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

1.5.7 · D2 · Physics › Rotational Mechanics › Perpendicular axis theorem — I_z = I_x + I_y — proof, restri

Step 1 — "Flat body" hota kya hai, aur hum usse rakhte kahan hain?

KIYA KYA: object ko is tarah rakha ki har particle ek flat sheet par ho.

KYUN: poora theorem ek baat par tika hai — object ka koi bhi hissa table se upar nahi nikal ta. Yeh hum pehli line se hi set kar lete hain taaki kabhi bhool na jayein.

PICTURE: Figure dekho. Black outline object hai. Red dot ek chhota sa mass piece hai jise hum track karenge — isko mass $d m$ ("mass ka ek tukda") bolte hain. Yeh sheet par baitha hai, isliye table ke upar iska height $0$ hai.

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

Woh $z = 0$ yaad rakho. Yeh woh ek seed hai jis se poora proof ugta hai.

Step 2 — Teen axes ek shared pin se guzaro

KIYA KYA: ek common point aur wahan teen perpendicular directions fix kiye.

EK SHARED PIN KYUN: agli steps mein hum teen integrals ko term by term add karenge. Add karna tabhi kaam karta hai jab saari teen measurements ek hi jagah se shuru ho. Koi bhi axis $O$ se hatao aur algebra bigad jaata hai (yeh hum Step 8 mein prove karte hain).

PICTURE: Do black axes flat pade hain; red $z$ -axis page se bahar nikalti hai. Yeh red axis "fidget-top ki tarah spin karo" wali axis hai — woh odd one out jo theorem apni do flat teammates se express karega.

Symmetry arguments in MOI bhi dekho — wahan bataya gaya hai ki $x, y$ directions ka choice aksar free kyun hota hai.

Step 3 — Spin karne ki cost kya hai? Moment of inertia define karo

KIYA KYA: continuous body ke liye $I$ ki definition likhi.

SQUARE KYUN, AUR YEH TOOL HI KYUN? Hume ek number chahiye jo bataye "yeh object spin karne mein kitna stubborn hai?" $r_{⊥}$ radius par speed $r_{⊥}$ ki tarah badhti hai, aur energy speed-squared ki tarah badhti hai — isliye difficulty mein har crumb ko $r_{⊥}^{2}$ se weight karna hoga, $r_{⊥}$ se nahi. Exponent ke peeche ka kyun yahi hai.

PICTURE: Red segment $r_{⊥}$ hai — crumb se axis tak ki sabse chhoti reach. Crumb ki cost hai (red segment ki length) $^{2} \times$ (uska mass).

Step 4 — Har axis ke liye perpendicular distance padho

Ab key geometric move: har axis ke liye, apne crumb ke liye $(x, y, z)$ par $r_{⊥}$ kya hai?

PICTURE: Red right triangle $z$ -axis tak ki doori dikhata hai. Crumb daayein $x$ aur aage $y$ hai; floor ke across straight-line reach red hypotenuse hai, aur Pythagoras uska square $x^{2} + y^{2}$ deta hai.

YEH STEP KYUN: yeh teen chhote Pythagoras facts proof ko jis geometry ki zaroorat hai woh poori hai. Baad mein sab kuch sirf bookkeeping hai.

Step 5 — Teen moments of inertia as integrals likho

KIYA KYA: Step 4 se har $r_{⊥}^{2}$ ko Step 3 ki definition mein plug kiya.

KYUN: ab teen "spin-difficulties" sab ek hi crumb coordinates $(x, y, z)$ se likhi hain. Kyunki yeh coordinates share karte hain, hum inhe compare aur combine kar sakte hain.

PICTURE: Teen panels, ek per axis, har ek same crumb aur woh distance dikhata hai jo woh axis "dekhti" hai. Notice karo ki har integrand do squared coordinates ka sum hai.

