Inertia tensor — principal axes, principal moments: Worked examples

2.1.22 · D3 · Physics › Analytical Mechanics › Inertia tensor — principal axes, principal moments

Yeh page parent topic ke liye drill-ground hai. Hum ek scenario matrix banate hain — ek checklist jo har tarah ki situation cover karti hai jo yeh topic exam mein daal sakta hai — phir examples karte hain jab tak har box tick na ho jaye. Agar "eigenvalue" ya "principal axis" jaisi phrases shaky lagti hain, toh Eigenvalues and eigenvectors aur Symmetric and orthogonal matrices kisi doosre tab mein khule rakho.

Formula Component formulas jo hum poori page reuse karenge (ek baar, pehle hi bata dete hain)

Point masses $m_{a}$ ke liye jo positions $r_{a} = (x_{a}, y_{a}, z_{a})$ par hain, inertia tensor ki entries hain: $I_{xx} = \sum_{a} m_{a} (y_{a}^{2} + z_{a}^{2}), I_{y y} = \sum_{a} m_{a} (x_{a}^{2} + z_{a}^{2}), I_{z z} = \sum_{a} m_{a} (x_{a}^{2} + y_{a}^{2}),$ $I_{x y} = I_{y x} = - \sum_{a} m_{a} x_{a} y_{a}, I_{x z} = - \sum_{a} m_{a} x_{a} z_{a}, I_{y z} = - \sum_{a} m_{a} y_{a} z_{a} .$ Isko aise padho: har diagonal entry total mass $\times$ us axis se squared-distance hai; har off-diagonal (product of inertia) mein minus sign hota hai. Continuous body ke liye $\sum_{a} m_{a} \to \int d m$ kar lo. Neeche sab kuch sirf inhi formulas ko kisi diye gaye layout ke liye evaluate karna hai.

Ek property jis par hum checks ke liye rely karenge: define karo $r_{a}^{2} = x_{a}^{2} + y_{a}^{2} + z_{a}^{2}$ (origin se mass $a$ ki squared distance). Toh trace — teen diagonal entries ka sum — hai $Tr I = I_{xx} + I_{y y} + I_{z z} = 2 \sum_{a} m_{a} r_{a}^{2},$ aur yeh $I_{1} + I_{2} + I_{3}$ (principal moments ka sum) ke barabar hai kisi bhi frame mein. Yeh har eigenvalue calculation par ek one-line sanity check deta hai.

The scenario matrix

Kisi bhi number se pehle, yeh poora case-space hai. Har worked example us cell ke saath tagged hai jo woh cover karta hai.

Cell	Case class	Isme kya tricky hai	Example
A	Masses on an axis (degenerate, $I = 0$ )	us line ke baare mein zero moment jis par woh hain	Ex 1
B	Off-diagonal negative, tilted $L$	product of inertia sign, non-parallel $L$	Ex 2
C	Off-diagonal positive (opposite quadrant layout)	$- \sum m x y$ ka sign quadrant ke saath flip karta hai	Ex 3
D	Full symmetry ( $I_{1} = I_{2} = I_{3}$ , spherical top)	har axis principal, isotropic	Ex 4
E	Continuous body + Parallel axis theorem shift	$\int d m$ aur axis centre se hokar nahin	Ex 5
F	Symmetric top ( $I_{1} = I_{2} \neq = I_{3}$ ) real word problem	do equal, ek different; frisbee	Ex 6
G	Fully asymmetric $3 \times 3$ , exam twist	genuine cubic, trace/det checks	Ex 7
H	Continuous box, 3 distinct + thin-rod limit	sab $I$ distinct, phir ek dimension $\to 0$	Ex 8

Matrix ke axes hain: off-diagonals ka sign (zero / plus / minus), eigenvalues ki degeneracy (sab distinct / do equal / sab equal), aur input type (point masses / continuous / limiting). Aath examples milke har woh combination hit karte hain jo exams mein aata hai.

Ex 1 — Cell A: masses sitting on an axis (the degenerate $I = 0$ case)

Forecast: abhi andaza lagao — $x$ -axis (jis line par masses hain) ke baare mein moment of inertia kya hai?

