Centre of mass — definition for system of particles: Worked examples

1.4.9 · D3 · Physics › Momentum & Collisions › Centre of mass — definition for system of particles

Yeh parent note on Centre of Mass ka ek deep-dive companion hai. Wahan humne formula derive kiya tha $r_{c m} = \frac{1}{M} \sum_{i} m_{i} r_{i}, M = \sum_{i} m_{i} .$ Yahan hum prove karne ke opposite karte hain: hum ise stress-test karte hain. Hum har tarah ka input dhundte hain jo formula ko mil sakta hai — positive coordinates, negative coordinates, zeros, equal masses, ek mass itna bada ki baaki sab ko nigal le, ek mass jo zero ho jaaye, aur finally ek real-world word problem aur ek exam trap. Agar tum yeh sab walk through kar sako, to koi bhi COM question tumhe surprise nahi kar sakta.

Scenario matrix

Point particles ke har COM problem ko inn cells mein se kisi ek mein rakha ja sakta hai. Aane wale examples par woh cell label hai jise woh cover karte hain, aur saath mein yeh sab cells cover ho jaati hain.

Cell	Kya tricky hai	Example
A. All-positive, 1-D	plain baseline, koi sign issue nahi	Ex 1
B. Mixed sign, 1-D	positions origin ke dono taraf — cancellation	Ex 2
C. Equal masses	weights cancel ho jaate hain → COM sirf plain midpoint/centroid hai	Ex 3
D. Degenerate: ek zero mass	ek particle kuch contribute nahi karta; formula survive karna chahiye	Ex 4
E. 2-D, mixed sign	$x$ aur $y$ alag alag karo, dono mein signs hain	Ex 5
F. Limiting: ek mass $\to$ bahut badi	COM dominant mass par collapse ho jaata hai	Ex 6
G. Real-world word problem	ek scene ko masses + positions mein translate karo	Ex 7
H. Exam twist: unknown dhundo	COM diya gaya hai, mass ya position solve karo	Ex 8

Ek rule jis par hum har cell mein lean karenge:

Cell A — All-positive, 1-D (baseline)

Worked example Ex 1: Do masses, dono right mein

$m_{1} = 2$ kg at $x_{1} = 1$ m, $m_{2} = 6$ kg at $x_{2} = 9$ m. $x_{c m}$ nikalo.

Forecast: $6$ kg ki mass teen guna bhaari hai aur $9$ m par baithi hai. Andaaza: COM $9$ se $1$ ke mukable mein kaafi zyada paas hona chahiye. Kahin $7$ ke aaspaas? Apna number likh lo.

Step 1 — Total mass. $M = 2 + 6 = 8$ kg. Yeh step kyun? Yeh denominator hai; kuch bhi position nahi ban sakta jab tak hum isse divide na karein.

Step 2 — Numerator banao (moment). $\sum m_{i} x_{i} = 2 (1) + 6 (9) = 2 + 54 = 56 kg\cdotm .$ Yeh step kyun? Har mass apni position ke liye " $m_{i}$ votes" daalta hai; hum saare votes add karte hain.

Step 3 — Divide karo. $x_{c m} = \frac{56}{8} = 7 m .$ Yeh step kyun? Division kg·m moment ko wapas metre position mein convert karta hai.

Verify: $7$ m, $1$ aur $9$ ke beech mein hai ✔, aur yeh $9$ par bhaari mass ki taraf jhuk raha hai ✔ ( $9$ se $2$ ki doori vs. $1$ se $6$ ki doori — ratio $2 : 6 = m_{1} : m_{2}$ ✔, see-saw balance). Forecast se match karta hai.

Figure — Centre of mass — definition for system of particles

Cell B — Mixed sign, 1-D (cancellation)

Yahan pehla asli trap aata hai: positions origin ke dono taraf. Negative $x$ par mass ka moment negative hota hai — yeh average ko left ki taraf kheenchta hai.

Worked example Ex 2: Ek mass origin ke left, ek right

$m_{1} = 4$ kg at $x_{1} = - 3$ m, $m_{2} = 1$ kg at $x_{2} = 6$ m. $x_{c m}$ nikalo.

Forecast: Bhaari mass ( $4$ kg) left mein $- 3$ par hai. To chahe $6$ kaafi right mein ho, bhaari side ko jeetna chahiye. Andaaza: ek negative $x_{c m}$ , size mein chhota. Karib $- 1.2$ ?

Step 1 — Total mass. $M = 4 + 1 = 5$ kg.

Step 2 — Moment, signs ke saath. $\sum m_{i} x_{i} = 4 (- 3) + 1 (6) = - 12 + 6 = - 6 kg\cdotm .$ Yeh step kyun? $- 3$ ka minus sign survive karna chahiye — yeh encode karta hai "yeh vote left mein point karta hai." Isse drop karna #1 sign error hai.

Step 3 — Divide karo. $x_{c m} = \frac{- 6}{5} = - 1.2 m .$

Verify: COM negative hai ✔ — yeh origin ke heavy-mass side par baitha hai, bilkul forecast ki tarah. Sanity: yeh $- 3$ aur $6$ ke beech mein hai ✔ aur $- 3$ (bhaari) ke zyada kareeb hai ✔.

