Prerequisites
Before studying this topic, review:
- Linear Momentum - Conservation principles
- Torque - Relationship with angular momentum
- Moment of Inertia - Rotational inertia
The Hook: Why Does a Figure Skater Spin Faster When Arms Are Pulled In?
Watch any figure skating championship - the skater starts spinning with arms outstretched, rotating slowly. Suddenly, they pull their arms in close to their body and - WHOOSH - they become a blur, spinning at incredible speeds!
No external force acts on them. No motor drives the spin. Yet they accelerate dramatically!
Similarly, in space movies like Interstellar, when spacecraft need to rotate, astronauts redistribute mass. Divers tuck into balls mid-air to complete multiple somersaults. Cats always land on their feet by twisting their bodies.
The mystery: How can you change your rotation speed without any external torque? What invisible quantity is being conserved?
The answer: Angular Momentum - one of nature’s most fundamental conservation laws!
The Core Concept
What is Angular Momentum?
Angular momentum (symbol: $\vec{L}$) is the rotational analog of linear momentum - it measures the “quantity of rotational motion” of a body.
Linear momentum ($\vec{p} = m\vec{v}$) = “quantity of straight-line motion” Angular momentum ($\vec{L} = \vec{r} \times \vec{p}$) = “quantity of rotational motion”
Just as linear momentum depends on mass and velocity, angular momentum depends on moment of inertia and angular velocity!
Definitions of Angular Momentum
For a particle:
$$\boxed{\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v}}$$Magnitude:
$$L = mvr\sin\theta$$where $\theta$ is angle between $\vec{r}$ and $\vec{v}$.
For a rigid body rotating about a fixed axis:
$$\boxed{L = I\omega}$$where:
- $L$ = angular momentum (kg·m²/s or J·s)
- $I$ = moment of inertia (kg·m²)
- $\omega$ = angular velocity (rad/s)
SI Unit: kg·m²/s (also written as J·s)
Direction: Along the axis of rotation (right-hand rule)
Interactive Demo: Visualize Angular Momentum
Explore how angular momentum is conserved in rotating systems.
Memory Tricks & Patterns
Mnemonic for Angular Momentum
Memory Trick: “Linear → p = mv | Rotational → L = Iω”
| Quantity | Linear | Rotational |
|---|---|---|
| “Mass” | $m$ | $I$ |
| Velocity | $v$ | $\omega$ |
| Momentum | $p = mv$ | $L = I\omega$ |
Pattern: Replace $m \to I$, $v \to \omega$, $p \to L$
The “Spin Faster, Pull Closer” Rule
For conservation of angular momentum (no external torque):
$$L = I\omega = \text{constant}$$ $$I_1\omega_1 = I_2\omega_2$$Key insight: $I \propto \frac{1}{\omega}$
- Decrease $I$ (pull mass closer) → Increase $\omega$ (spin faster)
- Increase $I$ (extend mass outward) → Decrease $\omega$ (spin slower)
Examples:
- Figure skater pulls arms in: $I \downarrow$ → $\omega \uparrow$ (spins faster)
- Diver extends body: $I \uparrow$ → $\omega \downarrow$ (slows rotation)
Relation Between Torque and Angular Momentum
Just as $\vec{F} = \frac{d\vec{p}}{dt}$ for linear motion:
$$\boxed{\vec{\tau} = \frac{d\vec{L}}{dt}}$$In words: Torque is the rate of change of angular momentum.
Proof:
For $L = I\omega$:
$$\frac{dL}{dt} = \frac{d(I\omega)}{dt} = I\frac{d\omega}{dt} = I\alpha = \tau$$(assuming $I$ is constant)
Key insight: Just as force changes linear momentum, torque changes angular momentum!
Conservation of Angular Momentum
The Law
If the net external torque on a system is zero, the total angular momentum remains constant.
$$\boxed{\text{If } \vec{\tau}_{ext} = 0, \text{ then } \vec{L} = \text{constant}}$$For a system:
$$L_{initial} = L_{final}$$ $$I_1\omega_1 = I_2\omega_2$$Conservation of angular momentum is one of the most powerful laws in physics!
