Parallel axis theorem — I = I_CM + Md² — proof: Visual walkthrough

1.5.6 · D2 · Physics › Rotational Mechanics › Parallel axis theorem — I = I_CM + Md² — proof

Step 0 — "Moment of inertia" ka matlab kya hai?

Squared kyun, sirf $r_{i}$ kyun nahin? Kyunki kisi cheez ko door se spin karne par ek bada circle sweep hota hai aur same turn-rate ke liye zyada fast move hota hai — dono effects multiply hote hain, ek ek factor of $r$ dete hain. Yahi $r^{2}$ tum Rotational kinetic energy mein phir miloge. Yaad rakho.

Figure — Parallel axis theorem — I = I_CM + Md² — proof

Figure dekho: axis vertical teal line hai, ek dana $m_{i}$ side mein baitha hai, aur burnt-orange arrow uski perpendicular distance $r_{i}$ hai — dane se axis tak ki sabse choti rope. Woh arrow, squared aur mass-weighted, sum ka ek term hai.

Step 1 — Do parallel axes: puri problem ek picture mein

HUM KYA CHAHTE HAIN: Axis 2 ke baare mein $I$ , given ki hum already $I_{C M}$ jaante hain (Axis 1 ke baare mein $I$ ).

KYUN KARNA HAI: Naye axis ke liye scratch se $I$ compute karna matlab poora sum dobara karna. Theorem hamare liye $I_{C M}$ reuse karne deta hai aur bas ek correction pay karte hain. Woh reuse hi pura payoff hai.

Dono teal dots woh jagah hain jahan axes page ko pierce karte hain (axes seedhe kaagaz se oopar tumhari taraf nikal rahe hain). Unke beech ka plum segment $d$ hai. Notice karo ek axis exactly balance point pe baitha hai — yeh condition Step 5 mein ek secret kaam kar rahi hai.

Step 2 — Smart coordinates chuno (origin CM par)

Maano doosra axis page ko point $(a, b)$ par pierce karta hai:

$a$ — naya axis CM se kitna door right mein hai,
$b$ — woh kitna oopar hai.

Kyunki $d$ CM $(0, 0)$ se $(a, b)$ tak ki straight-line distance hai, Pythagoras deta hai $d^{2} = a^{2} + b^{2}$

$a^{2} + b^{2}$ — horizontal-shift² plus vertical-shift²,
$= d^{2}$ — perpendicular gap, squared. Yeh chote plum triangle par sirf hypotenuse rule hai.

Orange grid CM par centered hai (isliye origin cross uski pe baitha hai). Naya axis $(a, b)$ par land karta hai; dotted right triangle $a$ , $b$ , aur plum hypotenuse $d$ dikhata hai.

Step 3 — Ek dane ke liye do distances likho

$a$ aur $b$ subtract kyun karte hain? Distance hamesha "grain minus jis cheez se measure kar rahe ho" hoti hai. Origin se tum $(0, 0)$ subtract karte ho (toh kuch nahi badlata); shifted axis se tum $(a, b)$ subtract karte ho. Har bracket grain se us axis tak ke right triangle ki ek leg hai, aur square-karke-add-karna phir se Pythagoras hai.

Wahi dana, do ropes: orange rope $r_{i, C M}$ CM-axis tak jaati hai, teal rope $r_{i}$ shifted axis tak jaati hai. Chote legs $(x_{i} - a)$ aur $(y_{i} - b)$ dotted draw kiye hain — woh formula ke andar ke brackets hain.

Step 4 — Sum mein daalo aur expand karo

KYA: naye axis ki distance ko $I = \sum_{i} m_{i} r_{i}^{2}$ mein daalo: $I = \sum_{i} m_{i} [(x_{i} - a)^{2} + (y_{i} - b)^{2}]$

EXPAND KYUN? Body brackets ke andar chhupi hai. Multiply out karne se "grain-only" pieces "shift-only" pieces se alag ho jaate hain taki hum purane doston ko pehchaan sakein: $I = \sum_{i} m_{i} [grain’s spread x_{i}^{2} + y_{i}^{2} cross terms - 2 a x_{i} - 2 b y_{i} + = d^{2} a^{2} + b^{2}]$

Ab ek sum ko chaar alag sums mein baanto (allowed hai — adding is adding, kisi bhi order mein): $I = (A) i \sum m_{i} (x_{i}^{2} + y_{i}^{2}) - 2 a (B) i \sum m_{i} x_{i} - 2 b (C) i \sum m_{i} y_{i} + (a^{2} + b^{2}) (D) i \sum m_{i}$

Term by term:

(A) $\sum m_{i} (x_{i}^{2} + y_{i}^{2})$ — har dane ki distance² CM-axis se, mass-weighted. Yeh literally $I_{C M}$ hai.
(B), (C) — mysterious cross terms; $a$ aur $b$ front mein nikale kyunki woh har dane ke liye same hain.
(D) $\sum m_{i} = M$ — bas total mass; aur $(a^{2} + b^{2}) = d^{2}$ .

