Rigid body — definition, degrees of freedom: Worked examples

1.5.1 · D3 · Physics › Rotational Mechanics › Rigid body — definition, degrees of freedom

Scenario matrix

Har rigid-body DOF sawaal neeche diye gaye cells mein se ek hai (ya unka blend). Har worked example us cell ke saath tagged hai jise woh cover karta hai.

Cell	Case class	Kya special hai	Example
A	Free body, 3D	kuch bhi constrained nahi → pure 6	Ex 1
B	Ek plane (2D) mein confined	ek translation + do rotations khatam	Ex 2
C	Fixed point (pivot)	saari translation khatam, saari rotation bachi	Ex 3
D	Fixed axis (hinge/axle)	sirf ek spin angle bachta hai	Ex 4
E	Degenerate geometry: collinear points	"3 non-collinear points" ka rule toot jaata hai	Ex 5
F	Zero/limit case: single particle & two-particle "body"	sabse chhota $N$ ; ek rotation gayab	Ex 6
G	Constraint-counting by subtraction	hidden constraint number nikalane ke liye DOF use karo	Ex 7
H	Real-world word problem	English ko constraints mein translate karo	Ex 8
I	Exam twist: composite / linked bodies	DOF joodo, linking constraints ghatao	Ex 9

Geometric cells (A, C, D, E) ke saath figures hain.

Ex 1 — Cell A: ek free brick space mein tumble karta hua

Worked example Ek kamre mein feka gaya brick

Statement: Ek solid brick ko is tarah feka gaya hai ki woh fly bhi karta hai aur tumble bhi karta hai freely. Kisi bhi instant mein ise pin karne ke liye kitne independent numbers chahiye?

Forecast: Padhne se pehle guess karo. (Trans + rot in 3D...)

Brick ka ek reference point locate karo. Uske liye $(x, y, z)$ chahiye → 3 numbers. Yeh step kyun? Ek rigid body shape mein frozen hoti hai, isliye jab ek baar ek point aur orientation pata ho, toh baaki har point follow karta hai. Shuruat us ek point ko place karke karo.
Us point ke around brick ko orient karo. 3D mein teen independent turning axes hain (roll, pitch, yaw) → 3 aur numbers. Yeh step kyun? Ek point fix karne ke baad brick ab bhi spin kar sakti hai; teen axes 3D mein har possible spin ko exhaust kar dete hain. Figure mein teen cyan rotation arcs dekho.
Joodo: $3 + 3 = 6$ .

Answer: 6 DOF.

Verify: Parent se "3 non-collinear points" method se cross-check karo: $3 + 2 + 1 = 6$ . ✔ Dono routes agree karte hain, yahi wajah hai ki yeh number trustworthy hai.

Ex 2 — Cell B: ek coin table par flat slide aur spin karta hua

Worked example Tabletop par confined ek coin

Statement: Ek coin flat padi hai aur table par kahin bhi slide kar sakti hai aur apne centre ke around spin kar sakti hai, lekin kabhi utha nahi sakti. DOF?

Forecast: Free case se har family mein ek kam freedom?

Centre ki position. Yeh table ke 2D plane mein rehti hai → $(x, y)$ → 2 numbers. Yeh step kyun? Plane mein confinement ek constraint hai: height $z$ fixed hai, ek translational DOF khatam (3 se 2).
Orientation. Ek plane mein sirf ek independent rotation axis hoti hai — vertical, seedha table ke bahar. Toh ek angle $θ$ → 1 number. Yeh step kyun? Coin ko kisi bhi in-plane axis ke around tipping karne se uska kuch hissa table se utha jaata, jo confinement forbid karti hai. Teen 3D rotations mein se do khatam; ek bachti hai.
Joodo: $2 + 1 = 3$ .

Answer: 3 DOF.