It explains:
- Why planets orbit the sun in ellipses (Kepler’s laws)
- Why pulsars (collapsed stars) rotate incredibly fast
- How figure skaters control their spin
- Why a cat always lands on its feet
- How helicopters work (need tail rotor to conserve angular momentum)
Remember: Even without external torque, internal forces can redistribute mass, changing $I$ and hence $\omega$!
Conditions for Conservation
- No external torque: $\tau_{ext} = 0$
- Isolated system: No external twisting forces
- Internal forces allowed: Can redistribute mass within system
Note: Internal torques (between parts of system) don’t affect total $L$ - they come in action-reaction pairs and cancel!
Important Cases and Applications
Case 1: Figure Skater / Ice Dancer
Initial: Arms extended, moment of inertia $I_1$, angular velocity $\omega_1$
Final: Arms pulled in, moment of inertia $I_2$ (smaller), angular velocity $\omega_2$
Conservation:
$$I_1\omega_1 = I_2\omega_2$$ $$\omega_2 = \frac{I_1}{I_2}\omega_1$$Since $I_1 > I_2$, we get $\omega_2 > \omega_1$ (spins faster!)
Case 2: Person on Rotating Platform
Setup: Person stands on frictionless rotating platform holding weights
When weights pulled in:
- Total $L$ conserved (no external torque)
- $I$ decreases (mass closer to axis)
- $\omega$ increases to compensate
Application: This is how astronauts change orientation in space!
Case 3: Planet in Elliptical Orbit
Kepler’s Second Law: A planet sweeps equal areas in equal times
Physics: Conservation of angular momentum!
$$L = mvr = \text{constant}$$At perihelion (closest to sun): Small $r$ → large $v$ (fast) At aphelion (farthest from sun): Large $r$ → small $v$ (slow)
Case 4: Collision with Rotating Object
Example: Bullet hits and embeds in rotating disc
Conservation of angular momentum:
$$L_{initial} = L_{final}$$ $$I_{disc}\omega_i + mvr = (I_{disc} + mr^2)\omega_f$$where $v$ is tangential component of bullet velocity.
Case 5: Two Discs Coupling
Setup: Disc 1 (rotating) placed on disc 2 (stationary). Friction causes them to rotate together.
Conservation:
$$I_1\omega_1 + I_2(0) = (I_1 + I_2)\omega_f$$ $$\omega_f = \frac{I_1\omega_1}{I_1 + I_2}$$Note: Kinetic energy is NOT conserved (friction is not conservative) but angular momentum IS!
Angular Momentum for Different Scenarios
1. Particle Moving in Straight Line
For particle of mass $m$, velocity $v$, at perpendicular distance $r$ from axis:
$$L = mvr$$Note: Even straight-line motion has angular momentum about an axis!
2. Particle in Circular Motion
For particle in circle of radius $r$ with speed $v$:
$$L = mvr$$(since $\vec{v}$ is always perpendicular to $\vec{r}$)
3. Rigid Body About Fixed Axis
$$L = I\omega$$This is the most commonly used form!
4. System of Particles
$$\vec{L}_{total} = \sum \vec{L}_i = \sum \vec{r}_i \times m_i\vec{v}_i$$Common Mistakes to Avoid
Mistake: Using wrong formula for the scenario
Truth:
- $L = I\omega$ for rigid body rotating about axis
- $L = mvr$ for particle (or point mass) at distance $r$ with perpendicular velocity $v$
Example: Particle moving in circle:
- Using $L = mvr$: Correct!
- Using $L = I\omega$ with $I = mr^2$: Also correct! (gives same answer)
Mistake: Assuming $KE_i = KE_f$ when $L_i = L_f$
Truth: Angular momentum conservation does NOT imply energy conservation!
Example: Figure skater pulls arms in
- $L$ conserved: $I_1\omega_1 = I_2\omega_2$ ✓
- $KE$ increases: $KE_2 > KE_1$ (skater does work pulling arms in!)
