Angular momentum in rotation about a fixed axis

A spinning ice skater pulls her arms in and, without any push from outside, suddenly whirls faster. A high diver curls into a tight tuck mid-air and the somersaults come quicker; she straightens out and the spin slows before she slices into the water. Nothing external is torquing either of them once they leave the ice or the board — so what quantity is being conserved that forces the spin rate to change the moment the body's shape changes? The answer is angular momentum, $\vec L = \vec r\times\vec p$ , the rotational twin of ordinary momentum $\vec p = m\vec v$ . This article builds angular momentum from the ground up: first for a single moving particle, where $\vec L=\vec r\times\vec p$ measures "how much turning" that particle carries about a chosen point, then for a rigid body spinning about a fixed axis, where the tidy result $L=I\omega$ takes over. We connect the two through the single most important relation in rotational mechanics — torque is the rate of change of angular momentum, $\tau_{net}=dL/dt$ — and show how, for a rigid body of fixed shape, that statement collapses to the familiar $\tau=I\alpha$ . Two fully worked numbers anchor the ideas: a particle gliding in a straight line still carries angular momentum about a point it never approaches, and a spinning disc, slowed by friction, hands us a clean way to time how long it takes to stop. By the end you should be able to look at any spinning or moving object, name the axis or point you are measuring about, and compute $\vec L$ two different ways — always landing on the same physical quantity. This sets up the very next idea in the chapter: what happens to angular momentum when no external torque acts at all.

Key takeaways

Angular momentum of a particle: $\vec L = \vec r\times\vec p$ , with magnitude $L = rp\sin\theta = mvr_\perp$ , where $r_\perp$ is the perpendicular distance from the axis to the line of motion.
Angular momentum of a rigid body about a fixed axis: $L = I\omega$ — the rotational analogue of $p=mv$ , built from moment of inertia $I$ instead of mass and angular velocity $\omega$ instead of velocity.
Torque drives the change: $\tau_{net} = \dfrac{dL}{dt}$ , the rotational analogue of Newton's second law $F=dp/dt$ ; this is the master relation, valid even when $I$ itself is changing.
For a rigid body of constant shape, $\tau_{net}=dL/dt$ reduces exactly to $\tau=I\alpha$ , so the two "laws" you learned separately are really one law seen two ways.
Angular momentum, like torque, is always quoted about a specific point or axis — the same particle has different $L$ about different origins, even though its motion never changes.
Internal forces cancel in a system's total angular momentum: for a system of particles, $\vec L_{total}=\sum_i \vec r_i\times\vec p_i$ , and only external torques can change it, because internal action-reaction pairs are equal, opposite and collinear.
Key idea: angular momentum is the rotational counterpart of linear momentum — $\vec L=\vec r\times\vec p$ for a particle and $L=I\omega$ for a rigid body about a fixed axis — and it changes only under the action of a net external torque, $\tau_{net}=dL/dt$ , which is exactly what makes conservation of angular momentum possible in the next topic.

The core insight

Angular momentum $\vec L=\vec r\times\vec p$ for a particle, or $L=I\omega$ for a rigid body about a fixed axis, is the rotational analogue of linear momentum, and it can be changed only by a net external torque, $\tau_{net}=dL/dt$ .

Where it fits in the syllabus

This section belongs to the CBSE Class 11 Physics unit System of Particles and Rotational Motion, immediately after "Rotational dynamics about a fixed axis" and immediately before "Conservation of angular momentum." The board syllabus asks for the definition $\vec L=\vec r\times\vec p$ for a particle, the rigid-body relation $L=I\omega$ , and the torque–angular-momentum relation $\tau_{net}=dL/dt$ , including its reduction to $\tau=I\alpha$ for a body of fixed shape rotating about a fixed axis. Derivations of moment-of-inertia formulas by integration, gyroscopic precession and rolling-energy subtleties stay outside this section's scope, in line with the chapter's stated keep-outs.

The physical picture

For a single particle of mass $m$ at position $\vec r$ (measured from a chosen origin), moving with linear momentum $\vec p = m\vec v$ , the angular momentum about that origin is defined as the vector product

\vec L = \vec r \times \vec p, \qquad L = rp\sin\theta = mvr_\perp

where: $\theta$ is the angle between $\vec r$ and $\vec p$ , and $r_\perp = r\sin\theta$ is the perpendicular distance from the origin to the line along which the particle moves — the same trick used for torque, $\tau = rF\sin\theta$ . A particle moving in a straight line at constant velocity still has a nonzero, and in fact constant, angular momentum about any point not on that line, because $r_\perp$ never changes even though $r$ and $\theta$ both do. The direction of $\vec L$ follows the right-hand rule for $\vec r\times\vec p$ and is perpendicular to the plane containing $\vec r$ and $\vec p$ . For a rigid body rotating about a fixed axis with angular velocity $\omega$ , every particle of the body moves in a circle about that axis, and summing $\vec r_i\times\vec p_i$ over all particles collapses — because the body is rigid and the axis fixed — to the compact result

