When the net external torque on a system is zero, the total angular momentum of the system is conserved: L_initial = L_final. Basis of figure-skater spin, planetary orbits, and rotating-bodies physics.
-- NCERT Class 11 Physics, Ch. 6, p. 121Conservation Angular Momentum
8 MCQs2 revision cards9-step worked example
Source: NCERT System of Particles and Rotational MotionPYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 26 Sep 2026
Conservation Angular Momentum, explained for NEET
Conservation of angular momentum is a direct consequence of Newton's second law for rotation: when the net external torque on a system is zero, the total angular momentum remains constant. Stated formally, if τ_ext = 0, then L = Iω = constant (NCERT Class 11 Physics Chapter 6, page 122).
The trap that costs marks: confusing conservation of angular momentum with conservation of rotational kinetic energy. These are different conservation laws with different conditions. When a body's moment of inertia changes through internal forces alone (no external torque), angular momentum L = Iω is conserved — but rotational kinetic energy KE = ½Iω² is NOT conserved. The internal forces do work, changing KE even as L stays fixed.
Why KE changes when L is conserved. Consider a spinning figure skater pulling her arms inward. No external torque acts, so L₁ = L₂, meaning I₁ω₁ = I₂ω₂. Since I₂ < I₁, the new angular velocity ω₂ > ω₁. Now check kinetic energy: KE₂ = L²/(2I₂) > L²/(2I₁) = KE₁. The kinetic energy increases — the skater's muscles did the work. The same logic applies to a collapsing star: as radius shrinks, I drops, ω rises, and rotational KE increases dramatically.
The deciding question for any problem: Is external torque zero? If yes → conserve L. Is there also no work done by any force? If yes → conserve KE too. In most NEET problems involving changing I, external torque is zero but internal work is done, so only L is conserved.
Watch out: if the problem gives a disc on a frictionless axle and a second disc is dropped onto it, this is an inelastic collision in rotation. L is conserved (no external torque from the axle along the rotation axis), but KE decreases — lost to friction between the surfaces during coupling.
Can you answer these Conservation Angular Momentum MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Under what condition is the angular momentum of a system conserved?
Show answer and why every option is right or wrong
Answer: A. Angular momentum is conserved when the net external torque is zero (NCERT Class 11 Physics Chapter 6, page 108). This is the rotational analogue of linear momentum conservation (zero net force), but the condition specifically involves torque, not force.
Why B is wrong: B is wrong because zero net external force conserves linear momentum, not angular momentum. A system can have zero net force but non-zero torque (e.g. a couple).
Why C is wrong: C is wrong because constant rotational KE is a separate condition. Angular momentum can be conserved even when KE changes (e.g. skater pulling arms in).
Why D is wrong: D is wrong because conservation of L does not require constant I. When I changes with zero external torque, ω adjusts so that Iω remains constant.
MCQ 2Direct ApplicationPractice
A skater spins at 2.0 rev/s with arms outstretched. She pulls her arms in, reducing her moment of inertia to half its value. Neglecting friction, her new rate of spin is:
Show answer and why every option is right or wrong
Answer: B. B is correct. With no external torque about the spin axis, angular momentum L = Iω is conserved. Halving I therefore doubles ω: 2 × 2.0 = 4.0 rev/s. Her kinetic energy ½Iω² actually doubles — the extra energy comes from the work her muscles do pulling her arms in.
Why A is wrong: A is wrong because 1.0 rev/s has ω changing in the same direction as I. Since Iω is constant, a smaller moment of inertia needs a LARGER angular speed.
Why C is wrong: C is wrong because 2.0 rev/s assumes the spin rate is fixed. It is the angular momentum that is conserved, not the angular velocity.
Why D is wrong: D is wrong because 8.0 rev/s makes ω go as 1/I². Conservation of Iω gives ω ∝ 1/I, so halving I only doubles ω.
MCQ 3Easy RecallPractice
A spinning figure skater pulls her arms inward. Which of the following quantities is conserved during this action?
Show answer and why every option is right or wrong
Answer: A. No external torque acts on the skater, so angular momentum L = Iω is conserved (NCERT Class 11 Physics Chapter 6, page 122). As I decreases (arms pulled in), ω increases, and KE increases due to work done by muscles.
Why B is wrong: B is wrong because angular velocity increases as I decreases — the whole point of the skater pulling arms in is to spin faster.
Why C is wrong: C is wrong because moment of inertia decreases when the skater's mass distribution moves closer to the rotation axis.
Why D is wrong: D is wrong because KE = ½Iω² changes when I changes. Since ω increases more than I decreases (KE = L²/2I, and I decreases), kinetic energy actually increases. (Trap: L vs KE conservation confusion.)
MCQ 4Direct ApplicationPractice
A uniform disc of moment of inertia I is spinning at angular velocity ω about its central axis on a frictionless axle. An identical disc, initially at rest, is gently placed on top of it. After the discs reach a common angular velocity, what is the final angular velocity?
