Torque = dL - dt: Exercises — Hinglish deep dive

1.5.11 · D4 · Physics › Rotational Mechanics › Torque = dL - dt

Yahan sab kuch parent note ke ek boxed law par tika hai: $τ_{ext} = \frac{d L}{d t}$ "Bahar se jo net twist milta hai woh equals hai spin-amount ke change hone ki speed." Hum iske do roop use karenge:

Constant $I$ : $τ = I α$ — dekho Torque and Angular Acceleration (tau = I alpha).
Zero torque: $L$ = constant — dekho Conservation of Angular Momentum.

Shuru karne se pehle symbols (kuch bhi assumed nahi):

Definition Har problem mein har character

$r$ (position vector) — pivot $O$ se us jagah tak ka arrow jahan force act karta hai; iska length $r$ distance hai, unit metre ( $m$ ).
$F$ (force) — ek seedha push ya pull, unit newton ( $N$ ). Dekho Newton's Second Law.
$T$ (tension) — ek taut string ke through lagne wali pulling force, $F$ ka ek type hai, unit newton ( $N$ ).
$g$ (gravitational field strength) — Earth ke paas gravity kitni strongly pull karti hai per kilogram, $g \approx 10 m/s^{2}$ ; ek mass $M$ neeche $M g$ weight feel karta hai.
$τ$ (torque) — twisting effort, unit newton-metre ( $N\cdotm$ ). $τ = r \times F$ ke liye dekho Cross Product.
$L$ (angular momentum) — "spin ki quantity," unit $kg\cdotm^{2} / s$ . Dekho Angular Momentum.
$I$ (moment of inertia) — kisi object ko spin karna kitna mushkil hai, unit $kg\cdotm^{2}$ . Dekho Moment of Inertia.
$ω$ (angular velocity) — spin rate, unit $rad/s$ .
$α$ (angular acceleration) — $ω$ kitni tezi se change hota hai, unit $rad/s^{2}$ .

Level 1 — Recognition

Exercise 1.1 (L1)

Ek wheel ka constant moment of inertia hai. Usp ar ek steady torque $τ = 3 N\cdotm$ act karta hai. Uska angular momentum $L (t)$ hai. $τ$ aur $L$ ko connect karne wali equation likho, aur words mein batao ki $3$ physically kya matlab rakhta hai.

Recall Solution 1.1

Law hai $τ = \frac{d L}{d t}$ . Yeh kya kehta hai: har second, angular momentum $3 kg\cdotm^{2} / s$ se badhta hai. Torque "spin" nahi hai — yeh woh rate hai jis par spin add hoti hai, bilkul waise jaise force linear momentum add hone ki rate hai (Newton's Second Law). $t$ seconds baad rest se shuru karke, $L (t) = 3 t$ (kg·m²/s).

Exercise 1.2 (L1)

Inme se kaunsa point $O$ ke baare mein kisi object ka angular momentum change kar sakta hai? (a) Ek force $F$ seedha $O$ ki taraf point karti hai. (b) Ek force $F$ us line ke perpendicular hai jo $O$ ko application ke point se jodhti hai. (c) Koi force nahi.

Recall Solution 1.2

Sirf (b).

(a) $r$ ke along point karti hai, to $r \times F = r F sin 0^{\circ} = 0$ — koi torque nahi.
(b) perpendicular hai, to $sin 9 0^{\circ} = 1$ deta hai maximum torque $r F$ — yeh $L$ change karta hai.
(c) koi force nahi ⇒ koi torque nahi ⇒ $L$ constant. Neeche diya figure dikhata hai ki sirf sideways part kyun twist karta hai.

Figure s01 — Pivot $O$ (navy dot) $r$ (navy) ke saath us point tak jahan force act karti hai. Orange arrow radial force hai ( $r$ ke along): yeh zero torque produce karta hai. Magenta arrow perpendicular force hai: yeh maximum torque $r F$ deta hai. Curved arrow CCW-positive sense mark karta hai.

Level 2 — Application

Exercise 2.1 (L2)

Ek disc jiska $I = 0.4 kg\cdotm^{2}$ (constant) hai, rest se ek constant torque $τ = 2 N\cdotm$ se spin up ki jaati hai. Angular acceleration $α$ aur $5 s$ baad angular velocity nikalo.

