Centre of mass — definition for system of particles: Question bank

1.4.9 · D5 · Physics › Momentum & Collisions › Centre of mass — definition for system of particles

Ek formula yaad rakho jis par neeche sab kuch tika hai:

Do pictures jo dimaag mein rakhni hain

Traps se pehle, do geometric words baar baar aate hain. Inhe ek baar picture ke saath samjho taaki ye kabhi trip na karein.

Figure — Centre of mass — definition for system of particles

Definition Bounding box & convex hull (seedhe shabdon mein)

Bounding box woh sabse chhotaa seedha rectangle hai (jo $x$ - aur $y$ -axes ke saath aligned ho) jo har particle ko apne andar rakhta hai. Iska left edge sabse chhote $x$ par hota hai, right edge sabse bade $x$ par, aur isi tarah top/bottom $y$ ke liye. Figure mein yellow rectangle dekho.
Convex hull woh rubber-band shape hai: imagine karo ki ek elastic band ko saare bahari particles ke around stretch karke chhod do taaki wo snap ho jaye. Ye sabse tight shape hai jisme koi dent nahi — figure mein blue polygon. COM (pink dot) hamesha dono ke andar hi hota hai. KYUN? Kyunki positions ka mass-weighted average kabhi bhi sabse daahine wale particle se zyaada daahine nahi ja sakta, na rubber band se bahar nikl sakta hai — averaging andar ki taraf kheenchti hai, bahar ki taraf nahi.

Doosri picture sabse simple possible case hai — do masses ek line par — jis par almost har trap secretly reduce hota hai:

Ek aur property jis par traps rely karte hain — aur jo beginners rarely notice karte hain:

True ya false — justify karo

Har claim aise likha hai jaise ek student maanta ho. True/false decide karo, phir reason do.

Centre of mass hamesha kisi ek particle par hota hai.

False — ye ek average position hai, isliye ye usually particles ke beech empty space mein hota hai (jaise do equal masses ke beech mein, jahan koi mass nahi hota).

Agar current COM position par aur particles add karo to COM shift hoga.

False — exactly

r_{c m}

par rakha gaya particle average ke same position contribute karta hai, isliye weighted average (aur COM) move nahi karta; sirf

M

badhta hai.

Agar system ki har mass ko double karo, to COM wahi rahega.

True — har

m_{i}

ko double karna numerator aur denominator dono ko

2

se multiply karta hai, jo cancel ho jaata hai; COM mass ratios par depend karta hai, absolute masses par nahi.

COM ko saare particles wale bounding box (sabse chhotaa seedha rectangle) ke andar hona chahiye.

True — coordinates ka weighted average kisi bhi axis par sabse bade se zyaada ya sabse chhote se kam nahi ho sakta, isliye ye charon edges ke andar rehta hai (upar yellow box dekho).

Same total mass wale do systems ka hamesha same COM hota hai.

False — COM is baat par depend karta hai ki har mass kahan hai, na sirf total par; do systems

M

share kar sakte hain phir bhi unka mass distribution aur COM bilkul alag ho sakta hai.

Ek particle ko sirf

x

-axis ke along slide karna

x_{c m}

change karta hai lekin

y_{c m}

ko nahi.

True — har axis ek independent weighted average hai, aur

y_{c m}

sirf

y_{i}

values use karta hai; particle ko bina uske

y_{i}

change kiye move karna

y_{c m}

nahi badal sakta.

Ek system ka COM particles ke convex hull (rubber-band shape) ke bahar ho sakta hai.

False — positive weights (masses) ke saath weighted average hamesha ek convex combination hoti hai, isliye ye rubber-band shape ke andar hota hai, kabhi bahar nahi niklta.

Agar sab masses equal hain, to COM positions ka plain geometric average hota hai.

True — equal weights numerator aur denominator mein cancel ho jaate hain, chodke

r_{c m} = \frac{1}{n} \sum r_{i}

, jo ordinary centroid hai.

Coordinate origin move karne par COM ka location change hota hai.

False — COM ek physical point hai masses ke beech fixed; origin ko

a

se shift karna uske coordinates ko

- a

se shift karta hai lekin actual point bilkul wahi rehta hai (do-origin figure dekho).

Error pakdo

Har item ek galat worked statement hai. Galti batao aur reasoning theek karo.

x_{c m} = \sum m_{i} x_{i}

, isliye

2

kg at

0

aur

3

kg at

5

m ka COM

15

m hai."

