Conservation Angular Momentum

8 MCQs2 revision cards9-step worked example

Source: NCERT System of Particles and Rotational MotionPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 7, page 6).

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.

Practice 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?

MCQ 2Easy RecallPractice

The SI unit of angular momentum is:

MCQ 3Easy RecallPractice

A spinning figure skater pulls her arms inward. Which of the following quantities is conserved during this action?

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?

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:

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?

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:

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.)

Quick recall before you leave

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.) ---

Before solving, remember these

Law

Conservation of angular momentum

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. 6

Formulas

8 formulas — click to collapse

Angular momentum

L = r \times p; L = I ω

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

Centre of mass of n-particle system

R_{c m} = \frac{\sum m _{i} r _{i}}{\sum m _{i}}

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

Valid when

System of point particles or rigid body
Inertial reference frame

Moment of inertia for common rigid bodies

I_{sphere} = \frac{2}{5} M R^{2}; I_{hollow} = \frac{2}{3} M R^{2}; I_{rod,c} = \frac{1}{12} M L^{2}; I_{ring} = M R^{2}; I_{disc} = \frac{1}{2} M R^{2}

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

Valid when

Uniform mass distribution
Rotation about symmetry axis (unless noted)

Parallel axes theorem

I = I_{c m} + M d^{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

Valid when

Both axes parallel
I_cm known about CM axis

Perpendicular axes theorem (planar)

I_{z} = I_{x} + I_{y}

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

Valid when

Body is planar (2D lamina)
All three axes intersect at one point

Rotational kinematic equations (constant alpha)

ω = ω_{0} + α t; θ = ω_{0} t + \frac{1}{2} α t^{2}; ω^{2} = ω_{0}^{2} + 2 α θ

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

Valid when

Constant alpha
Single rotation axis

Rotational kinetic energy

K E_{rot} = \frac{1}{2} I ω^{2}

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

Valid when

Rotation about fixed axis
I and omega about same axis

Torque (moment of force)

τ = r \times F

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

Valid when

Rigid body or extended object
r measured from chosen pivot/axis

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

7 items — click to collapse

Trap

Centre of mass: equal-distance default

Category: Overthinking

Student answers L/2 for two-particle CoM regardless of mass ratio.

When it triggers

Question gives two masses on rigid rod and asks for CoM distance.

How to avoid

R_cm from m1 = m2*L/(m1+m2). Heavier mass pulls CoM closer to it.

Trap

Angular momentum vs rotational KE conservation

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.

Trap

MOI coefficient swap (solid vs hollow)

Category: Similar Terms

Student confuses 2/5 (solid sphere) with 2/3 (hollow sphere) or 1/2 (disc) with 1 (ring).

When it triggers

Question gives a specific geometry and asks for I or radius of gyration.

How to avoid

Memorise: solid sphere 2/5, hollow sphere 2/3, disc/cylinder 1/2, ring/hoop 1, rod-centre 1/12, rod-end 1/3.

Trap

RPM vs rad/s confusion

Category: Unit Conversion

Student plugs rpm directly into formulas requiring rad/s. 1 rpm = 2π/60 rad/s.

When it triggers

Question gives ω in rpm and asks for kinematic quantities in SI units.

How to avoid

Convert: ω(rad/s) = (2π/60) × rpm. Always check units before substituting.

Mistake

Conserves rotational KE when only L is conserved (or vice versa) when MOI changes.

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).

Mistake

Confuses MOI coefficients (e.g. solid sphere 2/5 vs hollow sphere 2/3).

Root cause: concept gap

Correction

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).

Mistake