Rotational kinetic energy = ½Iω²: Question bank

1.5.9 · D5 · Physics › Rotational Mechanics › Rotational kinetic energy = ½Iω²

Intuition Is bank ko kaise use karein

Ye woh questions hain jo "maine $K = \frac{1}{2} I ω^{2}$ mein plug kiya" aur "main $K = \frac{1}{2} I ω^{2}$ ko samajhta hoon" ke beech fark karte hain. Neeche har jawab ek reason hai, verdict nahi. Right side cover karo, apni reasoning zaur se bolo, phir reveal karo. Agar tumhara kyun galat hai — chahe haan/nahi sahi ho — toh usse galat mark karo.

Symbols used (all built in the parent note): $K$ = kinetic energy (joules), $r$ (or $r_{i}$ ) = the perpendicular distance of a particle from the rotation axis (in metres) — woh ek geometric quantity jis par neeche sab kuch tika hai; $I = \sum_{i} m_{i} r_{i}^{2}$ = moment of inertia (axis ke baare mein mass ka spread, $kg\cdotm^{2}$ ), $ω$ = angular velocity (spin rate in radians per second), $M$ = total mass, $R$ = radius, $v_{c m}$ = speed of the center of mass.

True or false — justify

Ek body jiska center of mass stationary hai uski kinetic energy zero hoti hai.

False. Agar woh spin kare, toh axis par jo particles hain unke alawa har particle move kar raha hai, isliye

K = \frac{1}{2} I ω^{2} > 0

hoti hai chahe

v_{c m} = 0

ho. Poori body ka motion zaruri nahi — parts ka motion kaafi hai.

Equal mass aur equal

ω

wale do objects ki rotational KE hamesha equal hoti hai.

False.

K

depend karta hai

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

par (jahan

r_{i}

= distance from the axis), na ki sirf mass par. Usi mass ko bade

r

par spread karo toh

I

(aur

K

) badhta hai — ek hoop equal

M, R, ω

wali disc se double energy store karta hai.

ω

ko double karne se rotational kinetic energy double ho jaati hai.

False.

K \propto ω^{2}

hai, isliye

ω

double karne se

K

chaar guna ho jaata hai. Yahi square hai jis wajah se flywheels ko utni tez spin karte hain jitni material allow kare.

Fixed

ω

par moment of inertia double karne se rotational kinetic energy double ho jaati hai.

True.

K = \frac{1}{2} I ω^{2}

mein

I

linear hai. Sirf

ω

square hota hai —

I

pehli power par aata hai.

Bina slip kiye rolling karne wali body ki saari kinetic energy rotational hoti hai.

False. Uske paas dono hoti hain:

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

. Center of mass sach mein translate karta hai, isliye translational KE ek real, alag term hai.

Moment of inertia

I

ek object ki fixed property hai, jaise uska mass.

False.

I

depend karta hai chosen axis par. Same rod ka

I

center ke baare mein

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

hai lekin end ke baare mein

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

hai. Axis badlo,

I

badal jaata hai.

Rotational KE negative ho sakti hai agar body "ulti" spin kare.

False.

K = \frac{1}{2} I ω^{2}

mein

ω^{2} \geq 0

aur

I > 0

hai, isliye

K \geq 0

hamesha, spin direction chahe koi bhi ho.

ω

ka sign badlne se

ω^{2}

same rehta hai.

Ek sliding block aur ek uniform solid sphere (

I_{c m} = \frac{2}{5} M R^{2}

) same height se release hone par neeche same speed se pahunchte hain.

False. Sphere ki gravitational energy translation aur spin dono mein bant jaati hai. Us specific inertia ke liye algebra deta hai

v = 10 g h /7

, jo block ke

2 g h

se kam hai. Rotation mein energy store karna descent ko slow karta hai. (Alag shape, jaise hoop, alag aur aur bhi slow speed deta hai.)

Agar do flywheels same

ω

par same energy store karte hain, toh unka mass same hona chahiye.