Figure — Inertia tensor — principal axes, principal moments

Figure padho: do amber dots cyan horizontal line ( $x$ -axis) par exactly baithe hain. Notice karo ki kisi bhi dot aur us line ke beech koi perpendicular gap nahin hai — yahi visible zero gap neeche $I_{xx} = 0$ banayega.

$I_{xx} = \sum m (y^{2} + z^{2})$ compute karo. Dono masses ke liye $y = 0$ aur $z = 0$ hai, isliye har term $y^{2} + z^{2} = 0$ hai aur $I_{xx} = 0$ . Yeh step kyun? $I_{xx}$ mass ki squared-distance from the $x$ -axis measure karta hai; masses us axis par zero distance par hain — exactly wahi zero gap jo figure mein dikha.
$I_{y y} = \sum m (x^{2} + z^{2}) = m d^{2} + m d^{2} = 2 m d^{2}$ compute karo. Waisa hi $I_{z z} = 2 m d^{2}$ . Yeh step kyun? $y$ -axis se distance $x^{2} + z^{2} = ∣ x ∣ = d$ hai; $y$ ke baare mein spin karne par masses radius $d$ par swing karte hain.
Products of inertia: $I_{x y} = - \sum m x y = 0$ (sab $y = 0$ hain); waisa hi $I_{x z} = I_{y z} = 0$ . Yeh step kyun? Koi bhi product jisme ek coordinate uniformly zero ho, woh vanish karta hai.
Assemble karo. $I = 000 0 2 m d^{2} 0 00 2 m d^{2} = 00003600036 kg\cdotm^{2} .$ Yeh already diagonal hai, isliye principal moments $0, 36, 36 kg\cdotm^{2}$ hain aur principal axes sirf $x, y, z$ hain.

Verify: upar bataya gaya trace property use karo (har mass ke liye $r^{2} = x^{2} + y^{2} + z^{2} = d^{2}$ ke saath), trace $= 0 + 2 m d^{2} + 2 m d^{2} = 4 m d^{2} = 4 (2) (9) = 72$ , aur yeh $2 \sum m r^{2} = 2 (2 \cdot 9 + 2 \cdot 9) = 72$ ke barabar hona chahiye. ✓

Ex 2 — Cell B: off-axis pair, tilted $L$ (negative product)

Forecast: agar tum isko $x$ -axis ke baare mein spin karo, toh kya $L$ , $x$ ke along point karta hai? Pehle haan/nahin andaza lagao.

Figure padho: amber dots dashed cyan line $y = x$ ke aas-paas hain. White arrow spin $ω$ hai jo seedha $x$ ke along point kar raha hai; amber arrow resulting $L = (8 ω, - 8 ω, 0)$ hai jo fourth quadrant mein neeche point kar raha hai. White aur amber arrows ke beech visible angle hi is example ka poora point hai — $L$ , $ω$ se tilted hai.

Diagonal entries. Full definition $I_{xx} = \sum m (y^{2} + z^{2})$ se shuru karo. Har mass ka $z = 0$ hai (dono $x y$ -plane mein hain), isliye $z^{2}$ term drop ho jaati hai aur $I_{xx} = \sum m y^{2} = m a^{2} + m a^{2} = 2 m a^{2}$ . Waisa hi $I_{y y} = \sum m (x^{2} + z^{2}) = \sum m x^{2} = 2 m a^{2}$ , aur $I_{z z} = \sum m (x^{2} + y^{2}) = 4 m a^{2}$ . Yeh step kyun? Har diagonal apni axis se squared-distance hai; yahan sab masses ke liye $z = 0$ hai, isliye general formula in-plane terms tak collapse ho jaata hai, aur dono masses same $∣ x ∣ = ∣ y ∣ = a$ par hain.
Product $I_{x y} = - \sum m x y$ . Pehla mass: $x y = (a) (a) = a^{2}$ . Doosra: $(- a) (- a) = a^{2}$ . Sum $= 2 a^{2}$ , isliye $I_{x y} = - 2 m a^{2}$ . Yeh step kyun? Dono masses aise quadrants mein hain jahan $x$ aur $y$ ka sign same hai, isliye har $x y > 0$ — product pile hota hai, aur definition mein minus sign $I_{x y}$ ko negative banata hai.
$m a^{2} = 1 \cdot 4 = 4$ ke saath Assemble karo: $I = 2 m a^{2} 1 - 1 0 - 1 10 002 = 8 - 8 0 - 8 80 0016 kg\cdotm^{2} .$
$x$ ke baare mein spin karo: $ω = (ω, 0, 0)$ , isliye $L = I ω = (8 ω, - 8 ω, 0)$ . Yeh step kyun? Matrix times a column vector; $- 8$ off-diagonal ek $L_{y}$ component ko existence mein khींch laata hai — yahi figure mein amber arrow ka downward tilt hai.

