Visual walkthrough — Conservation of angular momentum — conditions

1.5.12 · D2 · Physics › Rotational Mechanics › Conservation of angular momentum — conditions

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Step 0 — Do arrows jinhe hum use kar sakte hain

Kisi bhi formula se pehle, humein agree karna hoga ki symbols kaunsi pictures represent karte hain.

Figure dekho. Amber dot pivot (origin) hai. Cyan arrow $\vec{r}$ object tak pahunchta hai; white arrow $\vec{p}$ dikhata hai woh kahan ja raha hai. Abhi tak kuch aur exist nahi karta — na torque, na cross product. Woh hum aage earn karenge.

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Step 1 — "Spin ki matra" ka matlab kya hai? $\vec{L}$ build karo

KYA. Hum chahte hain ek number-with-direction jo capture kare "yeh object pivot ke around kitna circle kar raha hai". Ise angular momentum $\vec{L}$ kehte hain.

Cross product kyun? Ek moving object tab hi pivot ke around circle karta hai jab uski motion ka kuch hissa $\vec{r}$ ke sideways ho. Agar woh straight pivot ki taraf (ya straight door) move kare, toh woh orbit hi nahi kar raha. Humein ek tool chahiye jo:

• zero de jab $\vec{r}$ aur $\vec{p}$ parallel hon (pure in/out motion, koi circling nahi),
• maximum de jab woh perpendicular hon (pure sideways motion, fastest circling).

Cross product $\vec{r}\times \vec{p}$ exactly yahi karta hai. Iski size hai

$|\vec{L}|=\underset{\text{kitna door hai}}{\underbrace{|\vec{r}|}}\underset{\text{kitni tezi se move kar raha hai}}{\underbrace{|\vec{p}|}}\underset{\begin{array}{c}\text{motion kitni}\\ \text{sideways hai}\end{array}}{\underbrace{\sin \theta }}$

$\theta$ do arrows ke beech ka angle hai. Jab woh line up karen, $\sin 0=0$ (koi spin nahi). Jab perpendicular hon, $\sin 90^{\circ }=1$ (full spin). Woh $\sin \theta$ hi reason hai ki cross product sahi tool hai — yeh automatically sideways part extract karta hai.

Figure dekho. Shaded parallelogram ka area equals $|\vec{r}||\vec{p}|\sin \theta$ — woh area hi $|\vec{L}|$ hai. $\vec{L}$ ki direction (woh chota curved amber arrow) circling ka sense hai: yahan counter-clockwise.

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Step 2 — Conservation ka matlab "change nahi hota" — toh time ke saath dekho

KYA. "Conserved" matlab $\vec{L}$ stable rehta hai jab clock tick karta hai. Change ki language mein, rate of change zero hoti hai.

Derivative kyun? Tool jo measure karta hai "koi cheez har instant kitni tezi se badal rahi hai" woh derivative $\frac{d}{dt}$ hai. "Kya $\vec{L}$ constant hai?" poochna exactly "kya $\frac{d\vec{L}}{dt}=0$ hai?" poochne jaisa hai. Toh hum $\vec{L}$ differentiate karte hain aur dekhte hain kya ise vanish karta hai.

Figure dekho. Do frozen snapshots ek tiny time $dt$ apart. Faint arrow $\vec{L}$ hai time $t$ par, bright wala $t+dt$ par. Unke tips ke beech ka amber gap $d\vec{L}$ hai — woh change jo humein zero tak chase karna hai. Flat line matlab conserved.

$\frac{d\vec{L}}{dt}=\frac{d}{dt}(\vec{r}\times \vec{p})$

Yahan $\frac{d\vec{L}}{dt}$ = spin-arrow ka drift rate har second. Agar yeh zero hai, $\vec{L}$ frozen hai — conserved.

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Step 3 — Product rule se change ko split karo

KYA. $\vec{L}$ do cheezon ka product hai jo dono change hoti hain: $\vec{r}$ (object move karta hai, toh position shift hoti hai) aur $\vec{p}$ (velocity change ho sakti hai). Toh product ke change ke do sources hain.

