Generalized momenta and generalized forces: Worked examples

2.1.7 · D3 · Physics › Analytical Mechanics › Generalized momenta and generalized forces

Prerequisites jo tum use kar sakte ho: Lagrangian Mechanics, Euler-Lagrange Equation, Generalized coordinates and constraints, D'Alembert's Principle, Angular momentum, Noether's Theorem.

Scenario matrix

Is topic ka har problem in cells mein se ek (ya zyada) hota hai. Neeche ke 8 worked examples us cell ke label ke saath hain jo wo cover karta hai, aur milke sab cells hit karte hain.

Cell	Kya cheez use alag banati hai	Covered by
A. Length coordinate	$q$ ek distance hai → $p_{j}$ ordinary momentum hai, $Q_{j}$ real force hai	Ex 1, Ex 5
B. Angle coordinate	$q$ ek angle hai → $p_{j}$ angular momentum hai, $Q_{j}$ ek torque hai	Ex 2, Ex 6
C. Sign of $Q_{j}$	force positive (driving) ya negative (restoring) ho sakti hai	Ex 2, Ex 4
D. Cyclic coordinate	$q$ $L$ mein absent hai → $p_{j}$ conserved (Noether)	Ex 3
E. Force hidden in $T$	koi potential nahi, phir bhi $\partial T / \partial q$ se generalized force aati hai	Ex 5
F. Zero / degenerate input	$\dot{θ} = 0$ , $r = 0$ , ya koi coordinate equilibrium par	Ex 4, Ex 7
G. Limiting behaviour	small-angle limit, large-mass limit, $ω \to 0$	Ex 6 (small angle), Ex 5 ( $ω \to 0$ )
H. Real-world word problem	figure-skater / merry-go-round ki kahani	Ex 7
I. Exam twist	do coupled coordinates, ek coupled cyclic momentum	Ex 8

Example 1 — Free particle in a line (Cell A)

Neeche ki figure coordinate fix karti hai: ek single number $x$ jo rail ke saath measure hota hai. Kuch bhi is par depend nahi karta ki hum origin kahan rakhte hain, aur theek isliye $x$ cyclic hai.

Figure — Generalized momenta and generalized forces

Worked example Statement

Ek particle jiska mass $m = 2 kg$ hai, $x$ -axis ke saath freely move karta hai, velocity $\overset{x}{˙} = 3 m/s$ ke saath. Koi force nahi lag rahi. $p_{x}$ aur $Q_{x}$ nikalo.

Forecast: Kya $p_{x}$ sirf ordinary momentum hai? Kya $Q_{x}$ bilkul zero hai?

Step 1 — $L$ likho. Koi potential nahi, toh $L = T - V = \frac{1}{2} m \overset{x}{˙}^{2} - 0 = \frac{1}{2} m \overset{x}{˙}^{2}$ . Yeh step kyun? Lagrangian hamesha $L = T - V$ hota hai; yahan $V = 0$ hai, toh $L$ pure kinetic energy hai.

Step 2 — Generalized momentum. $p_{x} = \frac{\partial L}{\partial x ˙} = m \overset{x}{˙} = 2 \cdot 3 = 6 kg\cdotm/s$ . Yeh step kyun? Yeh $p_{j}$ ki definition hai; length coordinate ke liye "mass factor" sirf $m$ hota hai, toh yeh ordinary momentum mein reduce ho jaata hai.

Step 3 — Generalized force. Upar ki boxed identity se, $Q_{x} = \frac{\partial L}{\partial x}$ . Kyunki $T = \frac{1}{2} m \overset{x}{˙}^{2}$ mein koi $x$ nahi hai, $\frac{\partial T}{\partial x} = 0$ , aur $V = 0$ , toh $Q_{x} = \frac{\partial L}{\partial x} = \frac{\partial T}{\partial x} - \frac{\partial V}{\partial x} = 0 - 0 = 0$ . Yeh step kyun? Generalized force hamesha $\partial L / \partial x$ ke barabar hoti hai (yahi $\overset{p}{˙}_{x} = \partial L / \partial x$ kehta hai). Yahan Lagrangian mein koi $x$ hai hi nahi, toh uska derivative zero ho jaata hai: koi force nahi, aur $x$ cyclic hai.

Verify: $p_{x}$ ke units: $kg \cdot m/s$ ✓ (ordinary momentum). $\overset{p}{˙}_{x} = Q_{x} = 0$ , toh momentum constant hai → particle sliding karta rahega, jaisa Newton kehta hai. ✓

Example 2 — Pendulum at a generic angle (Cells B, C)

Figure dekho: pivot height par dashed line humari potential-energy reference level hai ( $V = 0$ wahan); red arrow gravity torque hai jo $θ$ ko wapas zero par kheench raha hai.