Since $\omega_2 = \frac{I_1\omega_1}{I_2}$:
$$KE_2 = \frac{1}{2}I_2\left(\frac{I_1\omega_1}{I_2}\right)^2 = \frac{I_1^2\omega_1^2}{2I_2} = \frac{I_1}{I_2} \times \frac{I_1\omega_1^2}{2} = \frac{I_1}{I_2}KE_1$$Since $I_1 > I_2$, we get $KE_2 > KE_1$!
Mistake: Applying $L_i = L_f$ when external torque acts
Truth: Conservation requires $\tau_{ext} = 0$
Check for external torques:
- Gravity (if not through axis)
- Friction (if rotating on rough surface)
- Applied forces at a distance from axis
Remember: Internal forces (within system) don’t affect total $L$!
Mistake: Adding angular momenta without considering directions
Truth: Angular momentum is a vector! Use sign convention.
Convention:
- Counterclockwise rotation: $L$ positive (+)
- Clockwise rotation: $L$ negative (−)
Example: Two discs rotating in opposite directions couple together:
$$I_1\omega_1 + I_2(-\omega_2) = (I_1 + I_2)\omega_f$$Worked Examples
Level 1: Foundation (NCERT)
A particle of mass 2 kg moves with velocity 5 m/s along a straight line at perpendicular distance 3 m from an axis. Find its angular momentum about the axis.
Solution:
$$L = mvr = 2 \times 5 \times 3 = \boxed{30 \text{ kg·m}^2\text{/s}}$$A disc of mass 3 kg and radius 0.5 m rotates at 10 rad/s. Find its angular momentum.
Solution:
$I = \frac{MR^2}{2} = \frac{3 \times (0.5)^2}{2} = 0.375$ kg·m²
$$L = I\omega = 0.375 \times 10 = \boxed{3.75 \text{ kg·m}^2\text{/s}}$$A child of moment of inertia 2 kg·m² spins on a platform at 1 rad/s. They pull in their arms, reducing moment of inertia to 1 kg·m². Find the new angular velocity.
Solution:
Conservation of angular momentum:
$$I_1\omega_1 = I_2\omega_2$$ $$2 \times 1 = 1 \times \omega_2$$ $$\boxed{\omega_2 = 2 \text{ rad/s}}$$Angular velocity doubles!
Level 2: JEE Main
A person stands on a frictionless rotating platform with arms outstretched, rotating at 0.5 rad/s. The moment of inertia (person + platform) is 6 kg·m². They hold 2 kg weights at 1 m from axis. When weights are pulled to 0.3 m, find the new angular velocity.
Solution:
Initial moment of inertia:
$$I_1 = 6 + 2 \times 2 \times (1)^2 = 6 + 4 = 10 \text{ kg·m}^2$$(Each weight contributes $mr^2 = 2 \times 1^2 = 2$ kg·m²)
Final moment of inertia:
$$I_2 = 6 + 2 \times 2 \times (0.3)^2 = 6 + 4 \times 0.09 = 6.36 \text{ kg·m}^2$$Conservation:
$$I_1\omega_1 = I_2\omega_2$$ $$10 \times 0.5 = 6.36 \times \omega_2$$ $$\omega_2 = \frac{5}{6.36} = \boxed{0.786 \text{ rad/s}}$$Increase: $\frac{0.786 - 0.5}{0.5} \times 100\% = 57.2\%$ faster!
A disc of mass 5 kg and radius 0.2 m rotates at 20 rad/s. A bullet of mass 0.01 kg moving at 200 m/s tangentially hits the edge and embeds in it. Find the final angular velocity.
