Centre of mass — definition for system of particles: Visual walkthrough

1.4.9 · D2 · Physics › Momentum & Collisions › Centre of mass — definition for system of particles

Hum sirf yeh plain ideas use karenge, har ek pehle se samjhaaya hua:

ek position = ek number jo batata hai ki koi cheez line mein kitni door baithi hai (metres mein),
ek mass = koi cheez kitni "heavy" hai (kilograms, kg mein),
ek average = bilkul neeche precisely define kiya gaya hai.

Baaki sab kuch inheen ideas se grow karte hain.

Step 1 — Ek line, do dots, aur sawaal

KYA. Do particles ko ek horizontal line par rakho. Left waale ko mass $m_{1}$ kaho jo position $x_{1}$ par baitha hai, aur right waale ko mass $m_{2}$ position $x_{2}$ par. "Position" yahan sirf ek ruler reading hai metres mein — ek chosen zero mark (origin) se kitna right hai particle.

KYUN. Kisi bhi formula se pehle, humein fix karna hai ki hum dhundh kya rahe hain: ek single point $x_{c m}$ jo dono dots ko "represent" kare. Use dhundhne ki sabse simple jagah ek seedhi line hai, toh hum wahan se shuru karte hain aur baad mein dimensions add karte hain.

PICTURE. Figure mein, blue dot $m_{1}$ hai, orange dot $m_{2}$ hai, gray tick origin $0$ hai. Neeche green triangle ek pivot hai — imagine karo ki line ek stiff rod hai jo us pivot par raki hai. Humara poora sawaal yeh hai: pivot kahaan hona chahiye taaki rod balance kare?

Har symbol, jahan woh rehta hai:

$x_{1}$ — blue mass ki ruler reading (metres).
$x_{2}$ — orange mass ki ruler reading (metres).
$m_{1}, m_{2}$ — har ek kitna heavy hai (kg); dot size ke roop mein draw kiya gaya.
$x_{c m}$ — woh unknown pivot jo hum dhundh rahe hain.

Step 2 — "Sirf midpoint lo" ek trap kyun hai

KYA. Pehla tempting guess: pivot bilkul beech mein rakho, $\frac{x _{1} + x _{2}}{2}$ par — do positions ka plain average. Figure dikhata hai kya hota hai jab do masses unequal hain — rod heavy side ki taraf tip kar jaati hai.

KYUN. Balance turning effect ke baare mein hai, sirf distance ke baare mein nahi. Ek heavy mass bahut door aur ek light mass paas mein fir bhi balance kar sakti hai — toh sirf distance wala averaging galat hona chahiye jab masses differ karein. Humein upar ki definition se weighted average chahiye, har position ko uski mass se weight karte hue.

PICTURE. Left panel: equal masses → midpoint balance karta hai (rod level hai). Right panel: heavy orange mass → usi midpoint pivot pe orange side neeche tip karta hai (red arrow tip dikhata hai). Yeh picture parent ke "Common Mistakes" list mein har galti ka seed hai.

Step 3 — Lever law: mass × pivot-se-doori match karni chahiye

KYA. See-saw ki physics: ek mass balance karta hai agar ek side ka turning effect doosri side ke turning effect ke barabar ho. Turning effect = mass $\times$ pivot se uski doori. Toh balance ka matlab hai $m_{1} lever arm of m_{1} (x_{c m} - x_{1}) = m_{2} lever arm of m_{2} (x_{2} - x_{c m}) .$

Term by term:

$x_{c m} - x_{1}$ — pivot blue mass ke kitna right mein hai (uska lever arm).
$x_{2} - x_{c m}$ — orange mass pivot ke kitna right mein hai (uska lever arm).
har ek ko uski mass se multiply karne par do turning effects milte hain; balance ke liye zaruri hai ki woh equal hon.

KYUN yeh tool (product, sum nahi)? Kyunki ek see-saw ki rotate karne ki tendency dono se badhti hai — baccha kitna heavy hai aur kitna door baitha hai — kisi ek ko double karo toh effect double ho jaata hai. Sirf ek product $m \times (distance)$ "dono matter karte hain, multiplicatively" ko capture karta hai. Is product ka ek naam hai jise hum aage milenge: ek moment. (Dekho Weighted Average and Moments.)

PICTURE. Do shaded rectangles ka same area $= mass \times lever arm$ hai. Balance = equal areas. Jab orange mass heavier hoti hai, uska rectangle taller hota hai, toh uska arm shorter hona chahiye — pivot orange ki taraf slide karta hai taaki woh arm chhota ho jaaye. Tum dekh sakte ho pivot heavy mass ki taraf kheenchta hua.

