Unit 5 Rotation: Understanding and Using Angular Momentum

Angular Momentum

What angular momentum is (and why it’s more than “rotational momentum”)

Angular momentum is the rotational analogue of linear momentum. Linear momentum tells you “how hard it is to stop or redirect” translational motion; angular momentum tells you “how hard it is to stop or redirect” rotational motion about a chosen point (the origin) or about a chosen axis.

A key idea that often surprises students: angular momentum is not a property of an object “all by itself” the way mass is. It depends on what point you measure it about. The same moving particle can have different angular momentum values about different origins.

Angular momentum matters because it links directly to torque in the same way that linear momentum links to net force:

• Net force controls how linear momentum changes.
• Net torque controls how angular momentum changes.

That connection is what makes angular momentum so powerful on AP Physics C: it gives you a conservation law that can solve problems where forces are complicated, short-lived (impulses), or unknown.

Angular momentum of a particle about a point

For a particle at position vector \vec r (measured from your chosen origin) with linear momentum \vec p = m\vec v, the angular momentum about the origin is defined by the cross product

\vec L = \vec r \times \vec p

• \vec r is from the origin to the particle.
• \vec p is the particle’s linear momentum.
• Direction: given by the right-hand rule for the cross product.
• Magnitude:

L = r p \sin\theta

where \theta is the angle between \vec r and \vec p.

Interpreting the magnitude

The factor \sin\theta encodes a crucial physical point: only the component of momentum perpendicular to the radius contributes to angular momentum about that origin.

• If the particle moves directly toward or away from the origin, then \theta = 0 or \pi, so L = 0.
• If the particle moves perpendicular to \vec r (like circular motion about the origin), \sin\theta = 1 and L = rp.

Units

Angular momentum units are

\mathrm{kg\,m^2/s}

(also equivalent to \mathrm{N\,m\,s}).

Angular momentum for circular motion (a very common AP case)

Suppose a particle of mass m moves in a circle of radius r about the origin with speed v. Then \vec r is perpendicular to \vec v, so \sin\theta = 1 and

L = mvr

If the motion is described by angular speed \omega with v = r\omega, then

L = mr^2\omega

This is your first hint of the rigid-body formula L = I\omega, because for a single particle at radius r, the rotational inertia is I = mr^2.

Angular momentum of a rigid body about a fixed axis

For many AP Physics C: Mechanics problems, a rigid body rotates about a fixed axis (often through a pivot). You treat the object as made of many small masses m_i at distances r_i from the axis.

If the body rotates with angular speed \omega about that axis, each piece has speed v_i = r_i\omega and angular momentum magnitude L_i = m_i r_i^2 \omega about the axis. Adding contributions gives

L = I\omega

where I = \sum m_i r_i^2 is the moment of inertia about that axis.

Important subtlety: direction and when \vec L is parallel to \vec\omega

In the simplest AP setting (planar rotation about a principal axis), the angular momentum vector points along the rotation axis, same direction as \vec\omega by the right-hand rule, so you can safely use

\vec L = I\vec\omega

However, in more advanced 3D rotation, \vec L may not be parallel to \vec\omega unless you rotate about a principal axis. AP Physics C typically avoids full inertia-tensor complications, but it’s still worth remembering: the clean vector relationship assumes the standard “spin about a symmetry/principal axis” situation.

Total angular momentum of a system

For multiple particles, total angular momentum about a chosen origin is the vector sum

\vec L_{\text{tot}} = \sum_i \vec r_i \times \vec p_i

This is the quantity that appears in conservation laws. Even if internal forces between particles are large, they often do not change total angular momentum (the same spirit as internal forces not changing total linear momentum).

A useful decomposition for a system of particles (or an extended body) relates angular momentum about an origin O to angular momentum about the center of mass (CM):

\vec L_O = \vec r_{\text{CM}} \times M\vec v_{\text{CM}} + \vec L_{\text{CM}}

• M is total mass.
• \vec r_{\text{CM}} and \vec v_{\text{CM}} are CM position and velocity relative to origin O.
• \vec L_{\text{CM}} is angular momentum about the CM due to motion relative to the CM (this includes “spin”).

