Rigid body — definition, degrees of freedom: Visual walkthrough

1.5.1 · D2 · Physics › Rotational Mechanics › Rigid body — definition, degrees of freedom

Hum har idea zero se build karte hain. Pehle agree karte hain ki ek "coordinate" hota kya hai, phir yeh ki ek "constraint" kya remove karta hai, aur phir hum body ko ek point at a time assemble karte hain.

Step 1 — Ek single number ("coordinate") tumhe kya deta hai

KYA. Ek coordinate sirf ek number hai jo ek "kitni door?" wale sawaal ka jawab deta hai. Ek dot ko ek line par pin karne ke liye, tum ek number bolte ho: shuru se kitni door. Ek dot ko ek floor par pin karne ke liye, tumhe do chahiye: kitna right, kitna forward. Ek dot ko ek room mein pin karne ke liye, tumhe teen chahiye: right, forward, up.

KYUN. Space mein teen independent directions hain jo ek doosre se nahi ban sakte — tum "up" ko "right" aur "forward" combine karke reach nahi kar sakte. Har independent direction apna number maangta hai. Hum inhe $x$ , $y$ , $z$ kehte hain.

PICTURE. Neeche teen coloured rulers dekho. Blue dot "5 units right, 3 units forward, 4 units up" ke meeting point par baitha hai. Koi bhi ek ruler cover karo aur tum dot ko locate karne ki ability kho dete ho — proof ki teeno zaroori hain aur koi spare nahi hai.

$how far right (x, how far forward y, how far up z) \Rightarrow 3 free numbers$

Step 2 — Bahut saare free dots: freedoms bas add hoti jaati hain

KYA. Room mein $N$ dots rakho, har ek kahin bhi baith sakta hai, koi doosre ki parwah nahi karta. Woh numbers gino jo tumhe supply karne padte hain.

KYUN. "Independent" ka matlab hai ki dot 1 ki position choose karne se dot 2 ke baare mein kuch pata nahi chalta. Toh total count sirf 3 ko $N$ baar apne aap mein add karna hai.

PICTURE. Neeche teen free dots hain, har ek ka apna chhota $(x, y, z)$ tag hai. Unke beech koi strings nahi — poori freedom.

$total coordinates = per dot 3 \times number of dots N = 3 N$

Yahan $3 N$ hamaara starting budget of freedom hai. Aage jo bhi hoga woh is budget se kharcha karta hai — rules add karke jo dots ko ek doosre se baandhte hain.

Step 3 — Ek "distance rule" (constraint) kya remove karta hai

KYA. Ab dot A aur dot B ke beech ek invisible rigid stick glue karo: unka separation ek fixed length $d$ rehna chahiye. Woh single rule ek constraint hai.

KYUN yeh tool — ek equation hai, picture-erasure nahi. Ek constraint ek likhi hui equation hai jo coordinates ko link karti hai: $A aur B ke beech squared distance (x_{B} - x_{A})^{2} + (y_{B} - y_{A})^{2} + (z_{B} - z_{A})^{2} = fixed, ek baar choose kiya d^{2}$ Ek equation tumhe doosron ke terms mein ek unknown solve karne deti hai. Toh ek independent constraint exactly ek degree of freedom remove karta hai. Yeh master accounting rule hai: $DOF = (total coordinates) - (independent constraints)$

PICTURE. Dot A kahin bhi free hai. Dot B ab har jagah free nahi hai — stick use A ke centre se $d$ radius ki ek sphere ki surface par force karti hai. Woh "3D mein kahin bhi" (3 numbers) se "ek 2D skin par kahin bhi" (2 numbers) par aa gaya. Ek freedom kharcha hua.

