JEE Mains Physics · System of Particles and Rotational Motion

Angular Momentum and Its Conservation

A particle's angular momentum about a point is r × p; a rigid body's about its axis is Iω; with no external torque it stays constant, so a change in I forces a change in ω.

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Why this matters

Twenty-seven PYQs, twelve of them asking for a number, and four from 2026. Thirteen find a particle's angular momentum: along a straight line, in a circle (a conical pendulum among them), as a projectile, or from r × p in components. Fourteen conserve it: a disc dropped on a spinning disc, particles stuck on a ring, a shrinking earth, a moving body striking a rod.

Concept 1 of 2: Angular momentum of a particle

Angular momentum measures how much a moving particle sweeps around a chosen point. It is the momentum times the perpendicular distance from the point to the line of motion. A particle moving in a straight line at constant speed keeps the same angular momentum, because that perpendicular distance never changes.

Definition

$\vec L = \vec r \times \vec p = m(\vec r \times \vec v)$ ; its size is $mvd$ , where d is the perpendicular distance from the point to the line of motion.
Straight-line motion at constant velocity: L about any fixed point is constant.
Uniform circular motion: L about the centre is constant in size AND direction, $L = mvr$ .
Of A about B: use $\vec r_A - \vec r_B$ and $\vec v_A - \vec v_B$ . Two bodies on parallel lines a distance d apart: $L = m\,v_{\text{rel}}\,d$ .
Projectile, about the launch point, at the top: $L = m(u\cos\theta)H = \dfrac{mu^{3}\sin^{2}\theta\cos\theta}{2g}$ . Gravity's torque about that point is mgx, so L grows as the projectile moves out: $dL/dt = \tau$ .
If $\vec r \to -\vec r$ , then $\vec v$ and $\vec p$ reverse but $\vec L = \vec r \times \vec p$ does not.

Angular momentum of a particle, and of a projectile at its highest point

\vec L = \vec r \times m\vec v,\ |\vec L| = mvd \qquad L_{\text{top}} = \frac{mu^{3}\sin^{2}\theta\cos\theta}{2g}

Worked example

A 2 kg particle moves at 3 m/s in the +x direction along the line y = 4 m. Find its angular momentum about the origin, and about the point (0, 1) m.

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The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 2 · Q6Moderate

Example 1 · System of Particles and Rotational Motion · Angular Momentum and Its Conservation

Two cars A and B each of mass

10^{3}\text{ }kg

are moving on parallel tracks separated by a distance of 10 m , in same direction with speeds

72\text{ }km/h

and

36\text{ }km/h

. The magnitude of angular momentum of car A with respect to car B is

\_\_\_\_

J.s.

Distance to the point, not to the line

L = mvd uses the perpendicular distance from the point to the LINE of motion. The particle's straight-line distance from the point changes as it moves; d does not.

Angular momentum about a moving body

For one car about another, use the relative velocity. Using the first car's own speed gives its angular momentum about a fixed point beside the track.

A projectile's L is not conserved about the launch point

Gravity has a torque mgx about the launch point, so L changes along the path. Only about a point on the line of the force could L stay constant.

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Concept 2 of 2: Conservation of angular momentum

With no external torque about an axis, Iω about that axis cannot change. Add mass far from the axis, or spread out, and the body slows; pull in, and it speeds up. Kinetic energy is NOT conserved in these changes: sticking two spinning bodies together always loses energy.

Definition

Rigid body: $L = I\omega$ . No external torque: $I_1\omega_1 = I_2\omega_2$ .
A disc placed coaxially on a spinning disc: $\omega' = \dfrac{I_1\omega}{I_1 + I_2}$ . Two particles m stuck on the rim of a ring M: $\omega' = \dfrac{M}{M + 2m}\omega$ .
Earth with its mass fixed: $I \propto R^{2}$ , so the day length $T \propto R^{2}$ . Radius to $\tfrac{3}{4}$ : $24 \times \tfrac{9}{16} = 13.5$ h. Volume to $\tfrac{1}{64}$ means radius to $\tfrac{1}{4}$ : T to $\tfrac{1}{16}$ .
Kinetic energy $= L^{2}/2I$ : at fixed L, tripling I cuts the energy to a third.
Energy lost when two coaxial discs lock together: $\Delta K = \dfrac{I_1I_2(\omega_1 - \omega_2)^{2}}{2(I_1 + I_2)}$ .
A body striking a pivoted rod: L about the PIVOT is conserved through the impact (the pivot's force has no torque about itself). An impulse J at the end of a free rod: $J\tfrac{L}{2} = \tfrac{ML^{2}}{12}\omega$ , so $\omega = \dfrac{6J}{ML}$ .

Conservation of angular momentum, and energy lost on locking

I_1\omega_1 = I_2\omega_2 \qquad \Delta K = \frac{I_1I_2(\omega_1 - \omega_2)^{2}}{2(I_1 + I_2)}

Worked example

A thin ring of mass 2 kg and radius 0.5 m spins at 12 rad/s about its axis. Two beads of 0.5 kg each are stuck on its rim. Find the new angular speed and the kinetic energy lost.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 8 April 2024 · Q96Moderate

Example 2 · System of Particles and Rotational Motion · Angular Momentum and Its Conservation

A thin circular disc of mass

M

and radius

R

is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity

\omega

. If another disc of same dimensions but of mass

M/2

is placed gently on the first disc co-axially, then the new angular velocity of the system is :

Kinetic energy is not conserved when bodies stick

When a disc lands on a spinning disc, or beads stick to a ring, angular momentum is conserved but kinetic energy is lost. Equating the energies gives ω' = ω√(I₁/(I₁ + I₂)), which is wrong.

Day length goes with the radius squared

With fixed mass, I ∝ R², so T ∝ R². A volume change must first become a radius change: a sixty-fourth of the volume is a quarter of the radius, and T falls to a sixteenth.

About which point is L conserved?

For a body striking a rod on a fixed pivot, the pivot pushes during the impact, so linear momentum is not conserved. Angular momentum about the pivot is, because the pivot's force has no lever arm there.

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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Angular momentum of a particle
Angular momentum of a particle, and of a projectile at its highest point
$\vec L = \vec r \times m\vec v,\ |\vec L| = mvd \qquad L_{\text{top}} = \frac{mu^{3}\sin^{2}\theta\cos\theta}{2g}$
Conservation of angular momentum
Conservation of angular momentum, and energy lost on locking
$I_1\omega_1 = I_2\omega_2 \qquad \Delta K = \frac{I_1I_2(\omega_1 - \omega_2)^{2}}{2(I_1 + I_2)}$

Watch out for (6)

Distance to the point, not to the line
→ Angular momentum of a particle
Angular momentum about a moving body
→ Angular momentum of a particle
A projectile's L is not conserved about the launch point
→ Angular momentum of a particle
Kinetic energy is not conserved when bodies stick
→ Conservation of angular momentum
Day length goes with the radius squared
→ Conservation of angular momentum
About which point is L conserved?
→ Conservation of angular momentum

Test yourself on System of Particles and Rotational Motion

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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