Discuss the conservation of angular momentum with an example.

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Q. Discuss the conservation of angular momentum with an example.

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Concept understanding — Conservation of Angular Momentum

Conservation of Angular Momentum

The Intuition First

Imagine you're sitting on a spinning office chair with your arms stretched out. Someone gives you a gentle push, and you start rotating slowly. Now pull your arms in tight against your chest. What happens? You spin faster. Push your arms back out — you slow down again.

Nothing external pushed you to go faster or slower. The change came from inside — from how you arranged your mass relative to the axis of rotation.

That's the core idea: Angular momentum is a quantity that stays constant for a rotating system unless an external torque acts on it. When you pulled your arms in, you didn't change your angular momentum — you changed your distribution of mass, and your rotation speed had to adjust to keep the total constant.

The Precise Statement

\vec{L} = \text{constant} \quad \text{if} \quad \vec{\tau}_{\text{ext}} = 0

Where:

$\vec{L}$ is the angular momentum of the system
$\vec{\tau}_{\text{ext}}$ is the net external torque acting on the system

In words: The total angular momentum of an isolated system (no external torque) remains constant in both magnitude and direction.

What Is Angular Momentum?

For a point mass $m$ moving with velocity $\vec{v}$ at position $\vec{r}$ from a reference point:

\vec{L} = \vec{r} \times m\vec{v}

For a rigid body rotating about a fixed axis:

L = I\omega

Where:

$I$ = moment of inertia (how mass is distributed relative to the axis)
$\omega$ = angular velocity (how fast it spins)

Note

Moment of inertia $I$ depends on where the mass is, not just how much. Mass far from the axis gives larger $I$ ; mass close to the axis gives smaller $I$ .

Why It Works: The Physics

Newton's second law for rotation says:

\tau_{\text{ext}} = \frac{dL}{dt}

If $\tau_{\text{ext}} = 0$ , then $\frac{dL}{dt} = 0$ , so $L$ is constant.

Since $L = I\omega$ , if $I$ changes (you pull arms in), $\omega$ must change in the opposite way to keep $L$ the same:

I_1 \omega_1 = I_2 \omega_2

Smaller $I$ → larger $\omega$ (spin faster). Larger $I$ → smaller $\omega$ (spin slower).

Real-World Examples

Situation	What happens	Why
Ice skater pulling arms in	Spins faster	$I$ decreases, $\omega$ increases to keep $L$ constant
Diver tucking into a ball	Rotates faster in midair	Same principle — no external torque during flight
Cat falling upside-down	Twists body to land on feet	Changes $I$ of different body parts to rotate without external torque
Planet orbiting the Sun	Speeds up when closer, slows down when farther	Gravitational force is central (torque = 0), so $L$ is constant

Watch out

Angular momentum is a vector. Its direction matters too. If no external torque acts, the axis of rotation stays fixed in space. This is why a spinning gyroscope or a bicycle wheel resists being tilted.

Common Mistake to Avoid …

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Step 1. Relating torque to the rate of change of angular momentum.

For a rigid body rotating about a fixed axis with constant moment of inertia $I$ , the angular momentum is $L=I\omega$ and the torque-angular-acceleration relation is $\tau=I\alpha$ . Since $\alpha=\dfrac{d\omega}{dt}$ , the torque can be rewritten as

\tau=I\frac{d\omega}{dt}=\frac{d(I\omega)}{dt}=\frac{dL}{dt}.

So the net external torque on a body equals the rate of change of its angular momentum — the exact rotational counterpart of $\vec F=d\vec p/dt$ .

Step 2. The conservation law.

This relation immediately gives a conservation law: if the net external torque is zero, $\tau=0$ , then

\frac{dL}{dt}=0\quad\Longrightarrow\quad L=\text{constant}.

This is the law of conservation of angular momentum: in the absence of any external torque, a body's angular momentum about that axis never changes, however its own internal configuration (and hence its moment of inertia) might change.

Step 3. Writing it for an initial and final state.

Since $L=I\omega$ throughout, conservation of $L$ between an initial state (subscript $i$ ) and a later final state (subscript $f$ ) is written

I_i\omega_i=I_f\omega_f.

Step 4. The key physical consequence.

Because the product $I\omega$ is held fixed rather than either factor individually, if $I$ increases, $\omega$ must decrease correspondingly to keep the product the same, and if $I$ decreases, $\omega$ must increase — angular velocity and moment of inertia are always seen to move in opposite directions whenever $L$ is conserved.

Step 5. Everyday illustrations.