Step 4 — Pivot ke liye balance solve karo

KYA. Balance equation lo aur $x_{c m}$ isolate karne ke liye unwrap karo.

$m_{1} (x_{c m} - x_{1}) = m_{2} (x_{2} - x_{c m})$

Dono sides multiply karo (brackets khole): $m_{1} x_{c m} - m_{1} x_{1} = m_{2} x_{2} - m_{2} x_{c m}$

Har $x_{c m}$ ko left par gather karo, baaki sab right par: $m_{1} x_{c m} + m_{2} x_{c m} = m_{1} x_{1} + m_{2} x_{2}$

Left se $x_{c m}$ factor karo, phir divide karo: $x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}}$

KYUN. Humne parent ka formula sirf ek see-saw se derive kiya — abhi momentum ki zarurat nahi. Yeh bilkul weighted-average template hai weights $=$ masses ke saath. Box padhne par:

numerator $m_{1} x_{1} + m_{2} x_{2}$ — har mass apni jagah ke saath tagged, summed (yeh total moment hai).
denominator $m_{1} + m_{2}$ — total mass $M$ , jo moment ko wapas ek plain position mein convert karta hai.

PICTURE. Figure algebra ko ek slider ki tarah animate karta hai: jaise orange mass equal se heavy hoti jaati hai, $x_{c m}$ ki boxed value midpoint se $x_{2}$ ki taraf slide karti hai. Dashed midpoint line wahan ruki rehti hai taaki tum dekh sako ki true COM usse kaise alag hota jaata hai.

Step 5 — Do particles se $n$ tak: wahi balance, zyada dots

KYA. Teesra dot add karo, chautha, ... unme se $n$ . Balance idea nahi badalti: har mass ka moment sum karo, total mass se divide karo. $x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2} + \dots + m _{n} x _{n}}{m _{1} + m _{2} + \dots + m _{n}} = \frac{\sum _{i} m _{i} x _{i}}{\sum _{i} m _{i}} = \frac{1}{M} \sum_{i} m_{i} x_{i} .$

Compact symbols padhne par:

$\sum_{i}$ — "har particle $i$ par add up karo" (lambe $+ \dots +$ ka ek tidy shorthand).
$m_{i} x_{i}$ — ek particle ka moment: uski mass uski jagah ke saath tagged.
$M = \sum_{i} m_{i}$ — grand total mass.

KYUN sum $\sum$ ? Kyunki turning effects simply add hote hain — teen bachche ek see-saw par independently contribute karte hain, toh unke moments addition se pile up ho jaate hain. Koi naya cheez zaruri nahi hai; do-dot law untouched scale up hota hai. Yeh abhi bhi weighted average hai, ab $n$ numbers par.

PICTURE. Line par paanch alag-alag size ke dots; har ek ek area $m_{i} x_{i}$ ka stack contribute karta hai. Pivot wahan land karta hai jahan saare stacks balance hote hain — visibly bade dots ke cluster ki taraf kheecha hua.

Step 6 — Line se bahar: har axis par same idea, phir ek vector mein bundle

KYA. Real particles plane mein (ya space mein) rehte hain. Har particle ko readings ki ek pair do $(x_{i}, y_{i})$ — uski rightward aur upward ruler reading. Tab $x_{c m} = \frac{\sum _{i} m _{i} x _{i}}{M}, y_{c m} = \frac{\sum _{i} m _{i} y _{i}}{M}, z_{c m} = \frac{\sum _{i} m _{i} z _{i}}{M} .$ Teesra, $z_{c m}$ , full 3D ke liye hai (ruler reading "page se bahar"); yeh teesre axis par identical formula hai.

KYUN hum har axis ko alag kar sakte hain? Kyunki left–right balance karna aur up–down balance karna alag variables mein alag equations hain. Do-dot balance dekho: $x$ -equation $m_{1} (x_{c m} - x_{1}) = m_{2} (x_{2} - x_{c m})$ mein sirf $x$ 's hain — usme koi $y$ nahi, aur matching $y$ -equation mein koi $x$ nahi. Pivot ko sideways slide karna sirf $x$ -turning-effects ko affect karta hai aur har $y$ -term untouched rehti hai. Toh wahi weighted-average solution axis by axis kaam karta hai, aur teesra axis $z$ koi alag nahi hai — isi liye parent likh sakta tha "har axis independently handle hota hai."

