Generalized coordinates — choosing them, degrees of freedom: Worked examples

2.1.2 · D3 · Physics › Analytical Mechanics › Generalized coordinates — choosing them, degrees of freedom

Definition General

$D \times N$ setup (ek baar batao, har jagah use karo) Kisi bhi constraint se pehle, $D$ spatial dimensions mein rehne wale $N$ point particles ko locate karne ke liye $D \times N$ Cartesian numbers chahiye — har particle ke liye $D$ numbers. Isliye raw (unconstrained) count hai $raw = D N, n = D N - k .$

$N$ = particles ki sankhya.
$D$ = spatial dimensions ki sankhya ( $D = 2$ plane mein, $D = 3$ space mein).
$k$ = independent holonomic constraints ki sankhya — har independent constraint equation exactly ek number remove karti hai, isliye hum subtract karte hain. "Independent" (Ex 5) ka matlab hai ki koi bhi constraint baaki se derive nahi ki ja sakti.
$n$ = degrees of freedom = subtract karne ke baad bache hue generalized coordinates ki sankhya.

Jab bhi neeche " $2 N$ " ya " $3 N$ " dikhe, yeh bas yahi rule hai $D = 2$ ya $D = 3$ ke saath. Har baar $D$ explicitly batayenge taaki dimensionality guess na karni pade.

Scenario matrix

Kuch bhi count karne se pehle, aao territory ko map karein. Har degrees-of-freedom problem in cells mein se kisi ek mein aati hai. Har cell ek alag tarika hai jisme counting behave kar sakti hai — aur har ek ka ek worked example neeche hai.

Cell	Case class	Kya tricky hai	Example
A	Single particle, ek constraint (2D)	base case, sanity anchor	Ex 1
B	Kai particles, kai independent constraints	sahi $k$ subtract karna	Ex 2
C	Rigid body (finite $N$ trick)	$6$ use karo, $3 N$ nahi	Ex 3
D	Rheonomic (time-dependent) constraint	$t$ ek parameter hai, DOF nahi	Ex 4
E	Degenerate / dependent constraints	zaroorat se zyada equations likhna	Ex 5
F	Zero-DOF (fully pinned) limit	$n = 0$ ka matlab fixed hai, broken nahi	Ex 6
G	Non-holonomic (rolling) — formula fail	velocity constraint ≠ position equation	Ex 7
H	Real-world word problem	prose ko $N$ aur $k$ mein translate karna	Ex 8
I	Exam-style twist (coordinate pole)	ek achha coordinate ek khaas point par toot jaata hai	Ex 9

Is matrix ke axes hain: kitne particles (ek → kai → rigid body → ∞), kaunse type ki constraint (koi nahi → holonomic scleronomic → rheonomic → non-holonomic), aur degenerate limits (dependent constraints, zero DOF, coordinate singularities). Milaakar yeh nau cells poori jagah tile kar deti hain.

Recall Woh do words jo confuse nahi karni chahiye

Scleronomic constraint ::: time appear nahi karta: $f (r) = 0$ . Rheonomic constraint ::: time explicitly appear karta hai: $f (r, t) = 0$ ; phir bhi holonomic hai, phir bhi ek DOF remove karta hai.

Ex 1 — Cell A: single particle, ek constraint

Raw numbers count karo. Ek plane mein ek particle matlab $D = 2, N = 1$ , isliye raw $= D N = 2$ Cartesian numbers $(x, y)$ . Yeh step kyun? Hamesha sabse zyada wasteful honest description se shuru karo — parent ka rule "start raw".
Constraint ko ek equation ki tarah likho. Bead ring par rehti hai: $x^{2} + y^{2} = R^{2}$ . Yeh ek holonomic scleronomic constraint hai, isliye $k = 1$ . Yeh step kyun? Ek equation hame ek variable ko doosre ke terms mein solve karne deta hai, ek number kill karta hai.
Subtract karo. $n = D N - k = 2 - 1 = 1$ . Yeh step kyun? Har independent holonomic constraint exactly ek DOF remove karti hai.
Smart coordinate choose karo. Angle $φ$ use karo jahan $x = R cos φ, y = R sin φ$ . Yeh step kyun? Koi bhi $φ$ automatically $x^{2} + y^{2} = R^{2}$ satisfy karta hai: constraint bake in ho jaati hai aur disappear ho jaati hai.

