Inertia tensor — principal axes, principal moments: Question bank

2.1.22 · D5 · Physics › Analytical Mechanics › Inertia tensor — principal axes, principal moments

True or false — justify

L

is always parallel to

ω

for a rigid body.

False. Sirf ek principal axis ke saath (ya jab teeno principal moments equal hon) hi aisa hota hai. Off-axis pe angular momentum tilt ho jaata hai kyunki

I

mein non-zero products of inertia hote hain.

The inertia tensor is always a symmetric matrix.

True. Har off-diagonal term

- \sum m x_{i} y_{j}

hai, jo dono indices swap karne par bhi same rehta hai, isliye

I_{ij} = I_{j i}

hamesha — yahi guarantee karta hai real eigenvalues aur orthogonal axes (dekho Symmetric and orthogonal matrices).

Diagonalizing

I

changes the physical object.

False. Ye sirf tumhara coordinate description rotate karta hai. Body, uski mass distribution, aur invariants jaise

Tr I

aur

det I

bilkul untouched rehte hain.

If a coordinate axis is an axis of symmetry of the body, it is automatically a principal axis.

True. Symmetry force karti hai ki us axis se related products of inertia pairs mein cancel ho jaayein, isliye uske liye diagonalization ki zaroorat nahi.

Every rigid body has exactly three distinct principal axes.

False. Directions hamesha exist karti hain, lekin unique nahi honi zaroori: ek symmetric top mein ek special axis hoti hai aur equivalent directions ka poora ek plane hota hai; ek spherical top mein har axis principal hoti hai.

Principal moments of inertia can be negative.

False. Har ek

\sum m (distance from axis)^{2} \geq 0

hai, non-negative terms ka sum hai, isliye

I

ke eigenvalues

\geq 0

hote hain.

A principal moment of inertia can equal exactly zero.

True. Agar saari mass axis par hi ho (jaise ek line par point masses), to perpendicular distance zero hai, is axis ke baare mein

I = 0

milega.

Products of inertia carry a plus sign, matching the positive moments of inertia.

False. Products hain

I_{x y} = - \sum m x_{a} y_{a}

— derivation mein

- r (r \cdot ω)

term se ek minus sign aata hai. Sign galat karne se eigenvectors flip ho jaate hain.

The trace

I_{xx} + I_{y y} + I_{z z}

depends on which frame you compute it in.

False. Rotations trace preserve karti hain, isliye

I_{xx} + I_{y y} + I_{z z} = I_{1} + I_{2} + I_{3} = 2 \sum m r^{2}

har frame mein — eigenvalues ka ek badhiya sanity check.

Two masses on the x-axis have

I_{xx} = 0

but

I_{y y} = I_{z z} \neq = 0

True. X-axis se unki distance zero hai (isliye

I_{xx} = 0

), lekin woh y- aur z-axes se

x = \pm d

dur hain, jo

I_{y y} = I_{z z} = 2 m d^{2}

deta hai.

For a cube spun about a body diagonal,

L

tilts away from

ω

False. Ek uniform cube ek spherical top hai (

I_{1} = I_{2} = I_{3}

), isliye

I = I 1

aur

L = I ω

har axis ke liye parallel hai, diagonal bhi included.

I

is already diagonal in your chosen axes, those axes are principal.

True — bas agar off-diagonals exactly zero hon. Diagonal form ek principal frame ki definition hi hai.

Spot the error

I

is diagonal, so principal axes exist only because the body is symmetric."

Galat. Principal axes har rigid body ke liye exist karti hain kyunki

I

real aur symmetric hai (dekho Eigenvalues and eigenvectors); symmetry unhe sirf easily spot karne mein help karti hai, required nahi hai.

"The off-diagonal terms are tiny, so the axes are essentially principal."

Galat. Principal ka matlab off-diagonals exactly zero hain. "Small" products bhi

L

ko tilt karte hain aur rotational KE decomposition galat dete hain.

T = \frac{1}{2} (I_{xx} ω_{x}^{2} + I_{y y} ω_{y}^{2} + I_{z z} ω_{z}^{2})

in any frame."

Galat. Yeh clean diagonal KE formula sirf principal axes mein sahi hai. Ek general frame mein

T = \frac{1}{2} ω^{T} I ω

mein

I_{x y} ω_{x} ω_{y}

jaise cross terms bhi aate hain.

"Since

L = I ω

, doubling

ω

leaves

L

's direction unchanged."

Yeh actually sahi hai —

I

linear hai, isliye

ω

scale karne se

L

equally scale hota hai. Trap yeh sochna hai ki direction kabhi bhi magnitude par depend karta hai; woh sirf

ω

ki direction ka principal axes ke saath relation par depend karta hai.

