Torque = dL - dt: Question bank — Hinglish deep dive

1.5.11 · D5 · Physics › Rotational Mechanics › Torque = dL - dt

Shuru karne se pehle, aao har symbol aur har logical step ko dobara kamaaein, taaki is page par kuch bhi parent note ki yaad par na tike.

Definition

$\frac{d}{d t}$ ka matlab kya hai — ek arrow ke liye bhi? Symbol $\frac{d ( )}{d t}$ padhte hain "brackets mein jo cheez hai uski rate of change per second". Ek plain number jaise temperature ke liye ye uske time graph ki slope hai. Ek vector (ek arrow jisme length aur direction dono hain) ke liye iska matlab hai: abhi ka arrow lo, thodi der baad ka arrow lo, unhe tip-to-tip subtract karke chhota change-arrow nikalo, aur us tiny time se divide karo. Toh $\frac{d L}{d t}$ ek arrow hai jo batata hai ki $L$ kis taraf shift ho raha hai — ye $L$ ko badha sakta hai, ghata sakta hai, ya uski direction swing kar sakta hai (ek wobble) chahe uski length na bade. Yahi last case hai kyun $τ = d L / d t$ sirf spin-up nahi balki precession bhi describe karta hai. Ye picture yaad rakho: ek arrow ki derivative khud ek arrow hoti hai.

Recall Derivation chaar lines mein (taaki tumhe kabhi is page se bahar na jaana pade)

Ye woh steps hain jinka is page mein baar baar zikr hota hai. Step 1: $L = r \times p$ se shuru karo. Step 2: product rule se differentiate karo: $\frac{d L}{d t} = \frac{d r}{d t} \times p + r \times \frac{d p}{d t}$ . Step 3: pehla term hai $v \times m v = m (v \times v) = 0$ kyunki koi bhi arrow apne aap se cross kiya jaaye toh zero deta hai (parallel arrows, $sin 0 = 0$ ). Step 4: doosre term mein $\frac{d p}{d t} = F$ (Newton), aur $r \times F = τ$ . Result: $τ = \frac{d L}{d t}$ .

Formula

$τ = \frac{d ( I ω )}{d t}$ ko chain rule se unpack karna — $I α$ kahan se aata hai Fixed axis ke liye $L = I ω$ , aur $I$ khud time ke saath change ho sakta hai (arms andar aana). Product $I ω$ ko differentiate karo: $τ = \frac{d ( I ω )}{d t} = I α I \frac{d ω}{d t} + shape-change term ω \frac{d I}{d t} = I α + ω \frac{d I}{d t}$

Do terms kyun hain? Product rule kehta hai: (kuch $\times$ kuch-aur) ki rate-of-change = (pehla badal raha $\times$ doosra fixed) + (pehla fixed $\times$ doosra badal raha). Yahan $I$ aur $ω$ dono badal sakte hain.
$τ = I α$ kab milta hai? Sirf tab jab mass apni shape rakhta hai, $\frac{d I}{d t} = 0$ , jo doosra term khatam karta hai. Ye rigid-body-fixed-axis case hai Torque and Angular Acceleration (tau = I alpha) mein.
Zero torque wali skater: $τ = 0$ set karo, toh $I α = - ω d I / d t$ . Jab wo arms andar kheenchti hai, $d I / d t < 0$ , jo $α > 0$ force karta hai — wo speed up karti hai chahe koi external twist na ho. Ye $L$ ka conservation hai chain rule ke zariye dekha gaya.

True ya false — justify karo

TRUE ya FALSE: Ek force jo seedha pivot

O

ki taraf point karti hai woh

O

ke baare mein koi torque produce nahi karti.

True. Us line ke saath

r

aur

F

parallel hain, toh

θ = 0

aur

r F sin θ = 0

— twist karne ke liye koi perpendicular part nahi hai.

TRUE ya FALSE:

τ = I α

hamesha bilkul

τ = d L / d t

ke equivalent hai.

False. Ye sirf tab agree karte hain jab

I

constant ho. General chain-rule form hai

τ = I α + ω d I / d t

; agar

I

change karta hai (skater arms andar kheenchti hai) toh doosra term matter karta hai aur akela

I α

galat hai.

TRUE ya FALSE: Agar ek system ka angular momentum conserved hai, toh uski rotational kinetic energy bhi conserved hai.

False. Skater ko lo

ω_{1} = 2

rad/s par

I_{1} = 6

kg·m² ke saath; arms andar kheenchne par

I_{2} = 2

kg·m² tak

L = I ω

fixed rehta hai toh

ω_{2} = 6

rad/s. Tab

K E = \frac{1}{2} I ω^{2}

\frac{1}{2} (6) (2^{2}) = 12

J se badh kar

\frac{1}{2} (2) (6^{2}) = 36

J ho jaata hai — 3 ka factor. Uske muscles ne woh energy supply ki;

L

conserve karna energy conserve karne ke baare mein kuch nahi kehta.

