Inertia tensor — principal axes, principal moments: Exercises

2.1.22 · D4 · Physics › Analytical Mechanics › Inertia tensor — principal axes, principal moments

Jinpe hum rely karte hain unka quick reminder (sab parent mein define hain):

Intuition Teen facts jinpe hum rely karte hain — yahan sketch kiye hain, sirf cite nahi kiye

(A) Off-diagonals ka zero hona $\Leftrightarrow$ principal axes. Entry $I_{x y}$ measure karta hai ki $x$ ke baare mein ek spin kitna angular momentum $y$ -direction mein leak karta hai: $L_{x} = I_{xx} ω_{x} + I_{x y} ω_{y} + \dots$ . Agar purely $\overset{x}{^}$ ke baare mein spin (matlab $ω = (ω, 0, 0)$ ) se $L$ purely $\overset{x}{^}$ ke saath mile (yaani $L ∥ ω$ ), toh leaked components $L_{y} = I_{x y} ω$ aur $L_{z} = I_{x z} ω$ zero hone chahiye har $ω$ ke liye — forcing $I_{x y} = I_{x z} = 0$ . $\overset{y}{^}$ aur $\overset{z}{^}$ ke baare mein spins ke liye bhi yahi demand karo aur har off-diagonal zero ho jaata hai. Toh "teeno axes principal hain" aur "sab off-diagonals zero hain" literally same statement hai.

(B) Trace mein 2 ka factor. Diagonal entries add karo: $I_{xx} + I_{y y} + I_{z z} = \sum m [(y^{2} + z^{2}) + (x^{2} + z^{2}) + (x^{2} + y^{2})] = \sum m [2 x^{2} + 2 y^{2} + 2 z^{2}] = 2 \sum m r^{2},$ jahan $r^{2} = x^{2} + y^{2} + z^{2}$ . Har coordinate teen mein se do moments mein appear karta hai (jaise $x^{2}$ , $I_{y y}$ aur $I_{z z}$ mein aata hai par $I_{xx}$ mein nahi) — yahi double-counting exactly wahan se $2$ aata hai. Axes rotate karne se $\sum m r^{2}$ kabhi nahi badalta (origin se distance frame-independent hoti hai), isliye trace invariant hai.

(C) Clean kinetic-energy formula. Parent se, $T = \frac{1}{2} ω^{T} I ω$ . Principal frame mein $I$ diagonal hai, toh $ω = (ω_{1}, ω_{2}, ω_{3})$ ke saath: $ω^{T} I ω = (ω_{1}, ω_{2}, ω_{3}) I_{1} ω_{1} I_{2} ω_{2} I_{3} ω_{3} = I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2},$ jisse milta hai $T = \frac{1}{2} (I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2})$ . Cross terms $I_{ij} ω_{i} ω_{j}$ jo general formula ko clutter karte hain, sab zero ho jaate hain kyunki off-diagonals zero hain.

Level 1 — Recognition

Recall Solution L1·Q1

(a) Off-diagonal entries sab exactly zero hain, toh haan, lab axes already principal axes hain. (Fact (A) se: zero off-diagonals $\Leftrightarrow$ principal.) (b) Principal moments simply diagonal entries hain: $I_{1} = 5, I_{2} = 8, I_{3} = 8 kg\cdotm^{2}$ . (c) $L = I ω = (5 \cdot 3, 0, 0) = (15, 0, 0)$ . Yeh $ω = (3, 0, 0)$ ke parallel hai — haan. KYU: $\overset{x}{^}$ ek eigenvector hai, toh $L = I_{1} ω$ .

Recall Solution L1·Q2

Trace invariant (fact (B)) kehta hai $I_{1} + I_{2} + I_{3} = 2 \sum m r^{2}$ . Left side: $2 + 3 + 4 = 9$ . Right side: $2 \times 5 = 10$ . Kyunki $9 \neq = 10$ , data inconsistent hai. Kahin koi moment ya mass sum galat hai.

Level 2 — Application

Recall Solution L2·Q1

Dono masses ke liye $x = 0, z = 0, y = \pm b$ hai.

$I_{xx} = \sum m (y^{2} + z^{2}) = m b^{2} + m b^{2} = 2 m b^{2}$ .
$I_{y y} = \sum m (x^{2} + z^{2}) = 0$ (dono $y$ -axis par hain → usse zero distance).
$I_{z z} = \sum m (x^{2} + y^{2}) = 2 m b^{2}$ .
Products: har ek mein $x$ ya $z$ ka factor aata hai, yahan sab zero hain, toh $I_{x y} = I_{x z} = I_{y z} = 0$ . $I = 2 m b^{2} 00 000 00 2 m b^{2} .$ Already diagonal → principal moments $2 m b^{2}, 0, 2 m b^{2}$ ; principal axes hain $x, y, z$ . $0$ $y$ -axis ka hai kyunki masses usse par hain (uske baare mein spin karne se kuch sweep nahi hota).

