Visual walkthrough — Inertia tensor — principal axes, principal moments

2.1.22 · D2 · Physics › Analytical Mechanics › Inertia tensor — principal axes, principal moments

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Step 1 — Ek speck, ek circle mein ghoomta hua

KYA. Ek single tiny piece of the body imagine karo: mass $m$ ka ek dot, position $\vec{r}$ pe baitha hua. Position vector $\vec{r}$ ka matlab sirf yeh hai — "pivot point se dot tak khicha gaya arrow." Uske teen numbers $(x,y,z)$ batate hain ki dot kitna aage, kitna side mein, aur kitna upar baitha hai.

KYUN. Ek rigid body sirf laakhon aise dots ka collection hai jo ek doosre se chipke hain. Agar hum ek dot ka angular momentum samajh lein, toh end mein sab ko add kar denge. Sabse chote piece se banana — yahi poora trick hai.

PICTURE. Neeche, pivot woh black-dot origin hai. Pale-yellow arrow $\vec{r}$ hai. Body ghoom rahi hai, aur yeh ghoomna ek aur arrow se describe hota hai — $\vec{\omega }$ (chalk blue) — jo rotation ke axis ke saath-saath point karta hai. Uski length batati hai ki hum kitni tez ghoom rahe hain (radians per second); right-hand rule ke hisaab se tumhari curled fingers spin dikhati hain, tumhara thumb $\vec{\omega }$ hai.

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Step 2 — Speck kitni tez move karta hai? Cross product aata hai

KYA. Dot ki velocity hai $\vec{v}=\underset{\text{spin axis}}{\underbrace{\vec{\omega }}}\times \underset{\text{arm to the dot}}{\underbrace{\vec{r}}}.$ Symbol $\times$ cross product hai. Do arrows ke liye yeh ek teesra arrow output karta hai jo dono ke perpendicular hota hai, jiska length $|\vec{\omega }||\vec{r}|\sin \varphi$ hai jahan $\varphi$ dono ke beech ka angle hai.

KYUN yeh tool aur ordinary multiplication nahi? Hume yeh chahiye ki ek point ek axis ke around ghoomte waqt kis speed aur direction mein sweep karta hai. Axis se door ka point tez chalega; axis pe ka point (arm $\vec{\omega }$ ke parallel, toh $\sin \varphi =0$) bilkul nahi chalega. Cross product precisely woh operation hai jiska output zero ho jaata hai jab arm axis ke saath line up karta hai, aur perpendicular distance ke saath barhta hai — yahi circular motion ki geometry hai. Koi aur simple product yeh nahi karta.

PICTURE. Blue axis $\vec{\omega }$, yellow arm $\vec{r}$, aur pink velocity $\vec{v}$ ek right-handed corner banate hain. Notice karo ki $\vec{v}$ us circle ke tangent pe hai jis par dot chalata hai.

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Step 3 — Speck ka angular momentum

KYA. Ek dot ka angular momentum define hota hai: ${\vec{L}}_a=\vec{r}\times \left(m\vec{v}\right)=m\vec{r}\times (\vec{\omega }\times \vec{r}).$ Quantity $m\vec{v}$ (mass times velocity) ordinary momentum hai — "dot kitna motion carry kar raha hai." Ise $\vec{r}\times (\cdots )$ mein wrap karne se rotational version milta hai: momentum ko weighted kiya gaya hai ki woh kitna door hai aur kis turning sense mein act kar raha hai.

KYUN. $\vec{L}=\vec{r}\times (m\vec{v})$ angular momentum ki definition hai — yeh momentum ka rotational cousin hai, woh cheez jo constant rehti hai jab koi torque act nahi karta. Humne $\vec{v}=\vec{\omega }\times \vec{r}$ ko Step 2 se substitute kiya kyunki rigidity yeh relation force karti hai, aur ${\vec{L}}_a$ ko sirf $\vec{\omega }$ aur $\vec{r}$ se bani cheez mein badal deti hai.