Show answer and why every option is right or wrong
Answer: B. The frictionless axle exerts no torque along the rotation axis. By conservation of angular momentum: Iω = (I + I)ω_f, so ω_f = ω/2. Note that rotational KE is NOT conserved here — energy is lost to friction between the disc surfaces during coupling.
Why A is wrong: A is wrong because ω would mean the second disc acquired angular velocity without any transfer from the first. The combined moment of inertia doubled, so ω must halve to conserve L. (Trap: assuming KE conservation gives ω unchanged.)
Why C is wrong: C is wrong because ω/4 would correspond to I_total = 4I, which is not the case — two identical discs give I_total = 2I.
Why D is wrong: D is wrong because 2ω would violate conservation of angular momentum, giving L_f = 2I(2ω) = 4Iω > Iω = L_i.
MCQ 5Direct ApplicationPractice
A star collapses under gravity and its radius becomes half the original value. Assuming no mass is lost and no external torque acts, the ratio of the new angular velocity to the original angular velocity is:
Show answer and why every option is right or wrong
Answer: D. Model the star as a uniform solid sphere: I = (2/5)MR². When R → R/2, I_new = (2/5)M(R/2)² = I/4. By L conservation: Iω = (I/4)ω_new, so ω_new = 4ω. The ratio ω_new:ω = 4:1.
Why A is wrong: A is wrong because 1:4 would mean the star slows down when it shrinks. Since I decreases, ω must increase to keep L constant — the star spins faster, not slower.
Why B is wrong: B is wrong because 2:1 would correspond to I halving, i.e. R → R/√2. Since R → R/2, I drops to I/4, not I/2. (Trap: confusing linear vs quadratic dependence on R.)
Why C is wrong: C is wrong because 1:2 corresponds to I halving (R → R/√2). Here R halves, so I drops by a factor of 4 (since I ∝ R²), and ω quadruples.
MCQ 6Direct ApplicationPractice
A particle of mass m moves with velocity v along a straight line. What is its angular momentum about a point at perpendicular distance d from the line of motion?
Show answer and why every option is right or wrong
Answer: C. L = r × p. The magnitude is |r||p|sin θ = r·mv·sin θ. The perpendicular distance d = r sin θ, so L = mvd (NCERT Class 11 Physics Chapter 6, page 96). Angular momentum about a point is non-zero even for straight-line motion, provided the line does not pass through that point.
Why A is wrong: A is wrong because md/v has dimensions of kg·s, which does not match angular momentum [ML²T⁻¹].
Why B is wrong: B is wrong because mv/d has units of kg·(m/s)/m = kg/s, not kg·m²/s (angular momentum). There is no division by distance in the cross product formula.
Why D is wrong: D is wrong because angular momentum about a point depends on the perpendicular distance from that point to the line of motion. A particle moving in a straight line has constant, non-zero angular momentum about any point not on that line. (Trap: confusing L = 0 only when the line passes through the reference point.)
MCQ 7CalculationPractice
A disc of moment of inertia 4 kg·m² is spinning at 6 rad/s. A ring of moment of inertia 2 kg·m², initially at rest, is coaxially placed on the disc. After they reach a common angular velocity, the loss in rotational kinetic energy is:
Show answer and why every option is right or wrong
Answer: D. By L conservation: L = 4 × 6 = 24 kg·m²/s. Combined I = 4 + 2 = 6 kg·m². So ω_f = 24/6 = 4 rad/s. KE_i = ½(4)(6²) = 72 J. KE_f = ½(6)(4²) = 48 J. Loss = 72 − 48 = 24 J.
Why A is wrong: A is wrong — 12 J would result from halving the actual loss. The full calculation requires computing both KE_i and KE_f separately and taking their difference. (Trap: students who try to conserve KE get zero loss, which isn't even an option.)
Why B is wrong: B is wrong — 48 J equals the final KE, not the energy lost. The loss is KE_i − KE_f = 72 − 48 = 24 J.
Why C is wrong: C is wrong — 36 J corresponds to half of KE_i. The disc retains significant rotational energy at the reduced angular velocity.
MCQ 8CalculationPractice
A child of mass 30 kg stands at the edge of a merry-go-round of moment of inertia 600 kg·m² and radius 3.0 m, spinning at 2.0 rad/s. The child walks to the centre. What is the new angular velocity? (Treat the child as a point mass.)
Show answer and why every option is right or wrong
Answer: C. I_child at edge = mr² = 30 × (3.0)² = 270 kg·m². I_total,initial = 600 + 270 = 870 kg·m². At centre, r = 0, so I_child = 0. I_total,final = 600 kg·m². By L conservation: 870 × 2.0 = 600 × ω_f, so ω_f = 1740/600 = 2.9 rad/s.
Why A is wrong: A is wrong because 2.0 rad/s means no change in angular velocity. This would only be true if the child's moment of inertia were negligible, but 270 kg·m² is nearly half of the merry-go-round's 600 kg·m². (Trap: ignoring the child's contribution to I.)
Why B is wrong: B is wrong because 2.6 rad/s would need I_final = 870 × 2.0/2.6 ≈ 670 kg·m², leaving the child about 70 kg·m² at the centre; at r = 0 the child contributes nothing, so I_final = 600 kg·m² and ω_f = 2.9 rad/s.