TRUE ya FALSE: Ek object ko nonzero angular momentum rakhne ke liye rotate karna zaroori hai.

False. Ek particle seedha line mein glide karta hua

L = m v d

rakhta hai (

d

O

se line tak perpendicular distance), ek fixed nonzero value chahe kuch bhi spin na kare.

TRUE ya FALSE: Tum ek point ke baare mein torque compute kar sakte ho aur doosre point ke baare mein angular momentum aur phir bhi

τ = d L / d t

ki umeed rakh sakte ho.

False. Derivation ka Step 1

L = r \times p

se ek origin ke baare mein shuru karta hai; dono quantities ek hi point share karni chahiye (aur wo inertial ya centre of mass hona chahiye), warna equation meaningless hai.

TRUE ya FALSE: Agar ek body par net external torque zero hai, toh uski angular velocity

ω

constant rehni chahiye.

False. Zero torque

L = I ω

fix karta hai,

ω

ko nahi. Agar

I

change kare, toh

L

constant rakhne ke liye

ω

change karta hai.

TRUE ya FALSE: Ek rigid body ke andar internal forces uska total angular momentum change kar sakte hain.

False. Internal forces Newton's-third-law pairs hain jo joining line ke saath act karte hain; unke torques equal, opposite, aur collinear hote hain, toh unka sum zero hota hai. Sirf external torque

L

change karta hai.

TRUE ya FALSE: Term

\frac{d r}{d t} \times p

(jab tum

L

differentiate karte ho toh pehla term) sirf circular motion ke liye vanish hota hai.

False. Ye

v \times m v = m (v \times v) = 0

equals karta hai kisi bhi motion ke liye — kisi vector ka apne aap se cross product hamesha zero hota hai, chahe path koi bhi ho.

TRUE ya FALSE: Ek nonzero net torque ek aisi body par act kar sakta hai jiska angular momentum momentarily zero ho.

True.

τ = d L / d t

L

ki rate of change ke baare mein hai; jis instant ek body rest se spin karna shuru kare

L = 0

hai lekin

τ \neq = 0

— yahi to spinning shuru hone ka tarika hai.

TRUE ya FALSE: Pivot se distance double karne par hamesha ek given force ka torque double ho jaata hai.

False. Torque hai

r F sin θ

;

r

double karne par ye double hota hai sirf tab jab

r

aur

F

ke beech ka angle

θ

unchanged ho. Mass ko aisa move karo ki

F

zyada

r

ke along point kare aur extra distance kuch kaam nahi aata.

TRUE ya FALSE: Kyunki

τ = d L / d t

ek vector equation hai,

L

ke perpendicular ek torque

L

ki direction change karta hai bina uski length change kiye.

True.

L

ke right angles par ek chhota change-arrow

τ d t

add karne par

L

sideways swing ho jaata hai — uski length fixed rehti hai jabki wo rotate karta hai. Ye exactly ek spinning top ki wobble (precession) hai; planar

I α

intuition ise capture nahi kar sakta.

Error dhundho

Galti dhundho: "Skater apni arms andar kheenchne ke baad tezi se spin karti hai, toh angular momentum badh gaya."

Angular momentum conserved hai (

I ω

constant); wo nahi badha. Wo tezi se isliye spin karti hai kyunki

I

girta hai aur

ω

L

fixed rakhne ke liye badhna padta hai.

Galti dhundho: "Ek seedhi line mein move karte particle par koi net force nahi hai, toh kisi bhi point ke baare mein uska angular momentum zero hai."

Koi force nahi matlab

L

constant hai, zero nahi. Uska constant value

L = m v d

hai, jo motion ki line se hatke kisi bhi point ke liye nonzero hai.

Galti dhundho: "Kyunki

F = ma

, rotational law

τ = I α

hona chahiye — bas."

F = d p / d t

ka sach mein analogue

τ = d L / d t

hai.

τ = I α

constant

I

wala special case hai, bilkul jaise

F = ma

F = d p / d t

ka constant

m

wala case hai.

Galti dhundho: "Torque aur force ek hi direction mein point karte hain, kyunki torque bas rotational force hai."

Torque

r \times F

hai; right-hand rule se cross product dono

r

aur

F

ke perpendicular point karta hai (rotation ki axis ke saath), generally

F

ke along bilkul nahi.

Galti dhundho: "Ek torque kuch time tak act karne par utni kinetic energy add karta hai jitni angular momentum usne deliver ki."

Torque angular momentum deliver karta hai (

\int τ d t = Δ L

), jiske units energy se alag hain. Energy work

\int τ d θ

se aati hai; dono alag accountings hain.

Galti dhundho: "String tension pulley par koi torque nahi deti kyunki tension string ke along kheenchti hai, wheel ke around nahi."

String rim ko tangentially chhod ti hai, toh

r

(axle-se-rim)

T

ke perpendicular hai:

θ = 90°

sin 90° = 1

, aur

τ = R T

maximum hai, zero nahi.