PICTURE. Ab hamare paas ek double cross product hai. Board pe: $\vec{r}$ crossed into the (blue) $\vec{v}$ gives the (green) ${\vec{L}}_a$, jo axis ke direction mein nikalta hai — lekin dekho, yeh hamesha exactly $\vec{\omega }$ pe nahi baithega.

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Step 4 — Double cross product ko unfold karo (BAC–CAB)

KYA. Ek nested cross product $\vec{A}\times (\vec{B}\times \vec{C})$ ko hamesha aise rewrite kar sakte hain: $\vec{A}\times (\vec{B}\times \vec{C})=\vec{B}(\vec{A}\cdot \vec{C})-\vec{C}(\vec{A}\cdot \vec{B}),$ yeh BAC–CAB identity hai. Yahan $\vec{A}=\vec{r}$, $\vec{B}=\vec{\omega }$, $\vec{C}=\vec{r}$, jis se milta hai: ${\vec{L}}_a=m[\underset{\text{axis ke saath kheenchta hai}}{\underbrace{\vec{\omega }(\vec{r}\cdot \vec{r})}}-\underset{\text{dot ki taraf kheenchta hai}}{\underbrace{\vec{r}(\vec{r}\cdot \vec{\omega })}}]=m[\vec{\omega }{r}^2-\vec{r}(\vec{r}\cdot \vec{\omega })].$ Dot $\cdot$ yani dot product: $\vec{r}\cdot \vec{\omega }=x{\omega }_x+y{\omega }_y+z{\omega }_z$, ek single number jo measure karta hai ki do arrows kitna same direction mein point karte hain. Aur $\vec{r}\cdot \vec{r}=r^2$ arm ki length ka square hai.

KYUN yeh tool? Poora goal hai $\vec{\omega }$ ko openly bahar kheenchna taaki hum $\vec{L}$ ko "koi machine jo $\vec{\omega }$ pe act kar rahi hai" ki tarah dekh sakein. BAC–CAB woh ek identity hai jo nested product ko saaf $(\text{stuff})\cdot \vec{\omega }$ shape mein alag karti hai. Dot product isliye aata hai kyunki yeh natural "kitna aligned hai" number hai jo BAC–CAB se nikalta hai.

PICTURE. Do competing arrows: pehla term $\vec{\omega }{r}^2$ (blue) ${\vec{L}}_a$ ko axis ke saath rakhne ki koshish karta hai; doosra term $-\vec{r}(\vec{r}\cdot \vec{\omega })$ (pink) use dot ki taraf kheenchta hai. Dono ka sum (green) asli ${\vec{L}}_a$ hai — axis se tilted, jab bhi dot side mein hota hai.

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Step 5 — Ek component padho: ek matrix chupi hui hai

KYA. Chaliye sirf $\vec{L}={\sum }_a{\vec{L}}_a$ ka $x$-component likhte hain (sab dots ka sum). $r^2=x^2+y^2+z^2$ aur $\vec{r}\cdot \vec{\omega }=x{\omega }_x+y{\omega }_y+z{\omega }_z$ use karke: $L_x={\sum }_a{m}_a[{\omega }_x(x^2+y^2+z^2)-x_a(x_a{\omega }_x+y_a{\omega }_y+z_a{\omega }_z)].$ Group karo ki har term kaunsa $\omega$ multiply kar rahi hai: $L_x={\omega }_x\underset{I_{xx}}{\underbrace{\sum m(y^2+z^2)}}+{\omega }_y\underset{I_{xy}}{\underbrace{(-\sum mxy)}}+{\omega }_z\underset{I_{xz}}{\underbrace{(-\sum mxz)}}.$

KYUN. $L_x$ teen coefficients ke sum ke roop mein nikla, har ek $\vec{\omega }$ ke ek component times. Yahi exactly matrix times a vector ki top row hai. $L_y$ aur $L_z$ ke liye bhi same karo toh baaki do rows milti hain. Coefficients — mass ki pure geometry, andar koi $\vec{\omega }$ nahi — inertia tensor ki entries hain.