Figure 1 — kya dekhna hai: cyan circle wire hai (constraint $x^{2} + y^{2} = R^{2}$ ); amber dot bead hai; positive $x$ -axis se measure kiya gaya amber arc single coordinate $φ$ hai; white dashed radius dikhata hai ki $φ$ fix karne se bead completely fix ho jaati hai. Dekho ki bead slide karte waqt exactly ek cheez vary hoti hai — amber angle ka size.

Figure — Generalized coordinates — choosing them, degrees of freedom

Verify: Wापस substitute karo: $(R cos φ)^{2} + (R sin φ)^{2} = R^{2} (cos^{2} φ + sin^{2} φ) = R^{2} .$ ✓ Constraint har $φ$ ke liye hold karti hai, confirm karta hai ki $φ$ akela complete hai. Ek coordinate ⇒ $n = 1$ .

Ex 2 — Cell B: kai particles, kai constraints

Raw count. $D = 2, N = 3$ ⇒ raw $= D N = 6$ . Yeh step kyun? Teen independent points, har ek ke do numbers ( $D = 2$ ).
Rigid rods ko equations ki tarah list karo. Teen masses ke position vectors ke liye $r_{1}, r_{2}, r_{3}$ likho (bold- $r$ symbol upar box mein define hua hai). Sides $12, 23, 31$ fixed hain: $∣ r_{1} - r_{2} ∣ = a, ∣ r_{2} - r_{3} ∣ = b, ∣ r_{3} - r_{1} ∣ = c$ , jahan har $∣ r_{i} - r_{j} ∣$ do masses ke beech ki distance hai. Yeh $k = 3$ hai. Yeh step kyun? Har fixed distance ek holonomic constraint equation hai.
Independence check karo. Kya yeh teen sach mein independent hain? Ek triangle ke liye, haan — teen sides fix karne se shape fix ho jaati hai lekin tum ek length ko doosri do se derive nahi kar sakte (sivaaye degenerate collapse ke). Isliye teeno count hoti hain. Yeh step kyun? Parent warn karta hai: sirf independent constraints subtract hoti hain.
Subtract karo. $n = 6 - 3 = 3$ . Yeh step kyun? Rules-removed principle.
Interpret karo. Woh 3 DOF = triangle ke centre position ke liye 2 $(X, Y)$ + plane mein orientation angle $θ$ ke liye 1. Yeh "2D mein rigid body ka 3 DOF" rule se match karta hai (2 translation + 1 rotation). Yeh step kyun? Rigid-body rule ke against cross-checking count confirm karta hai.

Verify: Rigid-body shortcut se independent check: 2D plane mein ek rigid body ke $2 + 1 = 3$ DOF hote hain chahe kitne bhi particles ho. $6 - 3 = 3$ agree karta hai. ✓

Ex 3 — Cell C: rigid-body trick

$D N$ literally use MAT karo. $D = 3$ aur $N \to \infty$ ke saath rigid constraints bhi infinite ho jaati hain, aur $\infty - \infty$ meaningless hai. Yeh step kyun? Naive formula rigid bodies par choke karta hai — yahi woh trick hai jo parent flag karta hai.
Rigid-body shortcut use karo. Har finite rigid body ko exactly $6$ numbers chahiye: 3 yeh batane ke liye ki woh kahan hai (centre of mass $X, Y, Z$ ) aur 3 yeh batane ke liye ki woh kaise oriented hai. Yeh step kyun? Rigidity har particle ko pin kar deti hai jab tum body ke reference frame ki position + orientation jaante ho.
3 orientation numbers name karo. Yeh Euler angles $(α, β, γ)$ hain — dekho Rigid body kinematics — Euler angles. Yeh step kyun? Tumhe coordinates ko sirf count nahi karna, name bhi karna hoga.
Unki ranges batao aur unki singularity watch karo. Teen Euler angles freely nahi ghoomte: typically $α \in [0, 2 π)$ (pehli spin), $β \in [0, π]$ (tilt), $γ \in [0, 2 π)$ (final spin). Tilt extremes $β = 0$ ya $β = π$ par pehli aur teesri axes ek line mein aa jaati hain, isliye $α$ aur $γ$ body ko usi axis ke around ghuma rahe hain — tum unhe alag nahi bata sakte. Yeh gimbal lock hai: teen angle-coordinates mein se do momentarily merge ho jaate hain, exactly Ex 9 ke pole trap ki tarah. Yeh step kyun? Orientation coordinates ke edge cases hote hain bilkul position waalon ki tarah; unhe ignore karne se lagta hai DOF lost ho gayi.
DOF phir bhi $6$ hai gimbal lock ke bawajood. Gimbal lock chart $(α, β, γ)$ ka defect hai, body ka nahi: cube genuinely abhi bhi waahan 3 free rotations rakhta hai. Ex 9 ki tarah tum singular tilt ke paas ek different chart (ya quaternions) use karo. Yeh step kyun? Kabhi bhi ek toota hua coordinate label aur ek lost degree of freedom mein confuse mat karo.
Total. $n = 3 + 3 = 6$ . Yeh step kyun? Translation DOF plus rotation DOF.

