Angular momentum L = Iω (fixed axis), L = r × p (general): Visual walkthrough

1.5.10 · D2 · Physics › Rotational Mechanics › Angular momentum L = Iω (fixed axis), L = r × p (general)

Step 1 — "Position" aur "momentum" actually dikhte kaise hain

KYA HAI. Kisi bhi spinning se pehle, do arrows rakh do. Space mein ek dot choose karo aur use origin $O$ kaho — yeh aapki choice hai ki "woh center jahan se main sab kuch measure karunga." Ek moving ball kahin baithe hai; $O$ se seedha ball tak arrow kheecho. Woh arrow hai position $r$ . Ab ball par ek doosra arrow kheecho, jidhar woh ja rahi hai, jitna fast utna lamba — woh hai momentum $p = m v$ ( $m$ = mass in kg, $v$ = velocity in m/s).

KYUN. Angular momentum ek story hai in do arrows aur unke beech ke angle ki. Aur kuch nahi. Isliye pehle inhe clearly dekhna zaroori hai.

PICTURE. Magenta arrow hai $r$ ( $O$ se ball tak). Orange arrow hai $p$ (travel ki direction). Angle $θ$ (violet) unke beech — yahi poora khel hai.

Figure — Angular momentum L = Iω (fixed axis), L = r × p (general)

Step 2 — Motion ka kaun sa hissa actually "ghoomta" hai?

KYA HAI. Orange momentum arrow ko do aise arrows mein split karo jo milke usse bana saken: ek $r$ ke along point karta hai (seedha $O$ se door ya pass), aur ek $r$ ke perpendicular (sideways, $O$ ke around curl karta hua). Us split par basic right-triangle trig lagate hain: $p_{∥} = p cos θ, p_{⊥} = p sin θ .$

$p_{∥} = p cos θ$ — $r$ ke along wala piece. Jab $θ = 0$ (seedha bahar ja raha ho), $cos θ = 1$ : saari motion radial hai.
$p_{⊥} = p sin θ$ — $r$ ke across wala piece. Yahi akela piece ball ko $O$ ke around swing karta hai.

KYUN yahan $sin$ chahiye naa ki kuch aur. Humne ek specific sawaal pucha: "kitni motion origin ke around circle karti hai?" Right triangle ki sideways leg exactly iska jawab deti hai, aur right triangle mein angle ke opposite wali leg hoti hai (hypotenuse) $\times sin θ$ . Isliye $sin θ$ choose kiya gaya hai kyunki woh opposite (perpendicular) leg select karta hai — woh "around jaane wala" part.

PICTURE. Orange arrow ko dekhein ek faded radial piece (waste, $O$ se door point karta) aur ek bright perpendicular piece (asli swirl) mein tootte hue.

Step 3 — Lever arm: wahi swirl, ek doosre angle se dekhi

KYA HAI. Swirl ko isolate karne ke do equally correct tarike hain. $p$ ko trim karne ki jagah, poora $p$ rakho lekin pucho: $O$ se motion ki line kitni door miss karti hai? Momentum ko ek poori seedhi line mein extend karo aur $O$ se uss line par perpendicular daalo. Woh shortest distance hai lever arm $r_{⊥}$ : $r_{⊥} = r sin θ .$

$r$ — position arrow ki poori length.
$sin θ$ — dobara perpendicular leg pick karta hai, ab $r$ -triangle ki.
$r_{⊥}$ — path kitna $O$ ke paas se guzarta hai.

Magic yeh hai: $r p sin θ$ ko $p \cdot (r sin θ) = p r_{⊥}$ ya $r \cdot (p sin θ) = r p_{⊥}$ dono tarah padha ja sakta hai. Same number, do viewpoints.

KYUN. Seedha fly karne wale particle ke liye, $r_{⊥}$ constant rehta hai chahe $r$ swing kare — yahi baat free particles ko fixed angular momentum carry karna deti hai (Step 8 mein prove hoga).

PICTURE. Violet dashed segment hai $r_{⊥}$ , orange path $O$ ke kitna pass se guzarta hai.

Step 4 — Cross product: swirl ko vector banana

KYA HAI. Number $r p sin θ$ angular momentum ka size hai. Lekin spinning ka ek sense bhi hota hai — clockwise ya counter-clockwise — isliye $L$ ek poora arrow hona chahiye. Woh tool jo do arrows leta hai aur ek teesra arrow produce karta hai — dono ke perpendicular, length $r p sin θ$ ke saath — woh hai cross product: $L = r \times p, ∣ L ∣ = r p sin θ .$

$\times$ automatically $sin θ$ contain karta hai — radial part ko free mein khatam kar deta hai (jab $r ∥ p$ , $sin 0 = 0$ , to $L = 0$ ). Isliye nature ka formula cross product hai, plain multiply nahi.
Direction right-hand rule se: ungliyan $r$ ke along, $p$ ki taraf curl karo, thumb $L$ deta hai. Counter-clockwise swirl (page mein) $\Rightarrow$ $L$ page se bahar point karta hai.

