Equilibrium of rigid bodies — translational + rotational

1.5.18 · Physics › Rotational Mechanics

KYA hai equilibrium?

"Equilibrium" ka matlab "rest mein hona" nahi hai. Iska matlab hai zero acceleration — body constant velocity se move kar rahi ho sakti hai ya constant rate se spin kar rahi ho sakti hai. Hum static equilibrium (rest mein aur rest mein rehna) par focus karte hain.

KYO do conditions? (First principles se derivation)

Poori body ke liye Newton's second law (saare particles ka sum, internal forces Newton's 3rd law se cancel ho jaate hain):

$\sum {\vec{F}}_{\text{ext}}=M{\vec{a}}_{cm}$

Centre of mass ko koi acceleration na ho iske liye hume chahiye:

$\boxed{\sum {\vec{F}}_{\text{ext}}=0}$

Yeh translation fix karta hai. Lekin rotation ko change karta hai torque. Newton's law ka rotational analogue hai:

$\sum {\vec{\tau }}_{\text{ext}}=I\vec{\alpha }$

Koi angular acceleration nahi ke liye:

$\boxed{\sum {\vec{\tau }}_{\text{ext}}=0}$

Ek khoobsurat fact: pivot choose karna free hai

Proof. Point $A$ ke baare mein torque vs point $B$ ke baare mein torque (jahaan ${\vec{r}}_B={\vec{r}}_A-\vec{d}$, aur $\vec{d}=\vec{B}-\vec{A}$): ${\vec{\tau }}_B=\sum ({\vec{r}}_i-\vec{d})\times {\vec{F}}_i=\sum {\vec{r}}_i\times {\vec{F}}_i-\vec{d}\times \sum {\vec{F}}_i={\vec{\tau }}_A-\vec{d}\times \underset{=0}{\underbrace{\sum {\vec{F}}_i}}={\vec{\tau }}_A.$ Toh translational equilibrium satisfy hone par, net torque chosen axis se independent hai. ∎

Component form (kaam karne wale equations)

$xy$-plane mein coplanar (2-D) forces ke liye, teen scalar conditions hain:

$\sum F_x=0,\sum F_y=0,\sum {\tau }_z=0$

Sign convention: counter-clockwise torque = positive, clockwise = negative (consistent with $\hat{z}$ page se bahar).

Worked Example 1 — Seesaw (support ki distance nikalo)

Ek uniform plank, pivot uske centre par. Ek 40 kg ka bachha pivot se 2 m left mein baitha hai. 25 kg ka bachha right mein kahan baithega taaki balance ho? ($g=10$)

Step 1 — Translational? Pivot upar ki taraf normal force $N$ supply karta hai; woh vertical forces ko auto-balance karta hai, toh $\sum F_y=0$ satisfy ho jaata hai jab hum $N$ jaanein. Yeh step kyun? Hume question ke liye $N$ nahi chahiye, isliye hum is equation ka use nahi karenge.

Step 2 — Pivot ke baare mein torques lo. Kyun? Pivot reaction $N$ pivot se guzarti hai, isliye uska moment arm zero hai → woh gayab ho jaati hai. Smart axis choice!

$\sum \tau =0:(40)(10)(2)-(25)(10)(x)=0$ Yeh step kyun? Left bachha ek direction mein rotate karta hai (maano +), right bachha opposite direction mein (−).

Step 3 — Solve karo. $800=250x⟹x=3.2\text{ m}$ Yeh answer sensible kyun hai: halka bachha zyada door baithta hai — bada moment arm chhote weight ko compensate karta hai. ✓

Worked Example 2 — Leaning ladder (floor par friction)

$W$ weight, $L$ length ki ek uniform ladder, ek frictionless wall ke saath angle $\theta$ par tikki hai, rough floor par. Minimum coefficient of friction $\mu$ nikalo taaki woh slip na kare.

Forces: $W$ centre par; wall normal $N_w$ (horizontal, kyunki wall frictionless hai); floor normal $N_f$ (upar); floor friction $f$ (horizontal, wall ki taraf).

Step 1 — $\sum F_y=0$: $N_f=W$. Kyun? Sirf $N_f$ aur $W$ vertical hain.

Step 2 — $\sum F_x=0$: $f=N_w$. Kyun? Friction ko wall ke horizontal push ko rokna hoga.

Step 3 — Ladder ke paon ke baare mein torque lo. Yeh axis kyun? Yeh $N_f$ aur $f$ ko eliminate karta hai (dono paon par act karte hain → zero arm), sirf $W$ aur $N_w$ bachte hain.

Counter-clockwise positive lo. Weight horizontal distance $\frac{L}{2}\cos \theta$ par act karta hai; wall normal height $L\sin \theta$ par act karta hai: $N_w(L\sin \theta )-W\left(\frac{L}{2}\cos \theta \right)=0$ Yeh step kyun? Wall ka push ladder ko paon ke upar/paar rotate karne ki koshish karta hai; gravity usse neeche rotate karne ki koshish karti hai.

Step 4 — $N_w$ ke liye solve karo: $N_w=\frac{W\cos \theta }{2\sin \theta }=\frac{W}{2}\cot \theta .$

Step 5 — Combine karo: $f=N_w=\frac{W}{2}\cot \theta$ aur $N_f=W$. No-slip ke liye $f\le \mu N_f$ chahiye: $\mu \ge \frac{f}{N_f}=\frac{\frac{W}{2}\cot \theta }{W}=\frac{1}{2}\cot \theta .$ $\boxed{{\mu }_{\min }=\frac{1}{2}\cot \theta }$ Kyun sensible hai: zyada khadi ladder (bada $\theta$) → $\cot \theta$ chhota → kam friction chahiye. Near-flat ladder ko bahut bada $\mu$ chahiye. ✓

Worked Example 3 — Couple = pure rotation

Do forces $F=5$ N ek rod par act karti hain, opposite directions mein, $d=0.3$ m se separated.

$\sum \vec{F}=+F-F=0$ ✓ (force ka translational equilibrium), lekin $\sum \tau =F\cdot d=5\times 0.3=1.5$ N·m ≠ 0. Yeh kyun matter karta hai: body equilibrium mein nahi hai — woh angularly accelerate karti hai. Yeh steel-man proof hai ki tumhe genuinely doosri condition chahiye.

Flashcards

Connections

• Torque — force ka rotational analogue; doosri condition ke peeche ka engine.
• Centre of Mass — jahan weight effectively act karti hai.
• Newton's Laws of Motion — $\sum F=M{a}_{cm}$ pehli condition ke neeche hai.
• Moment of Inertia — $\sum \tau =I\alpha$ mein aata hai; zero net τ ⇒ zero α.
• Couple and Moment of a Couple — counter-example jo dikhata hai kyun do conditions chahiye.
• Static Friction — ladder/leaning problems mein holding force supply karta hai.
• Centre of Gravity vs Centre of Mass — toppling vs sliding analysis.

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