Angular Momentum Conservation Problems

What Angular Momentum Conservation Actually Means

Angular momentum is the rotational analogue of linear momentum. For a system with no external torque, the total angular momentum stays constant — even if the moment of inertia changes wildly.

This single rule explains why an ice-skater spins faster when she pulls her arms in, why a diver tucks tightly to flip more, why neutron stars spin thousands of times per second after collapse, and why a falling cat always lands feet-down. Every JEE/NEET aspirant should be able to spot conservation problems at first glance.

We will build the framework, work through worked examples graded by difficulty, and lock in the speed shortcuts that experienced rankers use.

Key Terms & Definitions

Angular momentum ( $\vec{L}$ ): For a rigid body, $\vec{L} = I\vec{\omega}$ , where $I$ is the moment of inertia and $\vec{\omega}$ is the angular velocity. For a point mass, $\vec{L} = \vec{r} \times \vec{p} = m\vec{r} \times \vec{v}$ .

Torque ( $\vec{\tau}$ ): The rotational analogue of force. $\vec{\tau} = d\vec{L}/dt$ , just as $\vec{F} = d\vec{p}/dt$ .

Moment of inertia ( $I$ ): How “spread out” the mass is from the axis. A skater with arms out has higher $I$ than the same skater with arms tucked.

Conservation condition: If $\vec{\tau}_\text{ext} = 0$ , then $\vec{L} =$ constant.

The Core Conservation Equation

$I_1 \omega_1 = I_2 \omega_2$

This equation is the workhorse for 90% of JEE/NEET problems. It assumes a rigid body changing shape (like an ice-skater) or two objects coupling/decoupling rotationally.

When the moment of inertia decreases, angular velocity must increase by the same factor to keep $L$ constant. The skater pulls in arms, $I$ drops, $\omega$ shoots up.

Why External Torques Often Vanish

Three situations where conservation applies cleanly:

Isolated systems in space (no contact with anything else)
Systems on smooth surfaces with vertical-only external forces (gravity and normal force have no torque about a vertical axis through COM)
Internal interactions only (like the skater’s muscles pulling arms in — these are internal torques and cancel)

Methods/Concepts (Step by Step)

Method 1: Identifying Conservation

Before plugging into formulas, ask: “Is there any external torque about the chosen axis?” If no, write $L_\text{before} = L_\text{after}$ and solve.

Method 2: Choosing the Right Axis

Sometimes a torque exists about one axis but not another. For a sphere falling through a hole and getting trapped, gravity provides torque about the contact point but not about the centre of mass. Always pick the axis where torques vanish.

Method 3: Handling Collisions and Coupling

When two rotating objects couple (clutch plates joining, putty hitting a rod), use angular momentum conservation about the rotation axis. Energy is generally NOT conserved in these inelastic interactions — only $L$ .

Solved Examples

Easy — Skater Pulling In Arms (CBSE/NEET level)

A skater spins at $2 \, \text{rev/s}$ with arms outstretched ( $I_1 = 5 \, \text{kg m}^2$ ). She pulls her arms in, reducing $I$ to $2 \, \text{kg m}^2$ . Find her new angular speed.

Conservation: $I_1 \omega_1 = I_2 \omega_2$ .

$5 \times 2 = 2 \times \omega_2$ , giving $\omega_2 = 5 \, \text{rev/s}$ .

She now spins 2.5× faster — exactly the famous “skater effect.”

The kinetic energy is NOT conserved here — it actually increases. The skater’s muscles do work pulling the arms in against centrifugal-like forces. Energy comes from her muscles, not from outside the system.

Medium — Putty Hits Rotating Rod (JEE Main level)

A rod of mass $M$ , length $L$ rotates about its centre at angular velocity $\omega_0$ . A piece of putty of mass $m$ falls vertically and sticks to the end of the rod. Find the new angular velocity.

Initial $L_\text{rod} = I_\text{rod}\omega_0 = (ML^2/12)\omega_0$ . Putty falling vertically has zero angular momentum about the rod’s centre.

After collision: $I_\text{new} = ML^2/12 + m(L/2)^2 = ML^2/12 + mL^2/4$ .

Conservation gives:

$\omega_\text{new} = \frac{(ML^2/12)\omega_0}{ML^2/12 + mL^2/4} = \frac{M\omega_0}{M + 3m}$

Hard — Disc and Insect (JEE Advanced level)

A horizontal disc of mass $M$ , radius $R$ rotates about a vertical axis through its centre with angular velocity $\omega_0$ . An insect of mass $m$ sits at the centre. The insect crawls slowly to the edge. Find the final angular velocity.

Initial $I_1 = MR^2/2$ (disc only — insect at centre has zero moment of inertia).

Final $I_2 = MR^2/2 + mR^2$ .

Conservation: $\omega_2 = \frac{(MR^2/2)\omega_0}{MR^2/2 + mR^2} = \frac{M\omega_0}{M + 2m}$ .

The disc slows down as the insect walks outward — the insect’s weight adds to $I$ , requiring lower $\omega$ to keep $L$ constant.