$L$ mein ek $y$ -component hai jabki humne sirf $x$ ke baare mein spin kiya — parallel nahin. Yahi Rigid body rotation hai jo tumhe kaatne aa jaata hai.

Verify: $2 \times 2$ block $8 (1 - 1 - 1 1)$ ke eigenvalues $8 (1 \mp 1) = 0$ aur $16$ hain; $I_{z z} = 16$ ke saath principal moments $0, 16, 16 kg\cdotm^{2}$ hain. Trace $= 8 + 8 + 16 = 32 = 2 \sum m r^{2} = 2 (1 \cdot 8 + 1 \cdot 8) = 32$ . ✓

Ex 3 — Cell C: opposite-quadrant layout product sign ko flip karta hai

Forecast: masses doosri diagonal ( $y = - x$ ) par move ho gaye. Andaza lagao ki $I_{x y}$ apna sign rakhta hai ya flip karta hai.

Figure padho: pichle wale se compare karo — amber dots ab dashed cyan line $y = - x$ (doosri diagonal) par hain, ek dot second quadrant mein aur ek fourth mein. Yeh mirror-image placement exactly wahi hai jo product of inertia ka sign flip karti hai, jaise figure mein amber caption batata hai.

Diagonal entries unchanged hain. $I_{xx} = \sum m y^{2} = 2 m a^{2}$ , $I_{y y} = 2 m a^{2}$ , $I_{z z} = 4 m a^{2}$ . Yeh step kyun? Squaring sign ko khatam kar deti hai: $(- a)^{2} = a^{2}$ , isliye masses Ex 2 ki tarah hi har axis se same distance par hain.
Product $I_{x y} = - \sum m x y$ . Pehla mass: $(a) (- a) = - a^{2}$ . Doosra: $(- a) (a) = - a^{2}$ . Sum $= - 2 a^{2}$ , isliye $I_{x y} = + 2 m a^{2}$ . Yeh step kyun? Ab masses wahan hain jahan $x$ aur $y$ ke opposite signs hain, isliye $x y < 0$ ; definition mein minus isko positive $I_{x y}$ mein flip kar deta hai.
Assemble karo: $I = 8808800016 kg\cdotm^{2} .$
Eigenvectors switch diagonal karte hain. $2 \times 2$ block $8 (1111)$ ka eigenvalue $0$ hai $\frac{1}{2} (1, - 1, 0)$ ke liye aur $16$ hai $\frac{1}{2} (1, 1, 0)$ ke liye. Yeh step kyun? Masses ab $y = - x$ ke along line up hain; woh line zero-moment principal axis hai (uske bahar unka koi extent nahin) — figure mein dashed line.

Verify: eigenvalues $0, 16, 16$ ; trace $= 8 + 8 + 16 = 32 = 2 \sum m r^{2}$ . Ex 2 se compare karo, sirf off-diagonal sign aur isliye $I = 0$ axis ki direction badli — geometry mirror, physics mirror. ✓

Ex 4 — Cell D: full symmetry, the spherical top

Forecast: is cube ke kitne principal axes hain — teen, ya infinitely many?

