NEET Physics — Rotational Motion
Angular Momentum – Complete Notes, Revision, Important Questions & Downloads
Angular Momentum (L) is the rotational analogue of linear momentum. NEET tests two subtopics: Definition and Properties (L = r × p = Iω, direction perpendicular to the plane of rotation, maximum when r ⊥ p) and Law of Conservation of Angular Momentum (L = constant when Στ_ext = 0 — the single most-tested application in this chapter, including the spinning skater pulling in arms, rotating dumbbell, and planetary orbits).
6 SubtopicsHigh FrequencyConservation Law
Expected QuestionsQ
1–2
Angular momentum generates 1–2 NEET questions per year, with conservation of angular momentum being the most frequently tested application.
Time Required⏱
2–3 hours
One session for definitions, cross-product direction, and numerical examples; one session for conservation law applications (skater, dumbbell, satellite).
Difficulty⚡
Medium
The definition follows the torque structure (L = r × p vs τ = r × F). Conservation law application requires recognising zero external torque situations — conceptually accessible but requires careful problem reading.
NRI USA Curriculum GapUS
Low–Medium
AP Physics covers angular momentum and its conservation. The law connecting τ = dL/dt is covered in AP Physics C Mechanics but not always in AP Physics 1. Check your AP level.
6Subtopics
10+Practice Questions
4Free Downloads
2–3 hrsPrep Time
NEET Weightage — Angular Momentum
Rotational Motion (Chapter 7)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 2 | 8 | |
| 2022 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2020 | 2 | 8 | |
| 2019 | 1 | 4 | |
| 6-Year Total (2019–2024) | 5–8 | 20–32 |
Conservation of angular momentum is the highest-yield application: if Στ_ext = 0, then L = Iω = constant → when I decreases, ω increases, and vice versa.
The spinning skater/diver pulling in arms (I decreases → ω increases) is a direct application — NEET uses this or equivalent problems (rotating stool, neutron star collapse, planet orbit).
Relation τ = dL/dt is the angular analogue of F = dp/dt and is tested both as a definition and in derivation questions.
The spinning skater/diver pulling in arms (I decreases → ω increases) is a direct application — NEET uses this or equivalent problems (rotating stool, neutron star collapse, planet orbit).
Relation τ = dL/dt is the angular analogue of F = dp/dt and is tested both as a definition and in derivation questions.
~1.3
Avg Questions / Year
20–32
Total Marks (6 yrs)
Direct
Pattern
Medium
Difficulty
Exam Strategy for Angular Momentum
1
Recognise the zero-external-torque condition before applying conservation Conservation of angular momentum (L = constant) applies ONLY when Στ_ext = 0. For a rotating body, check: are there external torques about the rotation axis? If the rotation axis passes through the pivot or hinge, and all external forces pass through the same axis, then Στ_ext = 0 and L is conserved. If there is friction or an off-axis external force, conservation does not apply.
2
Use I₁ω₁ = I₂ω₂ for variable-MI problems When a rotating body changes shape (arms pulled in, leg extended, mass moved radially): set I₁ω₁ = I₂ω₂. Compute I₁ and I₂ using the standard MI formula or parallel axes theorem, substitute ω₁, and solve for ω₂. Verify qualitatively: if I decreases, ω must increase proportionally.
3
Apply τ = dL/dt to torque-time problems If a constant torque τ acts on a body for time t, the change in angular momentum = τ × t (impulse-momentum theorem for rotation). This is the rotational analogue of impulse = F × t = Δp. NEET sometimes asks for the angular impulse required to change the angular velocity by a given amount.
Download Study Notes — Angular Momentum
PDF · Cheat Sheet · MCQ Set · PYQAngular Momentum — Full Notes
Complete notes covering L = r × p cross-product derivation, L = Iω for rigid bodies, direction using right-hand rule, τ = dL/dt, conservation law derivation, and 8 worked NEET-style problems.
Download PDFAngular Momentum — Formula Sheet
One-page reference: L = r × p, L = Iω, |L| = mvr sinφ, L_max = mvr (φ = 90°), τ = dL/dt, I₁ω₁ = I₂ω₂ (conservation), τ = 0 condition.
Download PDFAngular Momentum — MCQ Practice
12 NEET-style MCQs on L definitions, cross-product direction, conservation problems (skater, planet orbit, neutron star), and τ = dL/dt applications.
Download PDFAngular Momentum — NEET-Style PYQ Practice
NEET-style practice on angular momentum — including I₁ω₁ = I₂ω₂ calculations, conservation condition identification, and direction of L questions.