False. Equal

ω

par equal

K

se equal

I

milta hai, equal

M

nahi. Axis se door ek light ring, paas wali heavy disc ke barabar ho sakti hai.

Spot the error

ω = 60

deg/s, so

K = \frac{1}{2} I (60)^{2}

." — galti kahan hai?

ω

radians/s mein hona chahiye, kyunki derivation mein

v = r ω

use hua tha jo sirf radians mein valid hai. Pehle convert karo:

6 0^{\circ} / s = 60 \times π /180 \approx 1.05 rad/s

, phir usse square karo.

K = \frac{1}{2} (I ω)^{2}

." — kya galat hua?

Sirf

ω

square hota hai,

I

nahi: yeh

\frac{1}{2} I ω^{2}

hai, na ki

\frac{1}{2} (I ω)^{2}

. (Quantity

I ω

angular momentum hai — ek alag cheez.)

"Ball rolling kar rahi hai, isliye

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

." — kya missing hai?

Rotational term

\frac{1}{2} I_{c m} ω^{2}

. Rolling body translate bhi karta hai aur spin bhi; spin term drop karne se energy underestimate hoti hai aur speed overestimate.

"Center ke baare mein

I

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

hai, isliye main yahi use karunga chahe object rim par ek point ke baare mein spin kare." — ise fix karo.

Pehle Parallel axis theorem se axis shift karni hogi:

I = I_{c m} + M d^{2}

, jahan

d

do axes ke beech ki distance hai. Rim axis ke liye center value use karne se

M d^{2}

term miss ho jaata hai aur

I

undercount hota hai.

"Kai particles ki total KE pane ke liye main unki speeds average karunga aur

\frac{1}{2} M v_{a v g}^{2}

use karunga." — yeh kyun galat hai?

KE average speed se nahi banti; yeh sum

\sum \frac{1}{2} m_{i} v_{i}^{2}

hai. Kyunki speed square hoti hai, tez outer (large-

r

) particles disproportionately zyada contribute karte hain — pehle average karna woh

r^{2}

weighting destroy kar deta hai jo

I

ban jaata hai.

"Spin rate same hai, mass same hai, isliye dono shapes ki KE clearly equal hai." — kya ignore kiya gaya?

Mass ki distribution — har piece axis se kitni door baithe hai, jo

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

mein capture hoti hai. Same

M

aur

ω

lekin alag geometry matlab alag

I

, isliye alag

K

Why questions

\frac{1}{2} ω^{2}

ko particle-by-particle sum se bahar kyun nikala ja sakta hai?

Kyunki body rigid hai: har particle same time mein same angle sweep karta hai, isliye sabka ek common

ω

share hota hai. Ek constant factor sum se bahar aa jaata hai.

I

distance ke square ki tarah kyun badhta hai, na ki sirf distance ki tarah?

Axis se

r

door ek particle ki speed

v = r ω

hai, aur KE mein

v^{2} = r^{2} ω^{2}

aata hai. Energy mein yeh

r^{2}

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

mein survive karta hai.

Mass ko baahir move karne se object spin up karna mushkil aur given

ω

par zyada energy store karna dono kyun hota hai?

Dono effects bade

I

se aate hain: bada

I

matlab same angular acceleration ke liye zyada torque chahiye (

τ = I α

) aur same spin rate par zyada energy

\frac{1}{2} I ω^{2}

stored. Ek quantity dono ko govern karti hai.

Ball ke ramp se neeche aane ke liye

τ = I α

ki bajaye energy methods prefer kyun karte hain?

Energy conservation sirf start aur end states chahti hai, path mein varying friction aur geometry ko bypass karte hue.

τ = I α

solve karne ke liye har instant par forces track karni padti.

I

quote karne se pehle axis specify karna kyun zaroori hai?

Kyunki

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

har

r_{i}

ko us axis se distance measure karta hai; axis badlo toh har

r_{i}

badal jaata hai, isliye

I

(aur usse compute ki gayi koi bhi energy) badal jaati hai.