Off-diagonals symmetry se vanish karte hain. $I_{x y} = - \int x y d m = 0$ kyunki $(x, y, z)$ par har mass element ke liye ek mirror wala $(- x, y, z)$ par hai jo $x y$ cancel karta hai. Yeh step kyun? Symmetric and orthogonal matrices: body ki ek reflection symmetry corresponding products ko zero force karti hai.
Ek diagonal moment. $I_{z z} = \int (x^{2} + y^{2}) d m = \frac{M}{12} (L^{2} + L^{2}) = \frac{1}{6} M L^{2}$ . Yeh step kyun? Standard uniform-cube integral; $\frac{M}{12} L^{2}$ per axis box-inertia result hai.
Symmetry se teeno equal hain: $I_{xx} = I_{y y} = I_{z z} = \frac{1}{6} M L^{2} = \frac{1}{6} (6) (1) = 1 kg\cdotm^{2} .$
Toh $I = I 1$ . Phir $I ω = I ω$ har $ω$ ke liye. Yeh step kyun? Identity ka scalar multiple sab rotations ke saath commute karta hai, isliye centre se hoti har axis principal hai.

Verify: principal moments $1, 1, 1 kg\cdotm^{2}$ ; trace $= 3 kg\cdotm^{2}$ . ✓

Ex 5 — Cell E: continuous body + parallel axis shift

Forecast: kya end-axis moment centre wale se bada ya chhota hoga, aur kitna?

Centre moment by integration. Linear density $λ = M / ℓ$ ke saath, $I_{cm} = \int_{- ℓ /2}^{ℓ /2} x^{2} λ d x = λ \frac{x ^{3}}{3}_{- ℓ /2}^{ℓ /2} = \frac{M}{ℓ} \cdot \frac{ℓ ^{3}}{12} = \frac{1}{12} M ℓ^{2} .$ Yeh step kyun? Har element $z$ -axis se distance $∣ x ∣$ par hai; squared-distance $x^{2}$ hai; rod par integrate karo.
Number it: $I_{cm} = \frac{1}{12} (3) (4) = 1 kg\cdotm^{2} .$
Parallel axis theorem se end par shift karo. End centre se $d = ℓ /2 = 1 m$ door hai: $I_{end} = I_{cm} + M d^{2} = 1 + (3) (1)^{2} = 4 kg\cdotm^{2} .$ Yeh step kyun? Theorem $I = I_{cm} + M d^{2}$ centre value reuse karne deta hai re-integrate kiye bina. Yeh sirf centre of mass se bahar ki taraf kaam karta hai.

Verify: $I_{cm} = 1$ , $I_{end} = 4$ . End se direct integration $\int_{0}^{ℓ} x^{2} λ d x = \frac{1}{3} M ℓ^{2} = \frac{1}{3} (3) (4) = 4$ ✓ — theorem se match karta hai.

Ex 6 — Cell F: symmetric top (word problem, a frisbee)

Forecast: ek disc ka ek obvious axis hai (centre se, flat face ke perpendicular). Andaza lagao ki do in-plane moments equal hain ya different.

Setup confirm karo. Hamare chosen axes se disc $x y$ -plane mein hai aur sab jagah $z = 0$ hai, isliye perpendicular-axis theorem apply hota hai jisme $z$ -axis perpendicular wala hai. Yeh step kyun? Theorem sirf $x y$ -plane mein ek lamina ke liye valid hai; placement batana hi ise use karne ka license deta hai.
Perpendicular (spin) axis $I_{3}$ . Ek disc ke liye, $I_{3} = \frac{1}{2} M R^{2}$ . Yeh step kyun? Standard disc result; radius $r$ par har mass ring $r^{2} d m$ contribute karta hai.
In-plane axes $I_{1} = I_{2}$ perpendicular-axis theorem se: flat lamina ke liye $I_{3} = I_{1} + I_{2}$ , aur circular symmetry se $I_{1} = I_{2}$ , isliye $I_{1} = I_{2} = \frac{1}{2} I_{3} = \frac{1}{4} M R^{2}$ . Yeh step kyun? Disc flat hai, isliye perpendicular-axis theorem apply hota hai; symmetry phir $I_{3}$ ko do in-plane axes ke beech evenly split kar deti hai.
Numbers $M R^{2} = 0.2 \cdot 0.0225 = 0.0045$ ke saath: $I_{3} = \frac{1}{2} (0.0045) = 0.00225, I_{1} = I_{2} = \frac{1}{4} (0.0045) = 0.001125 kg\cdotm^{2} .$
Interpretation. Do equal, ek distinct: ek symmetric top. Perpendicular axis distinguished wala hai — wahan spin karo aur flight stable hai (isliye ek thrown frisbee face-flat spin karta hai, Euler's equations of rigid body motion ke according).

