Angular momentum L = Iω (fixed axis), L = r × p (general): Worked examples

1.5.10 · D3 · Physics › Rotational Mechanics › Angular momentum L = Iω (fixed axis), L = r × p (general)

Hum Cross Product, Linear Momentum, Moment of Inertia, Torque, Conservation of Angular Momentum, Rotational Kinetic Energy aur Kepler's Second Law ke ideas use karenge, jo sab parent topic note se seeded hain.

Scenario matrix

Is topic ke har problem ka ek case class hota hai. Last column us worked example ka naam deta hai jo use cover karta hai.

#	Case class	Kya special hai / kya trip kar sakta hai	Covered by
C1	Particle, general angle $θ$	$r sin θ$ use karna hoga, sirf $r$ nahi	Ex 1
C2	Degenerate: motion seedha toward/away from $O$	$θ = 0$ ya $18 0^{\circ} \Rightarrow L = 0$	Ex 2
C3	Straight-line motion, $L$ time mein constant	$r$ badalta hai par $r_{⊥}$ fixed	Ex 3
C4	Sign / direction (right-hand rule, $+ z$ vs $- z$ )	clockwise = negative	Ex 4
C5	Rigid body, fixed symmetry axis, $L = I ω$	sahi $I$ choose karo	Ex 5
C6	Conservation, $I$ shrinks, $ω$ grows	no external torque; energy conserved NAHI hoti	Ex 6
C7	Limiting behaviour: $r_{⊥} \to 0$ , aur $ω \to 0$	$L \to 0$ smoothly	Ex 7
C8	Real-world word problem (Kepler / planet)	equal-areas from $L$	Ex 8
C9	Exam twist: two particles / net $L$ with cancellation	vectors cancel kar sakte hain	Ex 9

Pehle woh geometry fix karte hain jo hum baar baar use karte rahenge.

Figure — Angular momentum L = Iω (fixed axis), L = r × p (general)

$∣ L ∣ = r p sin θ = p r_{⊥} p = m v$

Worked examples

Ex 1 — Cell C1: general angle

Worked example Particle ek slanted angle par

$m = 2 kg$ ka ek stone $O$ se $r = 4 m$ ki position par hai, $v = 5 m/s$ se move kar raha hai. $r$ aur $v$ ke beech angle $θ = 3 0^{\circ}$ hai. $O$ ke baare mein $∣ L ∣$ nikalo.

Forecast: compute karne se pehle guess karo — kya poora $r = 4$ use karna true $L$ se zyada estimate dega ya match karega? (Kyunki $sin 3 0^{\circ} = 0.5$ hai, true answer naive $r p$ ka half hai.)

Momentum $p = m v = (2) (5) = 10 kg m/s$ . Yeh step kyun? $L$ momentum se banta hai, raw velocity se nahi — mass matter karta hai (Linear Momentum).
Lever arm $r_{⊥} = r sin θ = 4 sin 3 0^{\circ} = 4 (0.5) = 2 m$ . Yeh step kyun? Sirf motion ka "going-around" part count hota hai; $sin θ$ use extract karta hai.
$∣ L ∣ = p r_{⊥} = (10) (2) = 20 kg m^{2} /s$ . Yeh step kyun? Yeh magnitude form $r p sin θ$ hai.

Verify: direct $r p sin θ = (4) (10) (0.5) = 20$ . ✓ Units: $kg m^{2} /s$ . ✓ Naive $r p = 40$ double hai — hamare forecast ko confirm karta hai.

Ex 2 — Cell C2: degenerate (radial motion)

Ex 3 — Cell C3: straight line, $L$ constant rehta hai

Worked example Free particle, koi force nahi

Ek puck ( $m = 2 kg$ , $v = 3 m/s$ ) ek straight line mein slide karta hai jiska $O$ se closest approach $r_{⊥} = 0.5 m$ hai. Koi forces act nahi kar rahe. $L$ nikalo aur dikhao ki puck ke move karne par yeh change nahi karta.

Forecast: $r$ continuously change karta rehta hai jab puck slide karta hai. Kya $L$ change hoga? Guess: nahi — kuch bhi ise twist nahi kar raha.

$p = m v = (2) (3) = 6 kg m/s$ . Yeh step kyun? Angular momentum momentum $p = m v$ se banta hai, raw speed se nahi — mass baked in hona chahiye (Linear Momentum).
$∣ L ∣ = p r_{⊥} = (6) (0.5) = 3 kg m^{2} /s$ . Yeh step kyun? Fixed lever arm use karo, kabhi nahi woh shrinking-then-growing $r$ .
Time-derivative: $\frac{d L}{d t} = r \times F$ . $F = 0$ ke saath, yeh $0$ hai. Yeh step kyun? Torque hi ek cheez hai jo $L$ change karta hai; koi force nahi means koi torque nahi.

Verify: jab puck move karta hai, $p$ (constant, koi force nahi) aur $r_{⊥}$ (ek fixed straight line ki geometry) dono same rehte hain, isliye $L = 3 kg m^{2} /s$ throughout. ✓ Constant, jaise forecast kiya tha.