Verify: Free 6 se shuru karo aur jo plane hata deta hai subtract karo: 1 translation ( $z$ ) aur 2 rotations (do tipping axes) hataata hai: $6 - 1 - 2 = 3$ . ✔

Ex 3 — Cell C: ek point par pinned rigid body

Worked example Apni tip par pinned ek spinning top

Statement: Ek rigid top floor ko ek single fixed point (uski tip) par touch karti hai (tip slide nahi karti). Kitne DOF?

Forecast: Translation chali gayi, rotation intact?

Translation khatam karo. Fixed point move nahi kar sakta, isliye reference point ke $(x, y, z)$ sab pinned hain → 0 translational DOF. Yeh step kyun? Ek fixed point ek saath teen constraints hain (ek per coordinate). $3 - 3 = 0$ .
Rotation rakho. Body us point ke around har direction mein swivel kar sakti hai → 3 rotational DOF. Yeh step kyun? Ek point pin karna sliding rokta hai lekin turning nahi; us point se guzarne wale teeno spin axes free rehte hain. Figure mein pinned tip ke around teen amber rotation arcs dekho.
Joodo: $0 + 3 = 3$ .

Answer: 3 DOF.

Verify: Free 6 se, ek fixed point exactly 3 translations hataata hai: $6 - 3 = 3$ . ✔ (Parent ke "fixed at a point → 3" row se match karta hai.)

Ex 4 — Cell D: apne hinges par ek door (fixed axis)

Worked example Ek swinging door

Statement: Ek door ek fixed vertical hinge line par latki hui hai. DOF?

Forecast: Yahan almost sab kuch nail down hai...

Hinge line ek fixed axis hai — body ki poori line frozen hai, sirf ek point nahi. Yeh step kyun? Ek axis fix karna ek point fix karne se zyada strong hai: yeh saari 3 translations rok deta hai aur un 2 rotations ko bhi rok deta hai jinki axes hinge nahi hain.
Jo bachta hai usse count karo. Sirf hinge axis ke around hi rotation allowed hai → ek swing angle $θ$ → 1 number. Yeh step kyun? Door ko na slide kar sakte ho na tip; figure mein single red arc hi ek akeli motion bachi hai.
Result: $DOF = 1$ .

Answer: 1 DOF.

Verify: Free 6 se, ek fixed axis 3 translations + 2 rotations $= 5$ hataata hai: $6 - 5 = 1$ . ✔ Ek axle par wheel jaisa hi. Yeh exactly single-angle world hai jahan saari rotational kinematics ( $θ, ω, α$ ) rehti hai.

Ex 5 — Cell E: degenerate case — teen collinear points

Worked example Kyun "3 points" secretly "3

non-collinear points" matlab rakhta hai Statement: Koi ek rigid rod ko fix karne ki koshish karta hai teen points thoka kar jo sab ek hi seedhi line par hain. Kya yeh rod ko freeze karta hai? Kaun se DOF bachte hain?

Forecast: Dhyan se — kya nails ki ek line rolling rokti hai?

Point 1 fix karo: uske liye 0 DOF bacha (yeh pinned hai) — yeh position pin kar deta hai. Yeh step kyun? Ek fixed point saari 3 translations hataata hai, bilkul Ex 3 jaisa.
Line par ek doosra point fix karo: yeh baaki 3 rotations mein se 2 hataata hai, do points se guzarne wali line ke around spin chodke. Yeh step kyun? Do fixed points ek axis define karte hain; sirf us axis ke around rotation bachta hai. Ab tak yeh fixed-axis door jaisa hi hai.
Ek teesra point fix karne ki koshish karo jo usi line par hai. Yeh koi nayi information nahi deta: yeh pehle se hi fixed axis par hai, isliye pehle do se uski position automatically force ho jaati hai. Yeh 0 aur DOF hataata hai. Yeh step kyun? "3 points ek body lock karte hain" rule ke liye teen points ek line par nahi hone chahiye, taaki teesra point leftover spin pin kar sake. Collinear points nahi kar sakte, kyunki woh sab us spin ke under invariant hain. Figure dekho: dashed axis ke around rotate karne par teeno amber dots bilkul wahin rehte hain.
Bacha hua DOF: line ke around spin $= 1$ .

