Rotational kinetic energy = ½Iω² — Hinglish notes

1.5.9 · Physics › Rotational Mechanics

WHAT hai yeh?

WHY matter karta hai? Energy methods forces aur torques ko sidestep kar dete hain. Agar tumhe sirf start aur end par speeds ki parwah hai (ek rolling ball, ek flywheel, ek yo-yo), toh conservation of energy step by step $τ = I α$ solve karne se kaafi tez hai.

HOW — Scratch se Derivation

Hum ise ek single point mass ke liye $K = \frac{1}{2} m v^{2}$ se build karte hain. Koi bhi formula assume nahi kiya gaya.

Step 1 — Body ko tiny pieces mein kato. Rigid body ko bahut saare particles $m_{1}, m_{2}, \dots$ ke roop mein imagine karo jo axis se $r_{1}, r_{2}, \dots$ distances par hain. Yeh step kyun? Total KE bas har piece ki KE ka sum hai — KE additive hoti hai.

$K = \sum_{i} \frac{1}{2} m_{i} v_{i}^{2}$

Step 2 — $v$ aur $ω$ ke beech rigid-body link use karo. Axis ke baare mein circular motion ke liye, radius $r_{i}$ par ek particle speed se move karta hai $v_{i} = r_{i} ω$ Yeh step kyun? Ek rigid body mein har particle same time mein same angle sweep karta hai, isliye sabhi ek ek $ω$ share karte hain. Linear speed sirf $r_{i}$ ke through alag hoti hai.

Step 3 — Substitute karo aur common factor bahar nikalo. $K = \sum_{i} \frac{1}{2} m_{i} (r_{i} ω)^{2} = \frac{1}{2} ω^{2} \sum_{i} m_{i} r_{i}^{2}$ Yeh step kyun? Kyunki $ω$ sabhi particles ke liye same hai, $\frac{1}{2} ω^{2}$ ek constant hai aur sum se bahar factor ho jaata hai.

Step 4 — Bache hue sum ko naam do. Quantity $\sum_{i} m_{i} r_{i}^{2}$ sirf is par depend karti hai ki mass axis ke baare mein kaise distribute hai. Hum ise moment of inertia kehte hain: $I \equiv \sum_{i} m_{i} r_{i}^{2} ⟹ K_{r o t} = \frac{1}{2} I ω^{2}$ Yeh step kyun? Yeh saari geometric cheez ko ek symbol $I$ mein isolate kar deta hai, ek aisa formula chhod kar jo bilkul $\frac{1}{2} m v^{2}$ jaisa lagta hai jisme $m \to I$ aur $v \to ω$ . Woh analogy hi poori baat hai.

Linear ↔ Rotational analogy

Linear	Rotational
mass $m$	moment of inertia $I$
velocity $v$	angular velocity $ω$
$K = \frac{1}{2} m v^{2}$	$K = \frac{1}{2} I ω^{2}$

Worked examples

Common mistakes (steel-manned)

Recall Feynman: 12-saal ke bachche ko explain karo

Ek merry-go-round imagine karo jo bachcho se bhari ho. Har bachcha ghoom raha hai, isliye har ek ke paas "moving energy" hai. Edge ke paas wale bachche tezi se zoom karte hain; middle ke paas wale mushkil se move karte hain. Total energy nikalne ke liye tum har bachche ki energy add karte. Har baar aisa karne ki jagah, hum note karte hain ki sab bachche ek saath same spin rate $ω$ par ghoomte hain. Hum "kitna heavy aur kitna spread out" ko ek number $I$ mein bundle kar dete hain. Tab spinning energy bas $I$ ka aadha times spin rate squared hai — bilkul waisi hi jaise ek moving truck ki energy uski mass ka aadha times uski speed squared hai. Bachcho ko aur bahar spread karo → spin karna mushkil → zyada energy store.

Active-recall flashcards

#flashcards/physics

Rotational kinetic energy define karo.

Ek rigid body ki KE jo axis ke baare mein spin kar rahi ho,

K = \frac{1}{2} I ω^{2}

ω

particle sum se bahar kyun factor hota hai?

Ek rigid body mein har particle same

ω

share karta hai, isliye

\frac{1}{2} ω^{2}

sum ke across constant hai.

Particles ke terms mein

I

kya hai?

I = \sum_{i} m_{i} r_{i}^{2}

— (mass × distance-from-axis²) ka sum.

\frac{1}{2} m v^{2}

ke rotational analogue mein

m

aur

v

ki jagah kya aata hai?

m \to I

aur

v \to ω

Bina slip kiye rolling karne wali object ki total KE?

K = \frac{1}{2} M v_{c m}^{2} + \frac{1}{2} I_{c m} ω^{2}

with

v_{c m} = R ω

\frac{1}{2} I ω^{2}

mein

ω

degrees mein hona chahiye ya radians mein?

Radians per second (kyunki derivation ne

v = r ω

use kiya).

Ek solid sphere ramp ke bottom par sliding block se slower kyun pahunchta hai?

PE ka kuch hissa rotational KE ban jaata hai, translation ke liye kam bachta hai:

v = 10 g h /7 < 2 g h

Hoop vs disc, same M,R,ω: kaun zyada rotational KE rakhta hai aur kyun?

Hoop (double), kyunki uski saari mass radius

R

par baithti hai, ek bada

I = \sum m r^{2}

deti hai.

Kya

I

chosen axis par depend karta hai?

Haan — jaise rod center ke baare mein

\frac{1}{12} M L^{2}

hai, end ke baare mein

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

Disc M=20kg, R=0.5m, ω=300 rad/s ki KE?

I = 2.5

K = \frac{1}{2} (2.5) (300)^{2} = 1.125 \times 1 0^{5}

Connections

Moment of inertia — formula mein $I$ ; mass distribution se compute hota hai.
Kinetic energy of a particle — $\frac{1}{2} m v^{2}$ jo humne sum kiya.
Rolling motion — translational + rotational KE combine karta hai.
Work-energy theorem (rotational) — $W = Δ K_{r o t}$ via $τ d θ$ .
Parallel axis theorem — energy calc se pehle $I$ ko off-center axes par shift karta hai.
Angular velocity — $ω$ ko rad/s mein define karta hai.
Conservation of energy — woh principle jo example 2 ko power karta hai.