Torque = dL - dt: Worked examples — Hinglish deep dive

1.5.11 · D3 · Physics › Rotational Mechanics › Torque = dL - dt

Hum sirf wahi tools use karte hain jo parent mein already build hain: angular momentum $\vec L=\vec r\times\vec p$, the cross product (jo ek vector ka perpendicular part doosre ke against measure karta hai), Newton's Second Law, moment of inertia $I$, aur Conservation of Angular Momentum. Baaki sab hum wahan pe re-explain kar denge.

Pehle, ek naya symbol: angular acceleration $α$

Parent ne $ω$ (angular velocity, "kitni tez spin ho rahi hai") aur $I$ (moment of inertia, "spin karna kitna mushkil hai") use kiya tha. Ab humein ek aur chahiye.

Yeh define ho gaya, toh parent ka shortcut $τ = I α$ (valid sirf tab jab $I$ constant ho) ab poori tarah spell out ho gaya hai: $L = I ω$ ko differentiate karo $I$ fixed rakh ke, toh milta hai $τ = I d ω / d t = I α$ .

Scenario matrix

Problems work karne se pehle, hum un tarah ke situations list karte hain jo yeh ek law produce karta hai. Har row ek "cell" hai — ek alag behaviour. Neeche ke examples mein se har ek us cell ke saath tagged hai jo woh cover karta hai.

Cell	Kya change hota hai	Sign / degeneracy	Kaun sa relation use karna hai
A	$ω$ badhta hai, $I$ fixed	positive torque (speed-up)	$τ = I α$
B	$ω$ ghatta hai, $I$ fixed	negative torque (braking)	$τ = I α$ , sign dekho
C	$ω = 0$ se shuru hota hai	torque zero se $L$ create karta hai	$τ = d L / d t$ , integrate karo
D	$I$ change hota hai, koi torque nahi	$τ = 0$ , $L$ conserved	$I_{1} ω_{1} = I_{2} ω_{2}$
E	$I$ aur $ω$ dono change hote hain	$d (I ω) / d t$ use karna hoga	full $τ = d L / d t$
F	force $r$ ke along point karta hai	degenerate: $τ = 0$	$r \times F$ , $sin θ$
G	straight-line motion, off-line point	spinning nahi phir bhi $L \neq = 0$	$L = m v d$
H	real-world word problem	two-body, string+pulley	Newton + $τ = I α$
I	exam twist: vector directions	sign of $L$ , right-hand rule	$τ = r \times F$

Ab hum cells A–I order mein cover karte hain.

Rotation ka sign padhna (cells A, B, I ke liye zaroori hai)

Neeche sab kuch ek convention use karta hai. Neeche wheel dekho.

Figure mein: cyan circle wheel ka rim hai; amber curved arrow chosen positive (CCW) sense dikhata hai; centre par chota cyan circle-with-a-dot ek vector ka standard symbol hai jo page se bahar point kar raha hai ( $+ \overset{z}{^}$ ) — yahi $ω$ , $α$ , aur $τ$ ki direction hai CCW quantities ke liye. White straight arrow right par ek tangential drive force hai; notice karo yeh rim ko sideways touch karta hai, centre ki taraf aim nahi kiya, jo exactly yehi allow karta hai ki yeh wheel ko twist kar sake.

Yeh ek rule hume speed up versus slow down track karne deta hai, kabhi bhi lost hue bina.

Cell A — Positive torque, constant $I$ (speed-up)

Worked example A · Belt se driven ek flywheel

Ek flywheel ka $I = 4 kg\cdotm^{2}$ hai. Ek belt uske rim par tangentially pull karti hai (radius $R = 0.5$ m) tension $T$ ke magnitude $20$ N ke saath CCW sense mein. Yeh rest se shuru hota hai. Angular acceleration $α$ aur $3$ s baad angular speed nikalo.

Forecast: Andaza lagao — $3$ s baad $ω$ 1, 7, ya 70 rad/s ke kareeb hoga?