Solution:
Initial angular momentum:
Disc: $L_1 = I_{disc}\omega = \frac{MR^2}{2}\omega = \frac{5 \times (0.2)^2}{2} \times 20 = 0.1 \times 20 = 2$ kg·m²/s
Bullet: $L_2 = mvr = 0.01 \times 200 \times 0.2 = 0.4$ kg·m²/s
Total initial: $L_i = 2 + 0.4 = 2.4$ kg·m²/s
Final moment of inertia:
$$I_f = I_{disc} + mr^2 = 0.1 + 0.01 \times (0.2)^2 = 0.1 + 0.0004 = 0.1004 \text{ kg·m}^2$$Conservation:
$$L_i = L_f$$ $$2.4 = 0.1004 \times \omega_f$$ $$\omega_f = \frac{2.4}{0.1004} = \boxed{23.9 \text{ rad/s}}$$Note: $\omega$ increased slightly (bullet added angular momentum in same direction)!
Disc A (moment of inertia 2 kg·m², angular velocity 10 rad/s) is dropped onto disc B (moment of inertia 3 kg·m², stationary). After friction acts, they rotate together. Find: a) Final angular velocity b) Loss in kinetic energy
Solution:
a) Final angular velocity:
Conservation of angular momentum:
$$L_i = L_f$$ $$I_A\omega_A + I_B \times 0 = (I_A + I_B)\omega_f$$ $$2 \times 10 + 0 = (2 + 3)\omega_f$$ $$\omega_f = \frac{20}{5} = \boxed{4 \text{ rad/s}}$$b) Loss in kinetic energy:
Initial KE:
$$KE_i = \frac{1}{2}I_A\omega_A^2 = \frac{1}{2} \times 2 \times 10^2 = 100 \text{ J}$$Final KE:
$$KE_f = \frac{1}{2}(I_A + I_B)\omega_f^2 = \frac{1}{2} \times 5 \times 4^2 = 40 \text{ J}$$Loss:
$$\Delta KE = 100 - 40 = \boxed{60 \text{ J}}$$60% of energy lost to heat/friction! (But angular momentum conserved)
Level 3: JEE Advanced
A uniform rod of mass $M = 2$ kg and length $L = 1$ m is pivoted at one end and hangs vertically at rest. A ball of mass $m = 0.5$ kg moving horizontally at $v = 10$ m/s strikes the rod at distance $d = 0.8$ m from pivot and sticks to it. Find the angular velocity just after collision.
Solution:
Initial angular momentum about pivot:
Rod: $L_{rod} = 0$ (at rest)
Ball: $L_{ball} = mvd = 0.5 \times 10 \times 0.8 = 4$ kg·m²/s
Total: $L_i = 4$ kg·m²/s
Final moment of inertia about pivot:
Rod: $I_{rod} = \frac{ML^2}{3} = \frac{2 \times 1^2}{3} = \frac{2}{3}$ kg·m²
Ball: $I_{ball} = md^2 = 0.5 \times (0.8)^2 = 0.32$ kg·m²
Total: $I_f = \frac{2}{3} + 0.32 = 0.667 + 0.32 = 0.987$ kg·m²
Conservation:
$$L_i = L_f$$ $$4 = 0.987 \times \omega_f$$ $$\omega_f = \frac{4}{0.987} = \boxed{4.05 \text{ rad/s}}$$A man of mass $m = 60$ kg stands at one end of a plank of mass $M = 40$ kg and length $L = 4$ m floating on frictionless ice. The plank is initially at rest. The man walks to the other end. How far does the plank move?
Solution:
Key concept: No external horizontal force → COM of system doesn’t move!
Initial COM position (taking left end as origin, man at left end):
$$x_{cm,i} = \frac{m \times 0 + M \times 2}{m + M} = \frac{40 \times 2}{100} = 0.8 \text{ m}$$Final configuration:
- Let plank move distance $x$ to left
- Man is now at distance $(4 - x)$ from original left end
- Plank center is at distance $(2 - x)$ from original left end
Final COM position:
$$x_{cm,f} = \frac{m(4-x) + M(2-x)}{m+M}$$COM doesn’t move:
$$x_{cm,i} = x_{cm,f}$$ $$0.8 = \frac{60(4-x) + 40(2-x)}{100}$$ $$80 = 240 - 60x + 80 - 40x$$ $$80 = 320 - 100x$$ $$100x = 240$$ $$\boxed{x = 2.4 \text{ m}}$$The plank moves 2.4 m in the opposite direction!