L = I\omega

where: $I=\sum m_ir_i^2$ is the moment of inertia about the same axis, and $\omega$ is the common angular velocity of every particle in the body. This is the exact rotational counterpart of $p=mv$ : mass $m$ is replaced by the "rotational mass" $I$ , and velocity $v$ is replaced by angular velocity $\omega$ . A large $I$ (mass spread far from the axis) means a body carries a large angular momentum even at a modest spin rate — this is precisely why a figure skater's outstretched arms, or a diver's extended body, correspond to large $I$ and, for the same $L$ , small $\omega$ . The bridge between force-based and torque-based mechanics is the statement that torque is the rate of change of angular momentum, exactly as force is the rate of change of linear momentum:

\tau_{net} = \frac{dL}{dt}

For a rigid body of constant $I$ rotating about a fixed axis, $L=I\omega$ gives $\dfrac{dL}{dt}=I\dfrac{d\omega}{dt}=I\alpha$ , so

\tau_{net} = I\alpha

recovering the familiar rotational equation of motion as a special case. The $\tau_{net}=dL/dt$ form is the more general and more powerful statement: it remains true even when $I$ itself changes (a skater pulling her arms in, a diver curling into a tuck), which is exactly the situation exploited in conservation of angular momentum, the next topic in this chapter.

Figure

A particle moving in a straight line past an origin, with the perpendicular distance from the origin to the line of motion marked, next to a spinning disc with its angular momentum vector drawn along the rotation axis

The units of angular momentum, $\text{kg}\,\cdot\,\text{m}^2/\text{s}$ , are identical to $\text{J}\,\cdot\,\text{s}$ (joule-second): $I\omega$ carries units $(\text{kg}\,\cdot\,\text{m}^2)(\text{rad/s})$ , and since the radian is a dimensionless ratio, this reduces cleanly to $\text{kg}\,\cdot\,\text{m}^2/\text{s}$ . Checking that a computed answer carries these units is a fast way to catch an arithmetic slip in an exam. It is also worth stating, without deriving, how angular momentum behaves for a system of particles. The total angular momentum of a system about a chosen point is the vector sum of the angular momenta of every constituent particle about that same point, $\vec L_{total}=\sum_i \vec r_i\times\vec p_i$ . The forces particles of a rigid body exert on each other to hold it together — internal forces — always occur in equal-and-opposite, collinear pairs, so their torques about any common point cancel exactly in the sum. Only external torques can change a system's total angular momentum. That single fact, stated here and used without proof, is what makes conservation of angular momentum possible: even as a skater's arms move in and $I$ changes, the internal muscular forces that pull the arms in exert zero net external torque, so $L$ stays fixed.

Derivation ATorque as the rate of change of angular momentum for a rigid body

The goal is to show, starting only from the definition $L=I\omega$ for a rigid body of constant moment of inertia rotating about a fixed axis, that the general relation $\tau_{net}=dL/dt$ reduces exactly to the familiar rotational equation of motion $\tau_{net}=I\alpha$ — establishing that these are not two separate laws but one law seen in two equivalent forms, and pinning down precisely which condition (constant $I$ ) is needed for the reduction to hold.

Step 1. Write down the angular momentum of the rigid body at any instant. About the fixed rotation axis, $L(t) = I\,\omega(t)$ , where $I$ is the body's moment of inertia about that axis and $\omega(t)$ is its instantaneous angular velocity. Because the axis is fixed and the body is rigid — its mass distribution about that axis never changes shape — $I$ is a constant number, not a function of time.

Step 2. Apply the general torque–angular-momentum relation, $\tau_{net}=\dfrac{dL}{dt}$ , which holds for any system regardless of whether $I$ is constant. Substituting the expression from Step 1, $\tau_{net} = \dfrac{d}{dt}\big(I\,\omega(t)\big)$ .

Step 3. Because $I$ is a constant, it can be pulled outside the derivative exactly as a constant mass is pulled outside $d(mv)/dt$ when deriving $F=ma$ from $F=dp/dt$ : $\tau_{net} = I\,\dfrac{d\omega}{dt}$ .

Step 4. Recognise $\dfrac{d\omega}{dt}$ as the angular acceleration $\alpha$ , giving $\tau_{net}=I\alpha$ — the rotational equation of motion used throughout the chapter. The derivation makes explicit the one condition under which this shortcut is valid: the body's shape (and hence $I$ about the given axis) must stay fixed while it rotates. If instead $I$ were allowed to change with time — a skater pulling in her arms, a rotating platform with a sliding mass — Step 3 would need the product rule, $\tau_{net}=I\dfrac{d\omega}{dt}+\omega\dfrac{dI}{dt}$ , and the simple $\tau=I\alpha$ would no longer capture the whole story; only the original $\tau_{net}=dL/dt$ remains exact in that case, which is exactly why the next topic in this chapter needs it.