Step 1. Axle ke baare mein torque: $τ = r \times T$ , magnitude $τ = R T sin 90° = 0.5 \times 20 = + 10 N\cdotm$ , direction $+ \overset{z}{^}$ (page se bahar). Yeh step kyun? Belt rim se tangentially nikalta hai, isliye $r ⊥ T$ , jo deta hai $sin 90° = 1$ — poori force twist karti hai. Right-hand rule se $r \times T$ page se bahar point karta hai, yaani positive.

Step 2. $I$ constant hai, isliye $τ = d L / d t = I α \Rightarrow α = τ / I = 10/4 = + 2.5 rad/s^{2}$ . Yeh step kyun? Ek rigid flywheel shape nahi badalta, isliye hum $I$ ko $d (I ω) / d t$ se bahar nikal sakte hain, jo chhodta hai $I d ω / d t = I α$ .

Step 3. Rest se constant $α$ : $ω = α t = 2.5 \times 3 = 7.5 rad/s$ . Yeh step kyun? Constant $I$ par constant torque constant $α$ deta hai, isliye " $v = a t$ " wala analogue $ω, α$ ke saath apply hota hai.

Verify: Units: $(N\cdotm) / (kg\cdotm^{2}) = kg\cdotm^{2} s^{- 2} / kg\cdotm^{2} = s^{- 2}$ ✓ (yahi hai rad/s²). Sign positive → CCW speed up hota hai, jaisa ek driving belt ko karna chahiye. Answer 7.5 rad/s guesses ke beech mein hai — moderate.

Cell B — Negative torque, constant $I$ (braking)

Worked example B · Friction flywheel ko rokti hai

Wahi flywheel ( $I = 4$ ) ab $ω_{0} = 7.5$ rad/s CCW spin kar raha hai. Ek brake constant torque lagata hai magnitude $6$ N·m motion ko oppose karte hue. Kitne time mein yeh rukega?

Forecast: Spin up hone mein lage 3 s se zyada ya kam time lagega?

Step 1. Brake CCW motion ko oppose karta hai, isliye $τ$ page ke andar point karta hai: $τ = - 6 N\cdotm$ (negative). Yeh step kyun? Positive $ω$ (page se bahar); ek torque jo isse reduce karta hai opposite direction mein point karna chahiye (page ke andar), isliye negative sign — yahi sign convention ka poora point hai.

Step 2. $α = τ / I = - 6/4 = - 1.5 rad/s^{2}$ . Yeh step kyun? $I$ phir se constant hai, isliye $τ = I α$ . Negative $α$ matlab $α$ $ω$ ko oppose karta hai — deceleration.

Step 3. $ω = 0$ par rukta hai: $0 = ω_{0} + α t \Rightarrow t = - ω_{0} / α = - 7.5/ (- 1.5) = 5 s$ . Yeh step kyun? Same kinematics, ab decelerating; wo time solve karo jab speed zero hit karti hai.

Verify: $t = 5$ s positive hai (acha — negative time ek impossible answer hota). Rukne mein spin up se zyada time lagta hai kyunki brake torque (6) drive torque (10) se weaker hai. Units: $(rad/s) / (rad/s^{2}) = s$ ✓.

Cell C — Torque zero se $L$ banata hai (rest se shuru, integrate karo)

Yeh trap hai "agar $ω = 0$ hai toh $L$ nahi hai, isliye torque kuch nahi karta." Galat — torque $L$ build karta hai.

Worked example C · Rest se ek time-varying torque

Ek disc ( $I = 2 kg\cdotm^{2}$ ) rest par hai. Ek torque $τ (t) = 3 t N\cdotm$ (time ke saath badhta hua, sab $+ \overset{z}{^}$ ke along) lagaya jaata hai. $t = 4$ s par $L$ aur $ω$ nikalo.

Forecast: Kyunki $τ$ badhta hai, kya tum expect karte ho ki $ω$ linearly badhega ya linearly se tez?

