Rolling KE = ½mv² + ½Iω² — Hinglish notes

1.5.14 · Physics › Rotational Mechanics

KYU translation + rotation mein motion split karein?

Scratch se Derivation (König's theorem)

Step 1 — Total KE ko sum ke roop mein likho. $K E = \sum_{i} \frac{1}{2} m_{i} v_{i}^{2} = \sum_{i} \frac{1}{2} m_{i} (v_{c m} + u_{i}) \cdot (v_{c m} + u_{i})$ Yeh step kyun? Speed-squared bas velocity ka khud se dot product hai; isse hum cleanly expand kar sakte hain.

Step 2 — Dot product expand karo. $K E = \sum_{i} \frac{1}{2} m_{i} (v_{c m}^{2} + 2 v_{c m} \cdot u_{i} + u_{i}^{2})$ Yeh step kyun? Standard $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ , lekin vectors ke saath, isliye middle term ek dot product hai.

Step 3 — Teeno terms handle karo.

Term 1: $\sum_{i} \frac{1}{2} m_{i} v_{c m}^{2} = \frac{1}{2} (\sum_{i} m_{i}) v_{c m}^{2} = \frac{1}{2} M v_{c m}^{2}$ . → Translational KE.
Term 2: $\sum_{i} m_{i} v_{c m} \cdot u_{i} = v_{c m} \cdot \sum_{i} m_{i} u_{i} = v_{c m} \cdot M u_{c m, r e l}$ . Lekin $u_{c m, r e l}$ cm ki velocity hai cm se dekhi gayi — jo zero hai! Isliye yeh term vanish ho jaati hai.
Term 3: $\sum_{i} \frac{1}{2} m_{i} u_{i}^{2}$ . cm ke around pure rotation ke liye, $u_{i} = ω r_{i}$ , isliye yeh ban jaata hai $\frac{1}{2} (\sum_{i} m_{i} r_{i}^{2}) ω^{2} = \frac{1}{2} I_{c m} ω^{2}$ . → Rotational KE.

Cross term kyun khatam ho jaata hai (key trick): center of mass apne aap se relative move nahi kar sakta. Woh zero hi wajah hai ki do energies cleanly alag ho jaati hain — yahi pura reason hai ki formula ek clean sum hai.

Rolling-without-slipping condition

$ω = v_{c m} / R$ substitute karo aur $I_{c m} = β M R^{2}$ likho (jahan $β$ ek shape number hai): $K E = \frac{1}{2} M v_{c m}^{2} + \frac{1}{2} (β M R^{2}) \frac{v _{c m}^{2}}{R ^{2}} = \frac{1}{2} M v_{c m}^{2} (1 + β)$

Shape	$I_{c m}$	$β = \frac{I _{c m}}{M R ^{2}}$	Rotational fraction $\frac{β}{1 + β}$
Ring/hollow cylinder	$M R^{2}$	$1$	$1/2$ (50%)
Disc/solid cylinder	$\frac{1}{2} M R^{2}$	$1/2$	$1/3$ (33%)
Solid sphere	$\frac{2}{5} M R^{2}$	$2/5$	$2/7$ (29%)
Hollow sphere	$\frac{2}{3} M R^{2}$	$2/3$	$2/5$ (40%)

Worked examples

Common mistakes

Flashcards

Total KE of a rigid body in plane motion (König's theorem)

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

König's derivation mein cross term kyun vanish hota hai?

Kyunki

\sum m_{i} u_{i} = M u_{c m, r e l} = 0

— cm ki velocity apne aap se relative zero hoti hai.

Rolling-without-slipping constraint

v_{c m} = ω R

(contact point instantaneously rest mein)

β

define karo aur uska role batao

β = I_{c m} / M R^{2}

; yeh energy split set karta hai aur yeh decide karta hai ki body ramp se kitni fast roll karti hai.

Rolling mein KE ka rotational fraction

\frac{β}{1 + β}

Rest se height h se rolling karne wali body ki speed

v = \frac{2 g h}{1 + β}

Solid sphere ke liye height h se

v

kya hai?

10 g h /7

Incline par race ka order: ring, disc, sphere

Sphere sabse pehle, phir disc, phir ring (sabse chhota β jeetta hai)

Kya winner mass ya radius par depend karta hai?

Nahi —

v \propto 1/ 1 + β

, M aur R dono cancel ho jaate hain.

Kya rolling KE ke liye

\frac{1}{2} I_{co n t a c t} ω^{2}

akele use kar sakte hain?

Haan, yeh parallel axis theorem se

\frac{1}{2} M v^{2} + \frac{1}{2} I_{c m} ω^{2}

ke equal hai. Double-add mat karo.

Recall Feynman: 12-saal ke bachche ko explain karo

Socho ek ball floor par roll ho rahi hai. Woh ek saath do kaam kar rahi hai: kamre ke across move karna aur top ki tarah spin karna. Har kaam mein energy lagti hai. "Moving" wali energy regular wali hai ( $\frac{1}{2} m v^{2}$ ). "Spinning" wali energy extra wali hai ( $\frac{1}{2} I ω^{2}$ ). Ball ke paas kitni total go-power hai yeh jaanne ke liye, dono kaamon ko add karo. Ring ka zyada weight edge par hota hai, isliye spinning mein zyada energy lagti hai — isliye ring ek downhill race mein solid ball se harti hai jo apni zyaadatar energy aage jaane ke liye rakhti hai.

Connections

König's Theorem — woh general parent result jiska yeh ek special case hai.
Moment of Inertia — $I_{c m}$ aur $β$ ki values provide karta hai.
Parallel Axis Theorem — isliye contact-point method same KE deta hai.
Rolling Without Slipping — $v = ω R$ constraint provide karta hai.
Conservation of Energy on Inclines — jahan yeh KE formula apply hota hai.
Angular Velocity and Angular Acceleration — $ω$ define karta hai.