Worked example Solution

Forecast: Kya $Q_{θ}$ yahan negative (restoring) aayega? Uske units kya honge?

Step 1 — Lagrangian. Kinetic aur potential energy ko combine karo: $L = T - V = \frac{1}{2} m ℓ^{2} \dot{θ}^{2} - (- m g ℓ cos θ) = \frac{1}{2} m ℓ^{2} \dot{θ}^{2} + m g ℓ cos θ$ . Yeh step kyun? $T = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}$ (speed $= ℓ \dot{θ}$ ). $V$ ke liye hum height $y = - ℓ cos θ$ pivot level se measure karte hain (figure mein dashed line, jahan $V = 0$ ), toh $V = m g y = - m g ℓ cos θ$ . $L = T - V$ mein ise subtract karne par sign flip ho jaata hai, isliye potential $L$ mein plus ke saath enter karta hai.

Step 2 — Generalized momentum. $p_{θ} = \frac{\partial L}{\partial θ ˙} = m ℓ^{2} \dot{θ} = 1 \cdot 4 \cdot 0.5 = 2 kg\cdotm^{2} / s$ . Yeh step kyun? Coordinate ek angle hai, toh uska conjugate angular momentum hai — units $kg\cdotm^{2} / s$ , na ki $kg\cdotm/s$ .

Step 3 — Generalized force (ek torque). Boxed identity se, $Q_{θ} = \frac{\partial L}{\partial θ} = \frac{\partial T}{\partial θ} - \frac{\partial V}{\partial θ}$ . Yahan $T = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}$ mein koi bare $θ$ nahi hai, toh $\partial T / \partial θ = 0$ aur poori force potential se aati hai: $Q_{θ} = - \frac{\partial V}{\partial θ} = - m g ℓ sin θ = - 1 \cdot 9.8 \cdot 2 \cdot sin 3 0^{\circ} = - 9.8 N\cdotm$ . Yeh step kyun? Hume dikhana hoga ki shortcut $- \partial V / \partial θ$ yahan legal kyun hai: kyunki $\partial T / \partial θ = 0$ hai, general $\partial L / \partial θ$ isme reduce ho jaata hai. Angle coordinate ⇒ $Q_{θ}$ ek torque hai; minus sign kehta hai gravity $θ$ ke badhne ko rok rahi hai — yeh restoring hai (Cell C, negative sign).

Verify: Units: $m ℓ^{2} \dot{θ} = kg\cdotm^{2} / s$ ✓; $m g ℓ sin θ = N\cdotm$ ✓. Sign: $θ = 3 0^{\circ} > 0$ par bob daayein hai, gravity use wapas baayein kheenchti hai → negative torque ✓.

Example 3 — Central force, cyclic angle (Cell D)

Figure polar coordinates dikhata hai: $r$ (distance) aur $θ$ (angle swept). Kyunki force seedha $r$ ke along point karti hai, poori picture ko centre ke baare mein spin karne se kuch nahi badalta — yahi rotational symmetry $θ$ ko cyclic banati hai.

Worked example Solution

Ek planet mass $m$ ek central potential $V (r)$ (sirf distance $r$ par depend) ke under plane mein move kar raha hai. Polar coordinates $(r, θ)$ use karo: $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2})$ . Dikhao ki $p_{θ}$ conserved hai aur, $m = 1$ , $r = 2$ , $\dot{θ} = 1.5$ ke liye, uski value compute karo.

Forecast: Kaun sa coordinate cyclic hai — $r$ ya $θ$ ? Conserved $p_{θ}$ physically kya represent karta hai?

Step 1 — Lagrangian. $L = T - V = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}) - V (r)$ . Yeh step kyun? Standard $L = T - V$ ; note karo $V$ mein koi $θ$ nahi kyunki force central hai.

Step 2 — Cyclic coordinate identify karo. $θ$ , $L$ mein sirf $\dot{θ}$ ke roop mein aata hai, kabhi $θ$ itself ke roop mein nahi. Toh $\frac{\partial L}{\partial θ} = 0$ . Yeh step kyun? Noether's Theorem: $L$ se absent ek coordinate ek symmetry signal karta hai (yahan rotational, "figure ko spin karo" ke roop mein visible), jo ek conservation law force karta hai.