Product rule kyun? Jab bhi koi quantity $A\times B$ ho aur dono $A,B$ move karen, total change hota hai "change in $A$, times old $B$" plus "old $A$, times change in $B$". Yeh product rule hai, aur yeh cross products ke liye bhi hold karta hai (order maintain karo, kyunki $\times$ commutative nahi hai).

$\frac{d\vec{L}}{dt}=\underset{\text{term A: }\vec{r}\text{ change ho raha hai}}{\underbrace{\frac{d\vec{r}}{dt}\times \vec{p}}}+\underset{\text{term B: }\vec{p}\text{ change ho raha hai}}{\underbrace{\vec{r}\times \frac{d\vec{p}}{dt}}}$

• $\frac{d\vec{r}}{dt}$ = position arrow kitni tezi se move karta hai = velocity $\vec{v}$.
• $\frac{d\vec{p}}{dt}$ = momentum arrow kitni tezi se change hota hai.

Figure dekho. Do panels: Term A $\vec{r}$ arrow (cyan) ko wiggle karta hai $\vec{p}$ hold karke; Term B $\vec{p}$ arrow (white) ko wiggle karta hai $\vec{r}$ hold karke. Unka combined effect $\vec{L}$ ka full change hai.

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Step 4 — Term A khatam ho jaata hai: koi vector apne aap ke around circle nahi kar sakta

KYA. Term A mein, $\frac{d\vec{r}}{dt}$ ko $\vec{v}$ se replace karo. Lekin $\vec{p}=m\vec{v}$, toh term A hai $\vec{v}\times m\vec{v}$.

Yeh zero kyun hai. Cross product do arrows ke beech ka sideways angle measure karta hai. Lekin $\vec{v}$ aur $m\vec{v}$ same direction mein point karte hain — unke beech angle $0$ hai, toh $\sin 0=0$.

$\frac{d\vec{r}}{dt}\times \vec{p}=\vec{v}\times (m\vec{v})=m\underset{=0}{\underbrace{(\vec{v}\times \vec{v})}}=0$

Figure dekho. Do arrows ek doosre ke upar flat pade hain. Woh jo "parallelogram" enclose karte, woh squash hokar ek line ban gaya — zero area — toh yeh term kuch contribute nahi karta. Term A gone.

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Step 5 — Term B torque ban jaata hai: survivor ko naam do

KYA. Term B mein, $\frac{d\vec{p}}{dt}$ woh rate hai jis par momentum change hota hai. Newton's second law kehta hai woh rate force hai: $\frac{d\vec{p}}{dt}=\vec{F}$. Toh term B hai $\vec{r}\times \vec{F}$.

Ise naam kyun den? $\vec{r}\times \vec{F}$ rotational push hai — ek force object ko pivot ke about kitna twist karta hai. Ise torque $\vec{\tau }$ kehte hain.

$\vec{r}\times \frac{d\vec{p}}{dt}=\vec{r}\times \vec{F}=\vec{\tau }$

Iski size hai $|\vec{\tau }|=|\vec{r}||\vec{F}|\sin \varphi$, jahan $\varphi$ $\vec{r}$ aur force ke beech angle hai. Bilkul $\vec{L}$ ki tarah, torque tab hi exist karta hai jab force ka sideways component ho — pivot ki taraf/door straight pull kuch twist nahi karta.

Figure dekho. White force arrow split hota hai ek cyan sideways part mein (twists — torque banata hai) aur ek grey along-$\vec{r}$ part mein (sirf in/out push karta hai — koi twist nahi). Sirf sideways part $\vec{\tau }$ mein survive karta hai.

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Step 6 — Master equation

Surviving pieces ko saath rakho: term A $=0$, term B $=\vec{\tau }$.

Conservation ke roop mein padhna: ${\vec{\tau }}_{ext,net}=0\Rightarrow \frac{d\vec{L}}{dt}=0\Rightarrow \vec{L}=\text{constant}$

Figure dekho. Ek balance beam: torque $\vec{\tau }$ ek pan mein $\vec{L}$ ki "drift speed" set karta hai doosre pan mein. Torque pan empty ⟹ $\vec{L}$ pointer pin ho jaata hai — frozen — conserved. Torque aur uske translational twin Conservation of linear momentum se compare karo.