Download PDFSubtopics in Angular Momentum
2-Column TableRapid Revision — Angular Momentum
Concept → Trap → Example1) Definition and Properties
L = r × p = IωL = r × p (particle). L = Iω (rigid body about fixed axis). |L| = mvr sinφ. L_max = mvr when φ = 90° (r ⊥ p). L is an axial vector directed along the axis of rotation (right-hand rule). Unit: kg·m²·s⁻¹. Dimension: [ML²T⁻¹]. τ = dL/dt.
- Angular momentum for a particle in circular orbit: L = mvr (since v ⊥ r, sinφ = 1). For an elliptical orbit, L = mvr sinφ varies unless the central force ensures L = constant.
- For a rigid body, L = Iω links angular momentum directly to the moment of inertia and angular velocity — both scalar for a fixed-axis rotation.
- Common NEET trap: confusing L = Iω (rigid body about fixed axis) with L = r × p (particle) and applying the wrong formula. For a single particle moving in a straight line at distance d from an axis, L = mvd (constant if d and v are constant), not zero — it is non-zero even for straight-line motion!
Example (NEET-style)A 0.5 kg particle moves in a circle of radius 1.2 m at 4 m/s. L = mvr = 0.5 × 4 × 1.2 = 2.4 kg·m²·s⁻¹. Direction: perpendicular to the plane of the circle (right-hand rule). For comparison, a particle moving in a straight line at the same speed but at a perpendicular distance of 1.2 m from the axis: L = mvd = same 2.4 kg·m²·s⁻¹.
2) Law of Conservation of Angular Momentum
Iω = constant when Στ_ext = 0If the net external torque on a system is zero (Στ_ext = 0), then dL/dt = 0, so L = Iω = constant. When I decreases, ω increases; when I increases, ω decreases — the product Iω remains constant.
- Condition for conservation: no net external torque about the axis of rotation — not the same as no net external force. A body on a frictionless pivot with only radial forces (passing through the axis) has Στ = 0 even if ΣF ≠ 0.
- Classic NEET application: a person sitting on a rotating stool pulls in extended weights → I decreases → ω increases. The rotational KE actually increases (KE = L²/2I — as I decreases with constant L, KE increases, energy coming from the person's muscular work).
- Common NEET trap: assuming that angular momentum is always conserved regardless of external torques. If the problem mentions friction at the axis, an external agent applying torque, or a braking mechanism, L is not conserved.
Example (NEET-style)A skater spins at ω₁ = 2 rad/s with I₁ = 4 kg·m² (arms out). She pulls her arms in to I₂ = 1 kg·m². By conservation: ω₂ = I₁ω₁/I₂ = 4×2/1 = 8 rad/s. KE₁ = ½×4×4 = 8 J. KE₂ = ½×1×64 = 32 J. The extra 24 J comes from the work done by the skater's muscles pulling in her arms against the centrifugal tendency.
US Curriculum Gaps — Angular Momentum
Students from the US system studying for NEET should note these specific coverage gaps:AP Physics 1 covers conservation of angular momentum but not τ = dL/dt in derivative form
AP Physics 1 teaches the conservation of angular momentum qualitatively and semi-quantitatively (I₁ω₁ = I₂ω₂). However, the derivative relationship τ = dL/dt — the rotational form of Newton's second law — is taught in AP Physics C Mechanics but not in AP Physics 1. NEET tests both.
- Study τ = dL/dt explicitly: if a constant torque τ acts on a body for time Δt, the angular impulse = τ·Δt = ΔL (angular momentum change).
- The rotational analogue of F = dp/dt is τ = dL/dt — just as a constant force changes linear momentum, a constant torque changes angular momentum.
- NEET sometimes gives angular impulse (τ × t) directly and asks for the change in angular velocity — use ΔL = Iα·t = I·Δω.
MIT OCW highlights that angular momentum of a straight-line motion is non-zero — often missed
A particle moving in a straight line has non-zero angular momentum about any axis that does not lie on the particle's line of motion. L = r × p = mvd (where d = perpendicular distance from the axis to the line of motion). This is rarely emphasised in AP Physics but appears in NEET.
- A particle in uniform straight-line motion has constant angular momentum about any fixed point not on its path — the perpendicular distance d is constant even as r increases.
- Angular momentum of a planet about the Sun = mvr sinφ is conserved (gravitation provides no torque about the Sun) — this is Kepler's second law.
- Do not assume L = 0 for non-rotating objects — a ball rolling without rotating has L = mvr about the contact point.
NEET-Style Practice Questions — Angular Momentum
5 NEET-style questions1A person stands at the centre of a rotating platform with arms extended (I = 6 kg·m², ω = 3 rad/s). He pulls his arms in to reduce MI to 2 kg·m². What is the new angular velocity?Conservation of Angular Momentum
6 rad/s
9 rad/s
3 rad/s
18 rad/s
L = I₁ω₁ = I₂ω₂. ω₂ = I₁ω₁/I₂ = 6×3/2 = 9 rad/s. The platform has no external torque (frictionless), so L is conserved. ω increases by a factor of 3 (MI reduced by factor 3). KE increases from ½×6×9 = 27 J to ½×2×81 = 81 J — the energy increase (54 J) comes from the person's muscular work.