Check: Man walks 4 m relative to plank, but only $4 - 2.4 = 1.6$ m relative to ice.
A satellite of mass $m$ orbits Earth at radius $r_1$ with speed $v_1$. At one point, it fires thrusters tangentially to change orbit to radius $r_2$ (larger ellipse). Find the speed at radius $r_2$.
Solution:
Conservation of angular momentum (thruster fires tangentially, so no torque about Earth’s center):
$$L_1 = L_2$$ $$mv_1r_1 = mv_2r_2$$ $$v_2 = \frac{r_1}{r_2}v_1$$ $$\boxed{v_2 = \frac{r_1}{r_2}v_1}$$If $r_2 > r_1$: $v_2 < v_1$ (slower at larger radius)
Example: $r_1 = 7000$ km, $v_1 = 7.5$ km/s, $r_2 = 10000$ km
$$v_2 = \frac{7000}{10000} \times 7.5 = \boxed{5.25 \text{ km/s}}$$Note: This is Kepler’s Second Law in action!
Comparison: Linear vs Angular Momentum
| Concept | Linear | Angular |
|---|---|---|
| Momentum | $\vec{p} = m\vec{v}$ | $\vec{L} = I\vec{\omega}$ or $\vec{r} \times \vec{p}$ |
| Newton’s 2nd Law | $\vec{F} = \frac{d\vec{p}}{dt}$ | $\vec{\tau} = \frac{d\vec{L}}{dt}$ |
| Conservation | If $\vec{F}_{ext} = 0$, then $\vec{p}$ = const | If $\vec{\tau}_{ext} = 0$, then $\vec{L}$ = const |
| Collision | $\sum p_i = \sum p_f$ | $\sum L_i = \sum L_f$ |
| Unit | kg·m/s | kg·m²/s |
Real-World Applications
Figure Skating Spins: Pull arms in → decrease $I$ → increase $\omega$ (conservation)
Helicopter Design: Main rotor creates angular momentum → need tail rotor to cancel it (conservation)
Pulsars (Neutron Stars): Collapsed stars spin incredibly fast (small $I$, large $\omega$ to conserve $L$)
Cat Righting Reflex: Cats change $I$ of front/back independently to flip while conserving total $L = 0$
Bicycle Stability: Spinning wheels have angular momentum → resist tilting (gyroscopic effect)
Planetary Orbits: Planets move faster when closer to sun (Kepler’s 2nd law = conservation of $L$)
Olympic Diving: Tucked position → small $I$ → fast spin; extended → large $I$ → slow spin
Quick Revision Box
| Concept | Formula | Key Point |
|---|---|---|
| Angular momentum (particle) | $\vec{L} = \vec{r} \times \vec{p}$ | “Rotational momentum” |
| Angular momentum (rigid body) | $L = I\omega$ | Like $p = mv$ |
| Torque relation | $\vec{\tau} = \frac{d\vec{L}}{dt}$ | Like $\vec{F} = \frac{d\vec{p}}{dt}$ |
| Conservation | If $\tau_{ext} = 0$: $L$ = const | $I_1\omega_1 = I_2\omega_2$ |
| Direction | Right-hand rule | Along rotation axis |
| Unit | kg·m²/s or J·s | - |
Memory tricks:
- “Spin faster → Pull closer” (decrease $I$ → increase $\omega$)
- “No external twist → $L$ conserved”
- Angular momentum is rotational momentum (just like torque is rotational force)
Decision Tree: Solving Angular Momentum Problems
Step 1: Identify if angular momentum is conserved
- Check: Is $\tau_{ext} = 0$?
- Yes → Use $L_i = L_f$
- No → Use $\tau = \frac{dL}{dt}$
Step 2: Calculate initial angular momentum
- Rigid body: $L = I\omega$
- Particle: $L = mvr$ (perpendicular distance)
- System: $L_{total} = \sum L_i$
Step 3: Identify changes
- Does $I$ change? (mass redistribution)
- Does $\omega$ change?
- Are objects coupling/separating?
