Centre of mass — definition for system of particles: Exercises

1.4.9 · D4 · Physics › Momentum & Collisions › Centre of mass — definition for system of particles

Level 1 — Recognition

Exercise 1.1

Do particles: $m_{1} = 4$ kg at $x_{1} = 0$ , $m_{2} = 4$ kg at $x_{2} = 8$ m. $x_{c m}$ nikalo.

Recall Solution 1.1

WHAT we do: formula mein plug in karo. $x_{c m} = \frac{4 ( 0 ) + 4 ( 8 )}{4 + 4} = \frac{32}{8} = 4 m$ WHY the answer makes sense: masses equal hain, toh weights cancel ho jaate hain aur COM bilkul $0$ aur $8$ ka plain midpoint hai, jo ki $4$ m hai. ✔️

Exercise 1.2

Bina poori computation ke batao: $m_{1} = 1$ kg at $x = 0$ aur $m_{2} = 9$ kg at $x = 10$ m ke liye, kya $x_{c m}$ , $0$ ke paas hai ya $10$ ke paas?

Recall Solution 1.2

WHAT & WHY: COM hamesha bhaari mass ki taraf jhukta hai. Yahan $9$ kg $≫ 1$ kg, isliye $x_{c m}$ , $x = 10$ m ke paas hoga. (Poora number, agar chahiye toh: $x_{c m} = \frac{1 ( 0 ) + 9 ( 10 )}{10} = 9$ m — wakai $10$ ke kareeb hai.)

Level 2 — Application

Exercise 2.1

$m_{1} = 2$ kg at $x_{1} = 1$ m, $m_{2} = 3$ kg at $x_{2} = 4$ m, $m_{3} = 5$ kg at $x_{3} = 6$ m. $x_{c m}$ nikalo.

Recall Solution 2.1

Total mass: $M = 2 + 3 + 5 = 10$ kg. Numerator (the moment, units kg·m): $2 (1) + 3 (4) + 5 (6) = 2 + 12 + 30 = 44$ kg·m. $x_{c m} = \frac{44}{10} = 4.4 m$ WHY this looks right: $5$ kg (sabse bhaara) $6$ m par baitha hai aur average ko $4$ m waale beech wale particle se upar kheench raha hai — toh $4.4$ m bhaari side ki taraf lean karna exactly wohi hai jo hum expect karte hain. ✔️

Exercise 2.2

Ek plane mein teen particles: $m_{1} = 1$ kg at $(0, 0)$ , $m_{2} = 2$ kg at $(4, 0)$ , $m_{3} = 1$ kg at $(0, 4)$ . $(x_{c m}, y_{c m})$ nikalo.

Recall Solution 2.2

WHY split by axis: vector equation bas teen independent scalar averages hain (parent §3). $x$ aur $y$ ko alag-alag handle karo. $M = 1 + 2 + 1 = 4$ kg. $x_{c m} = \frac{1 ( 0 ) + 2 ( 4 ) + 1 ( 0 )}{4} = \frac{8}{4} = 2 m$ $y_{c m} = \frac{1 ( 0 ) + 2 ( 0 ) + 1 ( 4 )}{4} = \frac{4}{4} = 1 m$ COM at $(2, 1)$ m — triangle ke andar, $x$ -axis wali edge ki taraf khicha hua kyunki bhaari $2$ kg mass wahin baitha hai. Figure dekho. ✔️

Figure — Centre of mass — definition for system of particles

Level 3 — Analysis

Exercise 3.1

Do particles: $m_{1} = 2$ kg at $x_{1} = 0$ , aur ek mass $m_{2}$ at $x_{2} = 6$ m. Kis $m_{2}$ ke liye $x_{c m} = 4$ m hoga?