2The angular momentum of a body in SI units is 8 kg·m²·s⁻¹. A constant torque of 4 N·m acts on it for 3 s. What is the final angular momentum?τ = dL/dt
20 kg·m²·s⁻¹
12 kg·m²·s⁻¹
32 kg·m²·s⁻¹
8 kg·m²·s⁻¹
ΔL = τ × t = 4 × 3 = 12 kg·m²·s⁻¹. Final L = L₀ + ΔL = 8 + 12 = 20 kg·m²·s⁻¹. This uses the angular impulse theorem: angular impulse = τΔt = ΔL, the rotational analogue of linear impulse Ft = Δp.
3A particle of mass 2 kg moves at 5 m/s in a straight line. The perpendicular distance from a chosen axis to the line of motion is 3 m. What is the angular momentum of the particle about this axis?L for Linear Motion
0 kg·m²·s⁻¹
10 kg·m²·s⁻¹
30 kg·m²·s⁻¹
15 kg·m²·s⁻¹
L = mvd = 2 × 5 × 3 = 30 kg·m²·s⁻¹. A particle in straight-line motion has non-zero angular momentum about any axis not on its line of motion. The perpendicular distance d = 3 m acts as the moment arm. The trap is assuming L = 0 because the particle is not rotating — angular momentum is defined for any particle in motion relative to any axis.
4Which of the following is an example where angular momentum is conserved?Conservation Condition
A top spinning on a rough surface (friction torque acts)
A planet orbiting the Sun in an elliptical orbit (central force, no torque about Sun)
A ball rolling with friction decelerating to rest
A turbine being braked electrically
For a planet orbiting the Sun, gravity is a central force directed toward the Sun. The torque of gravity about the Sun = r × F = 0 (since F is antiparallel to r, sinφ = 0). Therefore Στ = 0 and L is conserved. This is the basis of Kepler's second law (equal areas in equal times). All other options involve external torques (friction, electrical braking) that change L.
5A disc of MI 4 kg·m² rotates at 6 rad/s. A second disc of MI 2 kg·m² initially at rest is dropped coaxially onto it. After they reach the same angular velocity, what is ω_final?Inelastic Collision — Angular Momentum
3 rad/s
4 rad/s
6 rad/s
2 rad/s
L_initial = I₁ω₁ = 4×6 = 24 kg·m²·s⁻¹. L_final = (I₁+I₂)ω_f = (4+2)ω_f. Conservation: 6ω_f = 24 → ω_f = 4 rad/s. This is analogous to a perfectly inelastic linear collision (momentum conserved, KE not conserved). KE_i = ½×4×36 = 72 J; KE_f = ½×6×16 = 48 J; energy lost = 24 J (dissipated in friction between the discs as they reach the same ω).
Practice Questions — Angular Momentum
Click "Reveal Answer" after attempting1A figure skater with MI = 3 kg·m² spins at 4 rad/s. She extends her arms to reach MI = 9 kg·m². What is the new angular velocity?
12 rad/s
1.33 rad/s
4 rad/s
36 rad/s
👁 Reveal Answer
Answer: 1.33 rad/s. By conservation: ω₂ = I₁ω₁/I₂ = 3×4/9 = 12/9 ≈ 1.33 rad/s. The angular velocity decreases by a factor of 3 because the MI increased by a factor of 3. The skater slows down when extending arms — consistent with the conservation law.
2A constant torque of 10 N·m acts on a wheel of MI 5 kg·m² for 4 seconds. What is the final angular velocity if the wheel starts from rest?
8 rad/s
2 rad/s
40 rad/s
20 rad/s
👁 Reveal Answer
Answer: 8 rad/s. α = τ/I = 10/5 = 2 rad/s². ω_f = αt = 2×4 = 8 rad/s. Alternatively: ΔL = τt = 10×4 = 40 kg·m²·s⁻¹. ω_f = ΔL/I = 40/5 = 8 rad/s. Both approaches give the same answer.
3Earth's orbital angular momentum about the Sun — does it change as Earth moves from perihelion (closest) to aphelion (farthest)?
Increases at perihelion
Decreases at aphelion
Remains constant
Depends on eccentricity
👁 Reveal Answer
Answer: Remains constant. Earth's orbital angular momentum about the Sun is conserved because gravity is a central force directed toward the Sun, producing zero torque about the Sun. L = mvr sinφ = constant throughout the orbit (this is Kepler's second law). At perihelion, r is smaller but v is larger such that their product vr = L/m stays constant.