Step 1. Fundamental form directly use karo: $\frac{d L}{d t} = τ (t) = 3 t$ . Yeh step kyun? $τ$ constant nahi hai, isliye $α$ bhi constant nahi hai — hum $ω = α t$ use nahi kar sakte. Lekin $τ = d L / d t$ hamesha hold karta hai, aur hum isse integrate kar sakte hain.

Step 2. Rest se integrate karo ( $L (0) = 0$ ): $L (4) = \int_{0}^{4} 3 t d t = \frac{3}{2} t^{2}_{0}^{4} = \frac{3}{2} \times 16 = 24 kg\cdotm^{2} / s .$ Yeh step kyun? Torque-versus-time curve ke neeche ka area hi accumulated angular momentum hai — rotation ke liye "impulse" idea.

Step 3. $ω = L / I = 24/2 = 12 rad/s$ . Yeh step kyun? Constant $I$ ke saath, $L = I ω$ , isliye divide karne se speed recover hoti hai.

Verify: $L$ ke units: $(N\cdotm) (s) = kg\cdotm^{2} s^{- 1}$ ✓. $ω$ zero se badha — torque ne yeh sab manufacture kiya. Kyunki $τ \propto t$ , $L \propto t^{2}$ : linearly se tez, forecast se match karta hai.

Cell D — $I$ change hota hai, koi external torque nahi ( $L$ conserved)

Worked example D · Ek collapsing spinning ring

Ek ring $ω_{1} = 3$ rad/s par $I_{1} = 8 kg\cdotm^{2}$ ke saath spin kar rahi hai. Yeh contract hoti hai (koi external torque nahi) $I_{2} = 2 kg\cdotm^{2}$ tak. $ω_{2}$ nikalo.

Forecast: Shrink hone ke baad faster ya slower? Roughly kis factor se?

Step 1. Koi external torque nahi $\Rightarrow d L / d t = 0 \Rightarrow L$ constant. Yeh step kyun? $τ = 0$ ke saath fundamental law kehta hai $L$ change nahi ho sakta, chahe internal shape ho. Dekho Conservation of Angular Momentum.

Step 2. $I_{1} ω_{1} = I_{2} ω_{2} \Rightarrow ω_{2} = \frac{8 \times 3}{2} = 12 rad/s$ . Yeh step kyun? Sirf $L = I ω$ fixed hai; $ω$ ko ghatte hue $I$ ki compensate karne ke liye badhna hi padega.

Verify: $L_{1} = 8 \times 3 = 24$ , $L_{2} = 2 \times 12 = 24$ ✓ equal. $I$ ko 4× shrink karne se 4× speed up hua, jaise forecast kiya. (KE $\frac{1}{2} \cdot 8 \cdot 9 = 36$ J se $\frac{1}{2} \cdot 2 \cdot 144 = 144$ J ho gaya — internal work, conserved nahi.)

Cell E — Dono $I$ aur $ω$ change hote hain (full $d (I ω) / d t$ use karna padega)

Yahan $τ = I α$ illegal hai — yeh cell exist karta hai fundamental form ko force karne ke liye.

Worked example E · Ek spinning platform jisme mass gain ho rahi hai

Ek turntable ka moment of inertia linearly badhta hai jab usmein sand girta hai: $I (t) = (2 + t) kg\cdotm^{2}$ . Isse driven kiya jaata hai taaki uski speed bhi badhe: $ω (t) = (1 + t) rad/s$ . $t = 2$ s par zaruri torque nikalo.

Forecast: Zaroori torque naive $I α$ estimate se chota hoga ya bada?

Step 1. $L (t) = I (t) ω (t) = (2 + t) (1 + t)$ likho. Yeh step kyun? Dono factors time par depend karte hain, isliye hum kisi ko bhi bahar nahi nikal sakte; pehle $L$ banani padegi.

Step 2. Expand karo aur differentiate karo: $L = 2 + 3 t + t^{2}$ , isliye $τ = \frac{d L}{d t} = 3 + 2 t$ . Yeh step kyun? $τ = d L / d t$ ek hi correct relation hai jab $I$ vary karta hai; product rule automatically $I d ω / d t$ aur $ω d I / d t$ dono effects include karta hai.