Step 3 — Conserved momentum. Euler–Lagrange deta hai $\overset{p}{˙}_{θ} = \partial L / \partial θ = 0$ , toh $p_{θ} = m r^{2} \dot{θ}$ constant hai. Numerically $p_{θ} = 1 \cdot 2^{2} \cdot 1.5 = 6 kg\cdotm^{2} / s$ . Yeh step kyun? $\overset{p}{˙}_{θ} = 0$ matlab $p_{θ}$ kabhi nahi badlega — yeh Angular momentum conservation hai.

Verify: Units $kg\cdotm^{2} / s$ ✓. Physically $p_{θ} = L_{ang}$ — planet equal time mein equal areas sweep karta hai (Kepler's 2nd law), iska direct consequence. ✓

Example 4 — Degenerate input: pendulum at the bottom (Cells C, F)

Figure torque $Q_{θ} (θ) = - m g ℓ sin θ$ plot karta hai. Notice karo yeh $θ = 0$ par zero cross karta hai ek downward (negative) slope ke saath — red tangent line — jo ek stable equilibrium ki signature hai.

Worked example Solution

Ex 2 ka pendulum lo lekin equilibrium point $θ = 0$ par (seedha neeche latakta hua), momentarily at rest $\dot{θ} = 0$ . $p_{θ}$ aur $Q_{θ}$ nikalo.

Forecast: Dono zero? Ya unme se ek equilibrium par survive karta hai?

Step 1 — Momentum. $p_{θ} = m ℓ^{2} \dot{θ} = m ℓ^{2} \cdot 0 = 0$ . Yeh step kyun? $p_{θ} \propto \dot{θ}$ ; zero angular velocity ke saath zero angular momentum hoga. Yeh degenerate velocity input hai.

Step 2 — Force. $Q_{θ} = - m g ℓ sin (0) = 0$ . Yeh step kyun? $sin 0 = 0$ — bottom par gravity torque vanish ho jaata hai. Yeh exactly equilibrium condition hai (Cell F).

Step 3 — Interpret karo. $\overset{p}{˙}_{θ} = Q_{θ} = 0$ is instant par, toh abhi koi angular acceleration nahi. Lekin yeh ek stable equilibrium hai: $θ$ ko thoda positive nudge karo aur $Q_{θ} = - m g ℓ sin θ < 0$ push back karta hai (figure mein negative slope dekho). Yeh step kyun? Hum equilibrium ke paas $Q_{θ}$ ka sign check karte hain, sirf usp ar nahi, stability classify karne ke liye.

Verify: $θ = 0$ par: $sin θ = 0 \Rightarrow Q_{θ} = 0$ ✓. Slope $\frac{d Q _{θ}}{d θ}_{0} = - m g ℓ cos 0 = - m g ℓ < 0$ → restoring → stable ✓.

Example 5 — Bead on a rotating wire: force from $T$ (Cells A, E, G)

Figure spinning wire (angular rate $ω$ ) aur $r$ distance par bead dikhata hai. Red arrow outward "centrifugal" generalized force hai — yeh $\partial T / \partial r$ se aati hai, kisi potential se nahi.

Worked example Solution

Ek bead mass $m = 0.5 kg$ ek frictionless straight wire par slide karta hai jo origin ke baare mein constant angular rate $ω = 4 rad/s$ ( $θ = ω t$ ) par forced spin kar raha hai. Coordinate: $r$ . Toh $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} ω^{2})$ , $V = 0$ . $p_{r}$ aur generalized force $Q_{r}$ at $r = 1 m$ nikalo.

Forecast: $V = 0$ ke saath, kya $Q_{r}$ zero hai? Ya kisi aur jagah se force appear hoti hai?

Step 1 — Lagrangian. $L = T - V = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} ω^{2}) - 0 = \frac{1}{2} m \overset{r}{˙}^{2} + \frac{1}{2} m r^{2} ω^{2}$ . Yeh step kyun? Differentiate karne se pehle hume $L$ explicitly likhna hoga; $V = 0$ toh $L$ pure kinetic energy hai, lekin crucially isme $r$ hai ( $r^{2} ω^{2}$ term ke through).

Step 2 — Momentum. $p_{r} = \frac{\partial L}{\partial r ˙} = m \overset{r}{˙}$ . Yeh step kyun? $r$ ek length coordinate hai (Cell A), toh $p_{r}$ wire ke along ordinary linear momentum hai.