Recall Solution L2·Q2

$L = I ω = 2 m a^{2} 1 - 1 0 - 1 10 002 ω 00 = 2 m a^{2} (ω, - ω, 0) .$ Toh $L$ $(1, - 1, 0)$ ki taraf point karta hai jabki $ω$ $(1, 0, 0)$ ki taraf — parallel nahi. Angle, full norms ke saath (koi shortcuts nahi). Likho $L = 2 m a^{2} ω (1, - 1, 0)$ aur $ω = ω (1, 0, 0)$ . $L \cdot ω = (2 m a^{2} ω) (ω) [(1) (1) + (- 1) (0) + (0) (0)] = 2 m a^{2} ω^{2} .$ $∣ L ∣ = 2 m a^{2} ω 1^{2} + (- 1)^{2} + 0^{2} = 2 m a^{2} ω 2, ∣ ω ∣ = ω 1^{2} = ω .$ $cos θ = \frac{2 m a ^{2} ω ^{2}}{( 2 m a ^{2} ω 2 ) ( ω )} = \frac{2 m a ^{2} ω ^{2}}{2 m a ^{2} ω ^{2} 2} = \frac{1}{2} .$ Common factor $2 m a^{2} ω^{2}$ upar aur neeche cancel ho jaata hai — isliye hum use ignore kar sakte the — toh $θ = 4 5^{\circ}$ .

Neeche figure (s01): orange arrow $ω$ hai ( $x$ ke saath), magenta arrow $L$ hai (fourth quadrant mein $y = - x$ ke saath point karta hai). Dashed violet line mass line $y = x$ hai; notice karo $L$ uske across $ω$ se reflect hua hai, aur navy arc $4 5^{\circ}$ tilt mark karta hai — ek non-principal spin ka geometric signature.

Figure — Inertia tensor — principal axes, principal moments

Level 3 — Analysis

Recall Solution L3·Q1

$z$ -axis decoupled hai (iski column/row diagonal se bahar zero hai), toh $I_{3} = 6$ axis $\overset{e}{^}_{3} = (0, 0, 1)$ ke saath. Top-left block ke liye $det (4 - λ 1 1 4 - λ) = 0$ solve karo: $(4 - λ)^{2} - 1 = 0 \Rightarrow 4 - λ = \pm 1 \Rightarrow λ = 3 or 5.$ Eigenvectors: $λ = 5$ ke liye, $(4 - 5) x + y = 0 \Rightarrow y = x$ , axis $\frac{1}{2} (1, 1, 0)$ . $λ = 3$ ke liye, $y = - x$ , axis $\frac{1}{2} (1, - 1, 0)$ . Principal moments: $3, 5, 6$ . Check: trace $4 + 4 + 6 = 14 = 3 + 5 + 6$ . ✓

Recall Solution L3·Q2

$x = d, y = z = 0$ ke saath definitions se directly:

$I_{xx} = m (y^{2} + z^{2}) = 0$ .
$I_{y y} = m (x^{2} + z^{2}) = m d^{2}$ .
$I_{z z} = m (x^{2} + y^{2}) = m d^{2}$ .
Sab products zero hain (har ek ko ek zero coordinate chahiye). $I = 000 0 m d^{2} 0 00 m d^{2} .$ Toh $I_{z z} = m d^{2}$ hai. KYU yeh parallel axis se match karta hai: apne centre ke baare mein ek point mass ka $0$ hota hai; $z$ -axis ke perpendicular $d$ shift karne par $m d^{2}$ add hota hai, exactly theorem ka $I = I_{cm} + M d^{2}$ .

Level 4 — Synthesis

Recall Solution L4·Q1

$x$ -axis decouple hai: $I_{1} = 2$ aur $\overset{e}{^}_{1} = (1, 0, 0)$ . Block $(3113)$ : $(3 - λ)^{2} - 1 = 0 \Rightarrow λ = 2$ ya $4$ .

$λ = 4$ : $y = z$ , axis $\overset{e}{^}_{2} = \frac{1}{2} (0, 1, 1)$ .
$λ = 2$ : $y = - z$ , axis $\overset{e}{^}_{3} = \frac{1}{2} (0, 1, - 1)$ . Principal moments: $I_{1} = 2, I_{2} = 4, I_{3} = 2$ . Trace check: $2 + 3 + 3 = 8 = 2 + 4 + 2$ . ✓

Kinetic energy. Pehle lab frame mein, $T = \frac{1}{2} ω^{T} I ω$ : $I ω = (2 \cdot 1, 3 \cdot 2 + 1 \cdot 2, 1 \cdot 2 + 3 \cdot 2) = (2, 8, 8) .$ $T = \frac{1}{2} (1, 2, 2) \cdot (2, 8, 8) = \frac{1}{2} (2 + 16 + 16) = 17 J .$ Principal frame mein cross-check fact (C) use karke. $ω = (1, 2, 2)$ ke components har principal axis ke saath (dot product se project karo): $ω_{1} = (1, 2, 2) \cdot (1, 0, 0) = 1$ ; $ω_{2} = (1, 2, 2) \cdot \frac{1}{2} (0, 1, 1) = \frac{2 + 2}{2} = 22$ ; $ω_{3} = (1, 2, 2) \cdot \frac{1}{2} (0, 1, - 1) = \frac{2 - 2}{2} = 0$ . Har component ko apne moment ke saath pair karo: $ω_{2}$ ( $\overset{e}{^}_{2}$ ke saath) $I_{2} = 4$ carry karta hai, aur $ω_{3}$ , $I_{3} = 2$ carry karta hai: $T = \frac{1}{2} (I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2}) = \frac{1}{2} (2 \cdot 1^{2} + 4 \cdot (22)^{2} + 2 \cdot 0^{2}) = \frac{1}{2} (2 + 32 + 0) = 17 J . ✓$ Dono frames agree karte hain — energy ek scalar invariant hai.