Step 3. $t = 2$ par: $τ = 3 + 2 (2) = 7 N\cdotm$ . Yeh step kyun? Derivative mein interest ka instant plug karo.

Verify: Naive (galat) $I α$ deta $I (2) \cdot \overset{ω}{˙} = 4 \times 1 = 4$ N·m — bahut chota, kyunki yeh $ω d I / d t = 3 \times 1 = 3$ term ignore karta hai. Correct total $4 + 3 = 7$ ✓ match karta hai. Exactly yahi reason hai ki parent warn karta hai "fundamental form par trust karo."

Cell F — Force $r$ ke along: degenerate zero-torque case

Worked example F · Pivot ki taraf seedha push karna

Ek rod $O$ par hinged hai. Tum rod ke ek point par $r = 0.4$ m distance par $F$ force lagaate ho magnitude $15$ N ke saath, lekin force directly $O$ ki taraf point karta hai ( $r$ ke along). Torque nikalo.

Forecast: Zero, ya $0.4 \times 15 = 6$ N·m?

Step 1. Torque magnitude $τ = ∣ r \times F ∣ = r F sin θ$ , jahan $θ$ $r$ aur $F$ ke beech ka angle hai. Yeh step kyun? Cross product force ka sirf woh part rakhta hai jo $r$ ke perpendicular ho; $sin θ$ us perpendicular fraction ko measure karta hai.

Step 2. Force $O$ ki taraf point karta hai, isliye $F$ $r$ ke anti-parallel hai: $θ = 180°$ , $sin 180° = 0$ . Yeh step kyun? Hinge ki taraf push kiya hua kisi bhi sideways "twisting" component nahi hota — geometrically yeh kuch bhi rotate nahi kar sakta. (Same reasoning parent ke derivation mein $v \times p$ term ko kill karta hai.)

Step 3. $τ = 0.4 \times 15 \times 0 = 0 N\cdotm$ . Yeh step kyun? Zero perpendicular component ⇒ zero torque ⇒ $L$ mein koi change nahi.

Verify: Ek radial (central) force zero torque deta hai — wahi fact jo planetary orbits mein angular momentum conserve karta hai. Answer 0, 6 nahi, forecast trap pakad ke.

Cell G — Straight-line motion, off-line point (spinning nahi, phir bhi $L \neq = 0$ )

Figure mein: cyan horizontal line car ka straight path hai; amber dot origin par observation point $O$ hai; do white arrows $r$ hain jo do alag car positions par draw kiye gaye hain; amber vertical double-arrow fixed perpendicular distance $d = 3$ m mark karta hai. Key visual: jab car chalti hai $r$ swing aur lengthen karta hai, lekin uska $O$ par perpendicular projection — quantity $r sin θ$ — hamesha wahi $d$ rehta hai, isliye $L$ kabhi nahi badlta.

Worked example G · Ek car tumhare paas se guzarti hai

Ek $1200$ kg ki car constant $v = 10$ m/s par seedhi line mein chalti hai. Uska path tumse (point $O$ ) perpendicular distance $d = 3$ m par guzarta hai. $O$ ke baare mein uska angular momentum kya hai, aur kya yeh change hota hai?

Forecast: Zero (yeh tumhare around nahi ja rahi), ya kuch nonzero?

Step 1. Straight-line motion ke liye, $L = m v d$ jahan $d$ $O$ se line tak perpendicular ("closest approach") distance hai. Yeh step kyun? $L = r \times m v$ ka magnitude $m v (r sin θ)$ hai, aur $r sin θ$ exactly woh perpendicular distance $d$ hai — figure ka right triangle dekho.

Step 2. $L = 1200 \times 10 \times 3 = 36000 kg\cdotm^{2} / s$ . Yeh step kyun? Plug in karo; value fixed hai kyunki $m$ , $v$ , aur $d$ sab straight path ke along constant hain.