Recall Solution 2.1

Driving torque ko positive lo (yeh humara CCW-positive sense set karta hai). Step 1 — law ka kaun sa roop? $I$ constant hai, to $\frac{d L}{d t} = \frac{d ( I ω )}{d t} = I α$ . Kyun: constant $I$ ko derivative ke bahar khainch lo. Step 2 — angular acceleration: $α = \frac{τ}{I} = \frac{2}{0.4} = 5 rad/s^{2}$ (positive ⇒ CCW sense mein speed up hota hai). Step 3 — 5 s baad: rest se constant $α$ deta hai $ω = α t = 5 \times 5 = 25 rad/s$ .

Exercise 2.2 (L2)

Ek string radius $R = 0.15 m$ , $I = 0.3 kg\cdotm^{2}$ wale pulley ke upar wound hai. String rim se tangentially nikalta hai tension $T = 8 N$ ke saath. Torque aur angular acceleration nikalo.

Recall Solution 2.2

Step 1 — torque: string tangential hai to tension $T$ (ek force $F$ ) $r$ ke perpendicular hai, $θ = 9 0^{\circ}$ : $τ = R T sin 9 0^{\circ} = 0.15 \times 8 = 1.2 N\cdotm .$ Is driving torque ko positive lo. Step 2 — angular acceleration: $I$ constant ⇒ $τ = I α$ : $α = \frac{τ}{I} = \frac{1.2}{0.3} = 4 rad/s^{2} .$

Level 3 — Analysis

Exercise 3.1 (L3)

Mass $m = 2 kg$ wala ek particle straight line mein constant velocity $v = 3 m/s$ se move karta hai. Yeh line ek fixed point $O$ se $d = 4 m$ door hai. $O$ ke baare mein iska angular momentum nikalo, aur explain karo ki yeh constant kyun rehta hai jabki kuch bhi spin nahi karta.

Recall Solution 3.1

Step 1 — magnitude: $L = m v d = 2 \times 3 \times 4 = 24 kg\cdotm^{2} / s$ . $d$ kyun aur $r$ kyun nahi? $L = r \times m v = m v r sin θ$ , aur $r sin θ$ exactly perpendicular distance $d$ hai $O$ se line tak — figure dekho. Step 2 — constant kyun hai: koi force nahi ( $F = 0$ ) ⇒ $τ = r \times F = 0$ ⇒ $\frac{d L}{d t} = 0$ . Surprise: angular momentum ke liye circle ki zarurat nahi. Jaise particle slide karta hai, $r$ badhta hai lekin $θ$ chhota hota hai taaki $r sin θ = d$ locked rehe. Yeh product kabhi nahi badalta.

Figure s02 — Fixed point $O$ (navy) aur ek straight path (violet) perpendicular distance $d$ (orange) door. Particle ke do snapshots: near point par (magenta) $r$ chhota hai lekin $v$ se angle bada hai; far point par (navy) $r$ bada hai lekin angle chhota hai. Product $r sin θ = d$ — aur isliye $L = m v d$ — dono mein same hai, to $L$ constant hai.

Exercise 3.2 (L3)

Ek merry-go-round ( $I = 100 kg\cdotm^{2}$ ) freely $ω_{0} = 1.5 rad/s$ par spin karta hai. Mass $m = 30 kg$ wala ek bachcha (point ke roop mein treat karo) vertically uski edge par radius $R = 2 m$ par gir jaata hai. Nayi angular velocity nikalo.

Recall Solution 3.2

Step 1 — kaun sa roop? Bachcha mass add karta hai, $I$ change karta hai. Koi external horizontal torque nahi hai (drop vertical hai), to $L$ conserved hai. Step 2 — bachche ka added inertia: rim par ek point mass ka $I_{child} = m R^{2} = 30 \times 2^{2} = 120 kg\cdotm^{2}$ hai. Step 3 — $L$ conserve karo: $I_{0} ω_{0} = (I_{0} + I_{child}) ω_{1}$ $ω_{1} = \frac{100 \times 1.5}{100 + 120} = \frac{150}{220} \approx 0.682 rad/s .$ Ride slow ho jaati hai — mass bahar phailane se $I$ badhta hai, to $L$ fixed rakhne ke liye $ω$ girta hai.

