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

1.5.7 · Physics › Rotational Mechanics

Kya, Kyun, Kaise

KYA milta hai: ek shortcut. Do in-plane $I$ 's pata hain → perpendicular wala free mein mil jaata hai (ya ulta, aksar symmetry se).

KYUN sirf flat bodies tak restricted hai: ye secretly assume karta hai ki har particle ka $z = 0$ hai. Ek 3-D body yeh tod deti hai.

KAISE use karte hain: typically symmetry ke saath. Ek disc ke liye, $I_{x} = I_{y}$ hai (kisi bhi in-plane diameter se dekho to same dikhta hai), toh $I_{z} = 2 I_{x} \Rightarrow I_{diameter} = \frac{1}{2} I_{z}$ .

Derivation scratch se

Formula Step-by-step proof

Ek planar body lo jo $x y$ -plane mein hai. Teeno axes $x, y, z$ ek common origin $O$ se guzarti hain, $z ⊥$ plane.

Step 1 — Moment of inertia ki definition. Position $(x, y, z)$ par ek mass element $d m$ ke liye, kisi axis ke baare mein MOI hai $\int (perp. distance to axis)^{2} d m$ . Ye step kyun? MOI ko $\sum m r_{⊥}^{2}$ se define kiya jaata hai; hum bas isse ek continuous body ke liye integral ki tarah likhte hain.

Step 2 — Har axis tak distances.

$z$ -axis tak: perpendicular distance $^{2} = x^{2} + y^{2}$ .
$x$ -axis tak: perpendicular distance $^{2} = y^{2} + z^{2}$ .
$y$ -axis tak: perpendicular distance $^{2} = x^{2} + z^{2}$ .

Ye step kyun? Kisi point ki kisi axis se perpendicular distance baaki do coordinates use karti hai (Pythagoras us axis ke $⊥$ plane mein).

Step 3 — Teeno integrals likho. $I_{z} = \int (x^{2} + y^{2}) d m, I_{x} = \int (y^{2} + z^{2}) d m, I_{y} = \int (x^{2} + z^{2}) d m$

Step 4 — Planar restriction $z = 0$ apply karo. $I_{x} = \int y^{2} d m, I_{y} = \int x^{2} d m$ Ye step kyun? Yahi ek jagah hai jahan "flat body" condition enter hoti hai. Agar $z = 0$ nahi → koi theorem nahi.

Step 5 — $I_{x}$ aur $I_{y}$ jodo. $I_{x} + I_{y} = \int y^{2} d m + \int x^{2} d m = \int (x^{2} + y^{2}) d m = I_{z} ■$ Ye step kyun? Integral ki linearity hume combine karne deti hai, aur sum exactly $I_{z}$ integrand hai.

Figure — Perpendicular axis theorem — I_z = I_x + I_y — proof, restrictions

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Recall Pehle khud try karo, phir dekho

Theorem aur uski ek restriction batao.
Proof mein exactly kahan restriction use hoti hai?
$I_{z} = \frac{1}{2} M R^{2}$ se disc-about-diameter derive karo.

Answers: $I_{z} = I_{x} + I_{y}$ ek planar body ke liye; restriction = body ek plane mein lie karti hai ( $z = 0$ ). Step 4 mein $z = 0$ set karne par use hoti hai. Disc: $I_{z} = 2 I_{x} \Rightarrow I_{x} = \frac{1}{4} M R^{2}$ .

Flashcards

Perpendicular axis theorem state karo.

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

, ek planar body ke liye valid, jahan

x, y, z

ek common point se mutually perpendicular axes hain (

z ⊥

plane).

Proof mein kaun si ek condition zaruri hai?

Har mass element ka

z = 0

ho (body ek flat lamina hai).

Planar restriction kis step mein use hoti hai?

Jab

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

aur

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

mein

z = 0

set karte hain.

Theorem solid sphere ke liye kyun fail karta hai?

Sphere 3-D hai (

z \neq = 0

);

z^{2}

terms vanish nahi karte, toh sum ab

I_{z}

ke barabar nahi rehta.

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

diya ho toh disc ka MOI ek diameter ke baare mein?

\frac{1}{4} M R^{2}

(kyunki

I_{z} = 2 I_{dia}

I_{z} = M R^{2}

diya ho toh ring ka MOI ek diameter ke baare mein?

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

Rectangular plate ke liye sides ke terms mein

I_{z}

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

Kya teeno axes ko intersect karna zaroori hai?

Haan — teeno ko ek common point se guzarna chahiye.

Kya theorem hold karne ke liye

I_{x}

I_{y}

ke barabar hona zaroori hai?

Nahi; symmetry sirf ek convenience hai. Theorem tab bhi hold karta hai jab

I_{x} \neq = I_{y}

Point

(x, y, 0)

z

-axis se perpendicular distance²?

x^{2} + y^{2}

Recall Feynman: ek 12-saal ke bachche ko samjhao

Imagine karo ek flat coin table par rakhi hui hai. Isse ek fidget top ki tarah ghoomao (axis beech se upar nikalti hai) — ye " $z$ spin" hai. Ab isse end-over-end flip karo uske face par khinchi ek line ke baare mein — ye "in-the-plane spin" hai. Rule kehta hai: top-spin ki mushkil bas do flipping mushkilon ka sum hai — ek left-right flip ki, ek front-back flip ki. Ye tabhi kaam karta hai jab coin flat ho: dhaatu ka har tukda table par baitha hai, kuch upar nahi nikla. Ek mote ball ke liye, tukde har direction mein nikalte hain, aur ye neat adding trick toot jaati hai.

Connections

Parallel axis theorem — doosra shifting theorem; $I = I_{c m} + M d^{2}$ use karta hai, kisi bhi body ke liye kaam karta hai lekin offset parallel axes ke liye.
Moment of inertia — definition — $\int r^{2} d m$ jis par ye sab built hai.
Moment of inertia of standard bodies — disc, ring, rectangular plate ke results yahan use hue.
Radius of gyration — $I$ ki alternative packaging.
Symmetry arguments in MOI — kyun round/square plates ke liye $I_{x} = I_{y}$ hota hai.