Step 4: Apply conservation (if applicable)
$$I_1\omega_1 = I_2\omega_2$$Step 5: Solve for unknown ($\omega_2$, $I_2$, etc.)
Pro tip: Check if energy is conserved separately - usually it’s NOT when $L$ is conserved (except elastic collisions)!
Practice Problems
Foundation Level
A disc of radius 0.4 m and mass 2 kg rotates at 15 rad/s. Find its angular momentum.
A particle of mass 0.5 kg moves at 8 m/s at perpendicular distance 2 m from an axis. Find $L$ about the axis.
A rotating platform has $I = 5$ kg·m² and $\omega = 2$ rad/s. When a person sits on it, $I$ becomes 8 kg·m². Find new $\omega$.
JEE Main Level
A solid sphere of mass 3 kg and radius 0.1 m rotates at 50 rad/s. A thin spherical shell of same mass and radius (initially at rest) is dropped on it. Find final angular velocity when they rotate together.
A man stands at the center of a rotating platform (total $I = 4$ kg·m², $\omega = 3$ rad/s). He walks to edge 1 m away. If man’s mass is 60 kg, find new $\omega$.
A bullet of mass 20 g moving at 500 m/s hits a disc (mass 5 kg, radius 0.5 m, initially at rest) tangentially and embeds. Find: a) Final angular velocity b) Fraction of KE lost
JEE Advanced Level
A uniform rod of mass $M$ and length $L$ is hinged at one end. A particle of mass $m$ moving horizontally at speed $v$ strikes the free end and sticks. Find angular velocity just after collision.
Two discs A (mass $M$, radius $R$) and B (mass $2M$, radius $R/2$) rotate in opposite directions at $\omega_0$ and $2\omega_0$ respectively. They are brought into contact (coaxially). Find final angular velocities.
A satellite in circular orbit at height $h$ above Earth (radius $R$) fires thrusters to enter elliptical orbit with apogee at $2h$. Find the ratio of speeds at height $h$ before and after.
Connection to Other Topics
Prerequisites you should know:
- Linear Momentum - $L$ is rotational analog of $p$
- Torque - $\tau = dL/dt$ relationship
- Moment of Inertia - Needed for $L = I\omega$
What’s next:
- Rolling Motion - Uses both linear and angular momentum
- Gravitation - Planetary motion and Kepler’s laws
- Rotational Dynamics - Complete rotational motion analysis
Cross-chapter links:
- Conservation Laws - $L$ conservation is fundamental
- Collisions - Angular momentum in rotational collisions
- Circular Motion - $L = mvr$ for circular paths
- Satellites - Orbital angular momentum
Teacher’s Summary
Angular momentum is the “rotational analog of linear momentum” - just as $p = mv$ measures linear motion, $L = I\omega$ measures rotational motion
Two forms of angular momentum:
- Particle: $L = mvr\sin\theta$ or $\vec{L} = \vec{r} \times \vec{p}$
- Rigid body: $L = I\omega$ (most common in JEE)
Conservation is POWERFUL: If $\tau_{ext} = 0$, then $L_i = L_f$
- Explains skater spins, planetary motion, gyroscopes, helicopter tail rotors
- $I \downarrow$ → $\omega \uparrow$ (inverse relationship when $L$ conserved)
Torque-angular momentum relation: $\tau = \frac{dL}{dt}$ (like $F = \frac{dp}{dt}$)
Conservation ≠ Energy conservation:
- $L$ can be conserved while $KE$ changes (figure skater does work!)
- Always check separately for energy conservation
JEE loves:
- Figure skater / person on platform problems
- Disc coupling problems
- Bullet hitting rotating disc
- Collision problems with rotation
Direction matters: Use right-hand rule, watch for clockwise vs counterclockwise
“Linear momentum is to straight-line motion what angular momentum is to spinning motion - and both are conserved in their respective domains!”
This is a high-yield topic - 2-4 questions (8-16 marks) in JEE directly test angular momentum conservation. Master the skater/platform setup, disc coupling, and collision problems - they repeat every year!