"Because

L = I_{k} ω

along a principal axis, kinetic energy is

\frac{1}{2} I_{k} ∣ ω ∣^{2}

for any spin."

Generally galat. Yeh formula tabhi sahi hai jab

ω

sirf us ek principal axis ke saath point kare; general

ω

ke liye

\frac{1}{2} \sum_{k} I_{k} ω_{k}^{2}

sum karna padega.

"The eigenvectors of

I

point in the direction of

L

Galat. Eigenvectors woh special

ω

directions hain jinke liye

L

ω

ke saath align ho jaata hai; ye

ω

ki directions hain, kisi tilted

L

ki nahi.

Why questions

Why does

L

generally fail to be parallel to

ω

Off-diagonal products of inertia ka matlab hai ki

I

ω

ke alag components ko alag amounts se scale karta hai aur unhe mix karta hai, jisse output vector input se door rotate ho jaata hai.

Why are the principal axes exactly the eigenvectors of

I

L ∥ ω

" ka statement hai

I ω = λ ω

, jo eigenvalue equation hi hai — isliye principal axes aur eigenvectors literally ek hi cheez hain.

Why can a real symmetric

I

always be diagonalized by a rotation?

Spectral theorem: real symmetric matrices ke real eigenvalues aur mutually orthogonal eigenvectors hote hain, aur unse bana ek orthogonal (rotation) matrix ise diagonalize kar deta hai — dekho Symmetric and orthogonal matrices.

Why does the minus sign appear in the products of inertia?

Yeh

L_{a} = m_{a} [ω r_{a}^{2} - r_{a} (r_{a} \cdot ω)]

mein

- r_{a} (r_{a} \cdot ω)

term se aata hai; off-diagonal contributions sirf usi subtracted piece se aati hain.

Why is a cube a "spherical top" despite not being a sphere?

Iske teeno face-parallel principal moments symmetry se equal niklaate hain, isliye

I = I 1

; equal eigenvalues ka matlab hai inertia ellipsoid ek sphere hai aur har central axis principal hai.

Why is knowing the principal frame useful for Euler's equations?

Principal frame mein

I

diagonal hota hai, isliye torque–angular-momentum equations teen clean Euler equations mein decouple ho jaati hain sirf

I_{1}, I_{2}, I_{3}

ke saath — ek full matrix se kaafi simple.

Why does the parallel axis theorem not, by itself, give principal axes?

Yeh sirf moments/products ko ek parallel offset origin par shift karta hai; principal axes orientation ke baare mein hain, isliye shift karne ke baad bhi diagonalize karna padega (ya symmetry use karni padegi).

Edge cases

What are the principal moments of a single point mass at the origin?

Teeno zero hain:

r = 0

ke saath har

m r^{2}

term vanish ho jaata hai, isliye

I

zero matrix hai aur koi bhi axis trivially principal hai.

For point masses lying entirely on one line, how many principal moments are zero?

Exactly ek — us line ke baare mein moment (saari masses ke liye distance zero); do perpendicular moments equal aur non-zero hote hain.

If all three principal moments are equal, which axes are principal?

Sab.

I = I 1

sab kuch ke saath commute karta hai, isliye

I ω = I ω

har direction ke liye — ek degenerate (spherical) case.

If exactly two principal moments are equal (symmetric top), what is special about the eigenvectors?

Do equal eigenvalues ek poore plane ke eigenvectors share karte hain: us plane mein koi bhi orthonormal pair kaam karta hai, isliye wahan axes unique nahi hain — sirf alag teesri axis fixed hai.

What happens to

L

when

ω = 0

L = I \cdot 0 = 0

. Koi rotation nahi matlab koi angular momentum nahi, chahe mass distribution kitni bhi lopsided kyun na ho.

Does a flat lamina (all mass in the

z = 0

plane) satisfy any special relation among its moments?

Haan — perpendicular axis relation

I_{z z} = I_{xx} + I_{y y}

, kyunki

z = 0

I_{xx} + I_{y y} = \sum m (y^{2} + x^{2}) = I_{z z}

banta hai.

Can a non-symmetric-looking body still have three equal principal moments?

Haan. Equal moments ke liye sirf sahi mass distribution chahiye (jaise ek cube), visual sphericity nahi; "asymmetric dikhna" ek spherical inertia tensor ko forbid nahi karta.

Recall Har trap ki ek-line summary

$L = I ω$ linear hai; $I$ symmetric hai (real eigenvalues, orthogonal axes); products mein minus aata hai; principal ka matlab off-diagonals exactly zero; diagonal KE formula aur $L ∥ ω$ sirf principal frame mein sahi hain; aur degeneracies ( $I_{1} = I_{2} = I_{3}$ ya ek zero moment) genuine, common cases hain.