Conservation of angular momentum — conditions: Question bank

1.5.12 · D5 · Physics › Rotational Mechanics › Conservation of angular momentum — conditions

Shuru karne se pehle, teen simple-word anchors yaad rakho:

Angular momentum $L$ = "system kitna spin/orbit carry kar raha hai" — ek ghoomte hue body ke liye $L = I ω$ , jahan $I$ (Moment of inertia) batata hai mass kitna phaila hua hai, aur $ω$ ghoomne ki rate hai.
Torque $τ$ = "sideways twist" — $L$ ka rate-changer, bilkul jaisa force momentum ko badalta hai. Dekho Torque.
Master law: $\frac{d L}{d t} = τ_{e x t, n e t}$ . Koi external twist nahi ⟹ spin mein koi change nahi.

Sahi hai ya galat — justify karo

Har jawab mein reason hona chahiye, sirf haan/nahi nahi chalega.

Agar kisi system ki kinetic energy conserved nahi hai, toh uski angular momentum bhi conserved nahi ho sakti.

Galat — dono ki conditions alag-alag hain;

L

ko zero external torque chahiye jabki KE ke liye zero work chahiye, isliye ek skater ka

L

fixed rehta hai jabki uski KE badh jaati hai kyunki uski muscles kaam karti hain.

System ke parts ke beech internal forces total angular momentum ko badal sakte hain.

Galat — internal forces collinear third-law pairs mein aate hain, jinse torque

(r_{1} - r_{2}) \times f = 0

hota hai kyunki

f

(r_{1} - r_{2})

ke along hota hai, isliye ye cancel ho jaate hain. Dekho Newton's third law.

Angular momentum, agar conserved ho, toh har possible axis ke baare mein conserved hota hai.

Galat — ye sirf unhi axes ke baare mein conserved hota hai jahan external torque component zero ho; ek body ka

τ_{e x t} = 0

ek axis ke baare mein ho sakta hai aur doosre ke baare mein nonzero.

Sun ke around orbit karne wale planet ke liye, gravity Sun ke baare mein zero net torque deta hai.

Sahi — gravity ek central force hai jo seedha Sun ki taraf point karti hai, isliye

r ∥ F

aur

τ = r \times F = 0

Kisi given object ke liye

L

ki value same rehti hai chahe tum koi bhi origin chuno.

Galat —

L = r \times p

r

par depend karta hai, jo origin se measure hota hai, isliye alag-alag origins alag-alag

L

values dete hain.

Angular momentum conserved rehta hai chahe bade external forces act kar rahe hon, bas unka net torque zero hona chahiye.

Sahi — jo matter karta hai wo hai torque, force nahi; forces jo origin se guzarti hain (ya jinka twist cancel ho jaata hai)

L

ko unchanged chodh deti hain.

Ek rigid body jiska

L

constant hai, uska angular velocity

ω

bhi necessarily constant hoga.

Galat — agar mass redistribute ho aur

I

badle, toh

ω

badlega taaki

I ω

fixed rahe, jaise skater apni arms andar kheench leti hai.

Kepler ka equal-area law directly angular momentum conservation ka consequence hai.

Sahi — constant

L

ka matlab hai area sweep karne ki rate constant hai, isliye equal time mein equal areas sweep hote hain. Dekho Kepler's laws.

Galti dhundho

Neeche di gayi har statement mein ek flaw hai. Use ek ya do sentences mein identify karo.

"Skater jab apni arms andar kheenchti hai toh speed up ho jaati hai, isliye angular momentum badh gayi."

Galti —

L = I ω

constant rehta hai;

ω

sirf isliye badhta hai kyunki

I

kam hota hai, isliye product unchanged hai, badha nahi.

"Gravity planet ko kheechthi hai, isliye wo kaam karta hai aur planet ki angular momentum badal deta hai."

Galti —

L

badalne ke liye relevant quantity torque hai, force ya work nahi; central gravity zero torque deti hai, isliye

L

unchanged rehta hai.

"Hum

L

ka conservation tabhi use kar sakte hain jab system par koi external forces na hon."

Galti — tum tab bhi use kar sakte ho jab external forces exist karein lekin unka net torque zero ho (jaise origin se guzarne wali central forces).

"Putty turntable se chipak jaati hai, energy heat mein kho jaati hai, isliye turntable ka

L

kam ho gaya."

Galti — impact force internal hai aur gravity axis ke parallel hai, isliye axis ke baare mein

L

conserved hai; energy loss aur

L

conservation alag-alag matters hain.

"Kyunki

L = r \times p

, seedhi line mein move karne wale object ki angular momentum zero hoti hai."

Galti — jab tak line origin se na guzre,

r

aur

p

parallel nahi hote, isliye

L = r \times p \neq = 0

"Ek wheel ramp pe roll karte hue angular momentum conserve karta hai kyunki wo freely roll kar raha hai."

Galti — gravity contact-line axis ke baare mein nonzero torque produce karta hai, isliye us axis ke baare mein

L

conserved nahi hai; conservation claim karne se pehle axis choose karo.

"Internal muscular forces

ω

nahi badal sakti, kyunki internal forces system ko affect nahi karte."