Level 4 — Synthesis

Exercise 4.1 (L4)

Mass $M = 1 kg$ ek string se hang kar raha hai jo radius $R = 0.1 m$ aur moment of inertia $I = 0.02 kg\cdotm^{2}$ wale pulley par wound hai. Mass release kiya jaata hai aur girta hai. Girte hue mass ka linear acceleration $a$ nikalo. ( $g = 10 m/s^{2}$ lo, glossary mein define gravitational field strength.)

Neeche ka free-body picture compute karne se pehle har sign fix karta hai.

Figure s03 — Left: pulley (violet disc) string ke saath rim se nikalta hua; tension $T$ (magenta) tangentially pull karta hai, torque $T R$ deta hai jo pulley ko chosen positive sense mein (curved orange arrow) spin karta hai. Right: hanging mass $M$ weight $M g$ (orange, neeche) aur string tension $T$ (magenta, upar) ke saath; neeche positive choose kiya to $M g - T = M a$ . Link $a = R α$ dono diagrams ko jodhta hai.

Recall Solution 4.1

Do objects hain, to hum do laws likhenge aur unhe link karenge. Sign choices: mass ke liye, neeche positive hai; pulley ke liye, woh sense positive hai jismein girta hua string use turn karta hai. Yeh do choices $a = R α$ link se consistent banayi gayi hain (dono ek saath positive). Step 1 — mass (Newton): weight $M g$ neeche (positive), tension $T$ upar (negative): $M g - T = M a .$ Step 2 — pulley ( $τ = d L / d t = I α$ , $I$ constant): tension hi akela turning force hai; iska torque $+ T R$ positive sense drive karta hai: $+ T R = I α .$ Step 3 — link (string slip nahi karta): rim speed mass speed ke equal hai, to $a = R α \Rightarrow α = a / R$ (same sign — dono positive). Step 4 — substitute karo: $T R = I \frac{a}{R} \Rightarrow T = \frac{I a}{R ^{2}}$ . Step 1 mein daalo: $M g - \frac{I a}{R ^{2}} = M a \Rightarrow a = \frac{M g}{M + I / R ^{2}} .$ Step 5 — numbers: $I / R^{2} = 0.02/0.01 = 2 kg$ , to $a = \frac{1 \times 10}{1 + 2} = \frac{10}{3} \approx 3.33 m/s^{2} .$ $g$ se kam: pulley ki inertia giravat ko "rok" rahi hai.

Exercise 4.2 (L4)

Ek spinning space station ek ring hai jiska $I_{1} = 5000 kg\cdotm^{2}$ aur $ω_{1} = 0.4 rad/s$ par ghoom raha hai. Crew do arms extend karte hain, $I$ ko $I_{2} = 8000 kg\cdotm^{2}$ tak raise karte hain. $ω_{2}$ aur rotational kinetic energy mein change nikalo. Energy kahan gayi?

Recall Solution 4.2

Step 1 — koi external torque nahi (deep space) ⇒ $L$ conserved: $ω_{2} = \frac{I _{1} ω _{1}}{I _{2}} = \frac{5000 \times 0.4}{8000} = 0.25 rad/s .$ Step 2 — kinetic energies ( $K E = \frac{1}{2} I ω^{2}$ ): $K E_{1} = \frac{1}{2} (5000) (0.4)^{2} = 400 J, K E_{2} = \frac{1}{2} (8000) (0.25)^{2} = 250 J .$ Step 3 — change: $Δ K E = 250 - 400 = - 150 J$ . Energy giri. Kahan gayi: crew ki muscles ne bahar phenkti hui arms ko resist karte hue negative work kiya; energy rotation se nikal kar crew/heat mein chali gayi. $L$ fixed raha; energy ka rehna zaruri nahi tha.