Why D is wrong: D is wrong because 3.0 rad/s would require an initial total I of 3.0 × 600/2.0 = 900 kg·m², so the child's I at the edge would be 300 kg·m² — more than the 270 kg·m² that 30 kg at 3.0 m gives.
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Conservation Angular Momentum: quick recall before you leave
How do you solve a Conservation Angular Momentum question? A worked example
Pattern: Body's angular speed changes when its moment of inertia changes (NEET 2025 pattern).
- 1
Given
A uniform solid sphere (a model star) has initial angular velocity ω₁ = 1.0 rad/s. Due to gravitational collapse, its radius shrinks to one-third of its original value. No mass is lost and no external torque acts during the collapse.
- 2
Required
(a) The final angular velocity ω₂.
(b) The ratio of final to initial rotational kinetic energy, KE₂/KE₁. - 3
Concept
When external torque is zero, angular momentum is conserved: I₁ω₁ = I₂ω₂. However, rotational KE is NOT conserved because internal gravitational forces do work during the collapse.
- 4
Formula
• Moment of inertia of a solid sphere: I = (2/5)MR²• Conservation of L: I₁ω₁ = I₂ω₂• KE_rot = ½Iω² = L²/(2I)
- 5
Substitution
I₁ = (2/5)M R², I₂ = (2/5)M(R/3)² = (2/5)M(R²/9) = I₁/9.
From L conservation: I₁ω₁ = (I₁/9)ω₂. - 6
Calculation
ω₂ = 9ω₁ = 9 × 1.0 = 9.0 rad/s.
KE₁ = ½I₁ω₁² = ½I₁(1.0)² = 0.5 I₁.
KE₂ = ½I₂ω₂² = ½(I₁/9)(81) = ½ × 9 I₁ = 4.5 I₁.
KE₂/KE₁ = 4.5 I₁ / 0.5 I₁ = 9.
Note on exact constants: The fractions 2/5 and 1/9 are geometric constants of the sphere and the problem statement respectively. The angular velocity 1.0 rad/s is given as exact. These do not limit significant figures. The factor 9 in the final answer is exact. - 7
Final answer
(a) ω₂ = 9.0 rad/s. (b) KE₂/KE₁ = 9. The rotational kinetic energy increases ninefold — the gravitational potential energy released during collapse does this work.
- 8
Common trap
A common confusion is to assume KE is also conserved when L is conserved. If you set ½I₁ω₁² = ½I₂ω₂², you would get ω₂ = 3ω₁ (from ω₂ = ω₁√(I₁/I₂) = ω₁ × 3), which contradicts L conservation. The correct approach is to apply L conservation first (ω₂ = 9ω₁), then compute KE separately.
- 9
Similar NEET-style question
A ballet dancer with arms extended has moment of inertia 4.0 kg·m² and spins at 3.0 rev/s. She pulls her arms in, reducing her moment of inertia to 1.0 kg·m². Find her new spin rate and the ratio of her final to initial rotational kinetic energy. (Answer: 12 rev/s; KE ratio = 4.)
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What to remember before solving Conservation Angular Momentum questions
Which Conservation Angular Momentum formulas do you need for NEET?
1 formula — click to collapse
Angular momentum
For a particle: L = r x p. For a rigid body about its rotation axis: L = I omega. Vector quantity.
| Symbol | Quantity | SI Unit |
|---|---|---|
| L | angular momentum | kg*m^2/s |
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
Valid when
- Reference point/axis chosen
- I about same axis as omega
Where do students lose marks on Conservation Angular Momentum?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Similar Terms
Student conserves rotational KE when angular momentum is conserved (or vice versa). When I changes, L = Iω is conserved but KE = ½Iω² is NOT (it depends on I and ω together).
When it triggers
Question describes a body whose moment of inertia changes (skater pulling arms in, star collapsing).
How to avoid
L conservation requires zero external torque. KE conservation requires no work done — different criteria. When I changes via internal forces, L conserved, ω increases, KE increases.
Root cause: concept gap
Correction
L = Iω is conserved when external torque is zero. KE = ½Iω² is NOT conserved when I changes (since ω changes too). When skater pulls arms in, L conserved, ω increases, KE increases (work done by muscles).
More in System of Particles and Rotational Motion: 5 exam traps and mistakes · 7 formulas · 3 question patterns from its other lessons.
Conservation Angular Momentum questions from past NEET papers
1 question from NEET 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
1108 days
2100 days
3105 days
4115 days
NTA Answer: Option 1(final)
All 11 past-paper questions from System of Particles and Rotational Motion →
How does NEET ask about Conservation Angular Momentum?
1 recurring pattern from past papers — click to collapse
Body's angular speed changes when its moment of inertia changes (e.g. star collapses, skater pulls in arms). Apply L = I*omega = constant when no external torque.
Multi StepMedium
Common distractors
uses energy conservation instead
Confusing L conservation with KE conservation
Sources
NCERT refs: Class 11 Physics Chapter 6, p.122
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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