Perpendicular axis theorem — I_z = I_x + I_y — proof, restrictions: Question bank

1.5.7 · D5 · Physics › Rotational Mechanics › Perpendicular axis theorem — I_z = I_x + I_y — proof, restri

Shuru karne se pehle do words bilkul crisp hone chahiye:

Lamina / planar body = ek idealized flat sheet jahan har mass point ka $z = 0$ ho (zero thickness). Socho ek coin ko itna flatten kar do ki uski koi height hi na rahe.
Mutually perpendicular axes = $x$ , $y$ , $z$ axes sabhi $9 0^{\circ}$ pe milte hain, ek common point $O$ pe cross karte hain, aur $z$ sheet ke seedha bahar nikal raha hota hai.

True or false — justify karo

The theorem holds for a solid cube.

False — ek cube ka

z

mein real extent hai, isliye

I_{x}

aur

I_{y}

mein

z^{2}

ke terms kabhi vanish nahi hote, aur clean split

I_{z} = I_{x} + I_{y}

collapse ho jaata hai. Yeh sirf

z = 0

lamina ke liye kaam karta hai.

The theorem holds for a thin square plate of tiny but nonzero thickness.

Approximately true — ek real thin plate ka

z

thoda sa hota hai, isliye

z^{2} d m

terms negligible hain; lamina ki tarah treat karne par yeh essentially exact hai, lekin strictly yeh ek approximation hai, ideal law nahi.

For a uniform disc,

I_{z} = I_{x} + I_{y}

and also

I_{x} = I_{y}

True — disc planar hai isliye theorem apply hota hai, aur Symmetry arguments in MOI se har diameter equivalent hoti hai, jo

I_{x} = I_{y}

force karta hai; isliye

I_{z} = 2 I_{x}

The theorem requires

I_{x} = I_{y}

False — equal in-plane MOIs sirf ek symmetry convenience hai. Ek rectangular plate ka

I_{x} \neq = I_{y}

hota hai phir bhi

I_{z} = I_{x} + I_{y} = \frac{1}{12} M (a^{2} + b^{2})

satisfy karta hai.

Agar in-plane axes ko

3 0^{\circ}

tilt karo (abhi bhi ek doosre ke perpendicular, abhi bhi

O

se hote hue), toh theorem abhi bhi

I_{z}

deta hai.

True — same point se hote hue koi bhi do mutually perpendicular in-plane axes kaam karte hain; Pythagoras split

x^{'2} + y^{'2} = r^{2}

rotation-invariant hai, isliye sum unchanged rehta hai.

The theorem lets you find

I

about an axis that lies outside the plane but is not the

z

-axis.

False — yeh sirf teen special axes (

x

y

in-plane aur

z

out-of-plane) ko ek point se relate karta hai; ek slanted axis ke liye bilkul alag method chahiye.

For a flat L-shaped bracket (non-symmetric lamina), the theorem still holds.

True — requirement planarity hai, symmetry nahi; har mass point ka abhi bhi

z = 0

hai, isliye

I_{z} = I_{x} + I_{y}

odd shape hone ke bawajood hold karta hai.

I_{z}

is always the largest of the three MOIs for a lamina.

True — kyunki

I_{z} = I_{x} + I_{y}

aur

I_{x}, I_{y}

dono squares ke non-negative integrals hain,

I_{z} \geq I_{x}

aur

I_{z} \geq I_{y}

; yeh kisi ek ke equal tab hi ho sakta hai jab doosra zero ho.

Spot the error

"A solid sphere:

I_{x} = I_{y} = I_{z} = \frac{2}{5} M R^{2}

, so

I_{z} = I_{x} + I_{y} = \frac{4}{5} M R^{2}

. Contradiction — so MOI is undefined."

Error yeh hai ki theorem ek 3-D body pe apply kiya ja raha hai. Koi contradiction nahi hai; theorem simply apply nahi hota, kyunki sphere planar nahi hai.

I_{z} = \frac{2}{5} M R^{2}

apne aap mein correct hai.

"For a disc,

I_{z} = \frac{1}{2} M R^{2}

, so each diameter MOI is

\frac{1}{2} M R^{2}

as well."

Galat — pehle split karna padega:

I_{z} = I_{x} + I_{y} = 2 I_{dia}

, isliye

I_{dia} = \frac{1}{4} M R^{2}

. Do ka factor bhoolne se answer double ho jaata hai.

"The three axes can be parallel and offset like in the parallel axis theorem."

Nahi — yeh dono theorems ko confuse kar raha hai. Yahan axes mutually perpendicular hain aur ek point pe milte hain; kisi bhi axis ko offset karna Pythagoras relation

r_{⊥}^{2} = x^{2} + y^{2}

tod deta hai jo proof ko chahiye. Offsets ke liye Parallel axis theorem use karo.

"For a ring, mass is all at radius

R

, so

I_{x} = I_{y} = I_{z} = M R^{2}

Galat — sirf

I_{z} = M R^{2}

hai (saari mass central perpendicular axis se distance

R

par hai). Diameter ke baare mein mass varying perpendicular distances par hoti hai, jo

I_{dia} = \frac{1}{2} M R^{2}

deta hai

I_{z} = 2 I_{dia}

ke zariye.

"Since the theorem uses

x^{2} + y^{2}

, it must be about the origin only."

