Angular vs Linear Momentum: The Difference

One of them needs an address

Ask for the momentum of a $2.0$ kg puck sliding at $3.0$ m/s and there is one answer, $6.0\ \text{kg}\cdot\text{m/s}$, pointed the way the puck is going. Ask for its angular momentum and the question is incomplete until you say about what.

That is the difference, and everything else on this page follows from it.

The AP Physics 1 CED puts the two definitions in separate units and phrases them differently in a way that repays attention.

• EK 4.1.A.1: linear momentum is defined by the equation $\vec{p} = m\vec{v}$. EK 4.1.A.2 adds that momentum is a vector quantity and has the same direction as the velocity. Nothing in either statement mentions a location.
• EK 6.3.A.1: the magnitude of the angular momentum of a rigid system about a specific axis can be described with the equation $L = I\omega$. EK 6.3.A.2: the magnitude of the angular momentum of an object about a given point is $L = rmv\sin\theta$. Both statements name a reference before they name a formula.

Then the CED says it outright. EK 6.3.A.2.i: the selection of the axis about which an object is considered to rotate influences the determination of the angular momentum of that object. Change the axis, change the number, with the object doing exactly the same thing.

There is no counterpart statement for linear momentum, and there could not be. $m$ and $\vec{v}$ are properties of the object and of the reference frame; neither depends on where you stand within that frame. Two observers at rest in the same frame, one at the origin and one ten meters away, will always agree on a puck's linear momentum and will generally disagree on its angular momentum.

The CED even legislates the vocabulary. Topic 4.1 carries the boundary statement: unless otherwise stated, the general term "momentum" will refer specifically to linear momentum. So on an exam, bare "momentum" means the linear kind, and the angular kind always gets its adjective.

Angular vs linear momentum, side by side

Two rows carry most of the weight.

"Its inertia factor." Mass cannot change during a mechanics problem. Rotational inertia can, because it depends on how the mass is arranged relative to the axis, and rearranging is something a system can do to itself. That single asymmetry produces the whole family of skater and diver problems, which have no linear analogue at all. Mass vs rotational inertia works that difference in detail.

"Value for an object moving in a straight line." This is the fact that makes angular momentum click, and most treatments leave it out. It gets its own section next.

The case that separates them: a puck that is not rotating at all

A $2.0$ kg puck slides in a straight line at a constant $3.0$ m/s across frictionless ice. It is not spinning. It is not going round anything. It has angular momentum about almost every point in the room, and the CED says so explicitly.

EK 6.3.A.2.ii: the measured angular momentum of an object traveling in a straight line depends on the distance between the reference point and the object, the mass of the object, the speed of the object, and the angle between the radial distance and the velocity of the object.

That is the four-ingredient list for $L = rmv\sin\theta$, applied to something moving in a straight line. Take three reference points and compute:

Same puck, same instant, three answers, one of them zero. Meanwhile the linear momentum is $6.0\ \text{kg}\cdot\text{m/s}$ from all three vantage points and from every other.

Two things are worth pulling out of this.

The angular momentum about a fixed point is constant in time even though $r$ and $\theta$ both change. As the puck slides, its distance $r$ from the reference point grows and the angle $\theta$ between $\vec{r}$ and $\vec{v}$ shrinks, and the product $r\sin\theta$ is the perpendicular distance from the point to the line, which does not change at all. So $L$ holds still. It has to: no force acts on the puck, so no torque acts about any point, so by EK 6.3.C.2.i there is no angular impulse and no change in $L$.

"Rotating" is not a precondition. Angular momentum is not a measure of spin. It is a measure of motion around a chosen point, and something travelling past a point in a straight line is moving around it in exactly that sense. The reason a spinning wheel is the standard mental image is that $L = I\omega$ is the more common form in problems, not that spin is required.

Worked example one runs these numbers and confirms that $L = rmv\sin\theta$ gives $9.0\ \text{kg}\cdot\text{m}^2\text{/s}$ at two different moments in the puck's journey, with $r$ and $\sin\theta$ each different and the product the same.

The second asymmetry: where you push matters for one and not the other

Push an object once. The linear momentum it gains depends on how hard and for how long. The angular momentum it gains depends on that too, and also on where you pushed.

The first line is on all four AP equation sheets in one form or another. The second is EK 6.3.C.2.i, with the angular impulse defined at EK 6.3.B.1 as the product of the torque exerted and the time interval during which it is exerted, and $\tau = r_{\perp}F = rF\sin\theta$ printed on the AP Physics 1 and AP Physics 2 sheets.

The $r_{\perp}$ is the entire difference. Take a rod lying on frictionless ice and hit it with the same force for the same time in two places.

The rod's center of mass ends up moving at the same speed in both cases. That is not an approximation; Newton's second law for a system says the center of mass responds to the net external force and does not care where it is applied. What differs is that in the second case the rod is also spinning, and it carries more kinetic energy, because your hand travelled further while pushing the end than it would have pushing the center.