Ex 4 — Cell C4: sign aur direction

Worked example

$L$ kis direction mein point karta hai? $x y$ -plane mein ek particle $r = (3, 0, 0) m$ par hai aur momentum $p = (0, - 4, 0) kg m/s$ ke saath ( $- y$ direction mein move kar raha hai). $L = r \times p$ nikalo, sign ke saath.

Forecast: $r$ right ki taraf point karta hai, motion downward — right-hand rule se swirl clockwise hai. Guess $L$ page ke andar point karta hai ( $- z$ ), yani negative.

Components mein cross product: $L = (3, 0, 0) \times (0, - 4, 0) = (0 \cdot 0 - 0 \cdot (- 4), 0 \cdot 0 - 3 \cdot 0, 3 \cdot (- 4) - 0 \cdot 0) .$ Yeh step kyun? Cross Product ka determinant formula har component deta hai.
Simplify: $L = (0, 0, - 12) kg m^{2} /s$ . Yeh step kyun? Planar motion ke liye sirf $z$ -component bachta hai.

Verify: magnitude $= 12$ , direction $- \overset{z}{^}$ (page ke andar = clockwise = negative). ✓ Forecast se match karta hai. Compare: agar $p$ $+ y$ hota, toh $+ 12 \overset{z}{^}$ milta (page se bahar, counter-clockwise, positive). Sign spin direction encode karta hai.

Ex 5 — Cell C5: rigid body, $L = I ω$

Ex 6 — Cell C6: conservation, $I$ shrinks

Worked example Skater arms andar kheenchta hai

Ek skater $I_{1} = 4 kg m^{2}$ , $ω_{1} = 2 rad/s$ se start karta hai, phir arms $I_{2} = 1 kg m^{2}$ tak pull karta hai. $ω_{2}$ nikalo, aur kinetic energy mein change nikalo.

Forecast: koi external torque nahi → $L$ fixed. $I$ $4 \times$ drop karta hai, toh $ω$ $4 \times$ badh ke $8 rad/s$ hona chahiye. Energy? Muscles work karte hain, isliye KE increase honi chahiye.

Conservation: $I_{1} ω_{1} = I_{2} ω_{2}$ (Conservation of Angular Momentum). Yeh step kyun? Internal muscle forces koi external torque nahi lagate.
$ω_{2} = \frac{I _{1} ω _{1}}{I _{2}} = \frac{( 4 ) ( 2 )}{1} = 8 rad/s$ . Yeh step kyun? Unknown final spin ke liye conservation equation rearrange karo — dono sides ko $I_{2}$ se divide karo.
$K E_{1} = \frac{1}{2} I_{1} ω_{1}^{2} = \frac{1}{2} (4) (2)^{2} = 8 J$ ; $K E_{2} = \frac{1}{2} I_{2} ω_{2}^{2} = \frac{1}{2} (1) (8)^{2} = 32 J$ . Yeh step kyun? Rotational Kinetic Energy $= \frac{1}{2} I ω^{2}$ ; ise alag check karo kyunki yeh conserved NAHI hoti.

Verify: $L_{1} = (4) (2) = 8$ , $L_{2} = (1) (8) = 8$ — equal. ✓ $Δ K E = 32 - 8 = 24 J$ (muscles ne supply kiya). ✓

Ex 7 — Cell C7: limiting behaviour

Worked example Do limits jo zero dene chahiye

(i) Ek particle jiska $p = 10 kg m/s$ hai aur lever arm shrink ho raha hai $r_{⊥} \to 0$ . (ii) Ek disc jiska $I = 0.5 kg m^{2}$ hai aur spin slow ho raha hai $ω \to 0$ . Har case mein $L$ kis value ke paas jaata hai?

Forecast: dono limits $L \to 0$ deni chahiye — koi off-centre distance nahi, ya koi spin nahi.

(i) $L = p r_{⊥} = 10 r_{⊥}$ . Jab $r_{⊥} \to 0$ , $L \to 10 \times 0 = 0$ . Yeh step kyun? $O$ se hoke aim ki gayi motion mein koi angular momentum nahi hoti — Ex 2 ka smooth version.
(ii) $L = I ω = 0.5 ω$ . Jab $ω \to 0$ , $L \to 0$ . Yeh step kyun? Ek stationary body ka koi angular momentum nahi hota, chahe kitna bhi massive ho.

Verify: sample point $r_{⊥} = 0.01 m \Rightarrow L = 0.1$ ; $r_{⊥} = 0.001 \Rightarrow L = 0.01$ — $0$ ki taraf decrease ho raha hai. ✓ Similarly $ω = 0.1 \Rightarrow L = 0.05$ , $ω = 0.01 \Rightarrow L = 0.005$ . ✓ Dono limits continuous hain aur $0$ tak pahunchti hain.