Step 3. Kya $L$ change hota hai? Hume $O$ ke baare mein net torque check karna hoga, aur specifically uska component $L$ -axis ke along (yahan vertical, $\pm \overset{z}{^}$ ). Gravity ( $m g$ neeche) aur road ka normal force ( $N$ upar) real, nonzero forces hain, aur har ek ka $O$ ke baare mein nonzero torque ho sakta hai. Lekin dono vertically point karte hain (yahan page ke plane mein), isliye har ek ke liye $r \times F$ horizontal plane mein hota hai — iska $\overset{z}{^}$ ke along zero component hai. $L$ ke axis ke along, net torque zero hai, isliye $L_{z}$ (humara $36000$ ) change nahi hota. Yeh step kyun? "Koi horizontal force nahi" honest statement hai, "koi force hi nahi" nahi. Vertical forces torque lagaate hain, lekin sirf horizontal axes ke baare mein; yeh vertical angular momentum alter nahi kar sakte. Agar car constant height par bhi chalti hai, toh woh vertical torques balanced rehte hain aur use kabhi tilt nahi karte.

Verify: Jab car chalti hai, $r$ pehle ghatta hai phir badhta hai aur $θ$ change hota hai, lekin product $r sin θ = d$ exactly $3$ m rehta hai (figure dekho — har position same $d$ par project hoti hai). Isliye $L_{z}$ $36000$ par constant hai. Angular momentum ke liye circular motion zaruri nahi hai.

Cell H — Real-world word problem (string + pulley, two bodies)

Figure mein: cyan circle pulley hai (moment of inertia $I$ , radius $R$ ); uske right par white vertical line string hai; rim par amber arrow up tension $T$ hai jo torque produce karta hai; neeche cyan rectangle hanging block hai; block par, chota amber arrow up tension hai aur white arrow down gravity $m g$ hai; block ke side mein amber arrow uski downward acceleration $a$ mark karta hai. Dekho kaise tension rim par tangentially act karta hai — yahi torque mein $sin 90° = 1$ banata hai.

Worked example H · Girta hua block ek pulley ghuma raha hai

$m = 2$ kg ka ek block ek light string se hang kar raha hai jo $I = 0.1 kg\cdotm^{2}$ aur radius $R = 0.2$ m ke pulley ke around wound hai. Rest se release hone par, block gitta hai. Uski downward acceleration $a$ nikalo. ( $g = 10 m/s^{2}$ lo.)

Forecast: Free-fall $g = 10$ se zyada ya kam? Roughly kitna?

Step 1. Block (linear, Newton's Second Law): neeche ko positive lete hue, $m g - T = ma$ , jahan $T$ string tension ka magnitude hai. Yeh step kyun? Block par do vertical forces — gravity neeche, tension upar.

Step 2. Pulley (rotational): girta hua block right side par string unwind karta hai, pulley ko CW spin karta hai jaise draw kiya — us spin sense ko positive lo. Tension rim par tangentially act karta hai, isliye $τ = r \times T$ ka magnitude $τ = R T sin 90° = R T$ hai aur woh $+$ (CW spin axis) ke along point karta hai, yaani yeh rotation drive karta hai. Isliye $τ = I α$ deta hai $R T = I α$ jahan dono $τ$ aur $α$ is chosen sense mein positive hain. Yeh step kyun? $I$ constant hai, isliye $τ = d L / d t = I α$ . Positive spin sense fix karna baaki page ke saath har sign consistent rakhta hai; tension ek maatra torque hai (axle force $r = 0$ par act karta hai, isliye uska torque zero hai).

Step 3. Rolling-string constraint: $a = R α \Rightarrow α = a / R$ , same positive sense ke saath (block girna ↔ pulley humara positive way spin karna). Yeh step kyun? String slip nahi karti, isliye rim ki tangential acceleration block ki fall rate ke barabar hai, aur do positive directions ek saath tied hain.

Step 4. $α = a / R$ ko $R T = I α$ mein substitute karo: $T = I a / R^{2}$ . Step 1 mein daalo: $m g - \frac{I a}{R ^{2}} = ma \Rightarrow a = \frac{m g}{m + I / R ^{2}} .$ Yeh step kyun? $T$ aur $α$ eliminate karo ek unknown $a$ solve karne ke liye.