Figure chaar colored buckets sort karta hai: (A) orange mein = $I_{C M}$ , (D) teal mein = $M d^{2}$ , aur (B),(C) plum mein = "?" jo hum aage resolve karenge.

Step 5 — Cross terms gayab ho jaate hain (proof ka dil)

Physically iska matlab: " $\sum m_{i} x_{i} = 0$ " kehta hai CM ke right ka mass exactly left ke mass ko balance karta hai (yahi toh ek balance point hota hai). Toh cross terms cancel hote hain luck se nahin, balki "center of mass" ke meaning se.

Har dana $m_{i} x_{i}$ contribute karta hai: orange bars right (positive $x$ ) mein point karte hain, teal bars left (negative $x$ ) mein. Unhe stack karo aur woh ek flat zero line tak cancel ho jaate hain — cross terms ki visual death.

Step 6 — Bachne waale terms collect karo

(B) aur (C) gone, sirf (A) aur (D) bachte hain: $I = (A): CM ke baare mein real spread I_{C M} + (D): point mass M at distance d M d^{2}$ $I = I_{C M} + M d^{2}$

$I_{C M}$ — balance line ke baare mein body ki honest inertia,
$M d^{2}$ — woh inertia jo body ki hoti agar woh ek single dot of mass $M$ mein $d$ door baith ke simit jaati. (Yeh Radius of gyration mein dekho, jahan $I = M k^{2}$ mein $k$ ek "effective distance" hai.)

Yeh beautiful kyun hai: correction sirf itne par depend karta hai ki kitna door gaye, kabhi body ki shape par nahin. Kisi bhi body ko uske CM se $d$ door karo aur tum wahi $M d^{2}$ pay karte ho.

Step 7 — Edge aur degenerate cases (koi gap mat chhodna)

Plum curve $I (d) = I_{C M} + M d^{2}$ hai — ek parabola, CM ( $d = 0$ ) par bottom karta hua. Do sample axes A aur B curve par hain; unki heights subtract hoti hain, unke $d$ nahin hote.

Ek picture summary

Left se right padho: raw sum se shuru karo, origin CM par plant karo, chaar buckets mein expand karo, dekho ki do plum cross-term buckets zero ho jaate hain, aur jo bachta hai woh hai $I_{C M} + M d^{2}$ . Yeh puri derivation ek frame mein hai.

Recall Feynman retelling — poora walkthrough simple words mein

Kisi cheez ko spin karna utna mushkil hai jitna mass spin line se door baitha ho, squared count hota hua. Humne poocha: agar mujhe pehle se pata hai balance point ke baare mein spin karna kitna mushkil hai, toh $d$ door kisi parallel line ke baare mein kitna mushkil hoga? Humne balance point par centered ek grid rakha aur har dane ki naye line tak ki distance likhi. Jab humne multiply out kiya, chaar tarah ke terms aaye: (1) grains ka balance point ke baare mein spread — woh "easy" number jo hum pehle se jaante the; (2) do mixed terms; (3) total mass times gap squared. Mixed terms bas "balance point ke around mass kitna lopsided hai" hain — aur ek balance point ke around, by definition, lopsidedness zero hai. Left aur right exactly cancel karte hain. Toh woh gayab ho jaate hain. Jo bachta hai: easy number, plus mass-times-gap-squared — poori cheez ko distance $d$ par ek dot ki tarah swing karne ki mehnat. Home base sabse sasta hai; har kadam bahar jaane ki cost $M d^{2}$ hai.

Connections

1.5.06 Parallel axis theorem — I = I_CM + Md² — proof (Hinglish)
Moment of inertia — definition
Center of mass — definition and computation
Perpendicular axis theorem
Radius of gyration
Rotational kinetic energy
Physical pendulum