Step 4 — Body build karo: point 1 (the anchor)

KYA. Ab hum poori rigid body ko smartly assemble karte hain. Claim: ek poori rigid body ko space mein freeze karne ke liye, bas 3 points ko freeze karna kaafi hai jo ek straight line par nahi hain (teen non-collinear points). Unhe teen pin karo aur har doosra particle apni fixed distances ke zariye jagah par aa jaata hai. Hum unhe ek ek karke add karte hain aur count karte hain.

Point 1 pehle rakha jaata hai, abhi tak kuch nahi maanna.

KYUN. Koi earlier point nahi hai jisse fixed distance par hona ho, toh point 1 poore room mein freely ghoom sakta hai.

PICTURE. Ek single yellow anchor dot freely float karta hai, apne teen numbers ke saath tagged.

$point 1 deta hai x_{1}, y_{1}, z_{1} 3 DOF$

Step 5 — Point 2 (sphere par rehta hai) aur Point 3 (circle par rehta hai)

KYA. Point 2 ko point 1 se ek fixed distance par attach karo — woh 1 constraint hai. Phir point 3 ko point 1 aur point 2 dono se fixed distance par attach karo — woh 2 constraints hain.

KYUN, geometry-first.

Point 2: ek distance rule ⇒ woh point 1 ke around sphere par baithta hai ⇒ $3 - 1 = 2$ DOF (sphere ki surface par ghoomna).
Point 3: do distance rules. Point 1 se fixed distance use ek sphere par rakhta hai; point 2 se fixed distance use doosri sphere par rakhta hai. Do spheres ek circle mein cross karti hain. Circle par ek point ko sirf $1$ number chahiye (kitna around) ⇒ $3 - 2 = 1$ DOF.

PICTURE. Point 2 blue sphere par slide karta hai; point 3 pink circle par slide karta hai jahan do spheres milti hain. Freedom ko shrink hote dekho: room → sphere → circle.

$pt 1: free 3 + pt 2: on sphere 2 + pt 3: on circle 1 = 6$

Step 6 — Har doosra particle force ho jaata hai (0 new DOF)

KYA. Particle 4, 5, … $N$ tak laao. Har ek ki points 1, 2, aur 3 se ek fixed distance hai (rigid body mein saari distances frozen hain). Teen distances teen known, non-collinear points se ek point ko 3D mein uniquely pin karti hain.

KYUN — teen spheres ek point par milti hain. Point 1 se distance ⇒ ek sphere. Point 2 se distance ⇒ doosri sphere (woh ek circle mein milti hain). Point 3 se distance ⇒ teesri sphere jo us circle ko ek single spot par stab karti hai. Koi freedom nahi bachi: har extra particle ke liye 0 new DOF.

PICTURE. Teen spheres (teen anchors ke around) ek akele dot par intersect karti hain — particle 4 ke paas kahin aur jaane ki jagah nahi.

$DOF = 3 anchors se 6 + particles 4 \dots N 0 + 0 + \dots = 6 (kisi bhi N ke liye)$

Yeh us galti ka punchline hai ki "zyaada particles → zyaada DOF": rigidity count ko 6 par saturate kar deti hai. Dekho Moment of Inertia — wahan bataya hai ki woh "extra" particles, DOF-free hone ke bawajood, body ko spin karna kitna mushkil hai iske liye abhi bhi matter karte hain.

Step 7 — 6 ko 3 + 3 ki tarah padhna (slide vs turn)

KYA. 6 ko physically kya karte hain ke hisaab se split karo: 3 translations (body ko slide karo) aur 3 rotations (body ko turn karo).

KYUN. Body ka koi ek reference point lo — maano uska centre of mass. Us point ko place karne mein 3 numbers $(x, y, z)$ lagte hain: yeh hai kahan hai body (pure slide). Us point ko pin karne ke baad, body abhi bhi uske baare mein pivot kar sakti hai — $x$ , $y$ , $z$ axes ke baare mein spin karo: 3 angles ke liye kis taraf face kar rahi hai. Yeh teen turning freedoms Rotational Kinematics ka input hain.