Worked examples

Board level - angular momentum of a particle moving in a straight line

A particle of mass $0.5\ \text{kg}$ moves in the $xy$ -plane with velocity $4\ \text{m/s}$ along the $x$ -direction. At the instant considered, it is at position $(0,3)\ \text{m}$ , i.e. $3\ \text{m}$ from the origin along the $y$ -axis. Find its angular momentum about the origin.

Step 1. Identify the perpendicular distance. The velocity is directed along $x$ , so the particle's path is the straight line $y=3\ \text{m}$ . The perpendicular distance from the origin to this line is $r_\perp = 3\ \text{m}$ — this is the quantity that enters $L=mvr_\perp$ , not the instantaneous distance $r$ from the origin to the particle.

Step 2. Compute the momentum. $p = mv = (0.5\ \text{kg})(4\ \text{m/s}) = 2\ \text{kg}\,\cdot\,\text{m/s}$ .

Step 3. Compute the magnitude of angular momentum. $L = m v\, r_\perp = (0.5)(4)(3) = 6\ \text{kg}\,\cdot\,\text{m}^2/\text{s}$ .

Step 4. Fix the direction using the right-hand rule for $\vec r\times\vec p$ , with $\vec r=(0,3,0)\ \text{m}$ and $\vec p=(2,0,0)\ \text{kg}\,\cdot\,\text{m/s}$ : the cross product points along $-\hat z$ . So $\vec L = -6\hat z\ \text{kg}\,\cdot\,\text{m}^2/\text{s}$ , and the magnitude the exam wants is $L = 6\ \text{kg}\,\cdot\,\text{m}^2/\text{s}$ . Because $r_\perp=3\ \text{m}$ never changes as the particle glides along $y=3$ , $\vec L$ is constant throughout the motion — a useful check on the whole calculation.

The final answer is $L = 6\ \text{kg}\,\cdot\,\text{m}^2/\text{s}$ , directed along $-\hat z$ .

Competitive level - a decelerating spinning disc

A uniform disc of mass $4\ \text{kg}$ and radius $0.4\ \text{m}$ spins about its central axis at $\omega_0 = 15\ \text{rad/s}$ . A constant frictional torque of magnitude $0.8\ \text{N}\,\cdot\,\text{m}$ opposes the rotation. Find (a) the disc's initial angular momentum, and (b) the time taken for the disc to come to rest.

Step 1. Compute the moment of inertia of the disc about its central axis, using the standard (quoted, not derived) result for a disc: $I = \tfrac12 MR^2 = \tfrac12(4\ \text{kg})(0.4\ \text{m})^2 = \tfrac12(4)(0.16) = 0.32\ \text{kg}\,\cdot\,\text{m}^2$ .

Step 2. Compute the initial angular momentum using $L=I\omega$ : $L_0 = I\omega_0 = (0.32\ \text{kg}\,\cdot\,\text{m}^2)(15\ \text{rad/s}) = 4.8\ \text{kg}\,\cdot\,\text{m}^2/\text{s}$ . This answers part (a).

Step 3. Use $\tau_{net}=dL/dt$ in its average form, $\tau_{net}=\dfrac{L_{final}-L_0}{t}$ , to find the stopping time. The friction torque opposes the spin, so $\tau_{net}=-0.8\ \text{N}\,\cdot\,\text{m}$ , and $L_{final}=0$ once the disc is at rest: $-0.8 = \dfrac{0-4.8}{t}$ .

Step 4. Solve for $t$ : $t = \dfrac{4.8\ \text{kg}\,\cdot\,\text{m}^2/\text{s}}{0.8\ \text{N}\,\cdot\,\text{m}} = 6\ \text{s}$ . As a cross-check using $\tau=I\alpha$ : $\alpha = \tau/I = 0.8/0.32 = 2.5\ \text{rad/s}^2$ , and $t=\omega_0/\alpha = 15/2.5 = 6\ \text{s}$ — the two routes to $t$ , one through $L$ and one through $\alpha$ , agree exactly, confirming the equivalence proved in the derivation above.

The disc's initial angular momentum is $4.8\ \text{kg}\,\cdot\,\text{m}^2/\text{s}$ , and it comes to rest after $6\ \text{s}$ . Timing note: budget about 2 minutes for part (a) — it is a single substitution into $L=I\omega$ once $I$ is written down — and about 3 minutes for part (b), since a slip in the sign of the opposing torque, or in matching $L_{final}=0$ correctly, is the most common way marks are lost here; if time is short, jumping straight to the $\tau=I\alpha$ cross-check route is equally valid and marginally faster.