Galti — internal forces

I

ko badal kar $ω$ badal sakti hain (mass move karke); jo wo nahi badal sakti wo hai total

L

Why questions

Har ek ka underlying reason ke saath jawab do.

\frac{d L}{d t} = τ_{e x t, n e t}

mein sirf external torque kyun aata hai?

Kyunki internal third-law force pairs collinear hoti hain, unka combined torque

(r_{1} - r_{2}) \times f

vanish ho jaata hai, sirf external torques bachte hain.

Conservation condition derive karne ke liye hum

L

differentiate kyun karte hain?

Conservation ka matlab hai

L

time ke saath nahi badlta, yaani

\frac{d L}{d t} = 0

, isliye hum exactly ye dekhne ke liye wo derivative compute karte hain ki ye kab zero hoti hai.

Derivation mein pehla term

v \times m v

kyun vanish ho jaata hai?

Kisi bhi vector ka khud ke saath cross product zero hota hai kyunki dono vectors parallel hote hain, isliye

v \times v = 0

Skater ka

L

conserved rehta hai lekin uski rotational kinetic energy nahi — aisa kyun?

Koi external torque nahi hone se

L

fixed rehta hai, lekin uski muscles internal work karti hain mass ko andar kheenchne ke liye, jo Rotational kinetic energy

K E = \frac{1}{2} I ω^{2}

badha deti hai.

L

ka conservation apply karte waqt origin cleverly choose karna kyun help karta hai?

Kyunki

τ

origin par depend karta hai, aisa origin choose karna jahan

τ_{e x t} = 0

ho conservation use karne deta hai chahe doosre points ke baare mein wo fail kare.

Central force "classic" zero-torque case kyun hai?

Ek central force

r

ke along point karti hai (origin ki taraf/se), isliye

τ = r \times F = 0

kyunki parallel vectors ka cross product zero hota hai.

Planet ke liye

L = m v r

sirf perihelion aur aphelion par kyun likhte hain?

Wahan velocity radius ke perpendicular hoti hai, isliye

∣ r \times p ∣ = m v r

exactly hota hai; baaki jagah

v

mein ek radial component hota hai aur formula ko perpendicular part chahiye.

Torque ko angular momentum ka "pace-setter" kyun kaha jaata hai?

Kyunki ye

L

ke rate of change ke barabar hai, exactly wahi role nibhaata hai jo force linear momentum ke liye nibhaata hai Conservation of linear momentum mein.

Edge cases

Boundary aur degenerate scenarios — har ek ke baare mein reason karo.

Agar net external torque zero hai lekin ek single external force nonzero hai, kya

L

conserved hai?

Haan — conservation torque par depend karta hai, force par nahi; origin se guzarne wali nonzero force bhi zero torque aur constant

L

deti hai.

Agar ek particle exactly origin par baitha ho (

r = 0

), uski angular momentum kya hai?

Wo zero hai, kyunki

L = r \times p = 0 \times p = 0

chahe wo kitni bhi tez move kare.

Agar kisi body ka angular velocity

ω

zero ho, kya uski angular momentum necessarily zero hai?

Fixed axis ke around ghoomne wale rigid body ke liye,

L = I ω = 0

; lekin ek translating particle ab bhi orbital

L = r \times p

carry kar sakta hai ek off-line origin ke baare mein bina kisi spin ke.

Kya angular momentum direction mein conserved ho sakti hai lekin magnitude mein nahi, ya vice versa?

Truly constant

L

ke liye nahi — conservation ka matlab poora vector fixed hai, isliye magnitude aur direction dono constant hain; agar koi bhi badle, toh koi external torque act kar raha tha.

Planet ki ek poori orbit mein

L

ka kya hoga, given ki gravity central hai?

Ye poori orbit mein exactly constant rehta hai, kyunki torque har instant par zero hai, sirf average par nahi.

Agar external torque sirf vertical axis ke along zero hai, toh hum ab bhi kya conclude kar sakte hain?

Sirf vertical component

L_{z}

conserved hai;

L

ke horizontal components badal sakte hain kyunki unke torques nonzero hain.

Jab skater ka moment of inertia

I \to

bahut chhota ho jaaye (arms tight tuck mein),

ω

ka kya hoga?

ω \to

bahut bada ho jaata hai taaki

I ω

fixed rahe, isliye wo tezi se ghoomne lagti hai —

I_{1} ω_{1} = I_{2} ω_{2}

ka limiting behaviour.

Recall Yahan har trap ki ek-line summary

Is page par har trap inme se ek confusion hai: (1) $L$ ko KE se milana, (2) ye sochna ki internal forces total $L$ move karte hain, (3) bhool jaana ki conservation per-axis aur origin-dependent hai, ya (4) force aur torque ko confuse karna. Ye chaar theek karo aur topic tumhara hai.

Connections

Parent topic: conditions for conservation
Torque — wo twist jiska absent hona hi poori condition hai.
Moment of inertia — woh $I$ jo $ω$ ke saath trade karta hai.
Newton's third law — isliye internal torques cancel hote hain.
Central forces — zero-torque ka star example.
Kepler's laws — constant $L$ se equal-area law.
Conservation of linear momentum — translational twin.
Rotational kinetic energy — isliye $L$ fixed ≠ KE fixed.