Step 6 — Flatten karo: woh ek moment jahan $z = 0$ hota hai

Yeh poore proof ki heartbeat hai. Step 1 se, har crumb ka $z = 0$ hai. Ise $I_{x}$ aur $I_{y}$ mein daalo:

KIYA KYA: do $z^{2}$ terms delete kiye.

YEH THE STEP KYUN HAI: yeh woh ek line hai jahan "flat" use hota hai. Ise hatao aur $I_{x}, I_{y}$ mein apna $z^{2}$ ka baggage rehta hai — aur neeche wala clean sum kabhi nahi aata. Yeh yaad rakhna jab bhi koi solid ball par theorem try kare (Step 7).

PICTURE: Do $z^{2}$ boxes red se cross ho jaate hain; $y^{2}$ (for $I_{x}$ ) aur $x^{2}$ (for $I_{y}$ ) wale survivors glow karte hain.

Step 7 — Inhe add karo aur dekho $I_{z}$ kaise appear hota hai

KIYA KYA: do survivors ko add kiya. Integrals term by term add hote hain (ise linearity kehte hain — sum ka integral, integrals ka sum hota hai).

KYUN KAAM KARTA HAI: $x^{2} + y^{2}$ exactly woh $z$ -axis distance-square hai jo Step 4 mein tha. Do flat spins precisely woh do pieces carry karte hain jo top-spin ko chahiye.

PICTURE: Do black bricks jinpar $\int y^{2} d m$ aur $\int x^{2} d m$ likha hai, ek red brick mein slide ho jaate hain jispar $I_{z}$ likha hai. Yahi theorem hai, draw kiya hua.

Step 8 — Edge & degenerate cases (kabhi surprise mat lo)

Har woh case jo reader ko mil sakta hai, yahan dikhaya gaya hai.

PICTURE: Chaar mini-panels — ball (fails), offset axis (fails), skew axes (fails), thin plate (≈ works). Fail hone wale configurations red se mark hain.

Ek-picture summary

Upar sab kuch, compress kiya hua: crumb at $(x, y, 0)$ → teen Pythagoras distances → $z = 0$ ne $z^{2}$ terms ko maar diya → do flat integrals milke up-axis integral ban jaate hain.

Recall Feynman: poora walkthrough kisi dost ko sunao

Ek coin table par flat rakh do. Us par metal ka ek speck chuno. Kyunki coin flat hai, us speck ki koi height nahi — woh table par baitha hai. Ab teen sawaal pucho: coin ko top ki tarah spin karna kitna mushkil hai (axis upar nikli hai), usse front-to-back flip karna kitna mushkil hai, usse left-to-right flip karna kitna mushkil hai? Har "mushkilat" har speck par (distance-from-axis)² add karti hai. Up-axis tak ki doori, tabletop par Pythagoras se, $x^{2} + y^{2}$ hai. Flip-hardnesses mein $z^{2}$ bhi aana chahta tha height term ke roop mein — lekin har speck ki height zero hai, isliye woh terms gayab ho jaate hain. Jo bachta hai: ek flip $y^{2}$ count karta hai, doosra $x^{2}$ count karta hai. Inhe add karo aur exactly $x^{2} + y^{2}$ milta hai — top-spin hardness. To top = front-flip + side-flip. Aur yeh tabhi kaam karta hai jab coin flat ho: ek mote ball par specks upar chipke hain, height terms bachti hain, aur yeh saaf adding toot jaati hai.

Connections

Same idea in Hinglish
Parallel axis theorem — offset cousin ( $I = I_{c m} + M d^{2}$ ), kisi bhi body ke liye kaam karta hai.
Moment of inertia — definition — woh $\int r^{2} d m$ jo har step ke neeche hai.
Moment of inertia of standard bodies — disc, ring, plate results jo yeh unlock karta hai.
Radius of gyration — kisi bhi $I$ ko $I = M k^{2}$ ki tarah repackage karta hai.
Symmetry arguments in MOI — kyun $I_{x} = I_{y}$ aksar hota hai, jisse aap $I_{z}$ ko halve kar sako.