Bina slip kiye rolling karne wale wheel par friction net work kyun nahi karta?

Contact point ground ke relative instantaneously rest mein hota hai, isliye static friction force wahan zero displacement se act karta hai aur koi energy transfer nahi hoti — yeh sirf translation aur rotation ke beech motion redirect karta hai.

\frac{1}{2} I ω^{2}

formula deliberately

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

jaisa kyun banaya gaya hai?

Linear↔rotational analogy explicit karne ke liye:

m \to I

(motion ka resistance) aur

v \to ω

(motion ki rate). Parallel recognize karne se linear intuition directly reuse ho sakta hai.

Edge cases

ω = 0

par spin karne wali body ki rotational KE kya hai?

Zero.

K = \frac{1}{2} I (0)^{2} = 0

— koi spin nahi, koi rotational energy nahi, chahe

I

kitna bhi bada ho.

Ek point mass rotation axis par bilkul baithe hai. Woh kitni rotational KE carry karta hai?

Zero. Axis se uski distance

r = 0

hai, isliye

v = r ω = 0

; yeh

\sum m_{i} r_{i}^{2}

mein kuch contribute nahi karta. Sirf off-axis mass rotational energy store karta hai.

Ek idealized axis (ek geometric line) ki "zero radius" hoti hai. Kya axis par material ke paas bhi spin se KE hoti hai?

Nahi. Axis par points ka

r = 0

hota hai aur body ghoomne par woh apni jagah rahte hain, isliye woh koi rotational KE carry nahi karte chahe unke aaspaas ki body move kar rahi ho.

Kya same

ω

par ek bahut light object ek bhaari object se zyada rotational energy store kar sakta hai?

Haan, agar uska mass axis se door ho. Kyunki

K = \frac{1}{2} (\sum m_{i} r_{i}^{2}) ω^{2}

hai, bade

r

par ek chota mass bhi chote

r

par bade mass se zyada ho sakta hai.

Jab

ω \to \infty

K

ka kya hota hai, aur real flywheels ki limit kyun hoti hai?

K \to \infty

mathematically (

K \propto ω^{2}

), lekin real material pehle fail ho jaata hai — rim par baahri khichav (

\propto ω^{2} r

) eventually material ki strength se zyada ho jaata hai aur wheel burst ho jaata hai. Energy storage formula ki nahi, strength ki wajah se cap hoti hai.

Jis limit mein body ka saara mass axis par collapse ho jaaye, kisi bhi

ω

par uski rotational KE kya hogi?

Zero. Jab har

r_{i} \to 0

I \to 0

aur

K = \frac{1}{2} I ω^{2} \to 0

. Bina spread ke mass koi spin energy store nahi karta.

Ek body bina kisi rotation ke seedhi line mein translate kare. Kaun se KE terms apply hote hain?

Sirf translational,

\frac{1}{2} M v_{c m}^{2}

ω = 0

hone se rotational term vanish ho jaata hai — general rolling formula pure-translation case mein reduce ho jaata hai.

Ek wheel fixed axle ke baare mein spin kare apni jagah (center ka koi translation nahi). Kaun se KE terms apply hote hain?

Sirf rotational,

\frac{1}{2} I ω^{2}

v_{c m} = 0

hone se translational term drop ho jaata hai — pure-translation case ka mirror image.

Connections

Rotational kinetic energy = ½Iω² — woh formula jise ye traps probe karte hain.
Moment of inertia — kyun $I$ axis aur mass spread par depend karta hai.
Kinetic energy of a particle — woh $\frac{1}{2} m v^{2}$ pieces jo yahan sum hote hain.
Rolling motion — "sirf translational" trap ka source.
Parallel axis theorem — galat-axis $I$ ka fix.
Angular velocity — radians-not-degrees pitfall.
Conservation of energy — ramp reasoning ke peeche.
Work-energy theorem (rotational) — kyun rolling mein friction net work nahi karta.