An ice-skater or dancer spins comparatively slowly with the arms stretched out (large $I$ ); pulling the arms in close to the body sharply reduces $I$ , and since no external torque acts about the vertical spin axis once the spin is underway, $L$ stays fixed and $\omega$ increases visibly — the dancer spins faster. Similarly, a diver curls into a tight tuck in mid-air, deliberately reducing $I$ ; since gravity exerts no torque about the diver's own center of mass in free fall, $L$ is conserved throughout the dive, so the resulting increase in $\omega$ lets the diver complete extra somersaults before straightening out just before entering the water. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. While studying motion of an object under the gravitational effect of another object which of the following quantities is not conserved ?$
(a) Linear momentum
(b) Angular momentum
(c) Total mechanical energy
(d) None of these
›Reveal solutionSolution
Angular momentum and mechanical energy are conserved; linear momentum is not. Answer (A).

When a body moves under the gravitational attraction of another (central) body:

The force is central (directed along the line joining them), so torque about the centre is zero -> angular momentum is conserved.

Gravity is conservative, so total mechanical energy is conserved. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The motion of a particle under the central force is always confined to a plane. This is a consequence of$
(a) conservation of linear momentum
(b) conservation of angular momentum
(c) conservation of energy
(d) none of these
›Reveal solutionSolution

Motion under a central force stays in a plane because angular momentum is conserved. Answer (B).

A central force always points toward (or away from) a fixed centre, so its torque about that centre is zero (r and F are parallel/anti-parallel, r x F = 0).

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Kepler's second law is a consequence of$
(a) conservation of energy
(b) conservation of linear momentum
(c) conservation of angular momentum
(d) conservation of mass
›Reveal solutionSolution

Kepler's second law (the radius vector sweeps equal areas in equal times) is a direct geometric consequence of angular momentum being conserved in a planet's orbit around the Sun.

Gravitational force on a planet from the Sun always acts along the line joining them (a central force), so it produces zero torque about the Sun. With zero net torque, the planet's angular momentum L = m v r sin(theta) about the Sun stays constant throughout the orbit.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A particle undergoes uniform circular motion. The angular momentum of the particle remains conserved about:$
(a) any point inside the circle
(b) the centre point of the circle
(c) any point outside the circle
(d) the point on the circumference of the circle
›Reveal solutionSolution

Angular momentum in uniform circular motion is conserved only about the centre of the circle, because the centripetal force passes through that point and produces zero torque there.

For a particle moving in a circle of radius r with constant speed, the only force acting is the centripetal force F, which is always directed radially inward, straight at the centre of the circle.

Torque about any point O is tau = r' x F, where r' is the position vector from O to the particle. If O is the centre of the circle, r' is exactly along the same (or opposite) line as F, so r' x F = 0 at every instant -- zero torque, hence angular momentum about the centre stays constant (both in magnitude, m v r, and direction, perpendicular to the plane of motion).

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The motion of planets in the solar system is an example of conservation of$
(a) mass
(b) linear momentum
(c) angular momentum
(d) energy.
›Reveal solutionSolution

The gravitational force of the Sun on a planet always acts along the line joining the Sun and the planet (a central force), so it produces zero torque about the Sun.

Since $\vec\tau = \dfrac{d\vec L}{dt}=\vec r\times \vec F$ , and $\vec F$ is always anti-parallel to $\vec r$ (radial), $\vec\tau=0$ , so the angular momentum $\vec L$ of the planet about the Sun st …

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CBSE 2024Set ANNUAL1 mark
$Q. All planets move in ______ orbits with sun situated at one of the foci.$
›Reveal solutionSolution
[!TLDR]

Elliptical

Method

Kepler's first law of planetary m …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. The motion of planets in the solar system is an example of the conservation of$
(a) mass
(b) linear momentum
(c) angular momentum
(d) energy.
›Reveal solutionSolution

Planetary motion is a direct illustration of conservation of angular momentum, since gravity is a central force.

The Sun's gravitational pull on a planet always points from the planet towards the Sun, i.e. along the position vector $\vec r$ of the planet (measured from the Sun). Torque is $\vec\tau = \vec r \times \vec F$ ; since $\vec F$ is anti-parallel to $\vec r$ , this cross product is zero, so the torque on the planet about the Sun is always zero. Because $\vec\tau = d\vec L/dt = 0$ , the angular momentum $\vec L$ of the planet about …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. If without changing mass, radius of the earth is just half then decration of day will be:$
(a) 24 hours
(b) 12 hours
(c) 6 hours
(d) 4 hours
›Reveal solutionSolution

Halving Earth's radius shortens the day to 6 hours.

Earth spins with angular momentum L = Iω. With no external torque and unchanged mass, L is conserved.

Moment of inertia of a solid sphere I = (2/5)MR^2 ∝ R^2. If R → R/2, then I → I/4.

…

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Q.Discuss the conservation of angular momentum with an example.

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