Ek symbol mein bundle karna. Teen alag averages likhna tidy hai lekin clunky. Hum ek particle ki teen readings ko ek single object mein collect karte hain, uska position vector $r_{i} = (x_{i}, y_{i}, z_{i})$ — chhota arrow sirf matlab hai "yeh coordinates ka ek package hai, ek number nahi." Kyunki har coordinate same weighted average follow karta hai, saare teen ek line mein collapse ho jaate hain: $r_{c m} = \frac{1}{M} i \sum m_{i} r_{i} = (x_{c m}, y_{c m}, z_{c m}) .$ Toh $r_{c m}$ (arrow ke saath) koi naya idea nahi hai — yeh bilkul teen scalar averages $x_{c m}, y_{c m}, z_{c m}$ ek saath staple kiye hain. Yeh header formula hai, ab fully earned.

PICTURE. Teen particles ek plane mein (parent ka example: $1$ kg at $(0, 0)$ , $2$ kg at $(2, 0)$ , $3$ kg at $(0, 2)$ ). Hum har ek ko $x$ -axis par drop karte hain $x_{c m} = 0.67$ m ke liye, phir $y$ -axis par $y_{c m} = 1.0$ m ke liye. Jahan woh milte hain woh crosshair — vector $r_{c m}$ , triangle ke andar — COM hai.

Step 7 — Origin move karna: balance point parwah nahi karta

KYA. Ruler readings depend karti hain kahaan tumne zero mark lagaaya. Poore origin ko left mein $a$ amount se slide karo (toh har reading ban jaati hai $x_{i}^{'} = x_{i} - a$ ). $x_{c m}$ ka kya hota hai? $x_{c m}^{'} = \frac{\sum _{i} m _{i} ( x _{i} - a )}{M} = \frac{\sum _{i} m _{i} x _{i}}{M} - \frac{a \sum _{i} m _{i}}{M} = x_{c m} - a .$

Term by term:

$\sum_{i} m_{i} x_{i} - a \sum_{i} m_{i}$ — sum split karo; shift $a$ factor out ho jaata hai kyunki yeh har particle ke liye same hai.
$\frac{a \sum _{i} m _{i}}{M} = a$ — kyunki $\sum_{i} m_{i} = M$ hai, poora correction sirf $a$ hai.

KYUN yeh matter karta hai. Computed $x_{c m}$ exactly wahi $a$ se shift hota hai jitna baaki sab kuch. Yeh hai translation invariance: COM ek physical balance point hai jo particles se chipa hua hai, tumhare ruler ka artefact nahi. Ruler move karo, aur pivot ko label karne wala number uske saath move karta hai — lekin pivot wahi dots ke beech mein sama raha. (Yeh exactly woh reason hai kyun parent mein Step 3 "constant of integration ko origin fix karke absorb kar sakta tha.")

PICTURE. Wahi masses do baar draw ki gayi hain: ek baar origin left mein, ek baar origin right mein shift karke. Dono mein, green pivot same physical spot par baitha hai; sirf uska printed coordinate $a$ se change hota hai.

Step 8 — Edge cases: picture kabhi nahi tootni chahiye

KYA & KYUN. Ek achchi derivation har extreme mein survive karti hai. Teen check karne ke liye:

Equal masses. Weights cancel → weighted average plain midpoint ban jaata hai. Pivot dead centre baithta hai; dono rectangles symmetry se equal hain.
Ek mass dominate karti hai ( $m_{1} = 1$ kg at $0$ , $m_{2} = 99$ kg at $10$ ). Tiny mass barely kheenchti hai; $x_{c m} = \frac{1 ( 0 ) + 99 ( 10 )}{100} = 9.9$ m — giant ke thoda pehle.
Empty-space COM. Do masses $x = 0$ aur $x = 10$ par beech mein kuch nahi: COM $5$ m par baithta hai jahan koi particle exist nahi karta. Pivot ek geometric balance point hai, physical object nahi.

PICTURE. Teen mini-panels, ek har case ke liye, har ek pivot ko exactly wahan settle hote dikhaata hai jahan maths kehti hai — including teesre panel mein lonely pivot empty space mein float karta hua.

Step 9 — Kyun yeh exact point woh hai jise Newton love karta hai

KYA. Humne $x_{c m}$ ek see-saw se build kiya. Lekin parent ne wahi formula momentum se derive kiya. Pehle, momentum symbols ka plain-words matlab:

$v_{i}$ — particle $i$ ki velocity: yeh kitni tezi se aur kis direction mein move karta hai, yaani uski position change hone ki rate, $v_{i} = \frac{d r _{i}}{d t}$ (" $r_{i}$ har second kitna move karta hai").
$P$ — system ka total momentum, saari velocities ka mass-weighted sum: $P = \sum_{i} m_{i} v_{i}$ . Momentum hai "mass in motion" — Newton ke laws actually jise push karte hain.
$v_{c m}$ — COM point ki velocity, $v_{c m} = \frac{d r _{c m}}{d t}$ .