KYUN bilkul naya tool? Ek scalar counter-clockwise aur clockwise mein fark nahi bata sakta. Humein ek aisi cheez chahiye jo spin reverse hone par sign flip kare — axis ke along ek vector exactly yahi karta hai.

PICTURE. Green arrow $L$ seedha page se bahar nikalta hai; curved orange arrow woh circulation sense dikhata hai jise aapka right hand follow karta hai.

Step 5 — Ab ek poora rigid body spin karo: har particle axis ke around circle karta hai

KYA HAI. Ek real object lo (ek disc) jo ek fixed axis ke around spin kar raha ho — use $z$ -axis kaho, disc ke center se bahar nikalta hua. "Rigid" matlab shape kabhi deform nahi hoti, isliye har particle ek hi turning rate share karta hai — angular speed $ω$ (rad/s). Ek particle $i$ lo jo axis se perpendicular distance $r_{i}$ par hai. Woh $r_{i}$ radius ke circle par ride karta hai, aur uski speed hai $v_{i} = ω r_{i} .$

$ω$ — radians per second, sab particles ke liye identical (rigidity).
$r_{i}$ — particle $i$ axis se kitna door hai.
$v_{i} = ω r_{i}$ — bada circle $\Rightarrow$ faster particle, same $ω$ ke liye.

KYUN $v = ω r$ . Ek full turn mein ek particle circumference $2 π r_{i}$ cover karta hai jab $2 π$ radians sweep karta hai; distance ko time se divide karo aur $ω =$ radians/time use karo to $v_{i} = ω r_{i}$ directly milta hai.

PICTURE. Do sample particles — inner (magenta) aur outer (orange). Unke velocity arrows unke circles ke tangent hain; outer wala visibly lamba hai.

Step 6 — Axis ke baare mein ek particle ka angular momentum

KYA HAI. Particle $i$ ko $∣ L ∣ = r p sin θ$ mein feed karo uske momentum $p_{i} = m_{i} v_{i} = m_{i} ω r_{i}$ aur $θ = 9 0^{\circ}$ ke saath: $L_{i} = r_{i} p_{i} sin 9 0^{\circ} = r_{i} (m_{i} ω r_{i}) (1) = m_{i} r_{i}^{2} ω .$

Term by term, dekhte hain kaise build hota hai:

$r_{i}$ — lever arm (yahan yeh poora radius hi hai, kyunki motion perpendicular hai).
$p_{i} = m_{i} ω r_{i}$ — Step 5 ki speed substitute karo.
$sin 9 0^{\circ} = 1$ — swirl maximum hai; kuch waste nahi.
Result $m_{i} r_{i}^{2} ω$ — dhyan do ki $r_{i}$ do baar aata hai (ek baar lever arm se, ek baar speed se). Woh squared distance moment of inertia ka seed hai.

KYUN $r^{2}$ matter karta hai. Axis se distance double tarike se count hoti hai — door wali mass zyada penalized hoti hai. Isliye books ko arm's length par pakadne se aap itni dhheere spin karte ho.

PICTURE. Ek single particle ka $L_{i}$ arrow $+ z$ ke along, $r_{i}$ ke do roles label kiye: ek lever arm ke roop mein, ek speed ke andar chhupa hua.

Step 7 — Sab particles par sum karo: $L = I ω$ appear hota hai

KYA HAI. Har particle ka $L_{i}$ same direction mein point karta hai ( $+ z$ ke along), to unke sizes simply add ho jaate hain: $L = \sum_{i} L_{i} = \sum_{i} m_{i} r_{i}^{2} ω = (\sum_{i} m_{i} r_{i}^{2}) ω .$ $ω$ sab mein common hai (rigidity!), to woh factor out ho jaata hai. Bracket purely ek property hai mass kaise arrange hai ki — moment of inertia $I = \sum_{i} m_{i} r_{i}^{2}$ . Isliye: $L = I ω .$

$\sum_{i}$ — body ka har chunk add karo.
$m_{i} r_{i}^{2}$ — har chunk ka contribution; door wale chunks dominate karte hain.
$ω$ bahar nikala — tabhi valid jab har chunk ise share kare.
$I$ — collected bracket; ek baar measure karo, hamesha reuse karo (is axis ke liye).

