Angular momentum is defined as the moment of linear momentum about a point or axis (NCERT Class 11 Physics Chapter 7, page 5). 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:
The SI unit of angular momentum is:
For a rigid body rotating about a fixed axis, the angular momentum about that axis is given by:
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:
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:
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?
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:
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:
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. ---
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. 5For 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 |
The position of the centre of mass equals the mass-weighted average of particle positions. For continuous bodies use integral form.
| Symbol | Quantity | SI Unit |
|---|---|---|
| R_cm | CoM position | m |
| m_i | mass of i-th particle | kg |
| r_i | position of i-th particle | m |
Standard moments of inertia about the symmetry axis. For other axes use parallel/perpendicular axes theorems.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | mass | kg |
| R | radius | m |
| L | length | m |
| I | moment of inertia | kg*m^2 |
Moment of inertia about any axis = moment about parallel axis through CM + Md^2.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | MOI about given axis | kg*m^2 |
| I_cm | MOI about parallel CM axis | kg*m^2 |
| M | total mass | kg |
| d | perpendicular distance | m |
For planar lamina: MOI about axis perpendicular to plane = sum of MOI about two perpendicular in-plane axes through same point.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I_z | MOI perp to plane | kg*m^2 |
| I_x, I_y | MOI in plane | kg*m^2 |
Rotational analogues of linear kinematic equations under constant angular acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| omega | angular velocity | rad/s |
| alpha | angular acceleration | rad/s^2 |
| theta | angular displacement | rad |
| t | time | s |
Energy of rotation about an axis. Adds to translational KE for rolling bodies.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
Cross product of position vector and force vector. Magnitude r F sin(theta).
| Symbol | Quantity | SI Unit |
|---|---|---|
| tau | torque | N*m |
| r | position from pivot | m |
| F | force | N |
| theta | angle between r and F | rad |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student answers L/2 for two-particle CoM regardless of mass ratio.
Question gives two masses on rigid rod and asks for CoM distance.
R_cm from m1 = m2*L/(m1+m2). Heavier mass pulls CoM closer to it.
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).
Question describes a body whose moment of inertia changes (skater pulling arms in, star collapsing).
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.
Category: Similar Terms
Student confuses 2/5 (solid sphere) with 2/3 (hollow sphere) or 1/2 (disc) with 1 (ring).
Question gives a specific geometry and asks for I or radius of gyration.
Memorise: solid sphere 2/5, hollow sphere 2/3, disc/cylinder 1/2, ring/hoop 1, rod-centre 1/12, rod-end 1/3.
Category: Unit Conversion
Student plugs rpm directly into formulas requiring rad/s. 1 rpm = 2π/60 rad/s.
Question gives ω in rpm and asks for kinematic quantities in SI units.
Convert: ω(rad/s) = (2π/60) × rpm. Always check units before substituting.
Root cause: concept gap
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).
Root cause: concept gap
Memorise standard moments of inertia. Solid sphere has more mass near axis (smaller MOI = 2MR²/5); hollow sphere has all mass at radius R (larger MOI = 2MR²/3).
Root cause: unit error
Convert: ω(rad/s) = (2π/60) × ω(rpm). For example, 1200 rpm = 1200 × 2π/60 = 125.66 rad/s.
The angular acceleration of a body, moving along the circumference of a circle, is
forgets conversion rpm to rad s
Treats rpm as rad/s without 2*pi/60 conversion
uses equal distribution
Default to L/2 regardless of mass ratio
uses energy conservation instead
Confusing L conservation with KE conservation
swap solid hollow coefficients
Confusing 2/5 (solid sphere) with 2/3 (hollow sphere)
Test yourself on this topic with real past-paper questions:
Practice this topic →Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.