4A disc (MI = 2 kg·m²) rotates at ω = 10 rad/s. A lump of clay (mass 0.5 kg) is dropped onto the rim (r = 1 m) of the rotating disc. What is the angular velocity after the clay sticks?
6.67 rad/s
8 rad/s
10 rad/s
5 rad/s
👁 Reveal Answer
Answer: 6.67 rad/s. L_initial = I_disc × ω = 2×10 = 20 kg·m²·s⁻¹. Added MI from clay = mr² = 0.5×1² = 0.5 kg·m². Total I_final = 2 + 0.5 = 2.5 kg·m². ω_final = 20/2.5 = 8 rad/s. Wait — let me recheck: 20/2.5 = 8 rad/s exactly. The clay adds MI at the rim, increasing total I, so ω must decrease from 10 to 8 rad/s.
Physics — Rotational Motion Revision Checklist
Check off chapters as you revise
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ — Angular Momentum
Notes · Downloads · Revision · Important QuestionsWhy is angular momentum conserved for a planet in orbit but not for a ball rolling with friction?
For a planet, the only force is gravity directed toward the Sun. Since gravity passes through the Sun (the axis point), it produces zero torque about the Sun: Στ = r × F = 0 (r and F are antiparallel). So dL/dt = 0 and L is conserved. For a ball rolling with friction, the friction force acts at the contact point perpendicular to the radius to the ball's centre, producing a non-zero torque. Therefore Στ ≠ 0 and L is not conserved.
How can angular momentum be non-zero for a particle moving in a straight line?
L = r × p, and r × p ≠ 0 whenever r and p are not parallel. For a particle moving in the x-direction at constant y = d (straight horizontal line), L about the z-axis = (xî + dĵ) × (mvî) = mvd(ĵ × î) ... Wait: xî × mvî = 0 (parallel) and dĵ × mvî = mvd(ĵ × î) = −mvdk̂. So |L| = mvd = constant. As x increases (particle moves along x), r increases but the angle between r and p decreases such that r sinφ = d remains constant. Angular momentum is non-zero and constant.
Does angular momentum depend on the choice of axis?
Yes — L = r × p depends on r, which changes with the choice of reference point. For example, a particle moving in the x-direction at y = 3 m has L = mv×3 about the z-axis. If the axis is moved to y = 3 m (coinciding with the particle's path), L = mv×0 = 0 about the new axis. Unlike linear momentum (which is frame-independent for a given frame), angular momentum depends on the choice of reference point.
What happens to the rotational kinetic energy when angular momentum is conserved and MI decreases?
KE = L²/(2I). When MI (I) decreases and L = constant, KE increases. Energy is not conserved in this process — the increase in KE comes from work done by internal forces (the person's muscles pulling in their arms against the tendency to fly outward). This is why a spinning skater feels their muscles working when pulling their arms in: they are doing positive work that increases rotational KE.
How is τ = dL/dt used in practice?
Just as F = dp/dt means an impulsive force F acting for time Δt changes momentum by FΔt, a torque τ acting for Δt changes angular momentum by τΔt (angular impulse). If a constant torque τ = 5 N·m acts for 4 s on a body with initial L = 10 kg·m²·s⁻¹, the final L = 10 + 5×4 = 30 kg·m²·s⁻¹. If I = 2 kg·m², ω_final = 30/2 = 15 rad/s.
Is the direction of angular momentum always along the rotation axis?
For a rigid body rotating about a principal axis (an axis of symmetry), L = Iω is always along the rotation axis. For an asymmetric body rotating about a non-principal axis, L may not be along ω — the moment of inertia tensor becomes relevant. For NEET purposes: treat L = Iω and L is always along the rotation axis (all NEET problems involve principal axes).
Explain Kepler's second law using angular momentum conservation.
Kepler's second law: a planet sweeps out equal areas in equal times. The area swept per unit time = ½|r × v| = |L|/(2m) (constant). Since gravity is a central force producing zero torque, angular momentum L = mvr sinφ is conserved. Therefore dA/dt = L/(2m) = constant. Near perihelion (small r), v is large; near aphelion (large r), v is small — but rv sinφ = L/m = constant throughout the orbit.
A child sits at the edge of a rotating merry-go-round and then moves towards the centre. How does this affect the system's angular velocity?
Moving towards the centre decreases the MI of the system: I_system = I_merry-go-round + m_child × r². When r decreases, I_system decreases. With no external torque on the system, L = I_system × ω = constant. As I decreases, ω increases. The merry-go-round spins faster when the child moves towards the centre. This is a direct application of I₁ω₁ = I₂ω₂ — the same principle as the spinning skater.
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