This equation helps you reason about problems where something both translates and rotates.

Torque and the time rate of change of angular momentum

Just as net force is the time derivative of linear momentum, net torque is the time derivative of angular momentum.

For a particle, define torque about the origin as

\vec\tau = \vec r \times \vec F

Using Newton’s second law \vec F = d\vec p/dt and the product rule, you can show (for a particle)

\frac{d\vec L}{dt} = \vec\tau

This is one of the central “bridges” in rotational dynamics:

• If net torque is zero, angular momentum is constant.
• If there is a torque, angular momentum changes in the direction of that torque.

Angular impulse

Over a time interval, integrating the torque relation gives

\Delta \vec L = \int \vec\tau\,dt

The integral \int \vec\tau\,dt is the angular impulse. This is especially useful in collisions where a large force acts for a short time and you don’t know its detailed time dependence.

Choosing the origin (common source of mistakes)

Because \vec L = \vec r \times \vec p, the angular momentum depends on the origin. On exams, you must be intentional:

• If the problem involves a pivot, choosing the pivot as the origin is often strategic because the pivot force has zero lever arm and therefore produces zero torque about that point.
• If the problem asks about conservation of angular momentum, it will be “about a point” or “about an axis.” You must match that.

A frequent error is to claim “angular momentum is conserved” without specifying the origin. The correct statement is:

• Angular momentum is conserved about an origin if the net external torque about that origin is zero.

Worked example 1: Particle in uniform circular motion

A particle of mass m moves at speed v in a circle of radius r about the origin. Find its angular momentum magnitude about the origin.

Step 1: Identify vectors and angle. \vec r is radial, \vec p is tangential; they are perpendicular.

Step 2: Apply definition.

L = r p \sin\theta

With p = mv and \theta = 90^\circ:

L = r(mv)(1) = mvr

Interpretation: If the radius stays fixed and speed stays fixed, angular momentum is constant. A centripetal force can keep the particle in a circle without changing L because the centripetal force points along \vec r, making \vec\tau = \vec r \times \vec F = \vec 0.

Worked example 2: Rigid body about a pivot

A rigid disk rotates about its central axis with moment of inertia I and angular speed \omega. Find its angular momentum about that axis.

Step 1: Recognize fixed-axis rotation. The disk is rigid and rotating about a symmetry axis.

Step 2: Use the rigid-body relation.

L = I\omega

Direction: along the axis, right-hand rule with the rotation.

Notation and relationships (quick reference table)

Exam Focus

• Typical question patterns:
  • Compute \vec L of a particle about a point using \vec r \times \vec p, often with geometry to find the perpendicular component.
  • Use L = I\omega for a rotating object and connect it to changes in \omega when I changes.
  • Use \Delta \vec L = \int \vec\tau dt in collisions/short interactions where torque is easier than force.
• Common mistakes:
  • Treating angular momentum as origin-independent (it is not). Always state “about the pivot/origin.”
  • Using L = I\omega for situations that are not pure rotation about a fixed axis (e.g., a translating object without choosing the right origin).
  • Confusing torque with force: a large force can produce zero torque if its line of action passes through the chosen origin.

Conservation of Angular Momentum

The conservation law (what it is and when it applies)

Conservation of angular momentum means the total angular momentum of a system stays constant in time:

\vec L_{\text{tot}} = \text{constant}

This happens when the net external torque on the system about the chosen origin is zero:

\vec\tau_{\text{ext}} = \frac{d\vec L_{\text{tot}}}{dt}

So the condition for conservation is

\vec\tau_{\text{ext}} = \vec 0 \Rightarrow \vec L_{\text{tot}} = \text{constant}

The wording matters: it is net external torque. Internal forces (forces that parts of the system exert on each other) often produce equal and opposite torques that cancel in the total, so they do not change the system’s total angular momentum.