Level 5 — Mastery

Recall Solution L5·Q1

Cube ki symmetry se ( $x \to - x$ etc.), har product of inertia integral $\int x y d m$ $+$ aur $-$ contributions pair karta hai aur zero ho jaata hai, toh $I$ in axes mein diagonal hai. $\int x^{2} d m = \frac{1}{12} M L^{2}$ ki one-line derivation. Cube ki uniform density $ρ = M / L^{3}$ hai. Use $x$ ke perpendicular slabs mein slice karo; position $x$ par ek slab (jahan $- L /2 \leq x \leq L /2$ ) ka mass $d m = ρ L^{2} d x$ hai. Toh $\int x^{2} d m = ρ L^{2} \int_{- L /2}^{L /2} x^{2} d x = ρ L^{2} [\frac{x ^{3}}{3}]_{- L /2}^{L /2} = ρ L^{2} \cdot \frac{1}{3} \cdot \frac{L ^{3}}{4} = ρ \frac{L ^{5}}{12} = \frac{M}{L ^{3}} \frac{L ^{5}}{12} = \frac{1}{12} M L^{2} .$ Symmetry se $\int y^{2} d m = \int z^{2} d m = \frac{1}{12} M L^{2}$ bhi. Isliye $I_{z z} = \int (x^{2} + y^{2}) d m = \frac{1}{12} M L^{2} + \frac{1}{12} M L^{2} = \frac{1}{6} M L^{2},$ $I_{xx}, I_{y y}$ ke liye identically. Toh $I = \frac{1}{6} M L^{2} 1$ . Kyunki $I = I 1$ (ek scalar times identity hai), kisi bhi $ω$ ke liye, $I ω = I ω$ . Har direction ek eigenvector hai → har axis principal hai ("spherical top"). Diagonal ke baare mein spin karne par: $L = \frac{1}{6} M L^{2} ω$ , perfectly parallel $ω$ ke saath — koi wobble nahi, koi tilt nahi.

Neeche figure (s02): violet cube edges ke saath axes ke saath drawn hai; orange arrow $ω$ hai jo body diagonal $(1, 1, 1)$ ke saath point karta hai, aur magenta arrow $L$ hai jo exactly uske upar hai (sirf isliye chhota drawn hai taaki dono visible hon). Geometric message: spherical top ke liye koi special direction nahi hoti — $L$ kabhi $ω$ nahi chhodta, s01 ke tilted case ke bilkul opposite.

Recall Solution L5·Q2

Kyunki $x, y, z$ principal axes hain, $I$ diagonal hai: $I = 6000600010 .$ $L = I ω = (6 \cdot 2, 6 \cdot 0, 10 \cdot 5) = (12, 0, 50) .$ $ω = (2, 0, 5)$ ke parallel nahi hai kyunki $ω$ ke components alag moments wale axes ke saath hain ( $6$ vs $10$ ). Angle, full norms ke saath: $L \cdot ω = 12 \cdot 2 + 0 \cdot 0 + 50 \cdot 5 = 24 + 250 = 274,$ $∣ L ∣ = 1 2^{2} + 0^{2} + 5 0^{2} = 2644, ∣ ω ∣ = 2^{2} + 0^{2} + 5^{2} = 29 .$ $cos θ = \frac{274}{2644 29} \approx \frac{274}{51.42 \cdot 5.385} \approx \frac{274}{276.9} \approx 0.9895,$ jisse $θ \approx 8. 3^{\circ}$ milta hai. Ek chhota par nonzero tilt — ek symmetric top ka signature jo apne symmetry axis se hata ke spin ho raha hai. Yahi tilt exactly woh hai jo Euler's equations of rigid body motion mein precession drive karta hai.

Recap

Recall Self-test: kya tum yeh cold answer kar sakte ho?

$I$ ke off-diagonals principal axes ke liye exactly kya hone chahiye? ::: Exactly zero. Trace invariant relation (aur 2 kahan se aata hai)? ::: $I_{xx} + I_{y y} + I_{z z} = I_{1} + I_{2} + I_{3} = 2 \sum m r^{2}$ ; har coordinate square teen mein se do moments mein appear karta hai. $L ∥ ω$ kab hota hai? ::: Sirf tab jab $ω$ ek principal axis ke saath ho, ya jab sab principal moments equal hon (spherical top). $T$ ka formula jo sirf principal frame mein kaam karta hai? ::: $T = \frac{1}{2} (I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2})$ . Products of inertia ka sign convention? ::: Negative: $I_{x y} = - \sum m x y$ .