Parallel axis theorem — I = I_CM + Md² — proof

1.5.6 · Physics › Rotational Mechanics

Derivation from first principles

KAISE — smart coordinates choose karo. Dono axes ko $\overset{z}{^}$ ke along rakho. Origin ko CM par rakho, toh CM-axis $z$ -axis hai origin se guzarti hui. Dusri axis $x y$ -plane ko point $(a, b)$ par pierce kare, jahan $d^{2} = a^{2} + b^{2}$ .

Ek mass element position $(x_{i}, y_{i})$ par (hum $z_{i}$ ignore karte hain — $z$ -axis tak distance mein sirf $x, y$ use hote hain):

CM-axis tak Distance²: $r_{i, C M}^{2} = x_{i}^{2} + y_{i}^{2}$
Doosri axis tak Distance²: $r_{i}^{2} = (x_{i} - a)^{2} + (y_{i} - b)^{2}$

Ab dusri axis ke baare mein $I$ compute karo: $I = \sum_{i} m_{i} [(x_{i} - a)^{2} + (y_{i} - b)^{2}]$

Yeh step kyun? Humne sirf nayi axis location ko definition mein plug kiya. Squares expand karo: $I = \sum_{i} m_{i} [x_{i}^{2} + y_{i}^{2} - 2 a x_{i} - 2 b y_{i} + a^{2} + b^{2}]$

Terms ko cleverly group karo: $I = = I_{C M} i \sum m_{i} (x_{i}^{2} + y_{i}^{2}) - 2 a (⋆) i \sum m_{i} x_{i} - 2 b (⋆⋆) i \sum m_{i} y_{i} + (a^{2} + b^{2}) = M i \sum m_{i}$

Yeh step kyun? Pehla sum literally $I_{C M}$ hai. Aakhri sum total mass $M$ hai, aur $a^{2} + b^{2} = d^{2}$ . Beech ke do wali magic hain.

Therefore: $I = I_{C M} + M d^{2}$

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

Common mistake Steel-manned mistakes

Mistake A: "Main kisi bhi do parallel axes ko unki separation use karke $I_{2} = I_{1} + M d^{2}$ se connect kar sakta hoon." Kyun sahi lagta hai: Yeh symmetric aur tidy dikhta hai. Fix: $- 2 a \sum m_{i} x_{i}$ terms tabhi vanish hote hain jab origin CM ho. Toh exactly ek axis CM se guzarni chahiye. Do arbitrary axes ke liye, CM se hokar jaao (Example 3).

Mistake B: "Axes ke starting points ke beech ki koi bhi 'distance' use karo $d$ ke liye." Kyun sahi lagta hai: Koi bhi "distance" theek lagti hai. Fix: $d$ parallel axes ke beech perpendicular distance hai, equivalently CM se nayi axis tak. Dono ke ⟂ line ke along measure karo.

Mistake C: "Parallel axis theorem hamesha $I$ badhata hai, toh main isse ghatane ke liye bhi use kar sakta hoon." Kar sakte ho — lekin sirf CM ki taraf move karke, aur $I$ CM par minimum hota hai. Fix: Kyunki $M d^{2} \geq 0$ hai, CM-axis sabhi parallel axes mein sabse chhota $I$ deta hai. Isse parallel koi axis nahi hai jiska inertia chhota ho.

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tum ek hammer ghuma rahe ho. Use us jagah ke baare mein ghoomana jahan woh balance karta hai (uska centre point) sabse aasaan hai — kam se kam effort. Apna haath us balance point se door lo aur ghoomana mushkil hota jaata hai. Kitna mushkil? Maano poora hammer ek tiny dot mein shrink ho gaya balance point par, aur tum us dot ko $d$ radius ke circle mein swing karo — woh extra effort exactly $M d^{2}$ hai. Ise "easy" balance-point effort mein add karo aur tumhe nayi jagah ke liye effort milta hai. Bas itna hi!

Flashcards

Parallel axis theorem state karo.

I = I_{C M} + M d^{2}

, jahan

d

= parallel axes ke beech ⟂ distance, jinmein se ek CM se guzarti hai.

Woh ek condition kya hai jo theorem ko valid banati hai?

Ek axis center of mass se guzarni chahiye.

Proof mein cross terms

- 2 a \sum m_{i} x_{i}

kyun vanish hote hain?

Kyunki origin CM hai, isliye

\sum m_{i} x_{i} = \sum m_{i} y_{i} = 0

CM ki definition se.

Theorem mein

d

kya hai?

Do parallel axes ke beech perpendicular distance (= CM se nayi axis tak).

Ek given direction ke parallel sabhi axes mein, minimum I kaun deta hai?

Woh jo center of mass se guzarti hai (kyunki

M d^{2} \geq 0

Rod (mass M, length L) ka apne end ke baare mein

I

\frac{1}{3} M L^{2}

\frac{1}{12} M L^{2} + M (L /2)^{2}

se.

Disc (mass M, radius R) ka rim par perpendicular axis ke baare mein

I

\frac{3}{2} M R^{2}

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

se.

Do non-CM parallel axes ko kaise relate karte hain?

Dono ko alag-alag CM se compute karo aur subtract karo; kabhi directly unke beech theorem apply mat karo.

Connections

Moment of inertia — definition
Center of mass — definition and computation
Perpendicular axis theorem
Radius of gyration
Rotational kinetic energy
Physical pendulum

Derivation from first principles

Flashcards

Connections

Concept Map