Applying E-L equations to various systems: Question bank

2.1.6 · D5 · Physics › Analytical Mechanics › Applying E-L equations to various systems

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Parent note ne tumhe recipe sikhayi. Yeh page un jagahon par attack karta hai jahan recipe lagta hai kuch aur kehti ho lekin actually kuch aur kehti hai. Har item ek one-line reveal hai: prompt padho, answer chhupao, apni reasoning zyoor se bolo, phir uncover karo.

Shuru karne se pehle, ek convention jo har jagah use hoti hai, bilkul zero se explain ki gayi hai taaki kuch assume na ho:

$q$ = ek generalized coordinate — koi bhi ek number (ek angle, ek distance, kuch bhi) jo system ko locate karne mein madad karta hai. Har degree of freedom ke liye ek.
$\overset{q}{˙}$ = uska time-rate-of-change, "wo number abhi kitni tezi se badal raha hai".
$L = T - V$ jahan $T$ kinetic (chalti hui) energy hai aur $V$ potential (stored) energy hai.
$\frac{\partial L}{\partial q ˙}$ = $q$ ke conjugate generalized momentum; $\overset{q}{˙}$ ke alawa sab freeze karo aur dekho $L$ kaise respond karta hai. (Zyada detail mein 2.1.08-Generalized-momenta-and-the-Hamiltonian mein.)
Ek cyclic coordinate woh hai jo $L$ mein naam se appear nahi karta (sirf uska dot appear karta hai). Dekho 2.1.07-Cyclic-coordinates-and-conservation-laws.

True or false — justify karo

Har answer True/False se shuru hota hai, phir kyun — kabhi bare nahi.

Lagrangian hamesha system ki total energy

T + V

hoti hai.

False.

L = T - V

hai, ek difference hai. Sum

T + V

energy hai (aksar Hamiltonian);

L

ek alag bookkeeping object hai jiska stationary integral real path chunti hai.

Agar coordinate

q

cyclic hai, toh

q

khud time mein constant rehta hai.

False. Cyclic ka matlab hai

q

$L$ mein absent hai, isliye uska momentum

\partial L / \partial \overset{q}{˙}

constant hai — lekin

q

phir bhi change ho sakta hai (jaise central force mein

θ

steadily badhta hai jabki

L_{z}

fixed hai).

Aise coordinates chunna jo constraint ko respect karte hain, iska matlab hai constraint force koi kaam nahi karta.

True — yahi wajah hai kyun hum ise drop kar sakte hain. Ek ideal constraint force allowed motion ke perpendicular hota hai, isliye

F \cdot d r = 0

, aur yeh kabhi

T

V

mein enter nahi karta.

Euler–Lagrange equation sirf Cartesian coordinates mein kaam karti hai.

False. Iska pura selling point yahi hai ki ek hi form

\frac{d}{d t} \frac{\partial L}{\partial q ˙} - \frac{\partial L}{\partial q} = 0

kisi bhi coordinates mein hold karta hai — angles, arc-length, polar — kyunki yeh ek coordinate-free variational principle se aata hai.

Pendulum ke liye,

\ddot{θ} = - \frac{g}{ℓ} sin θ

simple harmonic motion hai.

Aam taur par False. Yeh sirf chote

θ

ke liye approximately SHM hai jahan

sin θ \approx θ

. Bade swings ke liye

sin θ

period ko amplitude ke saath badhata hai — genuine anharmonic motion.

L

ke partial derivatives ke andar, differentiate karne se pehle hum

\overset{q}{˙} = d q / d t

substitute kar sakte hain.

False.

\partial / \partial q

\partial / \partial \overset{q}{˙}

lete waqt,

q

aur

\overset{q}{˙}

ko independent slots ki tarah treat karo.

\overset{q}{˙} = d q / d t

ka link sirf bahari

d / d t

se restore hota hai.

V

mein constant add karna equations of motion ko change karta hai.

False.

V

ke sirf derivatives E–L mein enter karte hain (

\partial L / \partial q

), isliye constant shift cancel ho jaati hai. Sirf

V

ki shape matter karti hai, uska zero point kabhi nahi.

Error dhundho

Har line ek flawed move batati hai; reveal flaw ka naam aur fix deta hai.