PICTURE. Neeche ka rows-and-columns grid dikhata hai ki har coefficient apni slot mein kaise baitha. Diagonal boxes $I_{xx},I_{yy},I_{zz}$ (yellow) moments of inertia hain; off-diagonal boxes (pink) products of inertia hain, aur unme Step 4 ke $-\vec{r}(\vec{r}\cdot \vec{\omega })$ term se minus sign aata hai.

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Step 6 — Tilt, bilkul undeniable (worked degenerate + generic case)

KYA. Sabse chota off-axis body lo: do equal masses $m$ at $(a,a,0)$ aur $(-a,-a,0)$. Step 5 ke formulas mein plug karne par: $\mathbf{I}=2m{a}^2\left(\begin{array}{ccc}1& -1& 0\\ -1& 1& 0\\ 0& 0& 2\end{array}\right).$ Ab ise $x$-axis ke around spin karo, $\vec{\omega }=(\omega ,0,0)$: $\vec{L}=\mathbf{I}\vec{\omega }=2m{a}^2(\omega ,-\omega ,0).$ $\vec{\omega }$ $+x$ ke along point karta hai, lekin $\vec{L}$ fourth quadrant ki taraf — parallel nahi.

KYUN yeh dikhana? Pehle sab symbols the; yeh payoff hai jo tum literally dekh sakte ho. Off-diagonal $-1$ (ek product of inertia) ne $\vec{L}$ ko $\vec{\omega }$ se mod diya. Woh entry hatao aur arrows agree kar lenge.

PICTURE. Blue $\vec{\omega }$ $x$ ke along; green $\vec{L}$ mass line $y=x$ ki taraf neeche kheencha gaya. Masses (pink dots) us diagonal pe baithay hain — $\vec{L}$ wahan jhukta hai jahan mass actually hai.

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Step 7 — Fix: ghoomte raho jab tak tilt gayab na ho (principal axes)

KYA. "$\vec{L}$ parallel to $\vec{\omega }$" mathematically aise likha jaata hai: $\mathbf{I}\vec{\omega }=\lambda \vec{\omega },$ yani matrix sirf $\vec{\omega }$ ko ek number $\lambda$ se stretch karta hai, kabhi turn nahi karta. Yeh eigenvalue equation hai (dekho Eigenvalues and eigenvectors). Ek special $\vec{\omega }$ jo ise satisfy kare eigenvector hai; uska stretch factor $\lambda$ eigenvalue hai. Kyunki $\mathbf{I}$ real aur symmetric hai, teen mutually perpendicular aise directions hamesha exist karte hain — principal axes — aur unke $\lambda$'s principal moments $I_1,I_2,I_3$ hain.

KYUN yeh directions? Apna coordinate frame eigenvectors ke along point karne ke liye rotate karne par har off-diagonal box zero ho jaata hai. Step 6 ke example mein masses $y=x$ ke along hain; toh $\frac{1}{\sqrt{2}}(1,1,0)$ aur uska perpendicular $\frac{1}{\sqrt{2}}(1,-1,0)$ principal axes hain, jisme: $I=2m{a}^2(1-1)=0\text{mass line ke along},I=2m{a}^2(1+1)=4m{a}^2\text{iske across}.$ Inme se kisi ke along bhi spinning karo toh $\vec{L}\parallel \vec{\omega }$ milta hai — tilt gayi.

PICTURE. Same body, lekin ab blue $\vec{\omega }$ diagonal principal axis ke along rakha gaya hai; green $\vec{L}$ exactly uski upar pada hai. Frame rotate karne se $\mathbf{I}$ diagonalise ho gaya.

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Step 8 — Har case, ek board pe (edge cases)

KYA. Teen shapes saari possibilities exhaust karti hain, sorted by ki kitne principal moments equal hain.

