Angular momentum of a particle

A figure skater spins slowly with her arms flung wide, then pulls them tight against her chest — and suddenly she is a blur, spinning three or four times faster with no push from the ice at all. Nothing external touched her; no torque acted. Yet her rotation sped up dramatically. What quantity stayed exactly the same while everything else about her spin changed? The answer is angular momentum, and before it can be understood for a whole spinning skater, it has to be defined for the simplest possible object: one single particle, moving anywhere in space, watched from one chosen point. That single-particle definition, $\vec L = \vec r \times \vec p$ , looks unfamiliar the first time — a cross product of a position vector and a momentum vector — but it is built from ideas already known: position, momentum and the vector product from earlier in this chapter's torque discussion. This article isolates that one idea and works it hard. We define angular momentum of a particle precisely, connect it to torque through $\vec\tau = d\vec L/dt$ exactly as force connects to linear momentum, and show why the surprising fact about the skater — angular momentum staying constant when no external torque acts — is not a special trick of ice skating but a direct, general consequence of that one equation. Along the way we handle the single trickiest conceptual trap in the whole topic: that a particle moving in a straight line, never turning at all, can still carry nonzero angular momentum about a point that is not on its path. Get that one idea straight and the rest of the chapter — moment of inertia, $L = I\omega$ , conservation of angular momentum for orbits and spins — falls into place as a natural extension of this single-particle picture.

Key takeaways

Angular momentum of a particle about a point is defined as $\vec L = \vec r \times \vec p$ , where $\vec r$ is the position vector from that point and $\vec p = m\vec v$ is linear momentum.
Its magnitude is $L = rp\sin\theta = mvr\sin\theta$ , with $\theta$ the angle between $\vec r$ and $\vec p$ ; its direction follows the right-hand rule, perpendicular to the plane of $\vec r$ and $\vec p$ .
Angular momentum is always measured relative to a chosen point or axis — the same particle has different $\vec L$ about different origins.
Torque is the time rate of change of angular momentum, $\vec\tau = \dfrac{d\vec L}{dt}$ , exactly paralleling $\vec F = \dfrac{d\vec p}{dt}$ for linear motion.
If the net external torque about a point is zero, $\vec L$ about that point is conserved — the principle behind a skater pulling in her arms and a planet sweeping equal areas in equal times.
A particle moving in a straight line still has nonzero angular momentum about any point off that line, because $\vec r$ and $\vec p$ are not parallel there.
For circular motion, $\vec r \perp \vec p$ always, so $L = mvr = mr^2\omega$ — the bridge to the rigid-body result $L = I\omega$ used later in this chapter.
Key idea: a particle's angular momentum L = r × p about a chosen point is what torque changes (τ = dL/dt), so zero net torque locks L constant — the single-particle root of every rotational conservation law in this chapter.

The core insight

Angular momentum of a particle, $\vec L = \vec r\times\vec p$ measured about a chosen point, changes only under a net torque about that same point ( $\vec\tau = d\vec L/dt$ ), so it stays exactly constant whenever that torque is zero.

Where it fits in the syllabus

Angular momentum of a particle is a core definition inside the CBSE Class 11 chapter "System of Particles and Rotational Motion," sitting right after moment of a force (torque) in the prescribed sequence. The board expects the definition $\vec L = \vec r\times\vec p$ , the relation $\vec\tau = d\vec L/dt$ , and the resulting law of conservation of angular momentum, all stated and used without calculus-based proof — differentiating a vector cross product is shown only informally here, never demanded as a rigorous derivation on the CBSE board exam. This single-particle picture is deliberately built up because it is the seed from which $L = I\omega$ for a rotating rigid body — a separate, later topic in the same chapter — is generalised.