Cell C — Equal masses (weights cancel ho jaate hain)

Worked example Ex 3: Ek triangle ke corners par teen equal masses

Teen masses, har ek $m = 2$ kg, $(0, 0)$ , $(6, 0)$ , $(3, 6)$ par. COM nikalo.

Forecast: Equal masses equally vote karte hain — to mass drop out ho jaata hai aur COM plain centroid hota hai (corners ka average). Compute karne se pehle andaaza karo.

Step 1 — Cancellation note karo. Jab har $m_{i} = m$ ho, $x_{c m} = \frac{m x _{1} + m x _{2} + m x _{3}}{3 m} = \frac{x _{1} + x _{2} + x _{3}}{3} .$ Yeh step kyun? Common factor $m$ upar aur neeche dono mein aata hai aur cancel ho jaata hai — yahi wajah hai ki equal masses ordinary average dete hain.

Step 2 — $x$ 's ka average nikalo. $x_{c m} = \frac{0 + 6 + 3}{3} = \frac{9}{3} = 3$ m.

Step 3 — $y$ 's ka average nikalo. $y_{c m} = \frac{0 + 0 + 6}{3} = \frac{6}{3} = 2$ m.

Verify: COM $= (3, 2)$ m. Yeh triangle ke andar baitha hai ✔ (corners ka mass average kabhi unke convex hull ke bahar nahi ja sakta). Triangle ka centroid uske vertices ka average hota hai — match karta hai ✔.

Cell D — Degenerate input: ek zero mass

Agar kisi particle ki mass $0$ ho to? Formula toot-na nahi chahiye; zero-mass particle simply kuch contribute nahi karta.

Cell E — 2-D, dono axes mein mixed sign

Worked example Ex 5: Char particles, sab quadrants

$m_{1} = 1$ kg at $(2, 3)$ , $m_{2} = 2$ kg at $(- 4, 1)$ , $m_{3} = 1$ kg at $(0, - 5)$ , $m_{4} = 2$ kg at $(3, - 1)$ . COM nikalo.

Forecast: $x$ -votes: right ( $+ 2, + 3$ ) vs. left ( $- 4$ , weight $2$ ). Left mein ek heavy voter hai — expect karo $x_{c m}$ zero ke paas ya thoda negative. $y$ -votes mein $- 5$ par ek akela downward pull hai — expect karo $y_{c m}$ zero se neeche.

Step 1 — Total mass. $M = 1 + 2 + 1 + 2 = 6$ kg.

Step 2 — $x$ -moment (signs!). $\sum m_{i} x_{i} = 1 (2) + 2 (- 4) + 1 (0) + 2 (3) = 2 - 8 + 0 + 6 = 0 kg\cdotm .$ Yeh step kyun? Right aur left votes exactly cancel ho jaate hain — ek genuine mixed-sign cancellation.

Step 3 — $x_{c m}$ . $x_{c m} = \frac{0}{6} = 0$ m. Ek COM jo $y$ -axis par baitha hai.

Step 4 — $y$ -moment. $\sum m_{i} y_{i} = 1 (3) + 2 (1) + 1 (- 5) + 2 (- 1) = 3 + 2 - 5 - 2 = - 2 kg\cdotm .$

Step 5 — $y_{c m}$ . $y_{c m} = \frac{- 2}{6} = - \frac{1}{3} \approx - 0.333$ m.

Verify: COM $= (0, - 0.33)$ m. Yeh $y$ -axis par land kiya ( $x$ -votes cancel ho gaye) aur $x$ -axis se thoda neeche (net downward pull) ✔ — dono forecast se match karte hain. Yeh empty space mein hai, kisi particle ko touch nahi karta — COM ke liye bilkul legal hai.

Cell F — Limiting case: ek mass dominate kare

Worked example Ex 6: Ek boulder aur ek pebble

$m_{1} = 1000$ kg at $x_{1} = 0$ , $m_{2} = 1$ kg at $x_{2} = 50$ m. $x_{c m}$ nikalo, phir $m_{1} \to \infty$ imagine karo.

Forecast: Boulder $1000 \times$ bhaara hai aur $0$ par baitha hai. COM practically $0$ ke paas hona chahiye, $50$ ki taraf thoda sa.

Step 1 — Compute karo. $M = 1001$ kg, $x_{c m} = \frac{1000 ( 0 ) + 1 ( 50 )}{1001} = \frac{50}{1001} \approx 0.04995 m .$

Step 2 — Limit lo. Jaise jaise $m_{1} \to \infty$ , $1$ kg pebble ka vote negligible ho jaata hai: $x_{c m} = \frac{m _{1} ( 0 ) + 1 ( 50 )}{m _{1} + 1} m_{1} \to \infty \frac{50}{m _{1}} \to 0.$ Yeh step kyun? Yeh general truth dikhata hai: COM dominant mass par collapse ho jaata hai jab uska $M$ mein share $1$ ke paas pahunchta hai.