Body	Principal moments	$\vec{L}$ kya karta hai
sab alag $I_1\ne I_2\ne I_3$ (asymmetric top)	3 distinct axes	$\vec{L}\parallel \vec{\omega }$ sirf 3 axes pe
do equal $I_1=I_2\ne I_3$ (symmetric top)	axis ${\hat{e}}_3$ + koi bhi pair $\perp$	equal plane mein koi bhi axis bhi principal hai
sab equal $I_1=I_2=I_3$ (spherical top, e.g. cube)	har axis principal hai	$\vec{L}\parallel \vec{\omega }$ hamesha

KYUN yeh matter karta hai. Degenerate case $I_1=I_2=I_3$ ka matlab hai $\mathbf{I}=I1$ (identity times a number), toh $\mathbf{I}\vec{\omega }=I\vec{\omega }$ har $\vec{\omega }$ ke liye — ek cube sphere ki tarah behave karta hai. Aur zero case (Step 6 ki mass line, $I=0$): ek line pe bande masses us line ke around spinning ka koi resistance nahi offer karte, kyunki unme se koi off it nahi hai.

PICTURE. Teen mini-sketches — asymmetric brick, symmetric cylinder, cube — apne principal axes ke saath drawn.

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Ek-picture summary

Ek dot → uski velocity $\vec{\omega }\times \vec{r}$ → uska angular momentum $\vec{r}\times m\vec{v}$ → BAC–CAB use karke split karo → dots ke upar sum karne par geometry symmetric matrix $\mathbf{I}$ mein pack ho jaati hai → off-diagonals $\vec{L}$ ko tilt karte hain → eigenvectors (principal axes) par rotate karne se tilt hat jaati hai aur $\vec{L}=I_k\vec{\omega }$ wapas aa jaata hai.

Recall Feynman retelling (zor se bolo)

"Main spinning cheez ka ek tiny bit pakadta hoon. Kyunki poori body saath ghoomti hai, woh bit $\vec{v}=\vec{\omega }\times \vec{r}$ par whip around karta hai — agar axis se door hai toh tez, agar axis pe hai toh bilkul ruka. Uska spin-momentum hai $\vec{r}$ crossed with $m\vec{v}$. Jab main us double cross ko BAC–CAB se suljhata hoon, mujhe do tugs milti hain: ek momentum ko axis ke along rakhti hai, ek ise kheenchti hai jahan bhi mass baitha hai. Doosri wali hi troublemaker hai: woh $\vec{L}$ ko $\vec{\omega }$ se tilt karti hai. Jab main har bit ka sum karta hoon, saari geometry ek $3\times 3$ table $\mathbf{I}$ mein collapse ho jaati hai — bade diagonal numbers keh rahe hain 'is axis ke around spin karna mushkil hai,' off-diagonal minus-numbers keh rahe hain 'aur yeh sideways lean karega.' Leaning rokne ke liye main apne coordinate axes ko tab tak twist karta hoon jab tak woh off-diagonal numbers zero na ho jayein. Jin directions mein yeh hota hai woh principal axes hain — eigenvectors — aur wahan, khoobsurti se, $\vec{L}$ phir se $\vec{\omega }$ ke same direction mein point karta hai, exactly ek single point mass ki tarah."

Recall

$\vec{L}$ generally $\vec{\omega }$ ke parallel kyun nahi hota? ::: $-\vec{r}(\vec{r}\cdot \vec{\omega })$ term (off-diagonal products of inertia) $\vec{L}$ ko mass ki taraf, axis se door kheench deta hai. Kaunsa operation "$\vec{L}\parallel \vec{\omega }$" ko ek equation mein turn karta hai? ::: Eigenvalue equation $\mathbf{I}\vec{\omega }=\lambda \vec{\omega }$. Cube ke liye har axis principal kyun hai? ::: $I_1=I_2=I_3\Rightarrow \mathbf{I}=I1$, toh $\mathbf{I}\vec{\omega }=I\vec{\omega }$ sab $\vec{\omega }$ ke liye.