Common traps

Forgetting that angular momentum, like torque, depends on the choice of origin or axis — always state which point $\vec L$ (or which axis $L=I\omega$ ) is measured about, and never mix values found about two different origins.
Using $L=mvr$ with the wrong $r$ : it must be the perpendicular distance from the axis to the line of motion ( $r_\perp=r\sin\theta$ ), not the straight-line distance from the origin to the particle's current position.
Confusing $L=I\omega$ (valid for a rigid body rotating about a fixed axis) with the general particle definition $\vec L=\vec r\times\vec p$ — both are correct in their own context, but applying $L=I\omega$ to a single freely moving particle that is not part of a rotating rigid body is a category error.
Treating $\tau_{net}=dL/dt$ and $\tau=I\alpha$ as two independent laws to memorise separately, rather than recognising that the second follows from the first the moment $I$ is constant — this is exactly what the derivation above establishes.
Dropping the sign of an opposing torque when using $\tau_{net}=(L_{final}-L_0)/t$ , which flips the computed stopping time or angular acceleration to the wrong sign.

You should be able to…

Compute the angular momentum of a particle as $\vec L=\vec r\times\vec p$ , correctly identifying the perpendicular distance $r_\perp$ from the axis to the line of motion.
Compute the angular momentum of a rigid body about a fixed axis as $L=I\omega$ , given its moment of inertia and angular velocity.
State the general relation $\tau_{net}=dL/dt$ and derive $\tau_{net}=I\alpha$ from it for a body of constant moment of inertia.
Explain why angular momentum, like torque, must always be quoted about a specific axis or point.
Solve numerical problems in which a known torque changes a rigid body's angular momentum over a given time interval, including finding a stopping or starting time.
State, without proof, why only external torques can change the total angular momentum of a system of particles.

Ask the tutor: want to see angular momentum computed for the same object two different ways — once via $\vec r\times\vec p$ and once via $I\omega$ — to check they agree, or work through a few more torque-versus-time problems with the AI tutor?

Why does angular momentum depend on where I put the origin, when linear momentum p = mv does not?

Linear momentum is a property of the particle's motion alone — mass times velocity — with no reference point built in. Angular momentum is defined through $\vec r$ , the position vector from a chosen origin to the particle, so it is really describing the particle's motion relative to that origin, not an origin-free property of the particle by itself. Move the origin and $\vec r$ changes, so $\vec L=\vec r\times\vec p$ changes too, even though the particle's actual velocity and momentum haven't changed at all. This is exactly parallel to torque, $\vec\tau=\vec r\times\vec F$ , which is also always quoted about a specific point. The rule to remember: state the axis or origin every single time you write down an angular momentum.

If a particle is moving in a perfectly straight line, how can it have any angular momentum at all — it isn't rotating around anything?

Angular momentum does not require the particle to be moving in a circle; it only requires a chosen reference point that is not on the particle's line of motion. Picture standing off to the side of a straight road and watching a car drive past at constant speed: relative to where you stand, the car sweeps out an angle over time, even though it never turns. That sweeping is exactly what $\vec L=\vec r\times\vec p$ captures, and because the perpendicular distance from you to the road never changes, the car's angular momentum about you stays constant the whole time it drives past. Only once several particles, or a rigid body, are actually rotating about a fixed axis does $L=I\omega$ become the convenient shortcut.

Is torque equals I alpha wrong, then, since the article keeps saying tau_net = dL/dt is the real law?

Tau equals I alpha is not wrong — it is a special case of tau_net = dL/dt that holds exactly whenever the moment of inertia I stays constant while the body rotates about a fixed axis, which covers the great majority of Class 11 problems (a disc, a rod, a flywheel, all spinning rigidly). The derivation above shows the two are the same statement once I is pulled out of the derivative as a constant. The distinction matters only when I itself changes during the motion, such as a skater pulling in her arms — a situation this section deliberately sets aside for the next topic, conservation of angular momentum, where dL/dt = 0 (no external torque) does all the work even as I varies.

In the disc example, why did the torque have to be negative in the equation, and what would happen if I forgot the sign?

The torque was taken as negative because friction opposes the disc's existing spin — it is trying to reduce $L$ toward zero, not add to it. Writing $\tau_{net}=(L_{final}-L_0)/t$ with $\tau_{net}=+0.8\ \text{N}\,\cdot\,\text{m}$ by mistake would give $t=(0-4.8)/0.8=-6\ \text{s}$ , a negative time, which is a clear signal something is wrong with the sign convention, not a valid physical answer. In general, always fix a positive sense of rotation first (matching the direction of $\omega_0$ here), then assign every torque and every $L$ a sign relative to that same convention before substituting numbers — that single habit prevents almost every sign error of this kind.