Step 3 — Generalized force. Boxed identity se $Q_{r} = \frac{\partial L}{\partial r} = \frac{\partial T}{\partial r} - \frac{\partial V}{\partial r} = m r ω^{2} - 0 = m r ω^{2}$ . $r = 1$ par: $Q_{r} = 0.5 \cdot 1 \cdot 4^{2} = 8 N$ , outward point karta hua. Yeh step kyun? $V = 0$ ke bawajood, generalized force $\partial L / \partial r$ nonzero hai kyunki $T$ khud $r$ par depend karta hai (Cell E). Yeh centrifugal effect hai, aur ise legitimately $Q_{r}$ kehte hain — koi special " $F_{eff}$ " nahi — kyunki $Q_{r} = \partial L / \partial q$ definition hai.

Step 4 — Equation of motion & limiting check ( $ω \to 0$ ). Euler–Lagrange: $\overset{p}{˙}_{r} = Q_{r} \Rightarrow m \overset{r}{¨} = m r ω^{2}$ . Agar wire spinning band ho jaaye, $Q_{r} = m r \cdot 0 = 0$ — koi force nahi ek free bead ke liye (Cell G), Ex 1 recover karta hai. Yeh step kyun? Har formula ko apne limit mein known simpler case mein reduce hona chahiye; $ω \to 0$ force-free particle recover karta hai.

Verify: $Q_{r} = 0.5 \cdot 1 \cdot 16 = 8 N$ ✓ ; $ω = 0$ par, $Q_{r} = 0$ ✓. Units $kg \cdot m \cdot s^{- 2} = N$ ✓ (length coordinate → real force).

Example 6 — Small-angle pendulum torque (Cells B, G)

Figure exact torque $- m g ℓ sin θ$ (black) ko linear approximation $- m g ℓ θ$ (red) ke saath overlay karta hai. $θ = 0$ ke paas dono almost indistinguishable hain — isliye tiny swings simple harmonic hoti hain.

Worked example Solution

Ex 2 ke pendulum ke liye ( $m = 1$ , $ℓ = 2$ , $g = 9.8$ ), ek small angle $θ = 0.05 rad$ par restoring torque nikalo, aur exact value ko linear approximation $Q_{θ} \approx - m g ℓ θ$ se compare karo.

Forecast: $0.05$ rad tilt ke liye $sin θ$ kitna close hai $θ$ ke? $0.1%$ se kam error?

Step 1 — Exact torque. $Q_{θ} = - m g ℓ sin θ = - 1 \cdot 9.8 \cdot 2 \cdot sin (0.05)$ . Yeh step kyun? Angle coordinate ⇒ torque (Cell B); pehle full $sin$ rakhte hain.

Step 2 — Linearize karo. Small $θ$ ke liye, $sin θ \approx θ$ , jo deta hai $Q_{θ} \approx - m g ℓ θ = - 19.6 \cdot 0.05 = - 0.98 N\cdotm$ . Yeh step kyun? Small-angle limit (Cell G) pendulum ko simple harmonic oscillator mein turn karta hai; isliye tiny swings ke liye "SHM" hold karta hai.

Step 3 — Error size. Exact $- 19.6 sin (0.05) = - 0.979837 \dots$ vs approx $- 0.98$ . Relative error $\approx 0.084%$ . Yeh step kyun? Hume quantify karna hoga ki approximation kitni achhi hai, sirf assert nahi karna.

Verify: $∣ Q_{exact} - Q_{approx} ∣/∣ Q_{approx} ∣ \approx 0.00084 < 0.001$ ✓. Linear model kuch degrees se neeche excellent hai. ✓

Example 7 — Figure skater word problem (Cells F, H)

Figure dekho: red arrow angular momentum $p_{θ}$ hai — pehle aur baad mein SAME length.

Worked example Solution

Forecast: Woh faster spin karegi ya slower? Kya $\dot{θ}$ constant rehta hai (ek common trap)?

Step 1 — Conserved quantity identify karo. $θ$ cyclic hai (koi external torque nahi, $\partial L / \partial θ = 0$ ), toh $p_{θ} = I \dot{θ}$ conserved hai. Yeh step kyun? Yeh momentum $p_{θ}$ hai jo conserved hai, na ki velocity $\dot{θ}$ — yeh classic mistake hai.

Step 2 — Conservation apply karo. $I_{1} \dot{θ}_{1} = I_{2} \dot{θ}_{2} \Rightarrow \dot{θ}_{2} = \frac{I _{1} θ ˙ _{1}}{I _{2}} = \frac{4 \cdot 2}{1} = 8 rad/s$ . Yeh step kyun? $p_{θ}$ ko pehle aur baad mein equal karne se directly new rate mil jaata hai.

