Conservation of angular momentum

Watch an ice skater spin. Arms out, she turns slowly, almost lazily. She pulls her arms in against her chest and — without pushing off anything, without any outside hand touching her — she suddenly whirls so fast her outline blurs. Nothing pushed her. Nothing external touched her at all. Yet her spin rate roughly tripled in under a second, purely because she rearranged her own arms. That single trick is the visible fingerprint of one of the deepest conservation laws in mechanics. The rule behind it is almost insultingly simple to state: if no external torque acts on a system, its total angular momentum $L$ cannot change, no matter how the system twists, folds, or redistributes its own mass internally. The skater's muscles pull her arms inward and her moment of inertia $I$ drops sharply; because $L = I\omega$ must stay fixed, her angular speed $\omega$ has to shoot up to compensate. She is not adding spin — she is trading a large $I$ for a small one and letting $\omega$ absorb the difference, at the price of the muscular work she does pulling her arms in. This one idea reappears everywhere a rotating system is left alone by outside forces: a diver curling into a tuck to spin faster mid-air, a spinning platform speeding up as a person on it hugs a weight to their chest, a neutron star spinning ferociously fast because a slowly-rotating star collapsed to a fraction of its original size, and — reaching all the way into the Gravitation chapter — a planet sweeping out equal areas in equal times as it swoops closer to and farther from the Sun. This section builds the law from the torque–angular-momentum relation you already know, applies it with the same algebra you used for linear momentum conservation, and works through the CBSE board-level numbers step by step so the arithmetic becomes second nature before an exam ever asks for it.

Key takeaways

The law itself. If the net external torque on a system is zero, its total angular momentum $L$ stays exactly constant — in magnitude and direction — no matter how the system's own shape or mass distribution changes.
Where it comes from. It follows in one line from $\tau_{net} = dL/dt$ : set $\tau_{net} = 0$ and $dL/dt = 0$ immediately, so $L$ is constant.
The working formula. For a rotating rigid system with no external torque, $I_1\omega_1 = I_2\omega_2$ — a smaller moment of inertia forces a larger angular speed, and vice versa.
Internal forces cannot do it. Muscles, ropes pulled inward, or parts of a system pushing on each other are internal forces that occur in equal-and-opposite pairs and produce zero net external torque, so they can redistribute $I$ but never change total $L$ .
Energy is not automatically conserved. Kinetic energy $\tfrac12 I\omega^2$ generally changes when $I$ changes internally (it rises when the skater pulls her arms in), even while $L$ stays fixed — the two conservation laws are independent.
A gravitational cousin. Kepler's law of areas (a planet sweeps equal areas in equal times) is angular momentum conservation in disguise, because the Sun's gravity pulls along the line joining Sun and planet and produces zero torque about the Sun.
Key idea: with zero net external torque, L = Iω stays fixed, so a system that pulls its mass inward (I shrinks) must spin faster (ω grows) to compensate, exactly as I1ω1 = I2ω2 demands.

The core insight

Zero net external torque forces total angular momentum $L = I\omega$ to stay exactly constant, so any internal rearrangement that shrinks a system's moment of inertia $I$ must speed up its rotation $\omega$ in the fixed ratio $I_1\omega_1 = I_2\omega_2$ .

Where it fits in the syllabus

This is the capstone application of the CBSE Class 11 Physics unit System of Particles and Rotational Motion, sitting directly after torque and angular momentum about a fixed axis. The prescribed scope asks a student to state the law, connect it to $\tau_{net} = dL/dt$ , and apply $I_1\omega_1 = I_2\omega_2$ to systems whose moment of inertia changes with no external torque — this section stays exactly inside that boundary and does not touch gyroscopic precession or Euler's equations, both explicitly out of scope for Class 11. It also builds the bridge to Kepler's law of areas in the Gravitation chapter, which is the same conservation law applied to a planet under gravity's central pull.

The physical picture

Rotational motion has its own version of every linear-motion idea, and angular momentum is the rotational twin of ordinary momentum $p = mv$ . For a particle or system, the rate of change of angular momentum equals the net external torque applied:

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

where: $\tau_{net}$ is the net external torque about a chosen axis or point (units $\text{N}\,\cdot\,\text{m}$ ), $L$ is the total angular momentum about that same axis or point (units $\text{kg}\,\cdot\,\text{m}^2/\text{s}$ ), and $dL/dt$ is how fast $L$ is changing with time. Read physically, this says angular momentum only changes when something outside the system twists it — exactly the rotational parallel of $F_{net} = dp/dt$ for straight-line momentum. The moment something reaches in from outside and applies a torque, $L$ starts to change; the moment that outside torque vanishes, $L$ freezes at whatever value it held. Set the right-hand side to zero and the consequence is immediate:

\tau_{net} = 0 \quad\Rightarrow\quad \frac{dL}{dt} = 0 \quad\Rightarrow\quad L = \text{constant}

where: the condition $\tau_{net} = 0$ means the net external torque about the chosen axis vanishes — internal forces between parts of the same system do not count, because by Newton's third law they occur in equal-and-opposite pairs whose torques cancel exactly. This is the law of conservation of angular momentum: in the total absence of external torque about an axis, the total angular momentum about that axis is fixed for all time, however the system's own parts move relative to each other. For a rigid body, or a system that can redistribute its own mass while spinning about a fixed axis, angular momentum is $L = I\omega$ . Writing the conservation statement between an initial state 1 and a later state 2 with no external torque in between gives the single working equation used in almost every Class 11 problem on this topic:

I_1\omega_1 = I_2\omega_2

where: $I_1$ and $\omega_1$ are the moment of inertia and angular speed in the initial configuration, and $I_2$ , $\omega_2$ are the same quantities after the system has rearranged its own mass — both moments of inertia measured about the same fixed axis. Physically, this equation is a trade: whatever the system does to shrink $I$ , it must do the exact opposite to $\omega$ , and by the same factor, because their product is locked. Pull mass toward the axis and $I$ falls, so $\omega$ must climb to keep $L = I\omega$ unchanged; push mass outward and $I$ rises, so $\omega$ must fall.

Figure

Two states of a spinning skater on a turntable, arms out with a large moment of inertia and slow spin on the left, arms pulled in with a small moment of inertia and fast spin on the right, with the product of moment of inertia and angular speed marked equal on both sides

Two conditions must be checked before reaching for $I_1\omega_1 = I_2\omega_2$ : the torque must vanish about the specific axis or point being used, and the changes causing $I$ to shift must be internal to the system (mass moving within the system, not entering or leaving it, and no outside push or friction). If a bearing has friction, or an external hand grabs the spinning body, angular momentum is not conserved and the full relation $\tau_{net} = dL/dt$ must be used instead — there is no shortcut.

Derivation AAngular momentum conservation from the torque law, and the working equation I1ω1 = I2ω2

The goal of this derivation is to move from the general torque–angular-momentum relation, which every student already has from the previous section of this chapter, to the single practical equation used in board-level problems, $I_1\omega_1 = I_2\omega_2$ , while making explicit exactly which physical condition (zero net external torque, and internal redistribution of mass) licenses each step — because misapplying this shortcut to a system with an external torque is the single most common source of lost marks on this topic.

Step 1. Start from the defining relation between torque and angular momentum for any system, $\tau_{net} = \dfrac{dL}{dt}$ . This is exact and general; it says the total angular momentum of a system changes only in response to a net external torque, exactly as $F_{net} = dp/dt$ says ordinary momentum changes only in response to a net external force.

Step 2. Impose the physical condition that defines a conserved system: no net external torque acts, $\tau_{net} = 0$ . This can happen because no external force is present at all, because an external force acts exactly through the axis (giving it zero lever arm), or because internal forces between the system's own parts — muscles, ropes, contact forces between a person and a platform — exist strictly in equal-and-opposite Newton's-third-law pairs whose torques cancel and never appear in $\tau_{net}$ .

Step 3. Substitute $\tau_{net} = 0$ into the relation from Step 1: $0 = \dfrac{dL}{dt}$ . A quantity whose time derivative is identically zero cannot be changing, so $L$ must equal the same constant value at every instant: $L = \text{constant}$ . This is the law of conservation of angular momentum in its most general form, valid whether or not the system is rigid.

Step 4. Specialise to a system rotating about a fixed axis, where angular momentum takes the familiar rigid-body form $L = I\omega$ . Because $L$ is constant while the system freely redistributes its own mass between an initial configuration (moment of inertia $I_1$ , angular speed $\omega_1$ ) and a later one ( $I_2$ , $\omega_2$ ), the same constant value of $L$ must appear at both instants: $I_1\omega_1 = L = I_2\omega_2$ , giving the working equation $I_1\omega_1 = I_2\omega_2$ .

Step 5. Read the trade-off the equation enforces. Because the product $I\omega$ is pinned, $\omega_2 = \dfrac{I_1}{I_2}\,\omega_1$ : whenever internal rearrangement shrinks the moment of inertia ( $I_2 < I_1$ ), the ratio $I_1/I_2$ exceeds one and $\omega_2 > \omega_1$ — the spin speeds up. Whenever mass moves outward ( $I_2 > I_1$ ), the ratio falls below one and the spin slows down. Nothing about how the mass moved matters, only the ratio of the two moments of inertia — which is why the same equation handles a skater's arms, a diver's tuck, and a person walking on a rotating platform with no extra physics needed.

