Foundations — Physical pendulum — compound pendulum

1.6.7 · D1 · Physics › Oscillations & Waves › Physical pendulum — compound pendulum

Is page par kuch bhi assume nahi kiya gaya. Agar parent note ne koi symbol bina explanation ke likha tha, toh hum use yahan pehle ek picture se build karte hain. Upar se neeche padho — har block sirf wohi use karta hai jo us se upar wale block ne already define kiya hai.

1. "Rigid body" aur "pivot" ka matlab

Koi bhi maths se pehle, object ko khud picture karo.

Ek rigid body ek solid object hai jiske parts kabhi ek doosre ke relative move nahi karte — ek ruler, ek darwaza, ek wheel. Yeh jhoolte waqt apni shape maintain karta hai. Yeh ek floppy string ke bilkul ulta hai.

==Pivot $O$== woh fixed line hai (ek nail, ek rod, ek axle) jiske around body turn karti hai. Picture mein yeh upar black dot hai; body sirf iske baare mein rotate kar sakti hai — yeh slide away nahi kar sakti. Body ka har point $O$ par centred ek circle trace karta hai.

2. Angle $\theta$ — "kitna tilted hai" measure karna

Jab body shant hang hoti hai, uska balance-point pivot ke seedha neeche hota hai. Is resting position ko equilibrium kehte hain.

Isse ek taraf push karo aur yeh us vertical line ke saath ek angle banata hai. Is angle ko hum ==$\theta$== (Greek letter "theta") kehte hain, jo angular displacement hai.

• $\theta =0$ matlab seedha neeche hang karna (equilibrium).
• $\theta$ positive matlab ek taraf tilted, negative matlab doosri taraf — sign sirf tilt ki direction record karta hai.

3. Center of mass aur distance $d$

Har object mein ek special balance-point hota hai: center of mass (CM). Agar tum object ko exactly wahan support karo, yeh perfectly balance karega, aur gravity aise behave karti hai jaise saara weight is ek point par act karta hai.

Symbol ==$d$== pivot $O$ se CM tak ki straight-line distance hai.

Ek uniform ruler mein CM uske middle mein hota hai, isliye agar tum ek end par pivot karo, $d$ = half the length. Parent note mein disk ke liye, CM disk ka center hai, isliye rim par pivoting karne se $d=R$ milta hai.

4. Mass $m$ aur weight $mg$

• ==$m$== mass hai — object kitne stuff se bana hai, kilograms mein.
• ==$g$== gravity ke wajah se acceleration hai, lagbhag $9.8{\text{m/s}}^2$.
• Product ==$mg$== weight hai — woh downward force jo gravity exert karti hai. Yeh hamesha seedha neeche point karta hai, aur (§3 se) yeh CM par act karta hai.

5. Torque $\tau$ — ek force ki "twisting strength"

Ek force jo directly pivot ki taraf push karti hai woh kuch spin nahi kar sakti. Sirf sideways part twist karta hai. ==Torque $\tau$== (Greek "tau") measure karta hai ki ek force kitni strongly ek body ko axis ke baare mein twist karta hai.

$\tau =(\text{force})\times (\text{perpendicular distance from the axis to the force’s line})$

Hamare pendulum ke liye force $mg$ (neeche) hai aur CM pivot se $d$ distance par hai. $\theta$ se tilted hone par, us downward force ka perpendicular lever arm $d\sin \theta$ hai (CM ka pivot se horizontal offset). Toh

$\tau =-mgd\sin \theta .$

6. Moment of inertia $I$ — "rotational heaviness"

Yahan topic ka star hai. ==Moment of inertia $I$== measure karta hai ki body ki spin change karna kitna mushkil hai — mass ka rotational version. Iska size sirf is baat par depend nahi karta ki kitna mass hai, balki is baat par bhi ki woh mass axis se kitni door baith hai:

$I=\sum (\text{each bit of mass})\times (\text{its distance from the axis}{)}^2.$

Door wala mass bahut zyada count karta hai (distance squared hai), isliye wide spread mass wali body ko spin karna bahut mushkil hai.

Link Moment of inertia mein standard values ki full derivation hai ($\frac{1}{3}m{\ell }^2$ rod-about-end ke liye, $\frac{1}{2}m{R}^2$ disk ke liye, ...).

7. Radius of gyration $k$ — $I$ ko ek length mein package karna

Kabhi kabhi hum $I$ ko ek distance ke roop mein express karna chahte hain. ==Radius of gyration $k$== is tarah define hota hai ki

$I_{cm}=m{k}^2⟺k=\sqrt{\frac{I_{cm}}{m}}.$

Aise socho: "agar saara mass radius $k$ ki ek thin ring par squeeze kar diya jaaye, toh uski $I_{cm}$ same hogi." Yeh mass-spread ko ek single length ke roop mein repackage karta hai. Isse parent note $L_{\text{eq}}=k^2/d+d$ neatly rewrite kar pata hai aur minimum-period pivot $d=k$ par dhundh pata hai. Poori detail: Radius of gyration.

8. Angular acceleration aur rotational Newton's law

• $\frac{d\theta }{dt}$ = angle kitni fast change hota hai = angular velocity.
• $\ddot{\theta }=\frac{d^2\theta }{d{t}^2}$ = angular velocity kitni fast change hoti hai = angular acceleration $\alpha$ (do dots sirf matlab "rate of change, do baar").

Rotational Newton's second law torque ko spin-up se jodhta hai:

$\tau =I\ddot{\theta }.$

$F=ma$ se compare karo: torque $\tau$ force ka role play karta hai, moment of inertia $I$ mass ka role play karta hai, aur angular acceleration $\ddot{\theta }$ ordinary acceleration ka role play karta hai. Yahi woh equation hai jismein parent note torque feed karta hai.

9. SHM signature — $\ddot{\theta }=-{\omega }^2\theta$ "swing" kyun matlab karta hai

§5 aur §8 ko combine karo, phir small-angle shortcut $\sin \theta \approx \theta$ use karo (small tilts ke liye valid, radians mein):

$I\ddot{\theta }=-mgd\sin \theta \approx -mgd\theta \Rightarrow \ddot{\theta }=-\frac{mgd}{I}\theta .$

Koi bhi equation jo $\ddot{\theta }=-{\omega }^2\theta$ ki shape ki hai woh Simple Harmonic Motion hai: acceleration hamesha zero ki taraf wapas point karta hai, is baat ke proportion mein ki tum zero se kitna door ho. Yahi cheez kisi ko ek steady period ke saath oscillate karati hai. Dono forms ko match karne se milta hai

${\omega }^2=\frac{mgd}{I},T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{mgd}}.$

• ==$\omega$== ("omega") = angular frequency, cycle ke radians per second.
• ==$T$== = period, ek complete back-and-forth ke liye seconds. Yeh $T=2\pi /\omega$ se linked hain kyunki ek full cycle phase ke $2\pi$ radians span karta hai.

10. Har foundation topic ko kaise feed karta hai

Equipment checklist

Khud test karo — right side cover karo aur reveal karne se pehle zyech se jawab do.