Worked example two does this with numbers and shows the energy bookkeeping, which is where the result stops feeling like a trick.

The same asymmetry, read backwards, is why a free body diagram is not enough for a rotational problem. A force's contribution to $\Delta\vec{p}$ needs its magnitude and direction. Its contribution to $\Delta L$ needs those plus its line of action. How to calculate torque is the routine for the second part.

Conservation: same logic, different bookkeeping

Both quantities are conserved when the corresponding external influence is absent, and the CED states each separately.

For angular momentum, EK 6.4.A.2: any change to a system's angular momentum must be due to an interaction between the system and its surroundings, with EK 6.4.A.2.ii adding that a system may be selected so that the total angular momentum of that system is constant, and EK 6.4.A.1 defining the total as the sum of the angular momenta of the system's constituent parts about that axis.

The bookkeeping differs in one important way, and it is the row from the table above.

EK 6.4.A.2.iii: the angular speed of a nonrigid system may change without the angular momentum of the system changing if the system changes shape by moving mass closer to or further from the rotational axis.

Nothing like that sentence exists for linear momentum, because there is no way for a system to change its own mass. So:

The last row deserves a warning. A skater pulling their arms in conserves $L$ and gains kinetic energy, because $K = \frac{1}{2}I\omega^2$ with $L = I\omega$ fixed gives $K = \frac{L^2}{2I}$, which rises as $I$ falls. The energy comes from the muscular work done pulling the arms inward against the outward push they feel. Writing "angular momentum is conserved, so kinetic energy is conserved" is a specific and common way to lose a mark. Worked example three does the arithmetic.

The two conservation conditions are also independent of one another, which surprises people. A system can conserve one and not the other. Push a free rod at its end and its linear momentum changes while, about the center of mass, so does its angular momentum. Push a pivoted door and its angular momentum about the hinge changes while its linear momentum is being altered by the hinge as well. Check each condition on its own terms: net external force for $\vec{p}$, net external torque about your chosen axis for $L$.

What each of the four sheets prints

The four AP courses handle these two quantities with noticeably different equipment, and knowing which lines you will be handed is worth a few minutes.

Linear momentum is nearly identical across all four sheets. Every one prints $\vec{p} = m\vec{v}$ and a center-of-mass velocity $\vec{v}_{\text{cm}} = \frac{\sum\vec{p}_i}{\sum m_i} = \frac{\sum m_i \vec{v}_i}{\sum m_i}$. The algebra-based sheets give the rate law and impulse in difference form, $\vec{F}_{\text{net}} = \frac{\Delta\vec{p}}{\Delta t} = m\frac{\Delta\vec{v}}{\Delta t} = m\vec{a}$ and $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta\vec{p}$. The two Physics C sheets give the same content in calculus form, $\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$ and $\vec{J} = \int \vec{F}_{\text{net}}(t)\,dt = \Delta\vec{p}$.

Angular momentum is where the sheets diverge.

Read the middle two rows together. The algebra-based courses get the scalar $rmv\sin\theta$ form spelled out; the Physics C courses get the cross product $\vec{r} \times \vec{p}$ instead, which contains it, since the magnitude of a cross product is $AB\sin\theta$ and that identity is itself printed in the vectors table of the two Physics C sheets. Neither pair is missing content; they are given the same physics at two levels of machinery.

The direction question splits the courses too. Topic 6.3's boundary statement in AP Physics 1 reads: while AP Physics 1 expects that students can mathematically manipulate the magnitude of angular momentum using one-dimensional vector conventions, the direction of angular momentum and angular impulse is beyond the scope of the course. So an AP Physics 1 candidate handles $L$ with a sign for clockwise or counterclockwise and never with a right-hand rule. The Physics C sheets print $\vec{L}$ with a vector arrow and a cross product, so the direction is squarely in scope there.

You can read all of this off the formulas pages rather than taking it on trust.

Where the confusion costs a mark

• Quoting an angular momentum with no axis or point named. EK 6.3.A.2.i says the choice influences the value. An unaddressed $L$ is not a number yet.
• Saying an object moving in a straight line has zero angular momentum. It has zero angular momentum only about points on its line of motion. EK 6.3.A.2.ii spells out what it depends on everywhere else.
• Using $\text{kg}\cdot\text{m/s}$ for angular momentum. The unit is $\text{kg}\cdot\text{m}^2\text{/s}$. An extra meter comes in with the $r$, and a unit check catches most sign-of-trouble errors here.
• Concluding that conserved angular momentum means conserved kinetic energy. With $L$ fixed, $K = \frac{L^2}{2I}$, so pulling mass inward raises the kinetic energy. The extra energy is internal work.
• Applying conservation of angular momentum about one axis and then reading the answer as if it held about another. Conservation is a statement about a specific axis, because the torque condition is.
• Forgetting the $\sin\theta$ in $L = rmv\sin\theta$. It is the angle between the radial direction and the velocity, and it is $90^\circ$ only when the motion is perpendicular to the line joining the point to the object.
• Using the right-hand rule for $L$ in AP Physics 1. The Topic 6.3 boundary statement puts the direction of angular momentum and angular impulse outside the course. Use a sign.
• Assuming the linear impulse depends on where the force is applied. It does not. Only the angular impulse does.
• Writing bare "momentum" when you mean the angular kind. The Topic 4.1 boundary statement reserves the unqualified word for linear momentum.
• Treating $L = I\omega$ and $L = rmv\sin\theta$ as competing definitions. They are the same quantity for two different descriptions: a rigid system rotating about an axis, and a single object located relative to a point. For a point mass on a circle they agree, since $I = mr^2$ and $v = r\omega$ give $I\omega = mr^2\frac{v}{r} = rmv$, with $\sin\theta = 1$ because the velocity is perpendicular to the radius.