Answer: 1 DOF bachta hai — rod apni khud ki lambi axis ke around roll kar sakti hai. Yeh degenerate trap hai.

Verify: Do alag points ek fixed axis dete hain (Ex 4 logic): $6 - 5 = 1$ ; collinear teesra point $0$ add karta hai: $1 - 0 = 1$ . ✔

Ex 6 — Cell F: limit cases — ek particle aur do-atom "body"

Worked example Sabse chhote systems:

$N = 1$ aur ek diatomic molecule Statement: (a) 3D mein ek single point particle. (b) Ek diatomic molecule jise do point masses ek rigid bond se jude hue model kiya gaya hai. Dono ke DOF?

Forecast: (a) obvious. (b) famously 6 nahi hai — guess karo kyun.

(a) Single particle.

Ise place karo: $(x, y, z)$ → 3. Orient karne ke liye kuch nahi hai (ek point ki koi "facing" nahi hoti). Yeh step kyun? Orientation tabhi meaningful hai jab body ka kuch extent ho; ek mathematical point har angle se same lagta hai, isliye 0 rotational DOF.
Answer (a): 3 DOF.

(b) Diatomic (do point masses, rigid bond).

Pair ka translation: poore molecule ke reference point ko locate karo → 3.
Rotation: bond line 3D mein tumble kar sakti hai — lekin bond axis ke around rotation kuch bhi move nahi karta, kyunki dono masses points hain jo us axis par baithe hain. Yeh step kyun? Rotation axis par baitha point mass jab us axis ke around spin karte ho toh move nahi karta, isliye woh "rotation" physically undetectable hai aur DOF ke roop mein count nahi hota. Sirf do tumbling rotations (bond ke perpendicular axes) real hain → 2.
Joodo: $3 + 2 = 5$ .
Answer (b): 5 DOF.

Verify: (a) $3 = 3$ . ✔ (b) Full rigid body 6 hota; ek useless bond-axis spin hatao: $6 - 1 = 5$ . ✔ Yeh $5$ exactly woh number hai jo Equipartition Theorem ek diatomic gas ki specific heat mein feed karta hai. Full rigid body ke 6 se contrast karo.

Ex 7 — Cell G: constraint count wapas nikalao

Worked example Ek 5-particle rigid body ko saath mein kitne independent constraints rakhe hain?

Statement: Paanch point masses ek rigid body mein lock hain. Independent rigidity constraints ki sankhya nikalo.

Forecast: $(2 5) = 10$ nahi — yeh overcount karta hai. Asli number guess karo.

Total free coordinates: $3 N = 3 \times 5 = 15$ . Yeh step kyun? Lock karne se pehle, 5 particles mein se har ek ko 3 numbers chahiye.
Kisi bhi 3D rigid body ka DOF 6 hai (Ex 1 se). Yeh step kyun? Rigidity hamesha 6 par saturate hoti hai chahe $N$ kuch bhi ho.
Independent constraints $= coords - DOF = 15 - 6 = 9$ . Yeh step kyun? Master formula rearrange karo: constraints $=$ coords $-$ DOF.
Answer: 9 independent constraints (haalaanki $(2 5) = 10$ pair-distances hain — ek redundant hai).

Verify: $N = 5$ ke saath general rule $3 N - 6$ : $3 (5) - 6 = 9$ . ✔ Aur $(2 5) = 10 > 9$ , isliye exactly $10 - 9 = 1$ pair-distance redundant hai, jo parent ki pairs ko over-count karne ki warning se match karta hai.