Galti dhundho: "

τ = d (I ω) / d t

bas

I α

ke equal hai kyunki

ω

hi change ho raha hai."

Sirf tab jab

I

fixed ho. Product rule deta hai

I α + ω d I / d t

; doosra term ignore karna exactly wahi galti hai jo skater aur collapsing-star cases mein cheezein tod deti hai.

Why questions

τ = d L / d t

ke liye reference point inertial ya centre of mass kyun hona chahiye?

Accelerating frame mein har particle ek fake force

- m a_{O}

feel karta hai, jiska total torque

- (M R_{cm}) \times a_{O}

hai; ye vanish hota hai jab

O

inertial ho (

a_{O} = 0

) ya jab

O

centre of mass ho (

R_{cm} = 0

), aur sirf tabhi clean law bach ta hai.

Torque aur angular momentum mein ordinary multiplication ki jagah cross product kyun aata hai?

Sirf

r

ke perpendicular force ka component

O

ke baare mein twist kar sakta hai. Cross product ka

sin θ

automatically useless parallel part discard kar deta hai aur twisting part rakh leta hai.

τ = d L / d t

τ = I α

se "zyada fundamental" kyun kaha jaata hai?

Ye directly Newton's law se aata hai

I

ke baare mein koi assumption kiye bina, toh ye changing-

I

situations (skaters, collapsing stars) cover karta hai jahan dropped term

ω d I / d t

otherwise kho jaata.

τ = d L / d t

ek wobbling top describe kyun kar sakta hai jabki

τ = I α

nahi?

Kyunki

d L / d t

ek full vector rate hai: sideways torque

L

ki direction change karta hai (precession), jo genuinely 3D effect hai. Scalar

I α

sirf ek fixed axis ke baare mein speed up ya slow down track karta hai.

Ek free particle ka angular momentum constant kyun rehta hai chahe wo

O

se "door jaata" ho?

Jaise

r

lambi hoti hai, velocity se angle shrink hota hai taaki perpendicular distance

d

fixed rahe; isliye

L = m v d

kabhi nahi badlta bina torque ke.

Ek spinning system ki kinetic energy kyun badh sakti hai jabki uska angular momentum fixed rahe?

Kyunki

ω = L / I

, hum paate hain

K E = L^{2} / (2 I)

, toh fixed

L

par

I

aadha karne se

K E

double ho jaati hai;

I

mein teenguna giraawaT (jaise skater ki

6 \to 2

) use teen guna kar deta hai. Extra energy andar ki taraf mass kheenchne mein kiye kaam se aati hai.

Hum kyun kehte hain internal torques "on average" ki jagah "pairs mein" cancel hote hain?

Har Newton's-third-law force pair equal, opposite, aur ek hi line of action share karta hai, toh har pair individually exactly zero torque contribute karta hai — koi averaging ya approximation involve nahi hai.

Edge cases

Agar force exactly pivot par apply ki jaaye (

r = 0

) toh torque kya hoga?

Zero:

τ = r F sin θ = 0

kyunki

r = 0

. Hinge par seedha push karna door ko rotate nahi kar sakta chahe kitna bhi hard push karo.

Agar

L

conserved rahe aur

I \to 0

ho jaaye toh

ω

ka kya hoga?

ω = L / I

infinity ki taraf blow up karta hai — ek mathematical warning ki mass kabhi axis par fully collapse nahi ho sakta; real bodies ka minimum

I

hota hai.

Ek particle jo seedha point

O

ki taraf move kar raha hai, uska angular momentum kya hai?

Zero. Motion ki line

O

se gujarti hai, toh perpendicular distance

d = 0

aur

L = m v d = 0

— equivalently

r ∥ v

cross product ko vanish karta hai.

Agar do equal internal forces joining line ke saath act karein, toh wo net linear momentum aur net torque mein kitna contribute karte hain?

Dono zero hain: forces linear momentum ke liye cancel ho jaate hain, aur equal-opposite-collinear hone se unke torques bhi cancel ho jaate hain, isliye internal interactions kabhi total

p

ya total

L

change nahi karte.

Jis exact instant ek body torque ke under rest se shuru hoti hai,

L

ω

, aur

d L / d t

kya hain?

L = 0

aur

ω = 0

, lekin

d L / d t = τ \neq = 0

— body mein abhi tak koi spin nahi hai lekin wo ise acquire kar rahi hai, yahi

τ = d L / d t

ka poora point hai.

Jab force

r

ke parallel ho lekin body already fast spin kar rahi ho toh torque kya hoga?

Phir bhi zero:

sin θ = 0

ise khatam karta hai current spin se regardless. Existing

ω

ka koi bearing nahi hai ki ek given force torque produce karta hai ya nahi.

Genuinely 3D case mein jahan

τ

L

ke perpendicular ho, kya spin rate change hoti hai?

Nahi —

L

ki length (aur isliye spin rate) unchanged hai; sirf uski direction swing karti hai. Ye steady precession hai, kisi bhi planar

I α

treatment mein invisible.