Why conservation is so useful

On many rotation problems, the external forces are hard to model (contact forces, hinge forces, brief collision forces). Conservation of angular momentum lets you bypass those details.

A classic AP strategy is:

1. Pick a system (what objects are included?).
2. Pick an origin/axis about which you will apply conservation.
3. Argue that net external torque about that origin/axis is zero (or negligible during the interaction).
4. Set initial angular momentum equal to final angular momentum about that same origin/axis.

If you can justify step 3, the math often becomes straightforward even when forces are unknown.

Choosing the right axis: why “about the pivot” is a magic phrase

Suppose an object is attached to a frictionless pivot. The pivot force can be large, but it acts at the pivot point. If you choose the pivot as the origin, the lever arm is zero, so the pivot force produces no torque about that point.

That means for many pivot problems, the net external torque about the pivot can be zero (or at least the unknown pivot forces drop out), making angular momentum about the pivot the conserved quantity during brief events like collisions.

Be careful: external forces like gravity can still produce torque about the pivot, depending on the geometry. Often, problems rely on a short collision time so that gravity’s torque impulse is negligible during the collision.

Two major “families” of conservation problems

1) Changing moment of inertia (no external torque)

If there is no net external torque about the rotation axis, and the object’s moment of inertia changes (mass moves inward/outward), angular speed changes to keep L constant. For fixed-axis rotation where L = I\omega applies:

I_i\omega_i = I_f\omega_f

This is the physics behind a figure skater spinning faster when pulling in their arms. The key mechanism is not a mysterious “spin force” but conservation: decreasing I forces \omega to increase so that I\omega stays the same.

A common misconception is that “energy is conserved so it spins faster.” In reality, rotational kinetic energy usually changes:

K = \frac{1}{2}I\omega^2

If L is conserved, then \omega = L/I, giving

K = \frac{L^2}{2I}

So decreasing I increases K. The extra energy comes from internal work done by muscles (or internal forces) as the person pulls their arms in.

2) Collisions and sticking (angular momentum conserved, not kinetic energy)

In rotational collision problems (like a putty ball hitting a disk and sticking), angular momentum about a suitable axis is often conserved during the short collision, but kinetic energy is not.

The typical logic:

• During the collision, forces are large but act over a short time.
• External torque impulse about the pivot/axis is negligible.
• Therefore \vec L about that axis is conserved across the collision.

You then use energy ideas only after the collision if the subsequent motion is conservative (or you include energy losses explicitly).

Central forces and orbital angular momentum (a conceptual anchor)

A central force is a force that always points along the line connecting an object to a fixed point (the “center”), meaning it is parallel or antiparallel to \vec r. Gravity from a spherically symmetric mass and the force from an ideal spring anchored at the origin are common examples.

If \vec F is parallel to \vec r, then

\vec\tau = \vec r \times \vec F = \vec 0

So angular momentum about the center is conserved. This is the deep reason planets “sweep out equal areas in equal times” (Kepler’s second law): it’s a geometric consequence of constant orbital angular momentum when the net force is central.

AP Physics C may not require a full derivation of Kepler’s laws in this section, but it does expect you to understand and use the idea: zero torque about the center implies constant angular momentum, which tightly constrains motion.

Worked example 1: Skater pulls in arms (changing I)

A skater spins with initial moment of inertia I_i and angular speed \omega_i. They pull in their arms so their moment of inertia becomes I_f. Assume negligible external torque. Find \omega_f.

Step 1: Identify conservation law. With negligible external torque,

L_i = L_f

Step 2: Use L = I\omega for fixed-axis rotation.

I_i\omega_i = I_f\omega_f

Step 3: Solve.

\omega_f = \frac{I_i}{I_f}\omega_i

Check your intuition: if I_f < I_i, then \omega_f > \omega_i.

Common pitfall: claiming rotational kinetic energy stays constant. It generally does not; the skater does internal work.

