Visual walkthrough — Torque = dL - dt

1.5.11 · D2 · Physics › Rotational Mechanics › Torque = dL - dt

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Step 0 — Teen arrows jo hum use kar sakte hain

Kisi bhi formula se pehle, chaliye characters ko pictures ki tarah fix karte hain. Is page par sirf teen arrows aur ek point se sab kuch bana hai.

Arrows ke beech wala chhota "$\times$" hum Step 2 mein milenge — iska use tab tak nahi karenge jab tak humne draw nahi kar liya ki iska matlab kya hai. (Agar cross products naye hain, toh Cross Product dekho.)

Recall Cast check karo

Pivot se kaun sa arrow shuru hota hai? ::: $\vec{r}$, position arrow — yeh hamesha $O$ se particle tak jaata hai. $\vec{p}$ aur $\vec{v}$ pictures mein kaise alag hain? ::: Same direction; $\vec{p}$ bas $\vec{v}$ ko mass $m$ se scale karke lamba (ya chhota) kiya gaya hai. $\vec{r}\times \vec{p}$ kis taraf point karta hai jab dono board mein hon? ::: Board se bahar (right-hand rule, $\vec{r}$ ko $\vec{p}$ mein curl karo).

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Step 1 — "Angular momentum" $\vec{L}$ kaisa dikhta hai

Hum chahte hain ek single number-with-direction jo capture kare ki yeh moving particle $O$ ke around kitna circle kar raha hai. Do cheezein matter karni chahiye: yeh kitna fast move karta hai ($\vec{p}$) aur yeh kitna side se swing karta hai ($\vec{r}$). Woh tool jo do arrows ko "kitna sideways sweep karte hain" mein combine karta hai, woh hai cross product — isliye yeh enter karta hai, aur koi aur product kaam nahi karta (plain multiply se direction kho jaati hai; dot product alignment measure karta hai, jo exactly ulta hai jo hum chahte hain).

$\vec{L}=\underset{\text{from }O\text{ to particle}}{\underbrace{\vec{r}}}\times \underset{\text{momentum }m\vec{v}}{\underbrace{\vec{p}}}$

KYA: humne $\vec{L}$ ko position aur momentum ke cross product ke roop mein define kiya. KYUN: cross product ka $\sin {\theta }_{rp}$ automatically woh part throw kar deta hai jo motion $O$ ki taraf seedha aim hai (woh part kuch bhi circle nahi kar sakta) aur sirf sideways sweep keep karta hai. PICTURE: s02 mein shaded parallelogram — iska area exactly $|\vec{L}|$ hai. Wider swing ⇒ bada area ⇒ zyada angular momentum. Green dot with ⊙ dikhata hai ki $\vec{L}$ board se bahar point kar raha hai Step 0 ke right-hand rule se. Is arrow ki poori kahani ke liye Angular Momentum dekho.

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Step 2 — $\vec{L}$ ko change hote dekho: picture ko differentiate karo

Angular momentum tab badalta hai jab picture badlati hai. Dono arrows $\vec{r}$ aur $\vec{p}$ time ke saath move ho sakte hain, isliye $\vec{L}$ ke rate of change ko dono ke move hone ka account karna hoga. "Rate of change" ka tool hai derivative $d/dt$ — isliye yeh enter karta hai: hum pooch rahe hain "parallelogram per second kitni tezi se morph ho raha hai?"

$\frac{d\vec{L}}{dt}=\underset{\text{term A: }\vec{r}\text{ moved}}{\underbrace{\frac{d\vec{r}}{dt}\times \vec{p}}}+\underset{\text{term B: }\vec{p}\text{ changed}}{\underbrace{\vec{r}\times \frac{d\vec{p}}{dt}}}$

• $\frac{d\vec{r}}{dt}$ — position arrow per second kitna move karta hai. Lekin woh velocity hi hai: $\frac{d\vec{r}}{dt}=\vec{v}$.
• $\frac{d\vec{p}}{dt}$ — momentum arrow per second kitna change karta hai (isse hum Step 4 mein deal karte hain).

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Step 3 — Term A vanish ho jaata hai (self-cross-product picture)

Term A ko dhyan se dekho. Humne abhi kaha $d\vec{r}/dt=\vec{v}$, aur $\vec{p}=m\vec{v}$. Toh term A hai $\vec{v}\times m\vec{v}$ — ek arrow ka cross product apni hi scaled copy se.

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

KYA: humne term A delete kar diya. KYUN: velocity aur momentum hamesha same direction mein point karte hain, isliye yeh kabhi ek doosre ke "against sweep" nahi kar sakte. PICTURE: s04 mein flattened, area-zero parallelogram.

Sirf term B bachta hai: $\frac{d\vec{L}}{dt}=\vec{r}\times \frac{d\vec{p}}{dt}$

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Step 4 — Newton plug in karo: $d\vec{p}/dt$ force hai

Term B mein abhi bhi $d\vec{p}/dt$ chhupa hua hai. Momentum ka rate of change kya hai? Exactly yahi Newton's Second Law humein batata hai:

$\frac{d\vec{p}}{dt}=\vec{F}$

Ise seedha picture mein substitute karo:

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

• $\vec{F}$ — particle par push (momentum arrow ka rate of change).
• $\vec{r}\times \vec{F}$ — position crossed with force: yeh ek naya parallelogram hai, lever arm aur push se bana.

KYA: humne $d\vec{p}/dt$ ko $\vec{F}$ se replace kiya. KYUN: Newton's law woh ek fact hai jo "momentum kaise change hota hai" ko "acting force" mein turn karta hai ($m$ constant rakh kar). PICTURE: s05 parallelogram ko $\vec{p}$ ki jagah $\vec{F}$ ko second edge ke roop mein redraw karta hai.