The physical picture

Picture a particle of mass $m$ at some instant, located at position vector $\vec r$ from a chosen origin $O$ , moving with velocity $\vec v$ and therefore linear momentum $\vec p = m\vec v$ . The angular momentum of that particle about $O$ is defined as the vector (cross) product of its position and its momentum:

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

where: $\vec r$ is the position vector of the particle measured from the chosen point $O$ (not from anywhere else — the choice of $O$ matters), $\vec p = m\vec v$ is its instantaneous linear momentum, and $\vec L$ is the resulting angular momentum, a vector. Reading the physics: $\vec L$ is large when the particle is far from $O$ (large $r$ ), moving fast (large $p$ ), and moving nearly perpendicular to the line joining it to $O$ (large $\sin\theta$ ). If the particle happens to be moving directly toward or away from $O$ , so that $\vec r$ and $\vec p$ point along the same line, $\sin\theta = 0$ and $L = 0$ instantaneously — all its momentum is "radial," none of it circulating around $O$ . In magnitude form, using $\theta$ for the angle between $\vec r$ and $\vec p$ ,

L = rp\sin\theta = mvr\sin\theta

where: $r = |\vec r|$ , $p = |\vec p| = mv$ , and $\theta$ is measured between the position vector and the momentum vector at that instant. The direction of $\vec L$ , by the right-hand rule for the cross product, is perpendicular to the plane containing $\vec r$ and $\vec p$ — curl the fingers of the right hand from $\vec r$ toward $\vec p$ and the thumb gives the direction of $\vec L$ . This direction is a mathematical bookkeeping device (it tells you the sense of circulation, clockwise or counter-clockwise, and the axis about which that circulation happens) rather than a direction the particle physically travels in. A second display, equally central, links angular momentum to torque:

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

where: $\vec\tau$ is the net external torque acting on the particle, measured about the same point $O$ that $\vec L$ is measured about — the two must always share one common reference point. The physical reading is a direct parallel with Newton's second law in its momentum form, $\vec F = d\vec p/dt$ : just as unbalanced force is required to change linear momentum, unbalanced torque is required to change angular momentum. No torque about $O$ means no change in $\vec L$ about $O$ , however the particle's speed or direction may otherwise be varying.

Figure

A particle at position r from origin O moving with velocity v at angle theta to r, with L equal to r cross p pointing out of the page by the right hand rule

A key consequence follows immediately: if the net torque about $O$ is zero at every instant, then $\vec L$ about $O$ never changes — it is conserved. This is the law of conservation of angular momentum for a particle, exactly analogous to conservation of linear momentum whenever the net external force is zero. It is this single fact, applied to a spinning skater, a planet in orbit, or a rigid body made of many particles, that powers most of the interesting physics later in the chapter.

Derivation AWhy torque equals the rate of change of angular momentum

The goal is to show, starting only from the definition $\vec L = \vec r\times\vec p$ and the product rule for differentiating a cross product, that $\vec\tau = d\vec L/dt$ — and in doing so to see exactly why the term coming from the particle's own velocity vanishes, leaving only the torque of the external force. This is the algebraic heart of why angular momentum is conserved whenever torque is absent, and CBSE expects the flow of this argument (not a rigorous vector-calculus treatment) to be followed and reproduced conceptually.

Step 1. Write down the definition and differentiate both sides with respect to time, applying the product rule to the cross product exactly as it is applied to an ordinary product: $\dfrac{d\vec L}{dt} = \dfrac{d}{dt}(\vec r\times\vec p) = \dfrac{d\vec r}{dt}\times\vec p + \vec r\times\dfrac{d\vec p}{dt}$ .

Step 2. Identify the first term. Since $\dfrac{d\vec r}{dt} = \vec v$ and $\vec p = m\vec v$ , the first term becomes $\vec v\times m\vec v = m(\vec v\times\vec v)$ . The cross product of any vector with itself is zero (there is no plane between two parallel vectors, so $\sin 0° = 0$ ), so this whole term vanishes identically, for any particle, at any instant.

Step 3. Identify the second term. By Newton's second law, $\dfrac{d\vec p}{dt} = \vec F$ , the net force on the particle. So the second term is $\vec r\times\vec F$ , which is precisely the definition of torque, $\vec\tau = \vec r\times\vec F$ , about the same origin used for $\vec r$ .

Step 4. Combine the two results. With the first term gone and the second term identified as torque, $\dfrac{d\vec L}{dt} = 0 + \vec\tau = \vec\tau$ . This is the promised relation, $\vec\tau = \dfrac{d\vec L}{dt}$ , and it holds about any fixed point, provided $\vec L$ and $\vec\tau$ are both measured about that same point. Setting $\vec\tau = 0$ immediately gives $\dfrac{d\vec L}{dt} = 0$ , meaning $\vec L$ is constant — the law of conservation of angular momentum falls straight out of this one line.

