Visual walkthrough — Motion of centre of mass — external force determines a_CM

1.4.11 · D2 · Physics › Momentum & Collisions › Motion of centre of mass — external force determines a_CM

Kisi bhi symbol se pehle, chaar seedhe words jo hum baar baar use karenge:

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Step 1 — Balance point kahan hai?

KYA. Hum kuch dots ko alag alag masses ke saath ek line par rakhte hain aur poochhte hain: kaun si ek jagah par poora set balance karega, jaise ek see-saw?

KYU. Baad ke har step mein isi ek point ke movement ki baat hai. Isliye pehle yeh pin down karna zaroori hai ki woh point hai kya — mass-weighted average position, seedha midpoint nahi.

PICTURE. Figure mein halka dot (chhota $m$) baayein baitha hai, bhaara red dot (bada $m$) daayein baitha hai. Balance point (neeche kala triangle) bhaare wale ke paas baitha hai. Yeh nazdeeki hi "mass-weighted" ki poori kahaani hai.

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Step 2 — Fraction saaf karo

KYA. Definition ke dono sides ko $M$ se multiply karo taaki koi fraction na bache.

KYU. Agle steps mein hum rates of change lenge (dekho Step 3). $M$ neeche hone par woh messy ho jaata hai; $M$ ek constant hai, isliye usse aage rakhne se sab kuch clean rehta hai.

PICTURE. Ise is tarah socho ki balance-point arrow ko $M$ factor se scale up karo jab tak woh exactly sab chhote $m_i{\vec{r}}_i$ arrows ke tip-to-tail sum ke barabar na ho jaaye. Figure un weighted arrows ko stack karta hai aur dikhata hai ki single $M{\vec{R}}_{CM}$ arrow usi tip tak pahunchta hai.

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Step 3 — Move hote dekho (velocity)

KYA. Time ko chalne do. Har dot ki position arrow ${\vec{r}}_i$ badlti hai; uski rate of change uski velocity ${\vec{v}}_i$ hai. Hum Step 2 ki equation ke dono sides ki rate of change lete hain.

KYU — aur kyun yahi tool. "Time ke saath rate of change" wahi hai jo ek derivative measure karta hai: kitni tezi se ek arrow ki tip hilti hai. Hum ise use karte hain (aur positions subtract karne jaisi kisi cheez ko nahi) kyunki hum balance point ki instantaneous speed chahte hain, jo har instant par valid ho, time ke ek chunk mein average nahi. ${\vec{r}}_i$ ki "rate of change" ke liye ${\vec{v}}_i$ likhna exactly wahi hai.

PICTURE. Ab har dot ke peeche ek velocity arrow trail karta hai. Balance point ke peeche uska apna arrow ${\vec{V}}_{CM}$ trail karta hai. Figure dikhata hai ki red dot tezi se daayein ja raha hai, ek halka dot dheeray baayein drift kar raha hai, aur resulting ${\vec{V}}_{CM}$ bhaari motion ki taraf jhuk raha hai.

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Step 4 — Accelerate hote dekho (force aata hai)

KYA. Rate of change ek baar aur lo. Velocity ${\vec{v}}_i$ ki rate of change acceleration ${\vec{a}}_i$ hai; ${\vec{V}}_{CM}$ ki rate of change ${\vec{a}}_{CM}$ hai.

KYU. Hum woh force chahte hain jo balance point ko control karta hai. Force Newton's 2nd law ${\vec{F}}_i=m_i{\vec{a}}_i$ se acceleration se link karta hai — isliye hum dobara differentiate karte hain taaki accelerations (aur isliye forces) appear ho sakein.

PICTURE. Ab har dot ke paas ek force arrow hai (jo usse push karta hai). Figure in sab push-arrows ko dots par dikhata hai, aur ek single net push-arrow balance point par. Humne abhi pushes sort nahi kiye — woh Step 5 hai.

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Step 5 — Har push sort karo: andar vs bahar

KYA. Har dot par forces ko do piles mein sort karo: external (system ke bahar se — gravity, wall, haath) aur internal (dots ek doosre ko push karte hain — springs, strings, collisions).

KYU. Kyunki dono piles bilkul alag behave karti hain. Hum suspect karte hain (Step 6) ki internal pile khud cancel ho jaayegi. Yeh prove karne ke liye, pehle usse saaf alag karna zaroori hai.

PICTURE. System ke around ek dashed boundary draw karo. Boundary cross karne wale arrows = external (red mein rakhe). Poori tarah andar rehne wale arrows, dot-to-dot = internal (kale pairs). Figure ek internal pair ${\vec{f}}_{12}$ aur ${\vec{f}}_{21}$ label karta hai.

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Step 6 — Internal pile cancel ho jaati hai (dil ki baat)

KYA. Newton's Third Law use karo: dot $j$, dot $i$ ko ${\vec{f}}_{ij}$ se push karta hai, aur dot $i$, $j$ ko exactly ulta ${\vec{f}}_{ji}=-{\vec{f}}_{ij}$ push karta hai. Har internal push ka ek twin dusri taraf point karta hai.