Exam-Specific Tips

JEE Main / Advanced

JEE loves problems with putty-hits-rod, insect-on-disc, and platform-with-person setups. About 1-2 such questions per JEE Main shift since 2020. Master the three templates above and we cover most variations.

For Advanced, the twist usually involves combining conservation with energy methods or with translation (e.g., a wheel rolling and then jumping onto a turntable).

NEET

NEET tends to ask conceptual variants — “what happens to angular speed if…” rather than full numerical. Quick rule: if $I$ goes up, $\omega$ goes down, and vice versa.

CBSE Boards

Boards often ask for derivations: “Show that angular momentum is conserved in the absence of external torque.” Memorise the proof using $\vec{\tau} = d\vec{L}/dt$ .

Common Mistakes to Avoid

Mistake 1: Confusing angular momentum with angular velocity. They are related by $L = I\omega$ but are not the same. $L$ stays constant under no torque; $\omega$ generally changes.

Mistake 2: Using energy conservation instead of momentum conservation in inelastic rotational collisions. KE is NOT conserved in putty-stick problems.

Mistake 3: Forgetting to account for the falling object’s angular momentum about the chosen axis. If the putty falls with horizontal velocity, it carries angular momentum $mvr$ that must enter the equation.

Mistake 4: Computing moment of inertia about the wrong axis. Always use the parallel-axis theorem $I = I_\text{COM} + Md^2$ when the axis does not pass through the centre of mass.

Mistake 5: Treating contact friction as an external torque without checking. For pure rolling, friction provides the torque but does no work. For sliding, it dissipates energy AND provides torque.

Practice Questions

Q1. A child of mass $30 \, \text{kg}$ stands at the edge of a turntable of mass $60 \, \text{kg}$ , radius $2 \, \text{m}$ , rotating at $0.5 \, \text{rad/s}$ . The child walks to the centre. Find the new angular velocity.

$I_1 = (60)(2^2)/2 + (30)(2^2) = 120 + 120 = 240 \, \text{kg m}^2$ .

$I_2 = 120 + 0 = 120 \, \text{kg m}^2$ .

$\omega_2 = (240/120)(0.5) = 1.0 \, \text{rad/s}$ .

Q2. A solid sphere of radius $R$ rotates at $\omega_0$ . It melts and re-forms as a thin spherical shell of radius $R$ . Find the new angular velocity.

$I_\text{solid} = (2/5)MR^2$ , $I_\text{shell} = (2/3)MR^2$ .

$\omega_\text{new} = (2/5)\omega_0 / (2/3) = (3/5)\omega_0$ .

Q3. A bullet of mass $m$ moving horizontally with velocity $v$ hits and embeds in the end of a vertical rod (mass $M$ , length $L$ ) hinged at the top. Find the angular velocity of the rod just after impact.

About the hinge: $L_\text{bullet} = mvL$ .

After: $I = ML^2/3 + mL^2$ .

$\omega = mvL/(ML^2/3 + mL^2) = 3mv/[(M + 3m)L]$ .

Q4. Two discs of moment of inertia $I_1$ and $I_2$ rotating at $\omega_1$ and $\omega_2$ are brought into contact along their flat faces. Find the common final angular velocity.

$\omega = (I_1\omega_1 + I_2\omega_2)/(I_1 + I_2)$ .

Q5. A planet moves in an elliptical orbit. At perihelion (closest), its speed is $v_p$ and distance from sun is $r_p$ . Find its speed at aphelion (farthest, distance $r_a$ ).

Angular momentum about the sun: $mv_p r_p = mv_a r_a$ , so $v_a = v_p r_p / r_a$ .

This is Kepler’s second law in disguise.

FAQs

Is angular momentum always conserved? Only when external torque is zero about the chosen axis. Internal forces produce internal torques that always cancel pairwise.

What’s the difference between $L$ and $I\omega$ ? For a rigid body rotating about a fixed axis, they are equal. For more complex motion (like a precessing top), $\vec{L}$ may not be parallel to $\vec{\omega}$ and the relation involves a tensor.

Why does kinetic energy change but angular momentum stay constant? Internal forces can do work (skater’s muscles) but cannot exert net external torque. So energy is not conserved, but $L$ is.

Can a particle moving in a straight line have angular momentum? Yes — about any point not on its path. $L = mvd$ , where $d$ is the perpendicular distance from the point to the line of motion.

Why do helicopters need a tail rotor? The main rotor exerts torque on the body. Without a counter-torque from the tail, the body would spin in the opposite direction to conserve angular momentum.

What happens to angular momentum during a collision? External torques during a brief collision are usually negligible (impulsive forces dominate), so angular momentum about a fixed point is conserved across the collision.

Is the formula $L = I\omega$ valid for all rotations? It is valid when rotation is about a principal axis or about a fixed axis of symmetry. For non-symmetric rotations, $\vec{L}$ and $\vec{\omega}$ may not be parallel.

How is angular momentum quantised in atoms? In quantum mechanics, $L$ takes discrete values: $L = \sqrt{l(l+1)}\hbar$ , where $l = 0, 1, 2, \ldots$ . This is the basis of atomic structure.