Recall Solution 3.1

WHAT we do: $m_{2}$ ko unknown maano aur equation set up karo. $4 = \frac{2 ( 0 ) + m _{2} ( 6 )}{2 + m _{2}}$ Solve — dono taraf $(2 + m_{2})$ se multiply karo: $4 (2 + m_{2}) = 6 m_{2} \Rightarrow 8 + 4 m_{2} = 6 m_{2} \Rightarrow 8 = 2 m_{2} \Rightarrow m_{2} = 4 kg$ WHY sensible: $x_{c m} = 4$ m, $x_{1} = 0$ se zyada $x_{2} = 6$ m ke kareeb hai, toh $m_{2}$ bhaara wala hona chahiye — aur wakai $4 > 2$ kg hai. ✔️

Exercise 3.2

Ek COM origin par hai. Particle A: $3$ kg at $(- 2, 0)$ m. Particle B: $2$ kg at $(x_{B}, y_{B})$ . $(x_{B}, y_{B})$ nikalo.

Recall Solution 3.2

WHY this works: agar COM origin par hai, toh $\sum m_{i} x_{i} = 0$ aur $\sum m_{i} y_{i} = 0$ (numerators zero ho jaate hain, kyunki $x_{c m} = y_{c m} = 0$ aur $M \neq = 0$ ). x-equation: $3 (- 2) + 2 x_{B} = 0 \Rightarrow - 6 + 2 x_{B} = 0 \Rightarrow x_{B} = 3$ m. y-equation: $3 (0) + 2 y_{B} = 0 \Rightarrow y_{B} = 0$ m. Toh B at $(3, 0)$ m hai. WHY it looks right: dono masses ko origin ko balance karne ke liye same line par opposite sides par hona chahiye; halka $2$ kg mass zyada door baithta hai ( $3$ m vs $2$ m) apni halki hone ki bhaarpai karne ke liye. ✔️

Level 4 — Synthesis

Exercise 4.1

Ek $1.0$ m uniform rod jiska mass $2$ kg hai, uska COM middle mein hai, $x = 0.5$ m par. Ek $1$ kg ka point mass uske right end ( $x = 1.0$ m) par glue kiya gaya hai. Rod+mass system ka COM kahan hai?

Recall Solution 4.1

KEY idea (synthesis): ek extended body ko apne khud ke COM par ek single point mass se replace kiya ja sakta hai. Toh rod ko " $2$ kg at $x = 0.5$ m" samjho. (Yahi bridge hai Centre of Mass of Continuous Bodies ki taraf.) $x_{c m} = \frac{2 ( 0.5 ) + 1 ( 1.0 )}{2 + 1} = \frac{1.0 + 1.0}{3} = \frac{2.0}{3} \approx 0.667 m$ WHY it looks right: right end par mass glue karne se balance point rod ke apne centre ( $0.5$ m) se right ki taraf, added mass ki taraf shift hota hai. $0.667$ m exactly wohi rightward nudge hai. ✔️

Exercise 4.2

Ek square ke teen corners par chaar equal masses $m$ rakhe hain (side $a$ ): $(0, 0)$ , $(a, 0)$ , $(0, a)$ — aur fourth corner $(a, a)$ empty hai. COM nikalo.

Recall Solution 4.2

WHY equal masses simplify: jab saari masses $= m$ hon, toh $m$ 's cancel ho jaate hain aur $x_{c m}$ plain average ban jaata hai $x$ -coordinates ka (aur $y$ ke liye bhi yahi). Dhyan raho sirf teen particles hain. $x_{c m} = \frac{m ( 0 ) + m ( a ) + m ( 0 )}{3 m} = \frac{a}{3}, y_{c m} = \frac{m ( 0 ) + m ( 0 ) + m ( a )}{3 m} = \frac{a}{3}$ COM at $(\frac{a}{3}, \frac{a}{3})$ . WHY interesting: COM empty space mein hai — koi particle wahin nahi hai. Yah corner $(0, 0)$ ke kareeb padta hai jo do bhari hui edges ke beech "shared" hai. ✔️

Level 5 — Mastery

Exercise 5.1 (COM velocity, momentum se connect karna)

Do particles ek line par move kar rahe hain: $m_{1} = 2$ kg, velocity $v_{1} = + 3$ m/s; $m_{2} = 1$ kg, velocity $v_{2} = - 6$ m/s. COM velocity $v_{c m}$ nikalo, aur batao ki yeh total momentum ke baare mein kya kehta hai.