Worked examples

Board level - the skater on a stool

A student sits on a rotating stool with arms outstretched, holding a dumbbell in each hand, and spins at $\omega_1 = 2\ \text{rad/s}$ . The total moment of inertia of student, stool, and dumbbells about the rotation axis is $I_1 = 5\ \text{kg}\,\cdot\,\text{m}^2$ . The student then pulls the dumbbells in, reducing the total moment of inertia to $I_2 = 2\ \text{kg}\,\cdot\,\text{m}^2$ . Find the new angular velocity, assuming no external torque acts on the system.

Step 1. Identify that the system (student + stool + dumbbells) experiences no net external torque about the vertical axis — the stool's own bearing is treated as frictionless, and pulling the dumbbells inward is an internal muscular action. This licenses $I_1\omega_1 = I_2\omega_2$ .

Step 2. Substitute the given values: $(5\ \text{kg}\,\cdot\,\text{m}^2)(2\ \text{rad/s}) = (2\ \text{kg}\,\cdot\,\text{m}^2)\,\omega_2$ , so $10\ \text{kg}\,\cdot\,\text{m}^2/\text{s} = (2\ \text{kg}\,\cdot\,\text{m}^2)\,\omega_2$ .

Step 3. Solve for $\omega_2$ : $\omega_2 = \dfrac{10}{2} = 5\ \text{rad/s}$ . The student's angular velocity increases from $2\ \text{rad/s}$ to $\mathbf{5\ \text{rad/s}}$ — the angular momentum stays fixed at $10\ \text{kg}\,\cdot\,\text{m}^2/\text{s}$ throughout, and the more than twofold increase in spin rate is the direct, quantitative version of the "arms in, spin faster" effect.

Competitive level - a person walking to the rim of a platform

A circular platform of moment of inertia $I_p = 3\ \text{kg}\,\cdot\,\text{m}^2$ rotates freely at $\omega_1 = 4\ \text{rad/s}$ about a vertical axis through its centre. A person of mass $50\ \text{kg}$ , initially standing exactly at the centre (contributing negligible moment of inertia there), walks outward and stands at the rim, at radius $R = 1\ \text{m}$ . Find the platform's new angular velocity, and compare the system's rotational kinetic energy before and after.

Step 1. Find the initial total moment of inertia. With the person essentially on the axis, their contribution is $I_{person,1} \approx 0$ , so $I_1 = I_p + I_{person,1} = 3 + 0 = 3\ \text{kg}\,\cdot\,\text{m}^2$ .

Step 2. Find the final total moment of inertia. Treating the person as a point mass once at the rim, $I_{person,2} = mR^2 = (50\ \text{kg})(1\ \text{m})^2 = 50\ \text{kg}\,\cdot\,\text{m}^2$ , so $I_2 = I_p + I_{person,2} = 3 + 50 = 53\ \text{kg}\,\cdot\,\text{m}^2$ .

Step 3. Apply conservation of angular momentum — the person's walking forces act between the person and the platform, both part of the same system, so they are internal and produce no net external torque about the vertical axis: $I_1\omega_1 = I_2\omega_2 \Rightarrow (3)(4) = (53)\,\omega_2 \Rightarrow \omega_2 = \dfrac{12}{53} \approx 0.226\ \text{rad/s}$ .

Step 4. Compare kinetic energies to see that $L$ -conservation does not mean $KE$ -conservation. Initial $KE_1 = \tfrac12 I_1\omega_1^2 = \tfrac12(3)(4)^2 = 24\ \text{J}$ . Final $KE_2 = \tfrac12 I_2\omega_2^2 = \tfrac12(53)(0.226)^2 \approx 1.35\ \text{J}$ . Kinetic energy has dropped by roughly $22.6\ \text{J}$ — the opposite trend from the skater example, because here mass moves outward; the "lost" energy went into the muscular work the person did pushing against the platform's rotation while walking out. The platform's angular velocity falls to $\mathbf{\approx 0.226\ \text{rad/s}}$ — nearly eighteen-fold slower — since $I$ grew nearly eighteen-fold while $L = 12\ \text{kg}\,\cdot\,\text{m}^2/\text{s}$ stayed fixed. Timing note: this is a two-part numerical problem (find $\omega_2$ , then compare $KE$ ) worth about 4-5 minutes in a competitive setting — compute $I_2$ first and double-check the arithmetic ( $53$ , not $50$ , is the total), since a slipped $I_2$ propagates into both the angular velocity and the energy comparison.