Worked example 2: Inelastic rotational collision about a pivot

A small mass m moving at speed v hits the end of a light rod of length \ell pivoted at the other end and sticks. The impact point is at distance \ell from the pivot. Find the angular speed \omega immediately after the collision. Assume the rod is massless and external torque about the pivot during collision is negligible.

Step 1: Choose origin/axis. Use the pivot as the origin. The pivot force produces zero torque about the pivot.

Step 2: Compute initial angular momentum about the pivot. Just before impact, the mass has momentum magnitude p = mv. If it approaches perpendicular to the rod (typical setup), then \vec r is along the rod and \vec p is perpendicular, so

L_i = m v \ell

Step 3: Compute final angular momentum about the pivot. After sticking, the system rotates as a point mass at radius \ell. Its moment of inertia is

I_f = m\ell^2

Angular momentum after is

L_f = I_f\omega = m\ell^2\omega

Step 4: Apply conservation L_i = L_f.

mv\ell = m\ell^2\omega

So

\omega = \frac{v}{\ell}

What to notice: mass cancels here because both angular momentum and rotational inertia scale with m in the same way.

Common pitfall: using linear momentum conservation. The pivot provides an external impulse, so linear momentum of the mass is not conserved for the system; angular momentum about the pivot can be.

Worked example 3: Orbiting particle at changing radius (central force intuition)

A particle moves under a central force about the origin, so its angular momentum about the origin is constant. Suppose at some instant it is at radius r_1 with tangential speed v_{t1}, and later it is at radius r_2 with tangential speed v_{t2}. Relate these speeds.

Step 1: Use angular momentum magnitude for planar motion. For tangential component,

L = mrv_t

Step 2: Set conservation.

mr_1v_{t1} = mr_2v_{t2}

So

v_{t2} = \frac{r_1}{r_2}v_{t1}

Interpretation: if the particle moves closer to the center (smaller r), its tangential speed increases.

How conservation statements can fail (and how to diagnose it)

Conservation of angular momentum is not automatic; it rests on the torque condition.

Common situations where students incorrectly apply it:

• Gravity torque about a pivot is not negligible. If an object swings for a long time, gravity’s torque changes angular momentum. Conservation might still be usable during a brief collision, but not over the entire swing.
• Wrong choice of origin. Net external torque might be zero about one point but not about another. For example, a force might pass through the center of mass (zero torque about CM) but not pass through a corner point.
• Including the wrong system boundary. If you leave out an interacting object (like the Earth providing a gravitational torque, or a person applying an external twist), then what you call “external” changes.

A reliable check: explicitly ask “What external forces act, and do any have a lever arm about my chosen origin?” If yes, you likely have external torque.

Connecting conservation of angular momentum to other rotation tools

Angular momentum conservation often pairs with:

• Rotational kinematics after you find \omega (to get angles or times).
• Energy after a collision (to determine subsequent heights/speeds if motion is conservative).
• Torque and angular impulse when torque is not zero but acts for a known time.

The skill is choosing the right tool for the right time interval: conservation during short impulsive events, torque integration when torque is known, and energy when nonconservative work is absent (or quantified).

Exam Focus

• Typical question patterns:
  • “A mass sticks to a rotating disk/rod” or “a bullet embeds in a block” style problems: conserve angular momentum about the rotation axis (often a pivot) to find \omega right after impact.
  • “Person pulls in arms” or “radius changes under central force”: conserve L to relate initial and final speeds or angular speeds.
  • Mixed translation + rotation: choose an origin (often CM or pivot) and conserve angular momentum about that point during an impulse.
• Common mistakes:
  • Conserving angular momentum without checking external torque about the chosen origin (especially gravity torque about a pivot over non-negligible time).
  • Conserving kinetic energy through an inelastic collision; usually you conserve L, not K, during sticking.
  • Using the wrong lever arm or wrong component: only the perpendicular component of momentum contributes to L, and only perpendicular force components contribute to torque.