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Step 5 — Naye arrow ko naam do: woh hai torque $\vec{\tau }$

$\vec{r}\times \vec{F}$ quantity apna khud ka naam deserve karti hai kyunki yeh answer karti hai "yeh force particle ko $O$ ke around kitna twist karti hai?" Hum ise torque kehte hain.

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

PICTURE: s06 $\vec{F}$ ko ek pink piece mein split karta hai $\vec{r}$ ke along (spinning ke liye useless) aur ek blue piece $\vec{r}$ ke perpendicular (sirf yahi piece twist karta hai). $\sin {\theta }_{rF}$ sirf blue piece rakhta hai.

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Step 6 — Edge cases: kya law har scenario mein survive karta hai?

Ek law jis par trust kiya ja sake use weird inputs handle karne chahiye. Yahan sab hain, har ek picture ke roop mein.

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Step 7 — Ek particle se poore rigid body tak

Real objects mein billions of particles hote hain. Har ek ka angular momentum add karo: ${\vec{L}}_{\text{tot}}={\sum }_i{\vec{L}}_i$, isliye $\frac{d{\vec{L}}_{\text{tot}}}{dt}={\sum }_i{\vec{\tau }}_i$.

Ek fixed axis ke around spin karte rigid body ke liye, $\vec{L}=I\vec{\omega }$ ($I$ moment of inertia hai aur $\vec{\omega }$ angular velocity). Agar $I$ constant hai: ${\vec{\tau }}_{\text{ext}}=\frac{d(I\vec{\omega })}{dt}=I\frac{d\vec{\omega }}{dt}=I\vec{\alpha }$

Yeh special case hai $\tau = I\alpha$. Jab $I$ change hota hai (ek skater arms andar kheenchti hai), tumhe poora $d\vec{L}/dt$ rakhna hoga — yahi door hai Conservation of Angular Momentum ki taraf.

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Ek-picture summary

Yeh single figure saat steps compress karti hai: position arrow $\vec{r}$, momentum arrow $\vec{p}$ jo $\vec{L}$-parallelogram sweep karta hai, force $\vec{F}$ jo use twist karti hai, deleted self-cross-product, aur final boxed law $\vec{\tau }=d\vec{L}/dt$.

Recall Feynman: poora walkthrough plain words mein retell karo

Ek dot choose karo dekhne ke liye — use $O$ kaho. Us dot se ek chhoti si moving ball tak arrow draw karo; woh hai $\vec{r}$. Draw karo ball kahan ja rahi hai; use uske heft se scale karo; woh hai $\vec{p}$. Dono arrows ko ek slanted box mein glue karo — us box ka area ball ka "circling-amount" hai, uska angular momentum $\vec{L}$, aur woh kis direction point karta hai (board se bahar ya andar) right-hand rule se milta hai. Ab time tick hone do. Box do tarike se change ho sakta hai: position edge slide karti hai, ya momentum edge stretch karti hai. Sliding edge useless hai — woh momentum ke saath line up karti hai, toh box squash hokar kuch nahi ho jaata. Sirf stretching edge matter karti hai, aur momentum ko kya stretch karta hai woh hai ek push — ek force (jab tak ball ki mass nahi badlti). Position-arrow crossed with that push ek brand-new box hai jise hum torque naam dete hain: the twist. Toh poori cheez ek aise sentence mein collapse ho jaati hai jo tum dekh sako: tum jo twist apply karte ho woh equals karti hai circling-amount kitni tezi se grow karta hai. $\vec{\tau }=d\vec{L}/dt$. Straight-line ball with no push? Box kabhi nahi badlta — angular momentum still baitha rehta hai, chahe kuch bhi circle mein na ja raha ho.

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Active Recall

Recall Khud se rebuild karo

$\vec{L}$ ke derivative mein do terms kyun hain? ::: Product rule: dono $\vec{r}$ aur $\vec{p}$ time par depend karte hain, isliye har ek ek term contribute karta hai (parallelogram ki dono edges mein se har ek change ho sakti hai). Kaun sa term vanish hota hai aur kyun (picture ke roop mein)? ::: $\frac{d\vec{r}}{dt}\times \vec{p}=\vec{v}\times m\vec{v}$; parallel arrows, $\sin 0=0$, flattened zero-area box. $d\vec{p}/dt$ ki jagah kya aata hai, aur kis law se (aur kya hidden assumption hai)? ::: $\vec{F}$, Newton's second law $\vec{F}=d\vec{p}/dt$ se, jo is form mein tabhi valid hai jab mass $m$ constant ho. $\vec{\alpha }$ kya hai aur $\tau =I\alpha$ kaise appear hota hai? ::: $\vec{\alpha }=d\vec{\omega }/dt$, angular acceleration; $I$ constant hone par, $d(I\vec{\omega })/dt=Id\vec{\omega }/dt=I\vec{\alpha }$. Ek particle straight line mein drift karta hai, koi force nahi, $O$ ke 3 m side se guzarta hai $v=4$ m/s, $m=2$ kg par. Uska $\vec{L}$? ::: $L=mvd=2\cdot 4\cdot 3=24$ kg·m²/s, constant. $\vec{\tau }=\vec{r}\times \vec{F}$ zero kab hota hai real force hone ke bawajood? ::: Jab $\vec{F}$, $\vec{r}$ ke parallel ho (${\theta }_{rF}=0$), ek central push seedha $O$ ki taraf (ya usse).