Worked examples

Board level - angular momentum of a particle moving parallel to an axis

A particle of mass $2\ \text{kg}$ moves in the x-y plane with constant velocity $\vec v = (3\ \text{m/s})\hat i$ , while its position vector from the origin at the instant considered is $\vec r = (4\ \text{m})\hat j$ . Find its angular momentum about the origin.

Step 1. Find the linear momentum. $\vec p = m\vec v = 2\ \text{kg}\times(3\ \text{m/s})\hat i = 6\hat i\ \text{kg}\,\cdot\,\text{m/s}$ .

Step 2. Apply the definition $\vec L = \vec r\times\vec p$ , using $\hat j\times\hat i = -\hat k$ : $\vec L = (4\hat j)\times(6\hat i) = 24(\hat j\times\hat i) = 24(-\hat k) = -24\hat k\ \text{kg}\,\cdot\,\text{m}^2\text{/s}$ .

Step 3. State the magnitude and direction. The magnitude is $|\vec L| = 24\ \text{kg}\,\cdot\,\text{m}^2\text{/s}$ , and the direction $-\hat k$ means it points into the page. Notice the particle is travelling in a perfectly straight line (constant $\vec v$ , no force, no acceleration) — yet $\vec L$ about the origin is nonzero and, because there is no force at all here, it stays exactly $-24\hat k\ \text{kg}\,\cdot\,\text{m}^2\text{/s}$ forever. Answer: $\vec L = -24\hat k\ \text{kg}\,\cdot\,\text{m}^2\text{/s}$ , magnitude $24\ \text{kg}\,\cdot\,\text{m}^2\text{/s}$ , into the page.

Competitive level - angular momentum and angular velocity of a particle in circular motion

A particle of mass $0.5\ \text{kg}$ moves in a circle of radius $2\ \text{m}$ at a constant speed of $6\ \text{m/s}$ . Find its angular momentum about the centre of the circle, its angular velocity, and confirm the two routes to L agree.

Step 1. Note the geometry. For circular motion about the centre, the position vector $\vec r$ (pointing from centre to particle) and the velocity $\vec v$ (tangential to the circle) are always perpendicular, so $\theta = 90°$ and $\sin\theta = 1$ at every instant.

Step 2. Compute $L$ directly from $L = mvr\sin\theta = mvr$ : $L = 0.5\times6\times2 = 6\ \text{kg}\,\cdot\,\text{m}^2\text{/s}$ .

Step 3. Find the angular velocity from $v = r\omega$ : $\omega = v/r = 6/2 = 3\ \text{rad/s}$ .

Step 4. Cross-check using $L = mr^2\omega$ , the circular-motion form: $L = 0.5\times(2)^2\times3 = 0.5\times4\times3 = 6\ \text{kg}\,\cdot\,\text{m}^2\text{/s}$ , matching Step 2 exactly. Answer: $L = 6\ \text{kg}\,\cdot\,\text{m}^2\text{/s}$ , $\omega = 3\ \text{rad/s}$ , both routes agree. Timing note: this pattern (speed, radius, then either $L=mvr$ or via $\omega$ and $L=mr^2\omega$ ) should take well under two minutes in an exam; the only real risk is forgetting that circular motion guarantees $\theta=90°$ , so no sine factor needs computing — a trap students who over-generalise from the first example sometimes fall into by inserting a stray $\sin\theta$ where $\theta$ is already known to be $90°$ .

Common traps

Forgetting the reference point. Angular momentum has no meaning without stating what point or axis it is measured about; the same particle, at the same instant, has a different $\vec L$ about different origins.
"Straight line means zero angular momentum." A particle travelling in a perfectly straight line still has nonzero, constant $\vec L$ about any point that is not on that line, because $\vec r$ and $\vec p$ are not parallel there.
Mixing up torque and force conditions. Conservation of linear momentum needs zero net force; conservation of angular momentum needs zero net torque about the chosen point — these are related but distinct conditions, and one can hold without the other.
Sign or direction slips in $\vec r\times\vec p$ . Compute components carefully (or apply the right-hand rule consistently) — a sign error flips the whole answer from out-of-page to into-page.
Confusing $\theta$ with the angle of the path. $\theta$ in $L = mvr\sin\theta$ is the angle between the position vector $\vec r$ and the velocity $\vec v$ at that instant, not any angle of the particle's trajectory shape.