Ex 8 — Cell H: real-world word problem

Worked example Ek fixed workbench pakde robot forearm

Statement: Ek rigid robot forearm ek fixed shoulder joint (ek ball-and-socket jo saari rotations allow karta hai lekin sliding nahi) se bolt ki gayi hai, aur uska door wala end clamp kiya gaya hai taaki woh apni khud ki length ke around rotate na kar sake. DOF batao.

Forecast: Ball-and-socket = fixed point; phir ek aur clamp kuch remove karta hai.

Bolted shoulder ek fixed point hai → saari 3 translations hataata hai, 3 rotations chodta hai (yeh Ex 3 pivot situation hai) → ab tak 3. Yeh step kyun? "Bolted, slide nahi kar sakta" precisely ek fixed point hai.
Clamp arm ki apni lambi axis ke around rotation forbid karta hai → un 3 rotations mein se 1 hataata hai. Yeh step kyun? Ek extra independent constraint exactly ek DOF hatata hai; yahan yeh self-spin khatam karta hai, sirf do "point-in-a-direction" swings chodke.
Joodo: $3 - 1 = 2$ .

Answer: 2 DOF — forearm apni tip ko spherical patch par kahin bhi point kar sakta hai (do swing angles), lekin slide ya self-twist nahi kar sakta.

Verify: Free 6 se: fixed point $- 3$ , self-spin clamp $- 1$ → $6 - 3 - 1 = 2$ . ✔ (Physical picture ki sanity: do angles = pointing direction ka latitude + longitude, jo actually ek 2-parameter direction hai — ek point aur ek axis pin karne se consistent.)

Ex 9 — Cell I: exam twist — hinge se judi do rods

Worked example Composite body: ek pin se linked do rigid rods

Statement: Do alag rigid rods 3D mein freely float karti hain lekin ek end par ek pin se judi hain (woh us single point ko share karti hain, phir bhi freely angle kar sakti hain). Combined system ka total DOF?

Forecast: $6 + 6$ minus linking constraint — lekin ek shared point kitna remove karta hai?

Agar rods independent hoti toh DOF: har free rigid rod ka 6 hota, isliye $6 + 6 = 12$ . Yeh step kyun? Do disconnected rigid bodies apna DOF simply add karte hain (independent configurations).
Pin impose karo: dono rods ko space mein ek point share karna chahiye. Yeh 3 constraint equations hain (joint par unke $x$ , $y$ , $z$ match karne chahiye). Yeh step kyun? Do points ko coincide karvana 3 coordinates bandh karta hai; ek pin koi bhi relative rotation allow karta hai, isliye yeh sirf shared position constrain karta hai, orientation nahi.
Subtract: $12 - 3 = 9$ .

Answer: 9 DOF.

Verify: Independent check — linked system directly describe karo: rod 1 free (6), rod 2 ko apna pinned end rod 1 ke pinned end se chipkaye rakhna hai (yeh rod 2 ki position fix karta hai, 3 translations chale, 3 rotations bache): $6 + 3 = 9$ . ✔ Dono counts agree karte hain. Aise joint constraints ko systematically handle karna exactly wahi hai jo Constraints and Lagrangian Mechanics ke liye banaya gaya hai.

Recall One-line answer bank (reveal karne se pehle guess karo)

3D mein free rigid body? ::: 6 Ek plane mein confined rigid body? ::: 3 Ek point par pinned rigid body? ::: 3 Fixed axis par rigid body (door/wheel)? ::: 1 Teen collinear fixed points kya chodते hain? ::: 1 (line ke around spin) Single point particle? ::: 3 Diatomic point-mass molecule? ::: 5 N=5 rigid particles ke liye independent constraints? ::: 9 ( $3 N - 6$ se) Robot arm: fixed point + no self-spin? ::: 2 Pin se judi do free rods? ::: 9

Related building blocks: Centre of Mass (3 translational DOF ke liye natural reference point), Moment of Inertia aur Rotational Kinematics (jo rotational DOF ko dynamics se bharta hai), aur Rolling Motion (ek body jisme rolling constraint DOF aur trim karta hai).