Visual walkthrough — Physical pendulum — compound pendulum

1.6.7 · D2 · Physics › Oscillations & Waves › Physical pendulum — compound pendulum

Shuru karne se pehle, teen simple words jinhe hum baar baar point karte rahenge:

• pivot — woh fixed nail jiske around object ghoomta hai. Hum ise $O$ kehte hain.
• centre of mass (CM) — woh ek balance point jahan gravity effectively pull karti hai. Nail se uski distance ko hum $d$ kehte hain.
• ==tilt angle $\theta$== — object seedha neeche se kitna door swing hua hai.

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Step 1 — Body ko latkaao aur picture ko naam do

KYA. Ek rigid body (socho ek flat paddle) nail $O$ pe pin ki gayi hai. Apne aap chodne pe woh is tarah latkti hai ki uska CM nail ke seedha neeche ho — yeh equilibrium hai. Ab hum use angle $\theta$ se tilt karte hain.

KYUN. Har derivation ko pehle ek labelled picture chahiye. Agar hum $d$, $\theta$, aur $O$ ko point nahi kar sakte, to hum unke baare mein ek bhi honest equation nahi likh sakte.

PICTURE. Figure dekho. Teal dashed line "seedha neeche" (equilibrium) hai. Nail se CM tak ka orange arrow us line ke saath angle $\theta$ banata hai. Us single arrow ki length $d$ hai.

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Step 2 — Torque kyun, aur gravity actually kahan push karti hai

KYA. Gravity poori body ko pull karti hai, lekin ek aise body ke liye jo sirf $O$ ke around ghoom sakti hai, jo matter karta hai woh force nahi balki uska twisting effect hai — torque. Hum gravity ka torque $O$ ke around calculate karte hain.

KYUN. Body sideways nahi ja sakti; woh nailed hai. Woh sirf rotate kar sakti hai. Rotation ko govern karne wala law hai $\tau =I\alpha$ (rotational Newton), $F=ma$ ka twin. To hume twist chahiye, raw pull nahi. Yahi exact reason hai ki hum yahan force ki jagah torque use karte hain.

PICTURE. Gravity $mg$ seedha neeche CM se point karti hai (plum arrow). Torque = force $\times$ perpendicular lever arm — force ki line aur nail ke beech ka sideways gap. Woh gap orange bracket hai, aur geometry kehti hai yeh $d\sin \theta$ ke barabar hai.

$$\tau =\underset{\text{tilt ko oppose karta hai}}{\underbrace{-}}\underset{\text{gravity force}}{\underbrace{mg}}\underset{\text{lever arm}}{\underbrace{d\sin \theta }}$$

Minus sign decoration nahi hai: yeh kehta hai ki twist equilibrium ki taraf wapas point karta hai. Right tilt → torque left push karta hai. Ek restoring torque.

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Step 3 — Twist ko equation of motion mein convert karo

KYA. Is torque ko rotational Newton's law $\tau =I\alpha$ mein feed karo.

KYUN. $I\alpha$ hume batata hai ki body twist ke baad kaise respond karti hai. $\alpha$ angular acceleration hai — tilt-rate kitni tezi se change ho raha hai — jise hum $\ddot{\theta }$ likhte hain (time mein $\theta$ ka second rate of change). Response ko cause ke barabar set karne se motion milti hai.

PICTURE. Figure "cause" (gravity ka restoring torque, orange) ko "response" (body ki spin hone ki resistance, teal) ke saath stack karta hai, balance $\tau =I\ddot{\theta }$ pe milte hue.

$$\underset{\begin{array}{c}\text{inertia about }O\\ \text{(resistance to spin)}\end{array}}{\underbrace{I}}\underset{\text{angular accel.}}{\underbrace{\ddot{\theta }}}=\underset{\text{restoring torque}}{\underbrace{-mgd\sin \theta }}$$

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Step 4 — Straightening trick: small-angle approximation

KYA. Chhote tilts ke liye hum $\sin \theta$ ko $\theta$ se replace karte hain ($\theta$ radians mein).

