Angular momentum is defined as the moment of linear momentum about a point or axis (NCERT Class 11 Physics Chapter 6, page 96). For a point particle, L = r × p, where r is the position vector from the reference point and p is the linear momentum. For a rigid body rotating about a fixed axis, this simplifies to L = Iω, where I is the moment of inertia about that axis and ω is the angular velocity.
The trap that costs marks in NEET on this topic: confusing conservation of angular momentum with conservation of rotational kinetic energy. These are governed by different conditions. Angular momentum L = Iω is conserved when the net external torque on the system is zero. Rotational kinetic energy KE = ½Iω² is conserved only when no work is done — a different criterion entirely.
Consider a figure skater pulling their arms inward. No external torque acts, so L is conserved. As I decreases, ω must increase to keep Iω constant. But KE = ½Iω² = L²/(2I) — since I decreases, KE increases. The extra kinetic energy comes from the internal work done by the skater's muscles. Students who treat KE as conserved here get the wrong final angular speed.
The key test: when a problem describes a system whose moment of inertia changes (collapsing star, spinning platform with arms extended/retracted, two discs coupling), ask yourself — is external torque zero? If yes, conserve L, not KE. If the problem explicitly states no energy loss (e.g., elastic collision), then KE conservation applies separately.
Watch out: L is a vector quantity. For fixed-axis rotation, its direction is along the rotation axis (right-hand rule). NEET typically tests the scalar form L = Iω, but the vector nature matters when the axis itself can change.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Angular momentum is defined as:
Answer: A. By definition, angular momentum of a particle about a point is L = r × p, the cross product of the position vector and linear momentum (NCERT Class 11 Physics Chapter 6, page 96).
Why B is wrong: B describes the quantity Iα, which equals torque (τ = Iα), not angular momentum.
Why C is wrong: C describes impulse-angular (∫τ dt = ΔL), the change in angular momentum due to torque over time, not angular momentum itself.
Why D is wrong: D has no standard physical meaning; force × angular displacement mixes linear and angular quantities incorrectly.
The SI unit of angular momentum is:
Answer: B. L = Iω. The unit of I is kg·m² and ω is rad/s (rad is dimensionless), so [L] = kg·m²/s (NCERT Class 11 Physics Chapter 6, page 96).
Why A is wrong: A is the unit of linear momentum (p = mv), not angular momentum.
Why C is wrong: C is wrong because N·m·s² = kg·m/s² × m × s² = kg·m², which is not the unit of angular momentum (kg·m²/s).
Why D is wrong: D is the unit of energy (joules), not angular momentum.
For a rigid body rotating about a fixed axis, the angular momentum about that axis is given by:
Answer: C. For a rigid body about a fixed axis, L = Iω, where I is the moment of inertia about that axis and ω is the angular velocity (NCERT Class 11 Physics Chapter 6, page 96).
Why A is wrong: A conflates the point-particle formula (L = mvr sin θ for perpendicular distance) with the rigid-body form. For a rigid body, the correct quantity is I, not mr alone.
Why B is wrong: B equals torque (τ = Iα), not angular momentum.
Why D is wrong: D also equals torque for a point particle at distance r (since I = mr² gives mr²α = Iα = τ), not angular momentum.
A disc of moment of inertia 4.0 kg·m² rotates at 10 rad/s. A concentric disc of moment of inertia 2.0 kg·m², initially at rest, is dropped onto it and they rotate together. No external torque acts. The final angular velocity is:
Answer: D. No external torque → conserve angular momentum. L_i = I₁ω₁ = 4.0 × 10 = 40 kg·m²/s. After coupling, I_total = 4.0 + 2.0 = 6.0 kg·m². ω_f = L_i / I_total = 40/6.0 = 6.67 ≈ 6.7 rad/s (NCERT Class 11 Physics Chapter 6, page 96).
Why A is wrong: A results from incorrectly using ω_f = ω_i × I₁/I_total with I_total = 2 × I₁ = 8.0 kg·m², i.e., doubling I₁ instead of adding the second disc's MOI.
Why B is wrong: B results from inverting the ratio in angular-momentum conservation: ω_f = ω₁ × I_total/I₁ = 10 × 6.0/4.0 = 15 rad/s. Adding a disc increases I, so ω must fall, not rise (trap: inverted inertia ratio).
Why C is wrong: C assumes the angular velocity is unchanged, ignoring that the total moment of inertia increased — this violates L = Iω conservation.
A spinning star collapses under gravity, reducing its radius to half. Assuming no external torque and uniform density, the ratio of its new angular velocity to the original is:
Answer: A. Model the star as a solid sphere: I = (2/5)MR². When R → R/2, I_new = (2/5)M(R/2)² = I/4. Conservation of angular momentum: Iω = (I/4)ω_new → ω_new = 4ω. Ratio = 4. Note: the fraction 2/5 and integer 4 are exact geometric/mathematical factors and do not limit significant figures.
Why B is wrong: B comes from incorrectly assuming I scales linearly with R (I ∝ R), so halving R halves I and doubles ω. In reality I ∝ R², so halving R quarters I.
Why C is wrong: C implies ω is unchanged, which would require I to remain constant — contradicting the radius halving.
Why D is wrong: D would require I to decrease by a factor of 8, which would happen if I ∝ R³. Moment of inertia of a sphere scales as R², not R³.
A skater with arms extended has moment of inertia 4.0 kg·m² and rotates at 2.0 rev/s. She pulls her arms in, reducing her moment of inertia to 1.0 kg·m². Which statement is correct?
Answer: D. No external torque → L is conserved. New ω = 4.0/1.0 × 2.0 = 8.0 rev/s. KE_i = ½ × 4.0 × (2.0)² = 8.0 (arbitrary units). KE_f = ½ × 1.0 × (8.0)² = 32. KE increases by a factor of 4 — the muscles do internal work. This is the classic L vs KE conservation trap (NCERT Class 11 Physics Chapter 6, page 96).