"Polar coordinates mein

T = \frac{1}{2} m \overset{r}{˙}^{2}

Galat: tangential motion missing hai. Speed

v^{2} = \overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}

hai, isliye

T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2})

hai. Hamesha pehle

\dot{r}

v^{2}

banao.

"Atwood machine ke liye main equation mein string tension

T_{string}

ko force ki tarah include karunga."

Galat: inextensible string ek ideal constraint hai, jo ek coordinate

x

use karne mein pehle se baked in hai. Tension add karna ise double-count karta hai. Ise bilkul drop karo.

"Pendulum mein 2 DOF hain kyunki bob ek plane mein rehta hai."

Galat: rigid rod radius ko fix karta hai, ek coordinate kill kar deta hai. Sirf angle

θ

free hai ⇒ 1 DOF. Constraints ke baad independent freedoms gino.

L = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}

ke liye

\frac{d}{d t} \frac{\partial L}{\partial θ ˙}

barabar

m ℓ^{2} \dot{θ}

hai."

Galat: woh sirf

\partial L / \partial \dot{θ}

hai. Tumhe abhi bhi bahari

d / d t

dena baaki hai, jo

m ℓ^{2} \ddot{θ}

deta hai. E–L rule hai "pehle partial phir total-time-derivative".

"Incline par main normal force ko

V

mein add karunga taaki energy conserve ho."

Galat: normal force koi kaam nahi karta (slope ke perpendicular), isliye yeh kabhi

V

mein belong nahi karta. Slope ke along

s

use karna ise pehle se remove kar deta hai.

"Kyunki gravity incline block par act karti hai,

V = m g s

hai."

Sign aur geometry galat hai: block

s sin α

se neeche jaata hai, PE kho deta hai, isliye

V = - m g s sin α

hai. Actual height change track karo, path length nahi.

\partial L / \partial θ = - m g ℓ sin θ

isliye pendulum EOM hai

\ddot{θ} = - g ℓ sin θ

Galat: tumne

\frac{d}{d t} \frac{\partial L}{\partial θ ˙}

se aane wala

m ℓ^{2}

divide karna bhul gaye. Correct EOM hai

\ddot{θ} = - \frac{g}{ℓ} sin θ

"Centrifugal term

m r \dot{θ}^{2}

ek real applied force hai jo maine central-force problem mein add ki."

Galat: kisine ise add nahi kiya. Yeh

\partial T / \partial r

se automatically nikalta hai kyunki

T

mein

r^{2} \dot{θ}^{2}

hai. Yeh rotating polar coordinates use karne ka consequence hai, koi external push nahi.

Why questions

E–L method kyun tension aur normal forces ko bilkul ignore karne deta hai?

Kyunki woh ideal constraint forces permitted motion ke perpendicular hain aur zero work karte hain, aur independent coordinates ka achha chunav pehle se forbidden motion ko forbid kar deta hai — forces geometrically absorb ho jaate hain.

Hamen exactly ek generalized coordinate per degree of freedom kyun chahiye — na zyada, na kam?

Kam se system locate nahi ho sakta; zyada se redundant variables introduce hote hain jo constraints se bande hain, woh independence todta hai jo E–L assume karta hai. One-per-DOF minimal complete description hai.

L

se absent coordinate conserved quantity kyun produce karta hai?

Agar

\partial L / \partial q = 0

hai, toh E–L equation ban jaati hai

\frac{d}{d t} (\partial L / \partial \overset{q}{˙}) = 0

, isliye momentum

\partial L / \partial \overset{q}{˙}

ka zero time-derivative hai — yeh constant hai. Yeh apne simplest roop mein Noether's theorem hai.

Polar coordinates mein tangential kinetic term

\dot{θ}^{2}

nahi balki

r^{2} \dot{θ}^{2}

kyun hai?

Kyunki tangential speed radius times angular rate hai,

v_{⊥} = r \dot{θ}

. Kinetic energy speed squared use karti hai, isliye

v_{⊥}^{2} = r^{2} \dot{θ}^{2}

r

angle-rate ko real distance-rate mein convert karta hai.

Atwood acceleration sirf

(m_{1} - m_{2}) / (m_{1} + m_{2})

par depend kyun karta hai, tension par nahi?