Verify: $n = 6$ particle count se independent hai — rigid body ki defining property. Parent ke " $6$ for a rigid body" box se consistent; gimbal lock ise reduce nahi karta. ✓

Ex 4 — Cell D: rheonomic (time-dependent) constraint

Raw count. Ek bead ek plane mein: $D = 2, N = 1$ ⇒ raw $= D N = 2$ . Yeh step kyun? Base description.
DONO constraints likho.
- Angle forced: $θ (t) = ω t$ . Risky ratio form ki jagah, clean equation likho $x sin (ω t) - y cos (ω t) = 0$ (yeh bas yeh hai ki "point angle $ω t$ wali line par hai").
- Radius forced: $x^{2} + y^{2} = r_{0} - v t$ . Har ek mein $t$ explicitly hai — dono rheonomic hain lekin phir bhi holonomic. Isliye $k = 2$ . Yeh step kyun? Time-dependent equations bhi constraints count hoti hain; $t$ ek parameter hai jise hum vary nahi kar sakte.
Note karo kyun pole-free angle form use ki, $y / x = tan (ω t)$ nahi. Ratio form sirf wahan valid hai jahan $x \neq = 0$ ; yeh blow up karta hai (division by zero) jab bhi wire seedha upar point kare, yaani $ω t = \frac{π}{2}, \frac{3 π}{2}, \dots$ , jahan $tan (ω t)$ infinite hai. Form $x sin (ω t) - y cos (ω t) = 0$ ka koi pole nahi: yeh har angle ke liye finite aur correct rehta hai including vertical. Yeh step kyun? Hamen confirm karna hai ki jo constraint hum subtract karte hain woh har case mein valid hai, including degenerate $x = 0$ orientation — warna count un instants par silently fail kar deta hai.
Subtract karo. $n = 2 - 2 = 0$ . Yeh step kyun? Do independent equations do numbers remove karti hain.
Interpret karo. $n = 0$ matlab bead ki position fully determined function of time hai — bilkul koi freedom nahi: $x (t) = (r_{0} - v t) cos ω t, y (t) = (r_{0} - v t) sin ω t$ . Yeh step kyun? Zero DOF ka matlab "no motion" nahi — iska matlab motion prescribed hai, free nahi (Ex 6 se compare karo).

Verify: Kisi bhi $t$ par wapas plug karo: $x^{2} + y^{2} = ∣ r_{0} - v t ∣ cos^{2} ω t + sin^{2} ω t = ∣ r_{0} - v t ∣$ ✓, aur pole-free angle equation $x sin ω t - y cos ω t = (r_{0} - v t) (cos ω t sin ω t - sin ω t cos ω t) = 0$ ✓. Dono koi free coordinate bacha ke bina hold karti hain, confirm karta hai $n = 0$ .

Ex 5 — Cell E: degenerate / dependent constraints

Jo unhone likha list karo. $k_{claimed} = 4$ : teen sides + ek area. Yeh step kyun? Hamen independence audit karni hai, sirf equations count nahi.
Independence test karo. Ek triangle ki area uske teen side lengths se fully determined hoti hai (Heron's formula: $Area = s (s - a) (s - b) (s - c)$ jahan $s = \frac{a + b + c}{2}$ ). Isliye area equation teen side equations ka consequence hai, naya information nahi. Yeh step kyun? Parent ka rule: ek constraint jo logically doosron se follow kare woh DOF reduce nahi karta.
Recount karo. Independent constraints $k = 3$ . Isliye $n = 6 - 3 = 3$ , Ex 2 se match karta hai. Yeh step kyun? Sirf independent constraints subtract hoti hain.

