Angular Momentum ConservationJEE Main

Concept

With no external torque, angular momentum $L = I\omega$ is conserved. Pulling mass toward the axis shrinks $I$, so $\omega$ must rise — the skater spins faster. Kinetic energy is not conserved: her muscles do real work pulling her arms in against the centrifugal tendency.

Key formula

Derivation

$\tau_{ext} = dL/dt = 0 \Rightarrow L$ constant.

$KE = \tfrac12 I\omega^2 = \frac{L^2}{2I}$ — at fixed $L$, halving $I$ doubles the kinetic energy. The extra energy is exactly the work done by internal (muscular) forces, which can change KE but never L.

Scenarios to explore

• Angular Momentum — The skater pulls her arms in — Iω stays, KE rises.

Real-world applications

• Figure skating spins, diving tucks, gymnastics twists.
• Neutron stars: collapsing cores spin up to millisecond periods.
• Reaction wheels orient satellites without propellant.

JEE exam tips

• KE = L²/2I is the fastest route to energy comparisons at constant L.
• A dropped disc landing on a spinning disc: conserve L, then compute the (always positive) KE loss.
• Planets sweeping equal areas = angular momentum conservation in disguise (dA/dt = L/2m).

Common mistakes

• Conserving kinetic energy instead of angular momentum.
• Thinking internal forces can change L — only external torques can.
• Using L = mvr for extended bodies without integrating (that's for point masses).

Exam traps to avoid

• ΔKE > 0 when pulling in — energy INCREASES; the muscles supplied it.
• L is conserved about the specific axis with zero net torque — check the axis before conserving.