KYU. Ek number aur uska negative jodo → zero. Har pair ke liye yahi karo → poori internal pile $\vec{0}$ mein sum ho jaati hai. Yahi reason hai ki "internal fights balance point ko move nahi kar sakti", chahe kitni bhi violent ho.

PICTURE. Figure do dots par zoom karta hai jo spring se jude hain: dono force arrows equal length, opposite direction hain — tip-to-tail rakhne par kuch nahi bachta. Red "sum" arrow ek point tak chhota ho jaata hai.

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Step 7 — Master result

KYA. Step 6 ($\vec{0}$) ko Step 5 mein, phir Step 4 mein daalo. Sirf external pile bachti hai.

KYU. Yahi payoff hai: kisi bhi system ka balance point — chain, explosion, colliding cars — simple $\vec{F}=M\vec{a}$ follow karta hai sirf bahar ke forces use karke.

PICTURE. Sab internal clutter mita diya; ek single red external arrow total mass $M$ ke ek point ko drive karta hai. Messy many-body picture ek dot tak collapse ho jaati hai jo ek law follow karta hai.

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Step 8 — Edge cases (kabhi hanging nahi chhode)

KYA. Boundary scenarios check karo taaki koi reader kisi unseen case mein na phas jaaye.

KYU. Ek derivation jo tum trust karte ho usse apni extremes survive karni chahiye: zero external force, ek particle, zero mass.

PICTURE. Teen mini-panels: (a) ${\vec{F}}_{ext}=\vec{0}$ → balance point seedhi line mein constant speed se glide karta hai; (b) ek akela particle → ${\vec{R}}_{CM}$ wahi particle hai, formula ordinary $\vec{F}=m\vec{a}$ tak reduce ho jaata hai; (c) ek exploding shell → har fragment scatter karta hai (black) lekin balance point (red) usi parabola par rehta hai.

• Case (b) — sirf ek particle. ${\sum }_i{m}_i{\vec{r}}_i=m\vec{r}$, isliye ${\vec{R}}_{CM}=\vec{r}$. Balance point particle par baitha hai aur ${\vec{F}}_{ext}=m\vec{a}$ — ordinary Newton, koi surprise nahi.
• Case (c) — explosion / collision. Sab blast forces internal hain → cancel (Step 6). Sirf gravity (external) bachti hai → ${\vec{a}}_{CM}=\vec{g}$. CM kabhi apni parabola nahi chhodta.
• Rocket Propulsion ke baare mein ek note: wahan mass eject hota hai, isliye $M$ time ke saath badlta hai — Step 3 ka "spectator $M$" ab kaam nahi karta, aur tumhe exhaust ke saath carry hone wala momentum track karna padta hai. Isliye rockets ko ek alag treatment milti hai.

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Ek picture mein summary

Poora safar ek single frame mein: definition → do rates of change → forces split → internal pairs khatam → ek point, ek external force.

Recall Poore walkthrough ki Feynman retelling

Socho ek gang of friends space mein float kar rahi hai, kuch bhaari, kuch halke, springs se bande hue. Step 1–2: gang ka balance point dhundho — woh bhaare doston ke paas baitha hai, seedhe middle mein nahi. Step 3: time chalne do — balance point drift karta hai, aur uski drift gang ke total momentum ko total mass se divide karne ke barabar hai. Step 4: us drift ko speed up ya slow karo — uske liye forces chahiye. Step 5: forces ko sort karo "dost ek doosre ko dhakka dete hain" (andar) aur "koi bahar wala gang ko dhakka deta hai" (bahar). Step 6: har andar wala dhakka uska ek equal dhakka-waapas rakhta hai, isliye sab andar ka dhakka cancel ho jaata hai — gang chahe jitna bhi thrash kare, balance point ko koi fark nahi padta. Step 7: sirf bahar wale dhakke bachte hain, aur balance point ek poori mass ke lump ki tarah move karta hai, ${\vec{F}}_{ext}=M{\vec{a}}_{CM}$. Step 8: agar bahar se koi nahi dhakelta, balance point seedha coast karta rehta hai; ek akela dost apna khud ka balance point hai; aur ek exploding shell ka balance point usi parabola par ride karta hai jab pieces scatter hote hain.

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Connections

• Parent topic — Hinglish version
• Newton's Third Law — woh equal-and-opposite twin jo internal sum ko khatam karta hai (Step 6).
• Conservation of Linear Momentum — ${\vec{F}}_{ext}=\vec{0}$ wala case (Step 8a).
• Collisions (Elastic & Inelastic) — internal forces, CM ride karta hai through (Step 8c).
• Rocket Propulsion — jahan $M$ badlta hai aur Step 3 ka spectator trick toot jaata hai.
• Rigid Body Dynamics — solid body ka balance point same master law follow karta hai.