Verify: $a = 3, b = 4, c = 5$ (right triangle) ke saath, Heron deta hai $s = 6$ , Area $= 6 \cdot 3 \cdot 2 \cdot 1 = 36 = 6$ . Sides ho jaane ke baad area pin ho jaati hai — isliye koi nayi restriction carry nahi karta. Correct DOF $= 3$ , $2$ nahi. ✓

Ex 6 — Cell F: zero-DOF limit

Raw count. Plane mein do moving particles: $D = 2, N = 2$ ⇒ raw $= D N = 4$ . (Pivots $p_{1}, p_{2}$ fixed given points hain, particles nahi — yeh koi coordinates carry nahi karte.) Yeh step kyun? Sirf free masses coordinates carry karti hain; anchored points constants hain.
Teen independent rod equations list karo — exactly $k = 3$ , har ek ek genuine written equation.
- Rod pivot-1 → mass 1: $∣ r_{1} - p_{1} ∣ = ℓ_{1}$ (1 equation)
- Rod mass 1 → mass 2: $∣ r_{1} - r_{2} ∣ = ℓ_{2}$ (1 equation)
- Rod mass 2 → pivot-2: $∣ r_{2} - p_{2} ∣ = ℓ_{3}$ (1 equation) Isliye $k = 3$ , 4 nahi — koi fourth equation nahi hai, aur main ek invent nahi kar raha. Yeh step kyun? Honest accounting: har constraint jo subtract ki jaaye woh ek equation honi chahiye jo main actually likh sakun.
Subtract karo. $n = 4 - 3 = 1$ . Yeh step kyun? Teen independent holonomic constraints chaar mein se teen numbers remove karti hain.
Interpret karo — yeh one-DOF linkage hai, zero nahi. Equations (1) aur (3) har mass ko apne pivot ke around apne circle par confine karti hain; equation (2) unhe link karti hai. Generic pivot spacing ke liye yeh "four-bar-like" mechanism mein phir bhi ek freedom hai: $p_{1}$ par input rod ghoomaane se poori chain drive hoti hai. Isliye $n = 1$ . Yeh step kyun? Hamen woh number report karna hai jo honest count deta hai, chahe intuition ne "locked" guess kiya ho.

Common mistake Seductive "

$4 - 4 = 0$ " trap Yeh tempting hai ki frame ko "locked" declare karo aur $n = 0$ force karne ke liye ek phantom fourth constraint subtract karo. Lekin tum sirf woh equations subtract kar sakte ho jo tum likh sako. Yahaan sirf teen rods hain ⇒ $k = 3$ ⇒ $n = 1$ . Genuinely $n = 0$ tak pahunchne ke liye tumhe ek real fourth independent constraint add karni hogi, jaise step 5 mein.
Ab sach mein zero DOF tak pahuncho (pinned limit). Ek fourth rigid rod add karo, mass 1 se directly ek teesre fixed pivot $p_{3}$ tak: $∣ r_{1} - p_{3} ∣ = ℓ_{4}$ (equation 4). Mass 1 ab do fixed pivots ( $p_{1}$ aur $p_{3}$ ) se tied hai, isliye uski position pin ho jaati hai; equation (2) phir mass 2 ko pin kar deti hai. Ab $k = 4$ independent equations hain aur $n = 4 - 4 = 0.$ Yeh step kyun? $n = 0$ tabhi legitimate hai jab chaar honest independent equations present hoon. Yeh genuinely locked-truss limit deliver karta hai jo matrix ne promise kiya tha.
Padho $n = 0$ ka matlab kya hai. Zero DOF = poora structure rigidly fixed in place hai: vary karne ke liye koi number nahi bacha. Ex 4 se unlike (jahan $n = 0$ matlab time se prescribed motion), yahaan $n = 0$ matlab genuinely koi motion nahi. Yeh step kyun? Same number $n = 0$ ke do flavours hain — prescribed-by-time versus truly static — aur reader ko yeh alag batane mein capable hona chahiye.

Verify: Teen-rod chain deta hai $n = 4 - 3 = 1 \geq 0$ (ek working linkage). Four-rod version deta hai $n = 4 - 4 = 0$ (rigidly pinned). Agar koi count $n < 0$ return kare to tumne constraints over-count kiye hain (independence revisit karo, Ex 5). Dono results $\geq 0$ hain. ✓

Ex 7 — Cell G: jahan formula FAIL karta hai (non-holonomic)