Bilkul sahi nahi — theorem kisi bhi common point ke baare mein hold karta hai jahan se teeno perpendicular axes pass karti hain;

O

ko tum jahan chahein choose kar sakte ho, bas teeno axes wahan intersect karni chahiye.

I_{x} = \int (y^{2} + z^{2}) d m

and for a lamina that is

\int (y^{2} + 0) d m

, but I'll keep

z

just in case."

z

ko rakhna lamina ke liye galti hai — poora theorem

z = 0

set karne par hi based hai.

z^{2}

retain karne ka matlab hai tum ab planar body describe nahi kar rahe aur split fail ho jaati hai.

Why questions

Proof ke exactly ek step pe restriction kyun enter karti hai?

Kyunki sirf

I_{x}

aur

I_{y}

mein

z^{2}

terms hote hain;

z = 0

set karna ek hi move hai jo unhe khatam karta hai, aur yeh ek baar hota hai jab

I_{x} = \int y^{2} d m

I_{y} = \int x^{2} d m

likhte hain.

Do in-plane axes ek doosre ke perpendicular kyun hone chahiye?

Kyunki identity

x^{2} + y^{2} = r^{2}

(

z

-axis se perpendicular distance

^{2}

) Pythagoras hai, jo sirf orthogonal coordinate directions ke liye hold karta hai; non-perpendicular axes cross-terms dete hain jo clean sum ko kharab kar dete hain.

Theorem "come for free" kyun karta hai — disc ke diameter ke liye koi integration nahi?

Kyunki symmetry tumhe

I_{x} = I_{y}

de deti hai, aur theorem unhe known

I_{z}

se baandhta hai; do equations, ek unknown, isliye algebra integral ki jagah le leta hai. Dekho Moment of inertia of standard bodies.

I_{z}

I_{x}

aur

I_{y}

ka sum kyun hai, average kyun nahi?

Kyunki

z

-axis distance

^{2}

literally

x^{2} + y^{2}

ke barabar hai — un same squares ka sum jo

I_{y}

aur

I_{x}

banate hain — isliye integrate karne par addition milta hai, kabhi average nahi.

Theorem shape ki outline pe kabhi depend kyun nahi karta?

Kyunki derivation sirf coordinate identity

r^{2} = x^{2} + y^{2}

use karta hai jo

z = 0

wale har point par valid hai; lamina ki boundary kabhi appear nahi hoti, isliye koi bhi flat shape qualify karta hai.

Same theorem ko "backwards" padhke do in-plane MOIs se

I_{z}

kyun nikaal sakte hain?

Kyunki

I_{z} = I_{x} + I_{y}

ek equality hai, one-way recipe nahi; teeno mein se koi bhi do pata hono toh teesra determine ho jaata hai.

Edge cases

X-axis ke saath let hua ek thin straight rod (ek degenerate lamina).

I_{x}

kya hai?

I_{x} = \int y^{2} d m = 0

kyunki har point ka

y = 0

hai; phir

I_{z} = I_{y}

, yeh match karta hai ki rod ko apni length ke baare mein rotate karne mein kuch nahi lagta.

Origin par ek point mass. Teeno MOIs kya hain?

Sabhi zero — jab

x = y = z = 0

ho toh har axis se perpendicular distance zero hai, isliye

I_{x} = I_{y} = I_{z} = 0

aur sum

0 = 0 + 0

trivially hold karta hai.

x

-axis par do point masses,

O

ke baare mein symmetric. Kya theorem hold karta hai?

Haan — yeh abhi bhi ek lamina hai (

z = 0

). Yahan

I_{x} = 0

(

x

-axis par points) aur

I_{z} = I_{y}

, isliye

z

ke baare mein saari "spin difficulty" entirely

y

-axis term se aati hai.

Ek lamina jahan saari mass

y

-axis par hai. Teeno MOIs ko relate karo.

Phir saari mass ke liye

x = 0

hai, isliye

I_{y} = \int x^{2} d m = 0

aur

I_{z} = I_{x}

; out-of-plane aur in-line-perpendicular MOIs coincide karte hain.

Ek thin plate ki thickness

t \to 0

limit karne par —

I_{x} = \int y^{2} d m

mein error ka kya hota hai?

Drop kiya hua

\int z^{2} d m

term

t^{2}

ki tarah scale karta hai, isliye

t \to 0

hone par error quadratically vanish ho jaata hai aur lamina result exact ban jaata hai.

Ek lamina jo plane se thodi bahar muri hui hai (ek shallow dome). Kya theorem abhi bhi apply hota hai?

Nahi — jab points nonzero

z

le lete hain,

z^{2}

terms wapas aa jaate hain aur

I_{z} = I_{x} + I_{y}

fail ho jaata hai; body genuinely flat honi chahiye, sirf thin-and-curved nahi.

Connections

Parallel axis theorem — offset parallel axes ke liye sibling theorem; "Plus" vs "Plus a distance term" ko contrast karo.
Symmetry arguments in MOI — kyun discs aur rings ke liye $I_{x} = I_{y}$ hota hai, aur kyun symmetry yahan optional hai.
Radius of gyration — in MOIs ko ek equivalent single distance ki tarah recast karta hai.
Moment of inertia — definition — $\int r^{2} d m$ jis par upar saab kuch based hai.