Ex 8 — Cell C8: real-world (Kepler's second law)

Worked example Sun ke paas comet tezi se sweep karta hai

Ek comet aphelion (sabse door point) par $r_{a} = 5 \times 1 0^{11} m$ Sun se door hai, $v_{a} = 1 \times 1 0^{4} m/s$ se $r$ ke perpendicular move kar raha hai. Perihelion (closest point) par $r_{p} = 1 \times 1 0^{11} m$ . Perihelion speed $v_{p}$ nikalo.

Forecast: gravity $r$ ke along point karti hai (Sun ki taraf), isliye Sun ke baare mein uska torque zero hai → $L$ conserved. Chhota $r$ ka matlab bada $v$ hona chahiye. Guess karo $v_{p}$ $5 \times$ bada hoga.

Dono apsides par $v ⊥ r$ , isliye $L = m v r$ ( $sin 9 0^{\circ} = 1$ ke saath). Yeh step kyun? Closest/farthest points par motion purely tangential hoti hai.
Conservation: $m v_{a} r_{a} = m v_{p} r_{p}$ ; mass cancel ho jaata hai. Yeh step kyun? Central (radial) force ⇒ zero torque ⇒ constant $L$ — yahi Kepler's Second Law hai.
$v_{p} = \frac{v _{a} r _{a}}{r _{p}} = \frac{( 1 0 ^{4} ) ( 5 \times 1 0 ^{11} )}{1 \times 1 0 ^{11}} = 5 \times 1 0^{4} m/s$ . Yeh step kyun? Conservation equation ko unknown $v_{p}$ ke liye $r_{p}$ se divide karke solve karo, phir numbers plug in karo.

Verify: $L \propto v_{a} r_{a} = 1 0^{4} \cdot 5 \times 1 0^{11} = 5 \times 1 0^{15}$ aur $v_{p} r_{p} = 5 \times 1 0^{4} \cdot 1 0^{11} = 5 \times 1 0^{15}$ — equal. ✓ Speed $5 \times$ badi hai, jaise forecast kiya tha.

Ex 9 — Cell C9: exam twist, vector cancellation

Worked example Do particles ka net angular momentum

Particle A: $r_{A} = (0, 2, 0) m$ , $p_{A} = (3, 0, 0) kg m/s$ . Particle B: $r_{B} = (0, - 2, 0) m$ , $p_{B} = (3, 0, 0) kg m/s$ . $O$ ke baare mein total $L = L_{A} + L_{B}$ nikalo. (Exam trap: students magnitudes add kar dete hain aur sign bhool jaate hain.)

Forecast: A $O$ ke upar right move kar raha hai → ek taraf swirl karta hai; B $O$ ke neeche right move kar raha hai → opposite taraf swirl karta hai. Guess karo yeh cancel karke zero ho jaate hain.

$L_{A} = r_{A} \times p_{A} = (0, 2, 0) \times (3, 0, 0)$ . $z$ -component compute karo: $r_{x} p_{y} - r_{y} p_{x} = 0 \cdot 0 - 2 \cdot 3 = - 6$ . Toh $L_{A} = (0, 0, - 6)$ . Yeh step kyun? Planar motion → sirf $z$ bachta hai.
$L_{B} = (0, - 2, 0) \times (3, 0, 0)$ : $z = 0 \cdot 0 - (- 2) \cdot 3 = + 6$ . Toh $L_{B} = (0, 0, + 6)$ . Yeh step kyun? Flipped position sign swirl direction flip kar deta hai.
$L = L_{A} + L_{B} = (0, 0, - 6 + 6) = (0, 0, 0)$ . Yeh step kyun? Angular momentum ek vector hai — opposite swirls cancel ho jaate hain.

Verify: magnitudes $∣ - 6 ∣ = ∣ + 6 ∣ = 6$ hain, lekin signs opposite hain isliye vector sum $0$ hai. ✓ Magnitudes add karna ( $12$ ) trap hoga.

Recall

Recall Kaunsa formula kis cell ke liye?

$O$ ki taraf radial motion, $L$ kya hai? ::: Zero — $sin θ = 0$ , lever arm zero hai Straight-line free particle, kya $L$ time mein change karta hai? ::: Nahi — koi force nahi matlab koi torque nahi, $r_{⊥}$ aur $p$ fixed $x y$ -plane mein clockwise spin, $L_{z}$ ka sign? ::: Negative (page ke andar, $- \overset{z}{^}$ ) Skater arms andar kheenchta hai — kya energy conserved hoti hai? ::: Nahi, sirf $L$ hoti hai; muscles KE add karte hain Comet Sun ke paas tezi se move karta hai kyunki... ::: gravity radial hai → zero torque → $L = m v r$ constant Do equal particles opposite taraf swirl karte hain — net $L$ ? ::: Zero ho sakta hai; $L$ ek vector hai, signs matter karte hain

Connections

Cross Product — sign aur cancellation ke peeche ki machinery
Conservation of Angular Momentum — Ex 6, Ex 8
Kepler's Second Law — Ex 8 exactly yahi law hai
Torque — kyun free particles constant $L$ rakhte hain