Verify: $I_{3} = 0.00225$ , $I_{1} = I_{2} = 0.001125$ ; check karo $I_{1} + I_{2} = 0.00225 = I_{3}$ ✓ (perpendicular-axis theorem holds).

Ex 7 — Cell G: fully asymmetric tensor, exam twist

Forecast: ek axis already decoupled hai. Andaza lagao kaun sa, aur kya baaki do mix karte hain.

Block structure padho. Pehli row/column clean hai (sirf diagonal par $2$ ), isliye direction $\overset{n}{^}_{A} = (1, 0, 0)$ already ek principal axis hai moment $2$ ke saath. Yeh step kyun? Ek zero row aur column ka matlab hai $x$ mix nahin karta — $I (1, 0, 0)^{T} = (2, 0, 0)^{T} = 2 (1, 0, 0)^{T}$ , yahi $λ = 2$ ke liye eigenvalue equation hai.
$2 \times 2$ block $(3113)$ diagonalize karo. Characteristic equation $(3 - λ)^{2} - 1 = 0 \Rightarrow 3 - λ = \pm 1 \Rightarrow λ = 2$ ya $λ = 4$ . Yeh step kyun? $det (I - λ 1) = 0$ eigenvalue condition hai; block cubic ko isolated $λ = 2$ (step 1 se) aur is quadratic mein factorise karta hai.
Block ke eigenvectors. $λ = 4$ ke liye: $(3 - 4) v_{2} + v_{3} = 0 \Rightarrow v_{2} = v_{3}$ , direction $(0, 1, 1)$ . $λ = 2$ ke liye: $(3 - 2) v_{2} + v_{3} = 0 \Rightarrow v_{3} = - v_{2}$ , direction $(0, 1, - 1)$ . Normalise karo: $\overset{n}{^}_{B} = \frac{1}{2} (0, 1, - 1)$ (moment $2$ ) aur $\overset{n}{^}_{C} = \frac{1}{2} (0, 1, 1)$ (moment $4$ ). Yeh step kyun? Axis direction nikalne ke liye har root ke liye $(block - λ 1) v = 0$ solve karo, phir unit length mein normalise karo.
Collect aur order karo. Convention $I_{1} \leq I_{2} \leq I_{3}$ ke saath:
- $I_{1} = 2$ along $\overset{e}{^}_{1} = \overset{n}{^}_{A} = (1, 0, 0)$ ,
- $I_{2} = 2$ along $\overset{e}{^}_{2} = \overset{n}{^}_{B} = \frac{1}{2} (0, 1, - 1)$ ,
- $I_{3} = 4$ along $\overset{e}{^}_{3} = \overset{n}{^}_{C} = \frac{1}{2} (0, 1, 1)$ . Yeh step kyun? Har eigenvalue ko uske normalised eigenvector ke saath pair karna diagonalization complete karta hai; teen $\overset{e}{^}_{k}$ principal axes hain.

Verify: trace $= 2 + 3 + 3 = 8$ ko $I_{1} + I_{2} + I_{3} = 2 + 2 + 4 = 8$ ke barabar hona chahiye ✓. Determinant $= 2 \cdot (9 - 1) = 16$ ko $I_{1} I_{2} I_{3} = 2 \cdot 2 \cdot 4 = 16$ ke barabar hona chahiye ✓.

