hard it is to bring to a stop along a line. Angular momentum measures how much "rotational motion" an object has about a point or axis --- how hard it is to stop its rotation.
But angular momentum is richer than this, because it is geometric. It depends not just on how fast something moves and how much mass it has, but on where that mass is relative to the axis. A slow, massive flywheel with mass far from the axis can have enormous angular momentum. A fast, light particle can have large angular momentum about a distant point. The geometry --- the lever arm, the distance from the axis, the distribution of mass --- is woven into the definition.
This is why angular momentum conservation produces such dramatic effects. The skater's angular velocity doubles when she halves her moment of inertia, because angular momentum encodes both the distribution and the motion. Change one, and the other must adjust to keep the product constant.
In your own words: what does angular momentum tell you about a system?
Looking Ahead
You now have the definition of angular momentum for both particles ($\vec{L} = \vec{r} \times m\vec{v}$) and rigid bodies ($\vec{L} = I\vec{\omega}$). You know that it is the rotational counterpart of linear momentum, and you have seen a preview of conservation --- the skater's spin --- that hints at its power.
But how does angular momentum actually change? In Chapter 8, you learned that impulse $\vec{J} = \int \vec{F}\,dt$ changes linear momentum. The rotational analog should follow the same pattern: the accumulation of torque over time should change angular momentum. That quantity --- angular impulse --- is the subject of Section 11.4. It will formalize the link between torque and angular momentum change, setting the stage for the full conservation law in Section 11.5.