Intuition The one-line ideaAngular momentum L ⃗ \vec{L} L only when no net external torque twists the system . Just like linear momentum is conserved when no external force pushes, angular momentum is conserved when no external torque turns. Internal forces between parts of a system never change the total L ⃗ \vec{L} L
Definition Angular momentumFor a particle: L ⃗ = r ⃗ × p ⃗ \vec{L} = \vec{r} \times \vec{p} L = r × p r ⃗ \vec{r} r p ⃗ = m v ⃗ \vec{p} = m\vec{v} p = m v L = I ω L = I\omega L = I ω I I I moment of inertia and ω \omega ω conserved means L ⃗ \vec{L} L magnitude or direction over time.
We start from the rotational analogue of Newton's second law and derive the condition.
Step 1 — Define L ⃗ \vec{L} L
L ⃗ = r ⃗ × p ⃗ \vec{L} = \vec{r} \times \vec{p} L = r × p Why this step? Conservation means d L ⃗ d t = 0 \dfrac{d\vec{L}}{dt}=0 d t d L = 0
Step 2 — Apply the product rule for cross products.
d L ⃗ d t = d r ⃗ d t × p ⃗ + r ⃗ × d p ⃗ d t \frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt}\times\vec{p} \;+\; \vec{r}\times\frac{d\vec{p}}{dt} d t d L = d t d r × p + r × d t d p Why this step? L ⃗ \vec{L} L
Step 3 — Kill the first term.
d r ⃗ d t × p ⃗ = v ⃗ × ( m v ⃗ ) = m ( v ⃗ × v ⃗ ) = 0 \frac{d\vec{r}}{dt}\times\vec{p} = \vec{v}\times(m\vec{v}) = m(\vec{v}\times\vec{v}) = 0 d t d r × p = v × ( m v ) = m ( v × v ) = 0 Why this step? The cross product of any vector with itself is zero (v ⃗ ∥ v ⃗ \vec{v}\parallel\vec{v} v ∥ v
Step 4 — Identify torque.
d p ⃗ d t = F ⃗ ⇒ r ⃗ × d p ⃗ d t = r ⃗ × F ⃗ = τ ⃗ n e t \frac{d\vec{p}}{dt} = \vec{F} \quad\Rightarrow\quad \vec{r}\times\frac{d\vec{p}}{dt} = \vec{r}\times\vec{F} = \vec{\tau}_{net} d t d p = F ⇒ r × d t d p = r × F = τ n e t
Step 5 — The master equation.
d L ⃗ d t = τ ⃗ e x t , n e t \boxed{\frac{d\vec{L}}{dt} = \vec{\tau}_{ext,\,net}} d t d L = τ e x t , n e t
Why only EXTERNAL torque? For a system of particles, internal forces come in Newton's-third-law pairs acting along the line joining them. Each pair contributes torque r ⃗ 1 × f ⃗ + r ⃗ 2 × ( − f ⃗ ) = ( r ⃗ 1 − r ⃗ 2 ) × f ⃗ \vec{r}_1\times\vec{f} + \vec{r}_2\times(-\vec{f}) = (\vec{r}_1-\vec{r}_2)\times\vec{f} r 1 × f + r 2 × ( − f ) = ( r 1 − r 2 ) × f f ⃗ \vec{f} f along ( r ⃗ 1 − r ⃗ 2 ) (\vec{r}_1-\vec{r}_2) ( r 1 − r 2 ) = 0 = 0 = 0
L ⃗ \vec{L} L
No external force at all ⟹ certainly no external torque ⟹ L ⃗ \vec{L} L External forces exist but their net torque is zero — e.g. forces pass through the origin (central forces , like gravity in planetary orbit) so r ⃗ × F ⃗ = 0 \vec{r}\times\vec{F}=0 r × F = 0 Component-wise: if τ e x t \tau_{ext} τ e x t L L L about that axis is conserved even if other components aren't.