PICTURE. Left panel: teen straight arrows (slide along $x, y, z$ ). Right panel: teen curved arrows (roll, pitch, yaw) centre ke baare mein.

$6 = translation: x, y, z 3 + rotation: roll, pitch, yaw 3$

Step 8 — Degenerate cases: jab count girta hai

KYA. Extra pins (constraints) 6 se subtract karte hain. Teen classic cases, aur point-mass exception bhi.

KYUN & PICTURE. Har pin specific freedoms remove karta hai — unhe seedha geometry se padho.

Kaise pin karo	Kya khatam hota hai	DOF bacha	Reason (geometry)
Ek point fix karo	saare 3 slides	$6 - 3 = 3$	body sirf us point ke baare mein pivot kar sakti hai
Axle (ek line) fix karo	3 slides + 2 turns	$6 - 5 = 1$	sirf spin angle $θ$ bachta hai
Ek plane mein confine karo	1 slide ( $z$ ) + 2 turns	$6 - 3 = 3$	$(x, y)$ + ek in-plane angle
Diatomic (2 point atoms)	bond ke baare mein spin kuch nahi karta	$5$	$3$ slides $+ 2$ turns; teesra turn koi mass move nahi karta

Ek-picture summary

Ek figure, poori kahani: ek room (3 numbers) → ek sphere (−1) → ek circle (−1) → teen anchors 6 dete hain → 3 slides + 3 turns mein split, degenerate cases branching off ke saath.

Recall Feynman: walkthrough seedhe shabdon mein

Ek khaali room imagine karo. Usme ek marble girao. Mujhe batao ki woh kahan float kar raha hai toh tum teen cheezein bolte ho: kitna right, kitna forward, kitna high — 3 numbers.

Ab bahut saare marbles se ek stiff sculpture glue karo taaki andar kuch shift na ho sake. Mujhe har marble ke numbers nahi chahiye — main sirf teen acche tarah se spread marbles dekhta hoon. Pehla free hai: 3 numbers. Doosra pehle se ek fixed length par bandha hua hai, toh woh sirf pehle ke around ek invisible ball ki surface par slide kar sakta hai: 2 numbers. Teesra dono se bandha hua hai, toh woh us ring par stuck hai jahan do balls overlap karti hain: 1 number. Teen plus do plus ek hai chhah.

Har baaki marble unhe teeno se bandha hua hai, aur teen ropes teen spread-out points se ek marble ko exactly ek jagah par chhodte hain — toh millionth marble kuch nahi add karta. Isliye ek pebble aur ek planet dono ke chhah hote hain.

Aakhir mein, chhah ko do piles mein sort karo: teen poori cheez ko idhar udhar slide karne ke liye hain (left-right, forward-back, up-down) aur teen use jagah par turn karne ke liye hain (tilt, spin, roll). 6 = 3 slides + 3 turns. Sculpture ko wall pe nail karo aur tum kuch ropes kaatoge: ek spot par fixed rahne se 3 turns bachte hain; ek axle par stuck rahne se sirf 1 spin bachta hai. Har baar wohi recipe — chhah se shuru karo, jo pin kiya woh subtract karo.

Recall

Zero freedom-budget se, ek rigid body ke point 2 ko point 1 ke relative place karne ke liye kitne numbers chahiye? ::: 2 — woh point 1 ke around ek sphere par rehta hai ( $3 - 1$ ). Particle number 100 kyun 0 DOF add karta hai? ::: 3 non-collinear anchors se uski fixed distances use teen spheres ke single crossing point par rakhti hain — koi freedom nahi bachi. 6 ko physical piles mein slide karo. ::: 3 translations (slide along $x, y, z$ ) + 3 rotations (turn about $x, y, z$ ). Fixed axle DOF, aur kya remove hua? ::: 1 DOF; axle saare 3 slides aur 3 turns mein se 2 ko khatam karta hai ( $6 - 5$ ).