Error ye hai ki

M

se divide karna bhool gaya.

\sum m_{i} x_{i} = 15

ki units kg·m hain (ek moment), metres nahi;

M = 5

kg se divide karne par sahi

x_{c m} = 3

m milta hai.

1

kg mass at

0

aur

9

kg mass at

10

m ka COM midpoint,

5

m hai."

Error mass weighting ko ignore karna hai. Bhaari

9

kg mass average ko apni taraf kheenchta hai, deta hai

x_{c m} = \frac{0 + 90}{10} = 9

m, midpoint se bahut door.

y_{c m}

nikalne ke liye,

y

formula mein

x

-positions daalo."

Error axes mix karna hai.

y_{c m}

sirf

y_{i}

coordinates use karta hai; har axis ek alag scalar average hai, isliye

x

-values kabhi

y

calculation mein nahi aane chahiye.

"COM velocity

v_{c m} = \sum m_{i} v_{i}

hai."

Phir se

M

se divide karna chhoot gaya. Total momentum

P = \sum m_{i} v_{i}

hai, lekin

v_{c m} = P / M

; momentum khud velocity nahi hai.

"Kyunki particle 3 sabse bhaari hai, COM exactly particle 3 par baithta hai."

Error "heavy ki taraf jhukta hai" ko "heavy par baithta hai" samajhna hai. COM sirf sabse bhaari mass ki taraf shift hota hai; ye kisi particle par tabhi pahunchta hai jab baaki sab masses zero hon.

x = 100

m par ek massless particle COM ko

100

m ki taraf kheechta hai."

Error ye assume karna hai ki har particle average ko affect karta hai.

m_{i} = 0

wala particle numerator mein

0 \cdot x_{i} = 0

contribute karta hai aur

M

mein bhi

0

, isliye COM par uska koi effect nahi hota.

"L-shaped set of masses ka COM woh corner hai jahan dono arms milte hain."

Error ye assume karna hai ki COM kisi visually "central" particle par hota hai. COM actual positions ka mass-weighted average hai aur generally kisi bhi single particle par nahi hota, balkee L ke spread ke andar kahin hota hai.

Why questions

Ye formula ke peeche ki reasoning probe karte hain, arithmetic nahi.

Hum total mass

M

se kyun divide karte hain, particles ki sankhya

n

se kyun nahi?

M

se divide karna har mass ko utna contribute karta hai jitna woh bhaari hai (ek weighted average);

n

se divide karna ek

100

kg aur ek

1

kg mass ko equally important treat karta. Concretely,

2

kg at

0

aur

3

kg at

5

m deta hai

\frac{15}{5} = 3

m (mass-weighted), na ki

\frac{5}{2} = 2.5

m (plain average) — extra kilogram bhaari mass ko zyaada pull dilata hai.

COM ko momentum ke through kyun define karte hain, energy ke through kyun nahi?

Kyunki Newton's laws momentum par act karte hain (

F = d P / d t

), isliye COM ko

P = M v_{c m}

se anchor karna precisely wahi hai jo poore system ko

F_{e x t} = M a_{c m}

ki tarah ek particle jaise behave karne deta hai — dekho Newton's Second Law for a System of Particles. Energy ek scalar hai aur koi direction nahi rakhti, isliye wo kabhi pin down nahi kar sakti ki representative point kahan hai.

COM bhaari mass ki taraf kyun jhukta hai?

Kyunki har position sum mein uski mass se weight ki jaati hai; bada

m_{i}

apne

x_{i}

ko heavily multiply karta hai, weighted average ko us particle ki taraf kheenchta hai. See-saw figure socho: balance karne ke liye, pivot ko bade wale ke paas hona chahiye taaki unka chhota lever arm times bada weight chhote wale ke bade arm times chhote weight ke equal ho.

x

y

, aur

z

ko alag alag treat kyu kar sakte hain jab definition ek vector equation hai?

Ek vector equation component-by-component hold karti hai, aur mass weighting

m_{i}

har component mein same number hai. Isliye ek vector average cleanly teen independent scalar averages mein split hota hai —

x_{c m}

ko kabhi kisi

y

-value ki zarurat nahi padti, aur vice versa.

Jab koi external force na ho to COM velocity constant kyun rehti hai?