KYUN. $\sin \theta$ equation ko nonlinear banata hai — period ke liye koi clean formula nahi. Lekin graph dekho: chhote $\theta$ ke liye sine curve aur straight line $y=\theta$ almost ek doosre ke upar hain. Curve ko line se replace karna wahi hai jo ek hopeless equation ko SHM mein convert karta hai, woh ek oscillation jise hum exactly solve kar sakte hain.

PICTURE. Figure $\sin \theta$ (plum curve) aur $\theta$ (teal line) ko overlay karta hai. Woh origin ke paas chipke rehte hain aur roughly $20^{\circ }$ ke baad hi alag hote hain. Woh chipka hua region wahan hai jahan hamara formula honest hai.

$$I\ddot{\theta }=-mgd\theta ⟹\ddot{\theta }=-\underset{\text{ek positive constant}}{\underbrace{\frac{mgd}{I}}}\theta$$

Right side ka har symbol ek fixed number hai, to poori fraction sirf koi positive constant times $\theta$ hai, minus sign ke saath. Yeh baat yaad rakho.

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Step 5 — SHM pehchano aur period padhlo

KYA. $\ddot{\theta }=-\frac{mgd}{I}\theta$ ko master SHM equation $\ddot{\theta }=-{\omega }^2\theta$ se compare karo.

KYUN. "Acceleration $=-(\text{constant})\times (\text{position})$" form ki koi bhi equation simple harmonic motion hai — ek pure sinusoidal wobble. Constant ko ${\omega }^2$ naam diya jaata hai kyunki $\omega$ (angular frequency) directly timing set karta hai. Dono forms ko match karna hume $\omega$ free mein de deta hai.

PICTURE. Figure tilt $\theta (t)$ ko time mein cosine wave trace karte dikhata hai; ek poori wave = ek period $T$, aur steeper restoring (bada $mgd/I$) wave ko narrow kar deta hai.

Term by term match karte hue:

$${\omega }^2=\frac{mgd}{I}\Rightarrow \omega =\sqrt{\frac{mgd}{I}}\Rightarrow \boxed{T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{mgd}}}$$

Box padhte hue: bada $I$ (mass nail se door) → slow swing → lamba $T$. Bada $mgd$ (bhaari, ya CM zyada bahar → stronger restoring twist) → fast swing → chhota $T$. Formula ki shape hi physics hai.

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Step 6 — $I$ ko rewrite karo taaki hum actually compute kar sakein

KYA. $I$ ko "CM ke around spread" plus "CM se nail ki distance" mein split karo Parallel axis theorem use karke.

KYUN. Hum ek random nail ke baare mein $I$ rarely jaante hain, lekin hum hamesha table se $I_{cm}$ jaante hain. Parallel-axis theorem $I_{cm}$ ko pivot tak $m{d}^2$ add karke le jaata hai. $I_{cm}=m{k}^2$ likhna Radius of gyration $k$ introduce karta hai — woh single distance jo shape ko package karta hai.

PICTURE. Do dots: CM (uske compact spread $k$ ke saath) aur nail $d$ distance door. Theorem literally hai "CM ke around spread" + "shift $m{d}^2$".

$$I=\underset{=m{k}^2}{\underbrace{I_{cm}}}+\underset{\text{pivot tak shift}}{\underbrace{m{d}^2}}=m\left(k^2+d^2\right)$$

Box mein substitute karo — mass $m$ top aur bottom cancel ho jaata hai:

$$T=2\pi \sqrt{\frac{m(k^2+d^2)}{mgd}}=2\pi \sqrt{\frac{k^2+d^2}{gd}}\text{aur}{L}_{\text{eq}}=\frac{I}{md}=\frac{k^2}{d}+d$$

$L_{\text{eq}}$ us Simple pendulum ki length hai jo exactly same time mein swing karta hai — poori messy body ek number mein collapse ho gayi.

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Step 7 — Har edge case, taaki koi bhi swing surprise na kare

KYA. $d$ ko uski extremes tak push karo aur beech mein sweet spot dhundho.

KYUN. Ek formula jis par tum trust karo use apni limits survive karni chahiye. Hum $d\to 0$, $d\to \infty$ check karte hain, aur minimum dhundhte hain. Yeh "all cases" contract bhi complete karta hai.