Step 3 — Degenerate check ( $I_{2} = I_{1} = 4$ ). Toh $\dot{θ}_{2} = \frac{4 \cdot 2}{4} = 2 rad/s = \dot{θ}_{1}$ : inertia mein koi change nahi → rate mein koi change nahi (Cell F). Yeh step kyun? Ek sanity limit — agar woh kuch nahi karti, kuch nahi badalna chahiye.

Verify: $p_{θ}$ pehle $= 4 \cdot 2 = 8$ ; baad mein $= 1 \cdot 8 = 8$ ✓ conserved. Woh 4× faster spin karti hai ( $8$ rad/s vs $2$ rad/s) kyunki $I$ 4× drop hua; degenerate case correctly $2$ rad/s return karta hai. ✓ Trap note karo: velocity $\dot{θ}$ badla lekin momentum $p_{θ}$ nahi badla.

Example 8 — Exam twist: coupled cart–pendulum cyclic momentum (Cell I)

Figure cart (coordinate $x$ ) ko hanging bob (angle $θ$ ) ke saath dikhata hai. Poore apparatus ko rail ke saath left ya right slide karne se koi energy nahi badlati, toh $x$ cyclic hai chahe bob gravity feel karta ho $θ$ ke through.

Worked example Solution

Ek cart mass $M = 3 kg$ ek horizontal rail par freely slide karta hai (coordinate $x$ ); ek bob mass $m = 1 kg$ usse ek light rod par latakta hai, angle $θ$ . Koi horizontal external force nahi. Horizontal generalized momentum hai: $p_{x} = (M + m) \overset{x}{˙} + m ℓ \dot{θ} cos θ .$ Ek instant par $\overset{x}{˙} = 0.4 m/s$ , $ℓ = 2$ , $\dot{θ} = 0.3 rad/s$ , $θ = 6 0^{\circ}$ . $p_{x}$ compute karo aur batao yeh conserved kyun hai.

Forecast: Kya $x$ cyclic hai chahe pendulum clearly gravity feel karta hai? Kya survive karta hai?

Step 1 — $x$ ki cyclicity check karo. Gravity vertically act karti hai; potential $V = - m g ℓ cos θ$ sirf $θ$ par depend karta hai, $x$ par nahi. Kinetic energy $\overset{x}{˙}$ par depend karti hai lekin $x$ par nahi. Toh $\partial L / \partial x = 0$ → $x$ cyclic hai. Yeh step kyun? Sirf $L$ mein $x$ -dependence matter karta hai; bob ki gravity $θ$ ke through enter karti hai, $x$ ko ignorable chhodti hai. Isliye horizontal momentum conserved hai.

Step 2 — $p_{x}$ evaluate karo. $p_{x} = (3 + 1) \cdot 0.4 + 1 \cdot 2 \cdot 0.3 \cdot cos 6 0^{\circ} = 1.6 + 0.6 \cdot 0.5 = 1.6 + 0.3 = 1.9 kg\cdotm/s$ . Yeh step kyun? Sirf substitute karo; $cos θ$ weight bob ki motion ka horizontal par $\partial r / \partial q$ -style projection hai.

Step 3 — Meaning. $\overset{p}{˙}_{x} = \partial L / \partial x = 0$ , toh yeh $1.9 kg\cdotm/s$ tab tak fixed rehta hai jab $x$ aur $θ$ evolve karte hain, cart aur bob ke beech momentum exchange karte hue. Yeh step kyun? Conservation tumhe full coupled equations solve kiye bina cart ka recoil predict karne deta hai.

Verify: $p_{x} = 4 \cdot 0.4 + 2 \cdot 0.3 \cdot 0.5 = 1.6 + 0.3 = 1.9$ ✓. Units $kg\cdotm/s$ ✓ (length coordinate → linear momentum). Conserved kyunki $x$ $L$ mein absent hai. ✓

Recall Quick self-test

"No potential yet a force appears" — kaun sa cell hai? ::: Cell E — force $\partial T / \partial q$ mein chupi hoti hai (Ex 5, centrifugal term). Cyclic coordinate ke liye kya constant hota hai — velocity ya momentum? ::: Momentum $p_{j}$ (Ex 3, Ex 7). Velocity generally badlati hai. Angle coordinate ke liye $Q_{θ}$ ke units kya hain? ::: Torque, $N\cdotm$ (Ex 2). Skater problem mein woh faster kyun spin karti hai? ::: $p_{θ} = I \dot{θ}$ conserved hai; $I$ 4× drop hua toh $\dot{θ}$ 4× badh gaya (Ex 7).