Ex 8 — Cell H: general box (3 distinct moments) aur uska thin-rod limit

Forecast: teen different side lengths ke saath, andaza lagao ki koi do moments equal aate hain. Phir andaza lagao kaun sa moment collapse karta hai jab do sides wire ban jaate hain.

(a) The general box — Cell G-style but continuous, sab distinct.

Off-diagonals zero hain. Har coordinate plane ek symmetry plane hai, isliye $\int x y d m = \int y z d m = \int x z d m = 0$ ; face-parallel axes already principal hain. Yeh step kyun? Wahi reflection argument jo cube (Ex 4) mein tha, lekin unequal sides ka matlab diagonal entries alag hongi.
Numbers plug karo $\frac{M}{12} = 1$ ke saath: $I_{x} = b^{2} + c^{2} = 4 + 9 = 13, I_{y} = a^{2} + c^{2} = 1 + 9 = 10, I_{z} = a^{2} + b^{2} = 1 + 4 = 5 kg\cdotm^{2} .$ Yeh step kyun? Direct substitution. Teeno alag hain — yahi woh "three distinct eigenvalues, zero off-diagonals" cell hai jo matrix ne promise kiya tha.
Order karo: $I_{1} = 5 (z), I_{2} = 10 (y), I_{3} = 13 (x)$ — ek asymmetric top (sab moments distinct).

(b) The thin-rod limit.

$b \to 0, c \to 0$ bhejo (wire along $x$ mein collapse karo). $I_{x} = \frac{M}{12} (b^{2} + c^{2}) \to 0$ . Yeh step kyun? $I_{x}$ $x$ -axis se squared-distance hai; $x$ -axis par padi ek wire ki woh distance zero hoti hai — wahi degeneracy jo Ex 1 mein mili thi.
Do transverse moments. $I_{y} = \frac{M}{12} (a^{2} + c^{2}) \to \frac{M}{12} a^{2}$ aur $I_{z} = \frac{M}{12} (a^{2} + b^{2}) \to \frac{M}{12} a^{2}$ . Yeh step kyun? Jab $b, c \to 0$ toh sirf length $a$ bachti hai; dono perpendicular axes ab ek identical thin rod of length $a$ dekhte hain, isliye unhe same moment dena hi hoga — top symmetric ho gaya.
Conclusion. $M = 12, a = 1$ ke saath box ke three-distinct moments $(13, 10, 5)$ collapse ho kar ${0, 1, 1} kg\cdotm^{2}$ ban jaate hain: ek zero (wire ki apni line ke baare mein spin) aur do equal — exactly thin-rod result. Degeneracy limit se create hoti hai.

Verify: box moments $I_{x} = 13, I_{y} = 10, I_{z} = 5$ (sab distinct); limit mein $I_{x} = 0$ , $I_{y} = I_{z} = \frac{M}{12} a^{2}$ milta hai, jo $M = 12, a = 1$ ke saath $1$ hai. ✓

Recap

Recall Kaun si cell mein sign flip tha, aur kyun?

Cells B vs C ::: same distances (isliye equal diagonals) lekin masses same-sign vs opposite-sign quadrants mein, $I_{x y} = - \sum m x y$ ko $- 2 m a^{2}$ se $+ 2 m a^{2}$ mein flip karte hue aur $I = 0$ principal axis ko $y = x$ se $y = - x$ par rotate karte hue.

Recall Har "Verify" ne kaun sa invariant use kiya?

The trace ::: $Tr I = I_{1} + I_{2} + I_{3} = 2 \sum m r^{2}$ frame-independent hai, saath mein $det I = I_{1} I_{2} I_{3}$ — dono quick sanity checks hain jo diagonalization ke baad bhi survive karte hain.

Recall Kab

har axis principal hoti hai? Jab $I_{1} = I_{2} = I_{3}$ (Ex 4 cube) ::: tab $I = I 1$ aur $I ω = I ω$ sab $ω$ ke liye.

Scenario matrix ka har cell A–H ab worked hai. Agle, inhe Euler's equations of rigid body motion se connect karo yeh dekhne ke liye ki yeh principal moments spin ki time evolution ko kaise drive karte hain.