PICTURE. $T$ versus $d$: ek valley. Left wall ($d\to 0$) infinity tak jaata hai, right side ($d$ bada) steadily climb karti hai, aur floor $d=k$ pe baitha hai.

• Pivot CM pe, $d\to 0$: restoring torque $mgd\sin \theta \to 0$ — kuch bhi wapas nahi kheenchta → $T\to \infty$. Body oscillate nahi karti; woh kisi bhi orientation mein waise hi baith jaati hai.
• Pivot door, $d\to \infty$: $I\approx m{d}^2$ grows faster than $mgd$, to $T\approx 2\pi \sqrt{d/g}$ badhta rehta hai — yeh ek lamba, slow simple pendulum ban jaata hai.
• Minimum: $\frac{k^2}{d}+d$ minimize karo. Uski derivative $-\frac{k^2}{d^2}+1$ hai; ise zero set karne se $\boxed{d=k}$ milta hai.

$d=k$ pe: $L_{\text{eq}}=\frac{k^2}{k}+k=2k$, to

$$T_{\min }=2\pi \sqrt{\frac{2k}{g}}.$$

Pivot se $L_{\text{eq}}$ distance wala point centre of oscillation hai — wahan pivot karo aur tumhe same $T$ milega. Yeh reversibility wahi trick hai jiske peeche Kater's pendulum ka $g$ measure karna hai.

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Ek-picture summary

Ek image, poori chain: gravity ka twist $-mgd\sin \theta$ → rotational law $I\ddot{\theta }$ → $\sin \theta \approx \theta$ se straighten karo → SHM pehchano → $T=2\pi \sqrt{I/(mgd)}$ padho → $I=m(k^2+d^2)$ rewrite karo → equivalent length $L_{\text{eq}}=k^2/d+d$ apne minimum ke saath $d=k$ pe.

Recall Feynman retelling — walkthrough plain words mein

Ek paddle ko wall pe nail karo aur latkne do; uska middle nail ke seedha neeche baithta hai. Use thoda tilt karo. Gravity, middle ko kheenchte hue, use wapas swing karne ki koshish karti hai — aur woh kitna zor se kheenchti hai yeh depend karta hai ki middle kitna sideways slide hua, jo hai $d\sin \theta$. Woh sideways tug paddle ko nail ke around twist karta hai; twist torque hai. Paddle ek twist ka kaise jawab deta hai yeh uske moment of inertia $I$ se set hota hai (uska mass nail se kitna door baitha hai). "Twist causes turning" ko equation mein daalo aur tumhe wobble law milta hai. Chhote tips ke liye, "kitna slide hua" basically sirf angle hi hai, aur tab paddle ek perfect clock ki tarah rock karta hai — simple harmonic motion. Timing padh ke milta hai $T=2\pi \sqrt{I/(mgd)}$. Kyunki $I$ ko "mass apne middle ke paas kaise bunched hai" plus "woh middle nail se kitna door hai" mein split kiya ja sakta hai, period actually hai $2\pi \sqrt{(k^2+d^2)/(gd)}$. Exact middle pe nail karo aur woh swing hi nahi karega (kuch wapas nahi kheenchta); door nail karo aur woh lazily swing karta hai; beech mein kahin — $d=k$ pe — woh sabse tez swing karta hai.

Recall Quick self-test

Lever arm $\sin \theta$ kyun use karta hai aur $\cos \theta$ kyun nahi? ::: Kyunki gravity ki line aur nail ke beech ka sideways gap woh opposite side hai ek right triangle ki jiska hypotenuse $d$ hai aur angle $\theta$ hai; opposite/hyp $=\sin$. $\theta =0$ pe yeh vanish hona chahiye, aur $\sin 0=0$. Kaunsi ek approximation exact equation ko SHM mein convert karti hai? ::: $\sin \theta \approx \theta$ chhote $\theta$ ke liye (radians mein). Mass $m$ kahan jaata hai? ::: Woh cancel ho jaata hai — $T$ mass se independent hai, har gravity pendulum ki tarah. Kaun se $d$ pe period sabse chhota hai, aur wahan $L_{\text{eq}}$ kya hai? ::: $d=k$, jisse $L_{\text{eq}}=2k$ milta hai aur $T_{\min }=2\pi \sqrt{2k/g}$.