Common traps

Applying $I_1\omega_1 = I_2\omega_2$ when an external torque is actually present — friction at the axle bearing, someone pushing from outside the system, or an external force with a nonzero lever arm about the axis all break conservation, and the general relation $\tau_{net} = dL/dt$ must be used instead.
Confusing conservation of angular momentum with conservation of kinetic energy. $L = I\omega$ stays fixed with zero external torque, but rotational kinetic energy $\tfrac12 I\omega^2$ generally does not — it rises when $I$ shrinks (skater pulling arms in) and falls when $I$ grows (person walking outward), because internal muscular work is done in either direction.
Forgetting which axis the "zero torque" condition is being checked about — a system can have zero net torque about one axis while experiencing nonzero torque about a different axis through the same body, so the axis used for $I$ , $\omega$ , and $L$ must be identical throughout.
Leaving out part of the system's moment of inertia — when a person joins, leaves, or moves within a rotating system, every part's contribution to $I$ about the axis must be included on both sides of $I_1\omega_1 = I_2\omega_2$ , not just the platform's own $I$ .
Treating a person or mass moving toward the axis and one moving away from the axis as producing the same qualitative effect — they produce opposite effects on $\omega$ , and mixing up the direction of the $I$ change is a common sign error.

You should be able to…

State the law of conservation of angular momentum and the precise condition — zero net external torque about the chosen axis — under which it holds.
Derive the law in a few lines from $\tau_{net} = dL/dt$ , and explain why internal forces cannot violate it.
Apply $I_1\omega_1 = I_2\omega_2$ correctly to problems where a system's moment of inertia changes through internal mass redistribution, including multi-part systems like a person on a platform.
Explain, with a concrete numerical comparison, why rotational kinetic energy can rise or fall even while angular momentum stays exactly constant.
Identify situations where the law does not apply, because an external torque (friction, an outside push) is present.
Connect angular momentum conservation to Kepler's law of areas for planetary motion, in one clean line, without repeating the Gravitation chapter's derivation.

Ask the tutor: Want to see how this same law explains a neutron star's blistering spin rate, or try one more platform-and-person problem with different numbers? Ask the AI tutor for extra practice on conservation of angular momentum.

Why does the skater's kinetic energy increase when nothing external pushed her? Doesn't that break energy conservation?

It does not break energy conservation, because the skater is not an isolated energy system — her muscles do real mechanical work pulling her arms inward against the outward-flinging tendency of her own spinning mass (a bit like doing work against a centrifugal effect in the rotating frame). That muscular work is the source of the extra kinetic energy; it comes from the chemical energy stored in her body, not from nowhere. Angular momentum stays fixed because no external torque acts, but energy can still be added or removed internally by the work her muscles do — the two conservation statements track different quantities and are not required to agree.

If I stand on a frictionless rotating platform and just raise my arms straight up instead of pulling them in toward my body, does angular momentum conservation still apply?

Yes, the law itself always applies as long as there is no external torque — but whether your angular speed changes depends on whether your moment of inertia about the vertical rotation axis actually changes. Raising your arms straight up moves mass along the rotation axis, which barely changes your average distance from that axis, so $I$ stays roughly the same and $\omega$ barely changes. Pulling your arms in toward your chest moves mass much closer to the axis (reducing $I$ substantially), which is why that motion, not raising arms overhead, produces the dramatic skater speed-up.

How is this connected to Kepler's second law about planets sweeping equal areas in equal times?

They are the same law wearing different clothes. A planet orbiting the Sun feels gravity directed exactly along the line joining planet and Sun, so that force has zero lever arm about the Sun and therefore exerts zero torque about the Sun at every instant. With $\tau_{net} = 0$ about the Sun, the planet's angular momentum about the Sun is conserved throughout its orbit. It can be shown (using the geometry of area swept by a radius vector) that constant angular momentum about a point is exactly equivalent to sweeping equal areas in equal times about that point — which is precisely Kepler's second law, covered in the Gravitation chapter.

In the platform example, why is the person's own moment of inertia treated as mR squared instead of something more complicated?

Because the problem treats the person as a point mass concentrated at radius $R$ from the axis, which is the standard simplifying assumption at this level — a person's own body has some distributed mass and would strictly need its own moment of inertia formula, but that level of detail is outside the Class 11 syllabus (only $I$ for simple standard shapes is prescribed, without integration). Treating a person, or any object much smaller than the platform's radius, as a point mass at distance $R$ gives $I = mR^2$ and is accurate enough for board-level and most competitive-level numericals.