When they behave alike, and why that lulls you

The analogy between the two is real, exact in structure, and the source of the trouble.

That last chain is EK 6.3.C.2.ii verbatim, and the CED calls the rotational impulse-momentum theorem a direct result of the rotational form of Newton's second law for cases in which rotational inertia is constant. Every structural feature transfers. Both are conserved in isolated systems, both change only through an external agent acting over time, both can be read off the area under a graph, and EK 6.3.B.3 and EK 6.3.C.4 make the torque-against-time area the angular impulse just as the force-against-time area is the linear impulse.

Three settings hide the difference completely.

A single rigid body rotating about a fixed axis, with the axis given. The address problem is solved for you by the question, so $L = I\omega$ behaves exactly like $p = mv$ and the analogy is safe.

A rigid body whose shape cannot change. Then $I$ is as fixed as $m$ is and the two behave identically, including in collisions.

A problem phrased entirely in one language. Nothing forces you to notice that one quantity carries a hidden argument.

The difference reappears the moment any of these three holds:

• The problem names two candidate axes, or asks what happens if you measure about the other one.
• The system changes shape, so $I$ moves while $m$ cannot.
• Something travels in a straight line past a rotating system and you have to add its angular momentum to the total, which EK 6.4.A.1 requires as the sum over constituent parts about the chosen axis. A ball of putty thrown at the rim of a stationary turntable is the standard version, and it is a problem you cannot start until you have written down the axis.

The habit: write the axis down before you write the first equation. For linear momentum you never have to, which is why the omission feels harmless right up until it is not.

What the CED asks, and how the exam frames it

Linear momentum is AP Physics 1 Unit 4, Linear Momentum, weighted at 10 to 15 percent of the multiple-choice section. Topic 4.1, Linear Momentum, has one learning objective, 4.1.A: describe the linear momentum of an object or system. EK 4.1.A.1 gives $\vec{p} = m\vec{v}$, EK 4.1.A.2 says momentum is a vector with the same direction as the velocity, and EK 4.1.A.3 says momentum can be used to analyze collisions and explosions, with EK 4.1.A.3.i defining a collision as a model for an interaction in which the forces between the involved objects are much larger than the net external force, EK 4.1.A.3.ii licensing the object model because only the initial and final states are analyzed, and EK 4.1.A.3.iii defining an explosion as an interaction in which forces internal to the system move objects apart. Suggested skills for Topic 4.1 are 1.C, 2.B, 2.C and 3.B. Its boundary statement reserves the unqualified word "momentum" for the linear kind.

Angular momentum is Unit 6, Energy and Momentum of Rotating Systems, weighted at 5 to 8 percent of the multiple-choice section, a share it ties with Unit 7 and which no other AP Physics 1 unit goes below. Topic 6.3, Angular Momentum and Angular Impulse, carries three learning objectives: 6.3.A, describe the angular momentum of an object or rigid system; 6.3.B, describe the angular impulse delivered to an object or rigid system by a torque; and 6.3.C, relate the change in angular momentum of an object or rigid system to the angular impulse given to that object or rigid system. Suggested skills for Topic 6.3 are 1.B, 2.A, 2.D and 3.B. Topic 6.4, Conservation of Angular Momentum, adds learning objective 6.4.A, describe the behavior of a system using conservation of angular momentum, with suggested skills 1.B, 2.D, 3.A, 3.B and 3.C.

The two units are weighted very differently and the weighting understates the angular side's difficulty, because Unit 6 assumes all of Unit 5's torque and rotational inertia machinery on top of its own content.

AP Physics C: Mechanics covers the same two subjects in its own Unit 4, Linear Momentum, and Unit 6, Energy and Momentum of Rotating Systems, weighted at 10 to 20 percent and 10 to 15 percent respectively, with the vector and calculus treatment its equation sheet advertises.

For the routines: conservation of momentum and the impulse-momentum theorem cover the linear side, and how to calculate torque covers the piece the angular side needs first. The CED framing sits on Topic 4.1, Topic 6.3 and Topic 6.4. The momentum practice set and the torque and rotational motion practice set have problems on each side.