Configuration describe karo. Coin ko locate karne ke liye tumhe chahiye: contact point $(x, y)$ , heading angle $ϕ$ (kis direction mein point karta hai), aur spin angle $ψ$ (kitna rotate hua). Yeh 4 configuration numbers hain. Yeh step kyun? Yeh poochne se pehle ki koi constraint kuch reduce karta hai ya nahi, pehle complete list fix karni hogi un numbers ki jo object ko place karti hain — configuration matlab woh har number jo ise pose karne ke liye chahiye, yeh ignore karte hue ki woh wahan kaise pahuncha.
Rolling condition likho. Rolling-without-slipping velocities ko tie karta hai: $\overset{x}{˙} = R \dot{ψ} cos ϕ, \overset{y}{˙} = R \dot{ψ} sin ϕ$ . Inmen $\overset{x}{˙}, \overset{y}{˙}, \dot{ψ}$ hain — rates, positions nahi. (Dot ka matlab hai "time ke saath change ki rate".) Yeh step kyun? Yahi crux hai: constraint velocities par hai.
Ise ek position equation mein integrate karne ki koshish karo. Nahi kar sakte: koi function $f (x, y, ϕ, ψ) = 0$ nahi hai jiska derivative yeh deta ho. Constraint non-holonomic hai (dekho Constraints — holonomic vs non-holonomic). Yeh step kyun? $n = D N - k$ position equations subtract karta hai. Ek velocity constraint jo integrate nahi hogi woh position equation nahi hai.
Split conclude karo. Configuration space 4-dimensional rehta hai (coin ko clever manoeuvres se kisi bhi $(x, y, ϕ, ψ)$ tak le jaaya ja sakta hai), phir bhi har instant par sirf 2 velocity directions free hain. Isliye "DOF" split ho jaata hai: 4 accessible configurations, 2 instantaneous freedoms. Yeh step kyun? Yahi parent ki warning hai ki $n = 3 N - k$ "may not apply directly."

Figure 2 — kya dekhna hai: cyan dashed curve woh path hai jo contact point table par trace kar sakta hai; amber dot current contact $(x, y)$ hai; amber arrow heading $ϕ$ hai (woh direction jis taraf coin face karta hai); upar chhota white circle spin angle $ψ$ dikhata hai. Figure par caption text padho — woh key fact state karta hai: 4 configuration numbers exist karte hain, lekin unki sirf 2 velocities independently free hain, isliye rolling law ko position equation ki tarah subtract nahi kiya ja sakta.

Verify: Integrability test karo. Agar $\overset{x}{˙} = R \dot{ψ} cos ϕ$ $f = 0$ se aata, to $x - R ψ cos ϕ$ constant rehna chahiye — lekin $ϕ$ khud steer karte waqt change hota hai, isliye $R ψ cos ϕ$ clean total differential nahi hai. Koi integrating factor exist nahi karta ⇒ genuinely non-holonomic ⇒ formula ise subtract nahi karta. Configuration DOF $= 4$ . ✓

Ex 8 — Cell H: real-world word problem

Prose ko particles + constraints mein translate karo. Arm ke teen link endpoints ko shoulder ke beyond treat karo. Easier: recognize karo ki har joint ek free rotation angle hai. Yeh step kyun? Robot arms joint-by-joint count karne par cleanest hote hain.
Free angles directly count karo. Shoulder angle $θ_{1}$ , elbow angle $θ_{2}$ , wrist angle $θ_{3}$ — teen independent angles, har ek freely choosable. Yeh step kyun? Har free joint jo configuration change kare woh ek generalized coordinate hai.
$n = D N - k$ se cross-check karo. Plane mein teen moving endpoints: $D = 2, N = 3$ ⇒ raw $= D N = 6$ . Constraints: har link ki fixed length (3 equations), isliye $k = 3$ : $n = 6 - 3 = 3$ . Yeh step kyun? Do methods ka agree karna tumhara safety net hai.
Answer. $n = 3$ : tumhe teen motors chahiye, ek har joint par. Yeh step kyun? DOF = independent controls ki sankhya.

Verify: Dono methods $3$ dete hain. $J$ revolute joints wala serial planar arm $J$ DOF rakhta hai; yahaan $J = 3$ ⇒ $3$ . ✓ Units: angles dimensionless hain (radians) — valid generalized coordinates, parent ka confirm karta hai ki "coordinates ko lengths hona zaroori nahi."