KYUN yeh koi naya law nahi hai. Yeh literally $r \times p$ summed hai. $I ω$ woh hai jo general definition do conditions mein ban jaata hai: rigid (common $ω$ ) aur fixed symmetry axis (saare $L_{i}$ parallel, to magnitudes add hoti hain).

PICTURE. Kai chote $L_{i}$ arrows ek mote total $L$ mein stack ho rahe hain axis ke along.

Step 8 — Edge aur degenerate cases (kabhi surprise mat ho)

KYA HAI. Formula ke corners check karo taaki koi scenario tumhe off-guard na pakde.

Straight-line, force-free particle. Bilkul bhi spin nahi, phir bhi $L = p r_{⊥} \neq = 0$ off-line origin ke baare mein. Fly karte waqt $r$ swing karta hai lekin $r_{⊥}$ aur $p$ fixed rehte hain $\Rightarrow$ $L$ constant hai (no torque, conserved).
$O$ se seedhi motion ( $θ = 0$ ya $18 0^{\circ}$ ): $sin θ = 0 \Rightarrow L = 0$ . Pure radial motion mein koi angular momentum nahi hoti. ✔
Axis par particle ( $r_{i} = 0$ ): contributes $m_{i} \cdot 0 \cdot ω = 0$ . Center par mass "free" hai — yeh $L$ ya $I$ mein kuch add nahi karta.
$ω = 0$ (spin nahi kar raha): $L = I \cdot 0 = 0$ . Ek resting rigid body ki koi spin angular momentum nahi hoti, chahe shape koi bhi ho.
Spin reverse karo ( $ω \to - ω$ ): $L$ axis ke doosri taraf flip ho jaata hai — sign direction mein hota hai, exactly wahi jo cross product track karne ke liye bana tha (Step 4).

KYUN. Yeh sab ek hi $r p sin θ$ ki limits hain; inhe dekhna confirm karta hai ki formula har jagah gracefully degrade karta hai.

PICTURE. Char mini-panels: (i) straight-line ball with constant $r_{⊥}$ , (ii) radial motion $\Rightarrow L = 0$ , (iii) on-axis particle $\Rightarrow 0$ , (iv) reversed spin $L$ flip karta hua.

Ek-picture summary

KYA HAI. Ek frame poori journey ko stitch karta hai: left par single-particle triangle ( $r$ , $p$ , $θ$ , $r_{⊥}$ ), phir implication ka arrow, aur right par summed rigid body jo axis ke along $L = I ω$ deta hai.

Recall Feynman: walkthrough kisi dost ko sunao

Ek dot rakho aur use ghar kaho. Ghar se ek moving ball tak arrow kheecho (woh hai $r$ ), aur ek arrow dikhao jahan ball ja rahi hai (woh hai $p$ ). Ball ki motion ka sirf sideways hissa ghar ke around swirl karta hai — jo straight ghar ki taraf ya usse door ja raha hai woh kuch nahi karta. Us swirl ko hum measure karte hain (momentum) times (path ghar ko kitna miss karta hai), jo same hai (distance) times (sideways momentum), aur yeh $r p sin θ$ ke barabar hai. Yeh yaad rakhne ke liye ki woh kis taraf spin karta hai, hum use axis ke along ek arrow banaate hain — yahi cross product hai. Ab ek poori disc spin karo: har chhota bit apna circle ride karta hai, sab ek hi turning rate $ω$ par, har ek perfectly sideways chalte hue to kuch waste nahi. Har bit ka swirl hai uski mass times uski distance-squared times $ω$ . Sab add karo: $ω$ shared hai to woh bahar aa jaata hai, aur jo bacha — mass times distance-squared, summed — woh moment of inertia $I$ hai. To $L = I ω$ simply ek arab chote $r \times p$ 's ka sum hai, aur yeh tabhi itna clean hota hai jab object rigid ho aur ek achhi symmetric axis ke around spin kare.

Connections

Parent topic — woh full statement jise yeh page pictures mein draw karta hai
Cross Product — Step 4 ka tool jo $L$ ko vector banata hai
Moment of Inertia — Step 7 mein janam lene wala $\sum m_{i} r_{i}^{2}$
Linear Momentum — $p = m v$ , woh arrow jisse humne shuru kiya
Torque — $τ = d L / d t$ , isliye zero torque $L$ freeze karta hai (Step 8)
Conservation of Angular Momentum — free particle ka constant- $L$
Kepler's Second Law — planets ke liye wahi $p r_{⊥}$ constancy