Step 5. Numbers: $I / R^{2} = 0.1/0.04 = 2.5 kg$ , isliye $a = \frac{2 \times 10}{2 + 2.5} = \frac{20}{4.5} \approx 4.44 m/s^{2} .$

Verify: $a < g$ ✓ — pulley ki inertia block ko rok ke rakhti hai, jaise forecast kiya. Agar $I \to 0$ (massless pulley), $a \to g$ ✓, sanity limit. Units: $I / R^{2}$ ka $kg\cdotm^{2} / m^{2} = kg$ hai, isliye denominator ek mass hai ✓.

Cell I — Exam twist: $L$ ki direction right-hand rule se

Worked example I ·

$L$ kis taraf point karta hai? $m = 0.5$ kg ka ek particle position $r = (2, 0, 0)$ m par hai aur velocity $v = (0, 3, 0)$ m/s ke saath move kar raha hai. Origin ke baare mein $L$ nikalo, direction include karke.

Forecast: Kya $L$ $+ \overset{z}{^}$ (page se bahar) ya $- \overset{z}{^}$ ke along point karega?

Step 1. $L = r \times m v = m (r \times v)$ . Yeh step kyun? Angular momentum ki definition; Cross Product magnitude aur direction dono deta hai, isliye "kis taraf" bhi answer karta hai aur "kitna" bhi.

Step 2. $r \times v$ compute karo right-hand rule $\overset{x}{^} \times \overset{y}{^} = \overset{z}{^}$ use karke: $r \times v = (2 \overset{x}{^}) \times (3 \overset{y}{^}) = 6 (\overset{x}{^} \times \overset{y}{^}) = 6 \overset{z}{^} .$ Yeh step kyun? Sirf $r$ ka $x$ -component aur $v$ ka $y$ -component survive karte hain; unka product $\overset{x}{^} \times \overset{y}{^}$ se times $+ \overset{z}{^}$ ke along ek vector deta hai.

Step 3. $L = 0.5 \times 6 \overset{z}{^} = 3 \overset{z}{^} kg\cdotm^{2} / s$ . Yeh step kyun? Mass se multiply karo; direction $+ \overset{z}{^}$ hai, yaani page se bahar — CCW, humari sign convention se match karta hai.

Verify: Magnitude $= m v d = 0.5 \times 3 \times 2 = 3$ ✓ (yahan $d = 2$ kyunki $r ⊥ v$ ). Direction $+ \overset{z}{^}$ : right-hand fingers $r$ ( $+ x$ ke along) se $v$ ( $+ y$ ke along) ki taraf curl karo; thumb page se bahar point karta hai ✓. Agar $v$ $(0, - 3, 0)$ hota, toh same rule $- \overset{z}{^}$ (page ke andar) deta — hamesha right-hand rule se sign fix karo, intuition se nahi.

Recall Matrix across quick self-test

Kaun sa cell $τ = I α$ use karna forbid karta hai? ::: Cell E — $I$ change hota hai, isliye tumhe $τ = d (I ω) / d t$ use karna hoga. Ek radial force kya torque deta hai? ::: Zero (Cell F): $sin θ$ jab $θ = 0°$ ya $180°$ zero hota hai. Straight-line motion, off-line point — kya $L$ zero hai? ::: Nahi (Cell G): $L = m v d \neq = 0$ aur yeh $\overset{z}{^}$ -axis ke along constant rehta hai. Skater/collapsing ring $I$ ko 4× shrink karta hai bina torque ke — naya $ω$ ? ::: 4× faster (Cell D), kyunki $I ω$ fixed hai. Rest par kisi cheez par torque act kare toh kya hota hai? ::: Zero se $L$ build hota hai (Cell C) — $d L / d t = τ$ integrate karo.

Parent par wapas: Torque = dL/dt · related: Torque and Angular Acceleration (tau = I alpha).