Why A is wrong: A is wrong because KE is NOT conserved when I changes. KE = L²/(2I); since I decreased, KE increased. The skater's muscles do internal work. (trap: L vs KE conservation)
Why B is wrong: B is wrong for the same reason as C — L cannot increase without external torque. Additionally, saying both increase ignores the conservation law entirely.
Why C is wrong: C is wrong because angular momentum cannot increase without external torque. The skater's arm-pulling is an internal action.
Two identical discs, each with moment of inertia I, rotate about the same vertical axis. Disc 1 spins at ω and disc 2 spins at 3ω in the opposite direction. They are brought in contact and stick together. No external torque acts. The final angular velocity of the combined system is:
Answer: B. Take disc 2's direction as positive. L_total = I(3ω) + I(−ω) = 2Iω (positive = disc 2's direction). After coupling: I_total = 2I, so ω_f = 2Iω / 2I = ω in disc 2's direction. Angular momentum is a vector — direction matters.
Why A is wrong: A ignores the sign convention. If you add magnitudes (Iω + 3Iω = 4Iω) and divide by 2I, you get 2ω — but this incorrectly ignores that the angular momenta oppose each other.
Why C is wrong: C applies the same magnitude-addition error as A but correctly identifies the direction as disc 2's. The magnitude is wrong: the opposing angular momenta partially cancel.
Why D is wrong: D would be correct only if the two angular momenta were equal and opposite (both Iω in opposite directions). Since disc 2 has 3ω, the net angular momentum is nonzero.
A turntable of moment of inertia 0.50 kg·m² rotates freely at 4.0 rad/s. A ring of moment of inertia 0.50 kg·m² is dropped concentrically onto it. They reach a common angular velocity. The rotational kinetic energy lost in this process is:
Answer: C. Conserve L: 0.50 × 4.0 = (0.50 + 0.50) × ω_f → ω_f = 2.0 rad/s. KE_i = ½ × 0.50 × 4.0² = 4.0 J. KE_f = ½ × 1.0 × 2.0² = 2.0 J. Energy lost = 4.0 − 2.0 = 2.0 J. The lost energy goes to friction/heat during coupling.
Why A is wrong: A assumes rotational KE is conserved during the coupling. It is not — angular momentum is conserved, but KE is lost to friction between the surfaces. (trap: L vs KE conservation)
Why B is wrong: B results from an arithmetic error such as halving the loss again or miscalculating one of the KE terms.
Why D is wrong: D equals the initial KE (4.0 J), which would mean all energy is lost. The system still rotates at 2.0 rad/s, so KE_f = 2.0 J ≠ 0.
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Pattern: Conservation of angular momentum when moment of inertia changes (PYQ pattern: body with changing I, observed in NEET 2025).
Given
A solid disc of mass M = 2.0 kg and radius R = 0.40 m rotates at ω₁ = 6.0 rad/s about an axis through its centre. A ring of mass m = 1.0 kg and same radius R = 0.40 m, initially at rest, is coaxially placed on the disc. No external torque acts.
Required
Find (a) the final angular velocity, and (b) the change in rotational kinetic energy.
Concept
When net external torque is zero, angular momentum is conserved: L_i = L_f. Rotational KE is NOT conserved here because internal friction does work during coupling.
Formula
L = Iω (conservation: I₁ω₁ = I_total × ω_f)
I_disc = ½MR², I_ring = mR², I_total = I_disc + I_ring
KE = ½Iω²
Substitution
I_disc = ½ × 2.0 × (0.40)² = ½ × 2.0 × 0.16 = 0.16 kg·m²
I_ring = 1.0 × (0.40)² = 0.16 kg·m²
I_total = 0.16 + 0.16 = 0.32 kg·m²
L_i = I_disc × ω₁ = 0.16 × 6.0 = 0.96 kg·m²/s
Calculation
ω_f = L_i / I_total = 0.96 / 0.32 = 3.0 rad/s
KE_i = ½ × 0.16 × 6.0² = ½ × 0.16 × 36 = 2.88 J
KE_f = ½ × 0.32 × 3.0² = ½ × 0.32 × 9.0 = 1.44 J
ΔKE = 2.88 − 1.44 = 1.44 J (lost to friction)
Note: The fractions ½ (in I_disc = ½MR² and KE = ½Iω²) are exact geometric/mathematical constants and do not affect the significant-figure count.
Final answer
(a) ω_f = 3.0 rad/s
(b) Rotational KE lost = 1.44 J ≈ 1.4 J (2 significant figures, matching the given data)
The factor by which KE decreased: KE_f/KE_i = 1.44/2.88 = 0.50. KE halved — consistent with L²/(2I), since I doubled.
Common trap
Conserving KE instead of L would give: ½ × 0.16 × 36 = ½ × 0.32 × ω² → ω = √18 ≈ 4.24 rad/s — wrong. Angular momentum conservation is the correct principle when external torque is zero, regardless of whether energy is lost internally.
Similar NEET-style question
A uniform solid sphere of mass 3.0 kg and radius 0.20 m spins at 10 rad/s. A thin spherical shell of mass 2.0 kg and the same radius, initially at rest, is placed over it coaxially. They rotate together with no external torque. Find the final angular velocity and the fractional loss in rotational KE.
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Angular momentum of a particle about a point is L = r × p, where p is linear momentum. For a rotating rigid body about a fixed axis: L = I ω, where I is moment of inertia and ω is angular velocity. SI unit: kg·m²/s.
-- NCERT Class 11 Physics, Ch. 6, p. 107For 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 |
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