Centre of mass — definition for system of particles

1.4.9 · Physics › Momentum & Collisions

1. Centre of Mass kya hai?

WHAT matlab hai "mass-weighted" ka? Ek heavy particle average ko apni taraf kheenchta hai; ek light wala use barely shift karta hai. Yeh bilkul waise hai jaise class average mein har student ka score "weighted" hota hai us hisaab se ki course kitne credits ka hai.

2. Ise scratch se derive karna (WHY yahi formula, koi doosra nahi)

Hum chahte hain ek single point $r_{c m}$ jaise ki system behave kare jaise ek particle of mass $M$ wahan baitha ho. Sabse clean demand momentum pe hai.

Step 1 — System ka total momentum. $P = \sum_{i} m_{i} v_{i} = \sum_{i} m_{i} \frac{d r _{i}}{d t}$ Yeh step kyun? Momentum woh cheez hai jis pe Newton's laws act karti hain, toh hum definition ko usi se anchor karte hain.

Step 2 — Demand karo ki system ek mass $M$ ki tarah move kare $r_{c m}$ pe. Hum require karte hain: $P \equiv M v_{c m} = M \frac{d r _{c m}}{d t}$ Yeh step kyun? Yahi toh poora point hai: COM define kiya gaya hai taaki " $P = M v_{c m}$ " true ho.

Step 3 — Equate karo aur integrate karo. $M \frac{d r _{c m}}{d t} = \sum_{i} m_{i} \frac{d r _{i}}{d t} \Rightarrow M r_{c m} = \sum_{i} m_{i} r_{i}$ Yeh step kyun? Dono sides derivatives hain; derivatives ko match karne ka matlab hai quantities match hoti hain (constant of integration origin fix karke absorb ho jaata hai).

Step 4 — Solve karo. $r_{c m} = \frac{1}{M} i \sum m_{i} r_{i}$

Toh formula arbitrary nahi hai — yeh ek hi choice hai jo humein $P = M v_{c m}$ likhne deti hai.

3. Component form (HOW actually compute karte hain)

Figure — Centre of mass — definition for system of particles

4. Worked Examples

5. Common Mistakes (Steel-man + fix)

6. Active Recall

Recall Quick self-test (dekhne se pehle answer do)

COM ko ek sentence mein define karo. → Saare particles ki mass-weighted average position.
$M$ se kyun divide karte hain? → Moment $\sum m_{i} x_{i}$ (kg·m) ko position (m) mein convert karne ke liye.
Do equal masses ka COM kahan hota hai? → Bilkul midpoint pe.
Kya COM wahan ho sakta hai jahan koi mass na ho? → Haan (jaise ek ring ke centre pe).

Recall Feynman: ek 12-saal ke bacche ko samjhao

Soch ek see-saw pe alag-alag size ke bacche hain. Woh "balance point" jahan woh level baithega woh hai centre of mass. Ek bada baccha balance point ko apni taraf kheenchta hai, ek chota baccha use thoda kheenchta hai. Agar tum poore see-saw ko dhakka do, woh move karta hai jaise saare bacche ek lump hon jo bilkul us balance point pe baitha ho. Yahi woh magic point hai jo complicated cheezein simply behave karaata hai.

7. Forecast-then-Verify

Particles ke system ka centre of mass kya hota hai?

Mass-weighted average position,

r_{c m} = \frac{1}{M} \sum_{i} m_{i} r_{i}

COM formula mein total mass

M

se kyun divide karte hain?

Kyunki

\sum m_{i} x_{i}

ke units hain kg·m (ek moment);

M

(kg) se divide karne pe metres mein position milti hai.

COM ki woh defining property kya hai jo uska formula justify karti hai?

Yeh total momentum ko

M v_{c m}

banata hai, yaani system ek mass

M

ki tarah COM pe move karta hai.

Do equal masses ka COM kahan hota hai?

Bilkul unke midpoint pe.

Kya COM heavier ya lighter masses ki taraf jhukta hai?

Heavier mass ki taraf.

Kya COM empty space mein ho sakta hai?

Haan — jaise ek uniform ring of particles ke centre pe.

x_{c m}

ka component formula kya hai?

x_{c m} = \frac{\sum _{i} m _{i} x _{i}}{\sum _{i} m _{i}}

m_{1} = 2

kg at

0

aur

m_{2} = 3

kg at

5

m ke liye,

x_{c m}

nikalo.

x_{c m} = 15/5 = 3

Connections

Newton's Second Law for a System of Particles — leads to $F_{e x t} = M a_{c m}$ .
Conservation of Linear Momentum — COM velocity is constant when $F_{e x t} = 0$ .
Collisions — Elastic and Inelastic — analysed cleanly in the COM frame.
Centre of Mass of Continuous Bodies — replace $\sum$ by $\int$ .
Weighted Average and Moments — same maths as torque/lever balance.