Recall Solution 5.1

WHY the formula generalises: $x_{c m} = \frac{\sum m _{i} x _{i}}{M}$ ko time mein differentiate karo — masses constant hain, toh positions velocities ban jaate hain: $v_{c m} = \frac{\sum _{i} m _{i} v _{i}}{M}$ $v_{c m} = \frac{2 ( 3 ) + 1 ( - 6 )}{2 + 1} = \frac{6 - 6}{3} = 0 m/s$ WHAT it means: COM stationary hai. Kyunki total momentum $P = M v_{c m} = 3 \times 0 = 0$ , system ka momentum zero hai — dono particles ke momenta exactly cancel kar rahe hain. Yahi launching pad hai Conservation of Linear Momentum ke liye aur COM frame mein Collisions — Elastic and Inelastic analyse karne ke liye. ✔️

Exercise 5.2 (Mastery: "remove-a-piece" method)

Ek uniform square plate jiska side $2$ m aur mass $4$ kg hai, origin par centred hai (uska COM $(0, 0)$ par hai). Ek chhota square hole kaata jaata hai; nikale gaye piece ki mass $1$ kg hai aur uska apna centre $(0.5, 0)$ par tha. Baaki bachi plate ka COM nikalo.

Recall Solution 5.2

KEY synthesis idea — subtraction (negative mass trick): poori plate = (baaki plate) + (nikala gaya piece). Position ka momentum-bookkeeping matlab hai moments add hote hain: $M_{f u l l} x_{f u l l} = M_{r e m} x_{r e m} + M_{c u t} x_{c u t}$ Baaki part ka $x_{r e m}$ solve karo: $x_{r e m} = \frac{M _{f u l l} x _{f u l l} - M _{c u t} x _{c u t}}{M _{r e m}}$ Numbers: $M_{f u l l} = 4$ , $x_{f u l l} = 0$ ; $M_{c u t} = 1$ , $x_{c u t} = 0.5$ ; $M_{r e m} = 4 - 1 = 3$ kg. $x_{r e m} = \frac{4 ( 0 ) - 1 ( 0.5 )}{3} = \frac{- 0.5}{3} \approx - 0.167 m$ Symmetry se $y_{r e m} = 0$ . Toh baaki plate ka COM $(- \frac{1}{6}, 0) \approx (- 0.167, 0)$ m par hai. WHY it looks right: humne mass right side se ( $x = + 0.5$ ) kaata, toh jo bacha woh thoda left ki taraf balance karta hai. Negative answer confirm karta hai ki shift leftward hai. ✔️

Active Recall

Recall Kya tumne levels bhar ek pattern dhundha?

L1–L2: plug in — weights $m_{i}$ rakho, axes split karo.
L3: invert — moment equation set karke unknown mass ya position solve karo.
L4: compose — extended body ko uske apne COM par ek point se replace karo.
L5: differentiate ( $v_{c m}$ ) aur subtract (negative-mass hole). Ek formula, paanch moves.

Ladder of moves

recognise, apply, invert, compose, differentiate/subtract

Cut-out piece minus sign kyun carry karta hai?

Kyunki full = remaining + removed, toh removed moment subtract hota hai baaki part recover karne ke liye.

v_{c m}

ko momentum ke terms mein kya hai?

v_{c m} = \frac{\sum m _{i} v _{i}}{M} = \frac{P}{M}

Connections

Newton's Second Law for a System of Particles — Exercise 5.1, $F_{e x t} = M a_{c m}$ ka darwaza hai.
Conservation of Linear Momentum — zero COM velocity ka matlab zero total momentum hai.
Collisions — Elastic and Inelastic — COM frame (Exercise 5.1) collision algebra ko simplify karta hai.
Centre of Mass of Continuous Bodies — Exercises 4.1 & 5.2 "point at its own COM" bridge use karte hain.
Weighted Average and Moments — har solution ek weighted average / moment balance hai.