Verify: $0.04995$ m boulder se $\approx 5$ cm dur hai, essentially $0$ ✔. Limit $\to 0$ isse confirm karta hai. (Yahi reason hai ki Earth+person COM sabhi practical purposes ke liye Earth ke centre par hota hai.)

Cell G — Real-world word problem

Worked example Ex 7: Ek boat par do log

Ek $60$ kg ki aurat aage khadi hai aur ek $80$ kg ka aadmi ek halki boat ke peechhe. Boat $4$ m lambi hai; origin peechhe rakho, to aadmi $x = 0$ m par hai aur aurat $x = 4$ m par. (Boat ki apni mass negligible maano.) Do logon ka COM kahan hai?

Forecast: Aadmi ( $80$ kg) bhaara hai aur $0$ par hai; aurat ( $60$ kg) $4$ par. COM midpoint ( $2$ m) se aadmi ki taraf hona chahiye, yani $2$ m se kam. Andaaza $\approx 1.7$ ?

Step 1 — Scene translate karo. Masses ${80, 60}$ kg at positions ${0, 4}$ m. Yeh step kyun? Ek word problem sirf $(m_{i}, x_{i})$ ki disguise mein ek table hai; physics tabhi shuru hoti hai jab yeh table ban jaaye.

Step 2 — Total mass. $M = 80 + 60 = 140$ kg.

Step 3 — Moment aur divide karo. $x_{c m} = \frac{80 ( 0 ) + 60 ( 4 )}{140} = \frac{240}{140} = \frac{12}{7} \approx 1.714 m .$

Verify: $1.714$ m, $0$ aur $4$ ke beech mein hai ✔ aur midpoint $2$ m se kam hai, bhaare aadmi ki taraf jhuka hua ✔ — forecast se match karta hai. (Physical bonus: agar woh ab ek frictionless lake par jagah badal lein, to yeh COM wahi rehta hai — dekho Conservation of Linear Momentum — aur boat compensate karne ke liye slide karti hai.)

Cell H — Exam twist: unknown ke liye solve karo

Yahan COM diya gaya hai aur tumhe ek missing mass ya position dhundni hai. Wahi formula, ulta chalao.

Worked example Ex 8: Missing mass nikalo

$m_{1} = 3$ kg, $x_{1} = 0$ par baitha hai; ek unknown mass $m_{2}$ , $x_{2} = 8$ m par baitha hai. COM $x_{c m} = 6$ m par measure kiya gaya hai. $m_{2}$ nikalo.

Forecast: $6$ m par COM $8$ par mass ke zyada kareeb hai $0$ par mass ke mukable mein. Mass jितना bhaara, COM utna zyada us taraf jhukta hai — to $m_{2}$ bhaari hona chahiye, $3$ kg se zyada. Andaaza $\approx 9$ ?

Step 1 — Unknown ke saath COM equation likho. $6 = \frac{3 ( 0 ) + m _{2} ( 8 )}{3 + m _{2}} .$ Yeh step kyun? Hum $m_{2}$ ke alaawa sab kuch jaante hain; isse daalo aur yeh ek unknown mein ek equation ban jaati hai.

Step 2 — Denominator clear karo. $6 (3 + m_{2}) = 8 m_{2} \Rightarrow 18 + 6 m_{2} = 8 m_{2} .$ Yeh step kyun? Multiply out karne se fraction hat jaata hai taaki hum $m_{2}$ isolate kar sakein.

Step 3 — Solve karo. $18 = 2 m_{2} \Rightarrow m_{2} = 9 kg .$

Verify: Wapas plug karo: $x_{c m} = \frac{3 ( 0 ) + 9 ( 8 )}{3 + 9} = \frac{72}{12} = 6$ m ✔. Aur $m_{2} = 9 > 3$ kg — bhaara, jaise forecast tha ✔. See-saw check: distance ratio $6 : 2$ (har mass se COM tak) inverse mass ratio $m_{2} : m_{1} = 9 : 3 = 3 : 1$ ke barabar hona chahiye; aur $6 : 2 = 3 : 1$ ✔.

Matrix ka Recap

Recall Har cell kaunsa trap sikhata hai?

A — baseline, bhaari mass ki taraf jhuko. B — negative signs rakhna; COM negative ho sakta hai. C — equal masses cancel ho jaate hain → plain average. D — zero mass kuch contribute nahi karta; $M = 0$ undefined hai. E — $x$ aur $y$ alag alag karo; COM empty space mein ho sakta hai. F — dominant mass limit mein COM ko nigal jaata hai. G — word problems sirf $(m_{i}, x_{i})$ tables hain. H — COM diya gaya → unknown ke liye ulta solve karo.

Connections

Weighted Average and Moments — Ex 1 aur Ex 8 mein use kiya gaya see-saw / lever check moment balance hai.
Newton's Second Law for a System of Particles — jab COM locate kar lo, to uski motion track karo.
Conservation of Linear Momentum — boat problem (Ex 7) yahan continue hota hai.
Collisions — Elastic and Inelastic — yeh worked-example skills COM-frame analysis ko feed karti hain.
Centre of Mass of Continuous Bodies — yahan ke sums ko integrals se replace karo.