Common trapA straight-line particle is not "not rotating," and L is meaningless without stating the point it is measured about

The single biggest mark-loser on this topic is the belief that only particles moving in circles or curves can have angular momentum. The board-level worked example above disproves this directly: a particle travelling in a dead-straight line at constant velocity, feeling zero net force, still has a fixed, nonzero angular momentum about any point that does not lie on its line of motion — because $\vec r$ and $\vec p$ are not parallel for that point. Angular momentum is not a statement about whether the particle itself is "spinning" or curving; it is a statement about the particle's motion as seen relative to a chosen reference point. The second, equally costly habit is quoting a numeric value of $L$ without saying which point it was computed about — since $\vec r$ changes with the choice of origin, so does $\vec L$ , and an answer with no stated reference point is an incomplete answer.

You should be able to…

State the definition $\vec L = \vec r\times\vec p$ for a particle's angular momentum about a chosen point, and identify what $\vec r$ , $\vec p$ and the point of reference each mean.
Compute the magnitude $L = mvr\sin\theta$ and determine the direction of $\vec L$ using the right-hand rule.
Explain, following the four-step argument, why $\vec\tau = d\vec L/dt$ , and state the resulting law of conservation of angular momentum.
Recognise and correctly compute the angular momentum of a particle moving in a straight line about an off-line point.
Apply $L = mvr = mr^2\omega$ for a particle in circular motion, and connect this single-particle result to $L=I\omega$ for a rigid body.
Identify, given a physical situation, whether the net torque about the relevant point is zero, and if so state that $\vec L$ about that point is conserved.

Ask the tutor: If you are still unsure why a particle moving in a perfectly straight line can carry angular momentum, or want guided practice computing r cross p by components, ask the AI tutor to work through a few more examples with you.

Why does the direction of L not point in the direction the particle is moving?

That is a common first reaction, but $\vec L$ is not meant to describe where the particle goes — it describes the sense of circulation the particle's motion has about the chosen point. The cross product $\vec r\times\vec p$ is, by construction, always perpendicular to both $\vec r$ and $\vec p$ , so it points along the imaginary axis the particle would appear to be "orbiting" around at that instant, not along its actual velocity. Think of $\vec L$ as an axis-and-sense label (clockwise or counter-clockwise, and about which line) rather than a path direction, and it stops feeling strange.

If a particle moves in a straight line and never turns, how can it possibly have angular momentum?

Because angular momentum is defined relative to a point, not relative to the particle's own path. Picture a car driving straight down a highway and you standing beside the road, not on it. As the car approaches, passes, and recedes, the line from you to the car keeps changing direction even though the car itself never turns. That changing "sight line" is exactly what $\vec r\times\vec p$ captures. If you stood exactly on the highway (on the line of motion), $\vec r$ and $\vec p$ would be parallel and $L$ would be zero — but for any observer off that line, $L$ is nonzero and, since no force acts, stays perfectly constant.

Is L = r times p the same kind of multiplication as torque, tau = r times F?

Yes, exactly the same operation, the vector (cross) product, applied to a different pair of vectors — position with force for torque, position with momentum for angular momentum. That family resemblance is not a coincidence: it is exactly why differentiating $\vec L$ with respect to time produces $\vec\tau$ on the other side, as shown in the derivation above. Once the cross product is comfortable from the torque topic, angular momentum is really just "the same recipe applied to momentum instead of force."

Do I need to memorise the full derivation of tau equals dL by dt for the CBSE exam?

CBSE expects you to follow and be able to reproduce the flow of the argument — differentiate $\vec r\times\vec p$ , notice $\vec v\times m\vec v=0$ because a vector crossed with itself vanishes, and identify the remaining term as $\vec r\times\vec F=\vec\tau$ — but it is not treated as a rigorous, examiner-graded vector-calculus proof the way it might be at university. Knowing the four steps well enough to explain why the velocity term disappears is what actually earns marks; rote memorisation without understanding the vanishing term is the weak version of this answer.