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Step 7 — Sirf EXTERNAL torque kyun? (internal pairs cancel ho jaate hain)

KYA. Ek real system mein bahut saare particles ek doosre par pull karte hain. Kya woh internal pulls total $\vec{L}$ change karte hain? Hum check karte hain.

Woh cancel kyun hote hain. Newton's third law ke according, particle 1, particle 2 par $\vec{f}$ se pull karta hai aur particle 2, particle 1 par $-\vec{f}$ se — equal aur opposite, unhe join karne wali line ke along. Unke do torques same pivot ke baare mein add karo:

${\vec{\tau }}_{int}={\vec{r}}_1\times \vec{f}+{\vec{r}}_2\times (-\vec{f})=\underset{\text{joining line ke along}}{\underbrace{({\vec{r}}_1-{\vec{r}}_2)}}\times \vec{f}$

Lekin $\vec{f}$ bhi usi joining line ke along hai (yahi Central forces ka matlab hai). Do parallel arrows ⟹ cross product $=0$.

${\vec{\tau }}_{int}=0$

Figure dekho. Do internal forces (amber) exactly dashed line par lie karte hain jo particles ko link karti hai, toh $({\vec{r}}_1-{\vec{r}}_2)$ aur $\vec{f}$ parallel hain — parallelogram collapse ho jaata hai, zero torque. Sirf bahar se aane wale arrows total ko twist kar sakte hain. Dekho Moment of inertia ke liye ki kitne particles $L=I\omega$ mein combine hote hain.

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Step 8 — Edge & degenerate cases (kabhi hanging nahi chhodte)

Figure dekho. Ek planet ki char orbit positions (quadrants I–IV). Har ek par force arrow $\vec{r}$ line par lie karta hai — twist har jagah zero hai. Chahe close-and-fast ho ya far-and-slow, $\vec{L}$ kabhi nahi hilta.

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Ek-picture summary

Ek blueprint par poori derivation: $\vec{L}=\vec{r}\times \vec{p}$ se shuru karo → differentiate karo → product rule do mein split karta hai → term A khatam hota hai (parallel vectors) → term B torque hai → $\frac{d\vec{L}}{dt}={\vec{\tau }}_{ext}$ → torque zero set karo → $\vec{L}$ frozen. Internal pairs raaste mein cancel ho jaate hain, toh sirf outside twists matter karte hain.

Recall Feynman: plain words mein poora walk

Humne ek arrow draw kiya pivot se ek moving object tak ($\vec{r}$) aur ek arrow uski motion ke liye ($\vec{p}$). "Pivot ke around spin" ($\vec{L}$) sabse bada hota hai jab motion sideways ho aur zero hota hai jab straight in ya out ho — isliye humne sideways-measuring cross product use kiya. "Kya spin stable rehta hai?" poochne ke liye humne dekha yeh har instant kitni tezi se change hota hai. Woh change do pieces mein aaya. Pehla piece ne object se kaha apni khud ki velocity ke around circle karo — impossible, toh woh vanish ho gaya. Doosra piece ek force ki twist thi, jise humne torque naam diya. Toh spin ka change hai twist: $\frac{d\vec{L}}{dt}=\vec{\tau }$. Bahar se koi twist nahi ⟹ spin freeze ho jaata hai ⟹ conserved. Aur particles jo ek doosre par pulls lagate hain woh hamesha pairs mein cancel hote hain, toh sirf bahar se aane wale twists system ke liye matter karte hain. Woh ek sentence mein skater ka speed up karna, planet ka equal areas sweep karna, aur turntable par putty — yeh sab hain.

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Connections

• Parent · full conditions & worked examples
• Torque — master equation ka right-hand side.
• Moment of inertia — kaafi saare particles ka $\vec{L}$ kaise $L=I\omega$ banta hai.
• Newton's third law — internal torques cancel kyun hote hain (Step 7).
• Central forces — parallel-arrow zero-torque case (Step 8).
• Kepler's laws — constant $\vec{L}$ se equal areas.
• Conservation of linear momentum — is poori story ka translational twin.
• Rotational kinetic energy — conserved $L$ ka matlab conserved KE nahi hai.