Ex 9 — Cell I: exam twist — ek coordinate jo pole par break karta hai

DOF count karo (coordinate-free). 3D mein ek particle: $D = 3, N = 1$ ⇒ raw $= D N = 3$ ; ek constraint $x^{2} + y^{2} + z^{2} = ℓ^{2}$ ⇒ $k = 1$ ; isliye $n = 3 - 1 = 2$ . Yeh step kyun? DOF system ki property hai, koi bhi coordinates choose karne se pehle compute hoti hai.
Coordinates choose karo. Polar angle $θ$ (vertical se neeche) aur azimuth $ϕ$ (around): $x = ℓ sin θ cos ϕ, y = ℓ sin θ sin ϕ, z = - ℓ cos θ$ . Yeh step kyun? Yeh do angles length constraint automatically bake in kar dete hain.
Singularity dhundo. $θ = 0$ par (mass seedha neeche hang karta hai) ya $θ = π$ par (seedha upar), $sin θ = 0$ , isliye $ϕ$ ki har value same point deti hai. Azimuth $ϕ$ meaningless ho jaata hai — coordinate chart degenerate ho jaata hai. Yeh step kyun? Figure dekho: pole par amber $ϕ$ -circle ek single point mein shrink ho jaata hai, isliye $ϕ$ kuch bhi label nahi karta.
Trap resolve karo. DOF abhi bhi 2 hai har jagah. System mein genuinely do freedoms hain; sirf labelling $(θ, ϕ)$ poles par fail karta hai, physics nahi. Pole ke paas tum ek different chart use karoge. Yeh step kyun? Exam punchline: coordinate singularities choice ke artefacts hain, DOF count ke kabhi nahi.

Figure 3 — kya dekhna hai: cyan circle woh sphere of radius $ℓ$ hai jis par mass confined hai; white dashed vertical line axis hai; tilted amber ellipse woh circle hai jo azimuth $ϕ$ sweep karta constant latitude par; white dot angles $(θ, ϕ)$ par mass hai; top par amber star pole $θ = 0$ mark karta hai. Pole watch karo: woh amber $ϕ$ -circle wahan ek single point mein collapse ho jaata hai, visually dikhata hai ki $ϕ$ koi information carry nahi karta jabki DOF count $2$ rehta hai.

Verify: Constraint check: $x^{2} + y^{2} + z^{2} = ℓ^{2} sin^{2} θ (cos^{2} ϕ + sin^{2} ϕ) + ℓ^{2} cos^{2} θ = ℓ^{2} (sin^{2} θ + cos^{2} θ) = ℓ^{2}$ ✓ sab $θ, ϕ$ ke liye — including $θ = 0$ par, jahan answer $ℓ^{2}$ hai $ϕ$ ki parwah kiye bina, confirm karta hai ki $ϕ$ wahan redundant hai. DOF $= 2$ throughout. ✓

Recall

Recall Rapid-fire scenario recall

Fixed hoop par bead (2D) — DOF? ::: $1$ (angle $φ$ ). 3 masses ki rigid planar triangle — DOF? ::: $3$ ( $2$ translation $+ 1$ rotation). 3D mein free rigid body — DOF aur kyun $3 N$ nahi? ::: $6$ ; particle count infinite hai isliye position $3$ + orientation $3$ use karo. Gimbal lock kya hai, aur kya yeh DOF lower karta hai? ::: Euler tilt $β = 0$ ya $π$ par angles $α, γ$ same axis ke around ghoomte hain aur merge ho jaate hain; yeh ek chart defect hai, DOF $6$ rehta hai. Bead jo $r (t) = r_{0} - v t$ AND $θ = ω t$ par forced hai — DOF? ::: $0$ ; dono rheonomic constraints ek-ek remove karti hain; motion $t$ se fully prescribed hai. Triangle ke teen side-length constraints mein "area = const" add karne se $k$ kaise change hota hai? ::: Bilkul nahi — area sides se follow hoti hai (Heron), isliye yeh dependent hai aur count nahi hoti. Do fixed points par pinned three-rod chain — DOF? ::: $1$ ( $n = 4 - 3$ ); ek working linkage, locked nahi. $n = 0$ tak pahunchne ke liye ek fourth independent rod chahiye. Rolling upright disk — configuration DOF, aur kya $n = D N - k$ apply hota hai? ::: $4$ configuration numbers; formula rolling condition subtract nahi karta kyunki yeh non-holonomic hai (velocity constraint). Planar 3-joint robot arm — DOF? ::: $3$ (ek angle per joint). Spherical pendulum — DOF, aur pole par kya break hota hai? ::: $2$ ; azimuth $ϕ$ $θ = 0, π$ par meaningless ho jaata hai, lekin DOF $2$ rehta hai — sirf chart fail karta hai.

Connections

Parent: Generalized coordinates & DOF
Constraints — holonomic vs non-holonomic
Configuration space and phase space
Lagrangian mechanics — the Lagrangian L = T - V
Euler–Lagrange equations
Rigid body kinematics — Euler angles
Principle of virtual work and d'Alembert's principle