Level 5 — Mastery

Exercise 5.1 (L5)

Ek star jiska moment of inertia $I_{1} = 1.0 \times 1 0^{40} kg\cdotm^{2}$ hai, har $30$ days mein ek baar rotate karta hai. Yeh collapse ho kar neutron star ban jaata hai jiska $I_{2} = 2.0 \times 1 0^{38} kg\cdotm^{2}$ hai. Koi external torque nahi hone par, uska naya rotation period nikalo. Phir, $τ = d L / d t$ use karke explain karo (na ki $I α$ ) ki hum ise $τ = I α$ se solve kyun nahi kar sakte.

Figure s04 — Left: bada slow star ( $I_{1}$ bada, $ω_{1}$ chhota, period $T_{1}$ ke liye lamba arrow). Right: chhota neutron star ( $I_{2}$ chhota, $ω_{2}$ bada, period $T_{2}$ ke liye chhota arrow). Dono ke beech band par label hai $I ω = L$ = constant: jaise $I$ chhota hota hai spin $ω$ shoot up karta hai taaki product fixed rahe — koi torque ki zarurat nahi.

Recall Solution 5.1

Step 1 — spin ko $ω$ mein convert karo. Angular velocity hai $ω = \frac{2 π}{T}$ jahan $T$ period hai. Initial period $T_{1} = 30 days = 30 \times 86400 = 2.592 \times 1 0^{6} s$ . Step 2 — koi external torque nahi ⇒ $L$ conserved ⇒ $I_{1} ω_{1} = I_{2} ω_{2}$ . $ω = 2 π / T$ likhne par, $2 π$ cancel ho jaata hai: $\frac{I _{1}}{T _{1}} = \frac{I _{2}}{T _{2}} \Rightarrow T_{2} = T_{1} \frac{I _{2}}{I _{1}} .$ Step 3 — numbers: $T_{2} = 2.592 \times 1 0^{6} \times \frac{2.0 \times 1 0 ^{38}}{1.0 \times 1 0 ^{40}} = 2.592 \times 1 0^{6} \times 0.02 = 5.184 \times 1 0^{4} s \approx 0.6 days .$ To period 30 days se ghatakar lagbhag 14.4 hours ho jaata hai — collapse use dramatically spin up karta hai, skater ka astrophysical version jo arms andar kheench leta hai. Step 4 — $τ = I α$ kyun nahi? Collapse ke dauran $I$ enormously change ho raha hai. Formula $τ = I α$ secretly assume karta hai ki $I$ constant hai (taaki ise $d / d t$ ke bahar kheencha ja sake). Sachcha law $τ = \frac{d ( I ω )}{d t}$ mein ek extra $ω \frac{d I}{d t}$ piece hota hai. Yahan $τ_{ext} = 0$ , to $α$ chase karne ki jagah hum poore product $I ω = L$ ko constant rehte hue use karte hain. Yahi cheez $τ = d L / d t$ ko zyada general banati hai.

Exercise 5.2 (L5)

5.1 mein collapse ke dauran process $Δ t$ leta hai aur star ka $L$ constant rehta hai $L = I_{1} ω_{1}$ par. Ek student claim karta hai "internal gravitational squeeze ek bada torque exert karta hai use spin up karne ke liye." Ise $τ_{ext} = d L / d t$ use karke galat sabit karo, aur batao ki speed-up asal mein kis cheez se hota hai.

Recall Solution 5.2

Step 1 — pure star par law apply karo. Gravity yahan internal hai (star apne upar act kar raha hai). Wohi pair-cancellation argument jo parent derivation mein internal torques ko khatam karta hai, internal gravity bhi spin axis ke baare mein zero net torque contribute karti hai. Step 2 — to $τ_{ext} = 0 \Rightarrow d L / d t = 0$ . $L$ kabhi nahi badalta. Ise spin up karne ke liye koi torque nahi hai. Step 3 — asli wajah. $L = I ω$ fixed hai. Collapse $I$ ko chhota karta hai, to product constant rakhne ke liye $ω$ badh-na chahiye. Speed-up torque se nahi hai jo spin add kar raha ho — yeh wohi same spin ko chhote $I$ par redistribute karne se hai. "Torque $L$ add karta hai; $I$ chhota hona existing $L$ redistribute karta hai." Bilkul alag mechanisms hain.