Numerator net driving weight difference hai; denominator accelerate ho raha total mass hai. Tension ek internal constraint force hai jo ek baar single coordinate chunne ke baad kabhi appear nahi karta.

Differentiation ke dauran hum

q

aur

\overset{q}{˙}

ko independent variables kyun treat kar sakte hain jab physically

\overset{q}{˙} = d q / d t

hai?

Kyunki

L

possible paths ke pure space mein do independent slots ki function hai; physical relation sirf actual trajectory par hold karta hai, jise bahari

d / d t

reinstate karta hai. Variational calculus is separation ki demand karta hai.

Hamilton's principle (

δ S = 0

) deeper statement kyun hai, E–L sirf uska consequence hai?

Principle kehta hai ki true path action ko endpoints par fixed sab wiggles ke against stationary banata hai; arbitrary wiggles ke liye yeh demand karna aur parts mein integrate karna E–L yield karta hai. Dekho 2.1.05-Deriving-the-Euler-Lagrange-equation.

Edge cases

Atwood machine jahan equal masses

m_{1} = m_{2}

hain: EOM kya predict karta hai?

\overset{x}{¨} = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = 0

. Koi net weight difference nahi matlab koi acceleration nahi — system kisi bhi position par equilibrium mein hai. Sanity check pass.

Atwood machine jahan ek mass zero set hai, maan lo

m_{2} = 0

\overset{x}{¨} = \frac{m _{1} g}{m _{1}} = g

. Bacha hua mass free fall mein hai kyunki kuch oppose nahi karta — limiting case plain gravity reproduce karta hai.

Central-force particle jahan

\dot{θ} = 0

(purely radial launch): conserved

L_{z}

ka kya hota hai?

L_{z} = m r^{2} \dot{θ} = 0

hai aur zero rehta hai. Motion ek straight radial line hai; centrifugal term

m r \dot{θ}^{2}

vanish ho jaata hai, sirf

m \overset{r}{¨} = - V^{'} (r)

bachta hai, ek 1D problem.

Pendulum exactly

θ = π

se release kiya (seedha upar): EOM kya kehta hai?

\ddot{θ} = - \frac{g}{ℓ} sin π = 0

. Yeh equilibrium hai — lekin ek unstable wala; zameen ki tarah thodi si perturbation badhti hai,

θ = 0

point ki tarah stable nahi. Zero acceleration instability chhupata hai.

Incline angle

α \to 0

(flat ground):

\overset{s}{¨} = g sin 0 = 0

. Koi slope nahi, surface ke along koi gravitational pull nahi, isliye block accelerate nahi karta. Formula gracefully "kuch nahi hota" mein degrade ho jaata hai.

Incline angle

α \to 9 0^{\circ}

(vertical wall):

\overset{s}{¨} = g sin 9 0^{\circ} = g

. "Slope" ab ek cliff hai aur block free fall mein hai — tilted-coordinate result correctly vertical free fall reproduce karta hai.

Central force jahan

r \to 0

jabki

L_{z}

fixed aur nonzero hai: radial equation mein kya dominate karta hai?

Centrifugal term

m r \dot{θ}^{2} = L_{z}^{2} / (m r^{3})

blow up karta hai, ek infinite outward barrier ki tarah act karta hai. Yahi wajah hai koi particle angular momentum ke saath origin tak nahi pahunch sakta — "centrifugal wall".

Ek system jahan

T

explicitly time par depend karta hai (jaise ek wire forced rate

Ω t

par spin ho rahi ho): kya energy abhi bhi conserved hai?

Zaruri nahi.

L

mein explicit time-dependence time-translation symmetry todo hai, isliye Hamiltonian (energy) constant nahi ho sakta chahein E–L abhi bhi apply ho. Dekho 2.1.08-Generalized-momenta-and-the-Hamiltonian.

Recall Ek-line summary jo saath le jao

Recipe kabhi nahi badalti; traps sab is baare mein hain ki tumhe kya ignore karne ki permission hai (constraint forces, $V$ mein constant shifts) aur kya confuse nahi karna chahiye (cyclic ≠ frozen coordinate, partial vs total derivative, energy $T + V$ vs Lagrangian $T - V$ ).