Exercises — Rigid body — definition, degrees of freedom

Space mein ek point ko teen numbers se pin kiya jaata hai: $x$ ke along kitna, $y$ ke along kitna, $z$ ke along kitna. Inhe koi cheez aapas mein nahi baandhti, isliye teeno independent hain. $\text{DOF}=3$

$\text{DOF}=6=\underset{\text{translation}}{\underbrace{3}}+\underset{\text{rotation}}{\underbrace{3}}$ Teen numbers batate hain ki body kahaan hai ($x,y,z$ ke along slide); teen numbers batate hain ki woh kis taraf face kar rahi hai ($x,y,z$ ke baare mein turn).

Axle ko fix karne se teeno translations aur 3 mein se 2 rotations khatam ho jaate hain, sirf spin angle $\theta$ bachta hai. $\text{DOF}=1$

Step 1 — ek point locate karo. Coin ka centre ek 2D plane mein rehta hai, jiske liye $(x,y)$ chahiye → 2 numbers. Step 2 — orient karo. Ek plane mein sirf ek rotation axis hota hai (table ke upar seedha), isliye ek angle $\theta$. $\text{DOF}=2+1=3$ Figure s01 dekho: do arrows slide directions hain, curved arrow woh single spin hai.

Step 1 — kya fixed hai? Hinge line ek fixed axis hai: translation gone, aur 3 mein se 2 rotations gone. Step 2 — kya bacha? Sirf swing angle $\theta$. $\text{DOF}=1$ Axle wale wheel jaisi hi physics (L1.3).

Ek point fix karne se teeno translations hat jaate hain (pinned point move nahi kar sakta). Lekin body us point ke baare mein teeno rotational ways mein freely rotate kar sakti hai. $\text{DOF}=0_{\text{trans}}+3_{\text{rot}}=3$

(a) Har particle ko 3 numbers chahiye: $3N=3\times 4=12$. (b) Ek 3D rigid body ke 6 DOF hain, isliye $\text{constraints}=\text{coords}-\text{DOF}=12-6=6.$ (c) 4 particles mein distinct pairs ki sankhya: $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=\frac{4\cdot 3}{2}=6.$ $N=4$ ke liye yeh 6 pair-distances exactly independent hain — count match karta hai. ($N>4$ ke liye yeh sab independent nahi honge.)

Total pair equations: $\left(\genfrac{}{}{0ex}{}{8}{2}\right)=\frac{8\cdot 7}{2}=28.$ True independent constraints: $3N-6=3\cdot 8-6=18.$ Redundant (over-counted) equations: $28-18=10.$ Yahi wajah hai ki hum 3 points fix karte hain ($3+2+1=6$ DOF dete hain) instead of blindly saare pairs count karne ke.

Point 1: free → 3 numbers. Point 2: Point 1 se fixed distance → ek sphere par rehta hai → 2 numbers. Point 3: Points 1 aur 2 dono se fixed distance → ek circle par rehta hai → 1 number. Running total: $3+2+1=6$. Har agle particle ($4,5,\dots ,N$) ke in 3 non-collinear anchors se fixed distances hain, jo ise uniquely pin kar dete hain. Har ek 0 DOF add karta hai. Figure s02 mein sphere → circle → point ki freedom ki shrinking dikhayi gayi hai. $\text{DOF}=6\text{for all }N\ge 3.$

Translation: poora molecule 3D mein slide karta hai → 3. Rotation: yeh bond ke perpendicular do axes ke baare mein tumble kar sakta hai → 2. Jo ek missing hai: bond axis ke baare mein rotation karne par do point masses spin karte hain jo us axis par lie karte hain — kuch physically move nahi hota. Woh rotational DOF exist hi nahi karta. $\text{DOF}=3+2=5.$

Ek rigid diatomic mein $f=5$ active (quadratic) DOF hain: 3 translational + 2 rotational. Energy per molecule: $⟨E⟩=f\cdot \frac{1}{2}{k}_{B}T=\frac{5}{2}{k}_{B}T.$ Molar internal energy (Avogadro's number se multiply karo, $N_A{k}_B=R$): $U=\frac{5}{2}RT.$ Molar heat capacity at constant volume: $C_V=\frac{dU}{dT}=\frac{5}{2}R.$ Neighbouring idea: Equipartition Theorem.

Rolling constraint se pehle: ek plane mein rigid body ke 3 DOF hote hain — ground ke along position $x$, height $y$, aur spin angle $\theta$. Constraint 1 — ground par rehna: $y$ radius par fixed hai → 1 remove → $x,\theta$ bachte hain (2 DOF). Constraint 2 — no slipping: contact point instantaneously rest mein hota hai, $x$ aur $\theta$ ko $x=R\theta$ se baandhta hai → 1 aur remove. $\text{DOF}=3-1-1=1.$ Ek number (maano $\theta$, ya equivalently $x$) sab kuch bata deta hai. Dekho Rolling Motion.

Translation: $d$-space mein ek reference point place karne ke liye $d$ numbers chahiye. Rotation: ek rotation ek plane mein act karta hai, aur plane $d$ axes mein se 2 choose karke banta hai. Independent rotation planes ki sankhya hai $\left(\genfrac{}{}{0ex}{}{d}{2}\right)=\frac{d(d-1)}{2}.$ Isliye $\boxed{\text{DOF}(d)=d+\frac{d(d-1)}{2}=\frac{d(d+1)}{2}.}$ Check $d=2$: $\frac{2\cdot 3}{2}=3$ ✓ (coin se match karta hai, L2.1). Check $d=3$: $\frac{3\cdot 4}{2}=6$ ✓ (jaana-pehchaana answer).

Method 1 — physical angles. Rod A sirf apne ground pin ke baare mein swing kar sakta hai → 1 angle ${\theta }_A$. Rod B A ki tip ke hinge ke baare mein swing karta hai → 1 aur angle ${\theta }_B$. $\text{DOF}=2.$ Method 2 — bookkeeping. Ek plane mein har rigid rod ke 3 DOF hote hain, isliye do free rods = $6$.

• Ground pin, A ke ek end ko plane mein fix karta hai: 2 constraints.
• Hinge, A ki tip aur B ke end ko ek hi point par baandhta hai: 2 constraints. $\text{DOF}=6-2-2=2.✓$ Dono raaste 2 dete hain. General recipe ke liye dekho Constraints and Lagrangian Mechanics.

Start: free rigid body = 6 DOF (3 trans + 3 rot). Constraint: ek marked point ko ek line par rehna hai. Line par ek point ka sirf 1 free coordinate hota hai instead of 3, isliye translation part 3 se 1 ho jaata hai → 2 remove. Rotation: untouched — body phir bhi kisi bhi taraf face kar sakti hai → 3 bacha rehta hai. $\text{DOF}=(3-2)+3=1+3=4.$

DOF = coordinates − independent constraints. Free rigid body: $6$ in 3D, $3$ in 2D, $\frac{d(d+1)}{2}$ in $d$-D. Har fixed point/axis/plane/wire exactly uthne hi freedoms subtract karta hai jo woh kill karta hai — removed wale count karo.