Kyunki

F_{e x t} = M a_{c m}

, isliye zero external force ka matlab zero COM acceleration hai, yani constant

v_{c m}

— ye Conservation of Linear Momentum hai COM ke nazariye se. Internal forces action–reaction pairs mein aate hain jo sum mein cancel ho jaate hain, isliye wo COM ko kabhi nahi hila sakte.

COM frame mein collision analyse karna easier kyun hai?

COM frame mein total momentum zero hota hai, isliye do objects hamesha symmetrically approach aur separate karte hain, kai saare variables ko ek mein collapse karte hain — ye Collisions — Elastic and Inelastic ki backbone hai.

\sum

\int

se replace karne par solid body ka COM kyun milta hai?

Body ko

n

tiny chunks mein kaato, har ek ek "particle" hai jisme mass

Δ m_{k}

hai position

r_{k}

par; discrete formula deta hai

\frac{1}{M} \sum_{k} r_{k} Δ m_{k}

. Ab chunks ko chhota karo jab

n \to \infty

aur

Δ m_{k} \to 0

: tiny pieces ke sum ki wo limit exactly integral

\frac{1}{M} \int r d m

ki definition hai — dekho Centre of Mass of Continuous Bodies.

COM formula see-saw ke "moment" idea se identical kyun hai?

Dono weighted averages hain: COM ke baare mein torques balance karna

\sum m_{i} g (x_{i} - x_{c m}) = 0

ko rearrange karne par milta hai

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

, isliye Weighted Average and Moments same maths share karta hai. COM literally woh balance point hai jahan sab mass-moments cancel ho jaate hain.

Edge cases

Boundary aur degenerate scenarios jinhe formula phir bhi handle kare.

Single particle ka COM kahan hota hai?

Exactly us particle par — ek mass ke saath,

r_{c m} = \frac{m r}{m} = r

, isliye ek cheez ka average woh cheez khud hoti hai.

Uniform ring of particles ka COM kya hota hai?

Ring ka geometric centre, jo empty space hai — ek clean proof ki COM ka kisi mass ke saath coincide karna zaruri nahi.

Agar ek particle ki mass bina bound ke badhti rahe jabki baaki fixed rahen, to COM kahan jaata hai?

Woh us particle ki position ke paas jaata hai, kyunki limit mein

m_{1} \to \infty

wala term

m_{1} r_{1}

numerator aur denominator dono mein dominate karta hai, isliye

r_{c m} \to r_{1}

Kya do-particle system ka COM unhe milane wale segment ke bahar ja sakta hai?

Nahi — positive masses ke saath ye ek convex combination hai

r_{c m} = \frac{m _{1} r _{1} + m _{2} r _{2}}{m _{1} + m _{2}}

, jo hamesha do points ke beech aata hai, kabhi bahar nahi.

Agar saari masses equal hain aur ek point ke around symmetrically placed hain to COM kahan hoga?

COM exactly us symmetry point par hoga, kyunki equal weights ise centroid banate hain aur symmetry har displacement ko balance karti hai.

Agar do particles same position par hain, to COM kya hoga?

Wahi shared position — dono terms same jagah point karte hain, isliye weighted average woh jagah hai chahe mass ka split kuch bhi ho.

Kya system ki zero total mass formula mein koi sense banata hai?

Nahi —

M = 0

definition ko zero se divide kar deta hai; physically har real system ka positive total mass hota hai, isliye ye case excluded hai, answered nahi.

Recall Trap families ka one-line summary

Traps kuch flavours mein aate hain: (1) mass weighting $m_{i}$ drop karna; (2) $M$ se divide karna bhoolna; (3) axes mix up karna; (4) ye assume karna ki COM kisi particle par ya kisi "visually central" spot par hona chahiye; aur (5) ye sochna ki origin ya absolute mass scale matter karta hai. Paanchon ko guard karo aur COM kabhi surprise nahi karega.

Connections

Parent: COM definition — wo formulas jinhe ye traps test karte hain.
Newton's Second Law for a System of Particles — kyun momentum definition ko anchor karta hai.
Conservation of Linear Momentum — "no external force" wale edge cases.
Collisions — Elastic and Inelastic — jahan COM frame kaam aata hai.
Centre of Mass of Continuous Bodies — $\sum \to \int$ limit case.
Weighted Average and Moments — "why" items ke peeche see-saw analogy.