Intuition Why central forces conserveL ⃗ \vec{L} L r ⃗ × F ⃗ \vec{r}\times\vec{F} r × F sideways component of force to twist it — but gravity has none. Hence L ⃗ \vec{L} L is Kepler's 2nd law (equal areas in equal times).
Worked example 1) Spinning skater pulling arms inA skater spins with arms out: I 1 = 6 kg⋅m 2 I_1=6\ \text{kg·m}^2 I 1 = 6 kg⋅m 2 ω 1 = 2 rad/s \omega_1=2\ \text{rad/s} ω 1 = 2 rad/s I 2 = 2 kg⋅m 2 I_2=2\ \text{kg·m}^2 I 2 = 2 kg⋅m 2 ω 2 \omega_2 ω 2
Condition check: the ice is (nearly) frictionless and the pull is an internal muscular force → no external vertical-axis torque → L L L I 1 ω 1 = I 2 ω 2 I_1\omega_1 = I_2\omega_2 I 1 ω 1 = I 2 ω 2 Why this step? τ e x t = 0 \tau_{ext}=0 τ e x t = 0 L = I ω L=I\omega L = I ω ω 2 = 6 × 2 2 = 6 rad/s \omega_2 = \frac{6\times 2}{2} = 6\ \text{rad/s} ω 2 = 2 6 × 2 = 6 rad/s Check KE: K E = 1 2 I ω 2 KE=\tfrac12 I\omega^2 K E = 2 1 I ω 2 12 12 12 36 36 36 muscles did work , so energy is NOT conserved even though L L L
Worked example 2) Mass falling onto a rotating turntableTurntable I t = 0.5 kg⋅m 2 I_t=0.5\ \text{kg·m}^2 I t = 0.5 kg⋅m 2 ω 0 = 10 rad/s \omega_0=10\ \text{rad/s} ω 0 = 10 rad/s m = 0.2 m=0.2 m = 0.2 r = 0.5 r=0.5 r = 0.5
Condition check: gravity acts downward (parallel to axis) → no torque about vertical axis; impact force is internal → L L L I t ω 0 = ( I t + m r 2 ) ω f I_t\omega_0 = (I_t + mr^2)\,\omega_f I t ω 0 = ( I t + m r 2 ) ω f Why this step? New body's moment of inertia adds m r 2 mr^2 m r 2 ω f = 0.5 × 10 0.5 + 0.2 ( 0.5 ) 2 = 5 0.55 ≈ 9.09 rad/s \omega_f = \frac{0.5\times 10}{0.5 + 0.2(0.5)^2} = \frac{5}{0.55} \approx 9.09\ \text{rad/s} ω f = 0.5 + 0.2 ( 0.5 ) 2 0.5 × 10 = 0.55 5 ≈ 9.09 rad/s
Worked example 3) Planet in elliptical orbitAt perihelion the planet is r 1 r_1 r 1 v 1 v_1 v 1 r 2 , v 2 r_2,\,v_2 r 2 , v 2 τ ⃗ = 0 \vec{\tau}=0 τ = 0 L = m v r L=mvr L = m v r v ⃗ ⊥ r ⃗ \vec v\perp\vec r v ⊥ r m v 1 r 1 = m v 2 r 2 ⇒ v 1 r 1 = v 2 r 2 m v_1 r_1 = m v_2 r_2 \;\Rightarrow\; v_1 r_1 = v_2 r_2 m v 1 r 1 = m v 2 r 2 ⇒ v 1 r 1 = v 2 r 2 Why this step? Faster when close, slower when far — exactly Kepler's equal-area law.