Ab apne earned box $r_{c m} = \frac{1}{M} \sum_{i} m_{i} r_{i}$ ko time ke saath ek baar differentiate karo (masses constant hain): $M v_{c m} = M \frac{d r _{c m}}{d t} = \sum_{i} m_{i} \frac{d r _{i}}{d t} = \sum_{i} m_{i} v_{i} = P .$

KYUN dono approaches agree karni chahiye: mass-weighted average woh unique combination hai jiska time-rate $P / M$ ke barabar hai. Toh humara see-saw point automatically momentum point ban jaata hai — woh jagah jahan messy swarm mass $M$ ke ek particle ki tarah behave karta hai (leading to Newton's Second Law for a System of Particles, Conservation of Linear Momentum aur clean Collisions — Elastic and Inelastic frame).

PICTURE. Left: kaafi particles har ek ke paas ek chhota velocity arrow $v_{i}$ . Right: COM par mass $M$ ka ek bada dot jo summed arrow $P = M v_{c m}$ carry karta hai. Dono panels ke beech equals sign se equivalent declare kiye gaye hain.

Ek-picture summary

Upar ka sab kuch, compressed: line par alag-alag sizes ke dots, har ek ek moment $m_{i} x_{i}$ cast karta hai; ek pivot heavy cluster ki taraf kheecha hua; boxed formula upar; aur promise "yeh point ek mass $M$ ki tarah move karta hai."

Recall Feynman retelling — poora walkthrough plain words mein

Ek broomstick imagine karo jismein weights taped hain. Tum woh ek spot chahte ho jahan yeh tumhari ungali par balance kare. Ek heavy weight us balance spot ko apni taraf kheechti hai; ek light weight barely matter karta hai. Spot dhundhne ke liye, tum har weight ko tag karte ho jahan woh baitha hai — uski mass ko uski position se multiply karo — woh saare tags add karo, phir total weight se divide karo taaki tumhe wapas ek plain "kahaan" mile instead of ek "mass-times-kahaan." Woh "tagged sum ko total weight se divide karo" move bilkul ek weighted average hai. Yeh left–right aur up–down (aur in–out 3D ke liye) alag alag karo, kyunki apni ungali sideways slide karna ek up–down wobble fix nahi kar sakta — phir teen answers ko ek arrow $r_{c m}$ mein bundle karo. Poora ruler shift karo aur printed number change hota hai, lekin balance spot wahi weights ke beech mein sama raha. Jawaab weights ke beech khali hawaon mein bhi land kar sakta hai — yeh sirf ek balance point hai, koi cheez nahi. Aur yahan punchline hai: agar ab tum poore broomstick ko spinning room ke paar toss karo, woh balance point ek clean arc mein fly karta hai jaise poora broom ek dot ho wahan baitha hua. Isi liye hum ise invent karne ki taklif lete hain.

Recall Rapid self-check

Do masses ke liye line par balance law? ::: $m_{1} (x_{c m} - x_{1}) = m_{2} (x_{2} - x_{c m})$ — pivot ke baare mein equal moments. Ise $x_{c m}$ ke liye solve karo. ::: $x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}}$ . $x$ , $y$ , $z$ alag alag kyun split karo? ::: Har axis apne variable mein apna balance equation hai; sideways shift ek up–down imbalance theek nahi kar sakta. $r_{c m}$ mein arrow ka matlab kya hai? ::: Yeh teen scalar averages $x_{c m}, y_{c m}, z_{c m}$ ko ek package mein bundle karta hai. Agar tum origin ko $a$ se shift karo, $x_{c m}$ ka kya hota hai? ::: Yeh exactly $a$ se shift hota hai — physical balance point unchanged rehta hai (translation invariance). $1$ kg at $0$ , $99$ kg at $10$ : COM kahaan hai? ::: $9.9$ m — heavy mass ke paas hi. $P$ ko words mein define karo. ::: Total momentum, velocities ka mass-weighted sum $\sum_{i} m_{i} v_{i}$ .

Connections

Parent: Centre of mass — definition (Hinglish)
Weighted Average and Moments — see-saw "moment" $m_{i} x_{i}$ wahi maths hai.
Newton's Second Law for a System of Particles — Step 9 $F_{e x t} = M a_{c m}$ mein flow karta hai.
Conservation of Linear Momentum — $v_{c m}$ constant jab koi external force na ho.
Collisions — Elastic and Inelastic — COM frame mein sabse clean.
Centre of Mass of Continuous Bodies — $\sum$ ko $\int$ se swap karo jab dots ek smear ban jaayein.