Common mistake "If kinetic energy isn't conserved,L L L Why it feels right: we associate conservation laws together. The fix: they have different conditions. L L L τ e x t = 0 \tau_{ext}=0 τ e x t = 0 L L L
Common mistake "Internal forces can change the total angular momentum."Why it feels right: internal forces do change individual parts' motion. The fix: they always come in equal-opposite collinear pairs, so their torques cancel; net L ⃗ \vec L L
L L L Why it feels right: conservation sounds absolute. The fix: it's conserved only about axes where τ e x t = 0 \tau_{ext}=0 τ e x t = 0
Common mistake "Choosing a different origin doesn't matter."Why it feels right: physics shouldn't depend on bookkeeping. The fix: L ⃗ \vec{L} L τ ⃗ \vec\tau τ τ e x t = 0 \tau_{ext}=0 τ e x t = 0 L ⃗ \vec L L
Recall Feynman: explain to a 12-year-oldImagine spinning on a chair holding heavy books with arms stretched out. Pull the books to your chest and you suddenly whirl faster — nobody pushed you! That "amount of spin" (angular momentum) likes to stay the same unless something from outside twists you. Friction or someone grabbing the chair would be an outside twist that changes it. As long as nothing outside twists you, you keep the same spin: arms out = slow and wide, arms in = fast and tight.
Mnemonic Remember the condition"No twist ⟹ no shift." No external torque ⟹ no change in angular momentum. And TIPS : Torque Is the Pace-Setter — torque is the rate-changer of L L L p p p
Cover the answers. State the exact condition. Why don't internal forces matter? Why does the skater speed up but gain energy?
What is the exact condition for angular momentum to be conserved? The net external torque on the system about the chosen axis is zero.
What equation links torque and angular momentum? d L ⃗ d t = τ ⃗ e x t , n e t \frac{d\vec L}{dt}=\vec\tau_{ext,net} d t d L = τ e x t , n e t .
Why don't internal forces change total L ⃗ \vec L L They occur in collinear Newton's-third-law pairs, so their torques cancel:
( r ⃗ 1 − r ⃗ 2 ) × f ⃗ = 0 (\vec r_1-\vec r_2)\times\vec f=0 ( r 1 − r 2 ) × f = 0 .
Why is L ⃗ \vec L L Force passes through the origin, so
τ ⃗ = r ⃗ × F ⃗ = 0 \vec\tau=\vec r\times\vec F=0 τ = r × F = 0 since
r ⃗ ∥ F ⃗ \vec r\parallel\vec F r ∥ F .
Skater pulls arms in: what happens to ω \omega ω ω \omega ω increases (since
I ω I\omega I ω fixed); KE increases because muscles do work — KE is NOT conserved.
Can L L L Yes — only about axes where the external torque component is zero.
Is choosing the origin irrelevant for conservation of L L L No —
L ⃗ \vec L L and
τ ⃗ \vec\tau τ depend on origin; choose one where
τ e x t = 0 \tau_{ext}=0 τ e x t = 0 .
Which Kepler law is a direct consequence of angular momentum conservation? Kepler's 2nd law (equal areas swept in equal times).
Angular momentum L = r x p
Master eq dL/dt = tau ext net
L about an axis conserved
Intuition Hinglish mein samjhoDekho, angular momentum ka funda simple hai: L ⃗ = I ω \vec L = I\omega L = I ω r ⃗ × p ⃗ \vec r \times \vec p r × p net torque na lage. Jaise linear momentum tab conserve hota hai jab koi external force na ho, waise hi angular momentum conserve hota hai jab external torque zero ho. Master equation yaad rakho: d L ⃗ d t = τ ⃗ e x t \frac{d\vec L}{dt} = \vec\tau_{ext} d t d L = τ e x t L L L
Sabse important point: sirf external torque matter karta hai. System ke andar ke forces (jaise skater ke muscles) hamesha Newton ke third law ke pairs mein aate hain aur unke torque cancel ho jaate hain. Isiliye skater jab haath andar kheechti hai, I I I L = I ω L=I\omega L = I ω ω \omega ω L L L
Central force ka case bahut sundar hai. Planet par Sun ki gravity hamesha Sun ki taraf point karti hai, yaani r ⃗ \vec r r F ⃗ \vec F F r ⃗ × F ⃗ = 0 \vec r \times \vec F = 0 r × F = 0 L L L v r v r v r L i = L f L_i = L_f L i = L f
