Centre of mass — derivation for common shapes (rod, triangle, semicircle, hemisphere): Question bank

1.4.10 · D5 · Physics › Momentum & Collisions › Centre of mass — derivation for common shapes (rod, triangle

Prerequisites jo pehle dekh lene chahiye: Centre of mass — definition and system of particles, Integration as continuous summation, Symmetry arguments in physics, Centroid vs centre of mass vs centre of gravity.

True or false — justify

True/False items — har answer mein reason dena zaroori hai, sirf haan/nahi kabhi nahi.

Ek uniform rod ka COM hamesha rod ke andar hota hai.

True — ek seedhi uniform rod ke liye

0

aur

L

ke beech positions ka mass-weighted average

L /2

par aana hi chahiye, jo rod ka hi ek point hai.

Ek uniform L-shaped bent wire ka COM material ke bahar (empty space mein) ho sakta hai.

True — COM ek average position hai, body ka koi physical point nahi; concave/bent shapes ke liye yeh average uss region mein pad sakta hai jahan koi mass nahi, jaise "L" ke corner-gap mein.

Ek uniform body ke liye, centre of mass aur geometric centroid ek hi hote hain.

True — jab density constant ho toh woh

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

mein upar-neeche cancel ho jaati hai, aur sirf ek purely geometric average bachta hai, jo centroid ki definition hai (Centroid vs centre of mass vs centre of gravity).

Same radius

R

wali semicircular wire aur semicircular disc ka COM ek hi hota hai.

False — wire ka saara mass rim par hai (

\frac{2 R}{π} \approx 0.64 R

), jabki disc ka area centre ke paas zyada hai jo average ko neeche

\frac{4 R}{3 π} \approx 0.42 R

tak kheenchta hai. Alag

d m

distributions alag answers dete hain.

Ek triangular lamina aur ek triangular wire frame (teen rods) ka COM ek hi hota hai.

False — lamina ka COM vertex-average (medians ka intersection) hota hai, lekin frame ka COM teen edge midpoints ka mass-weighted average hota hai, jo generally alag point hota hai jab tak triangle equilateral na ho.

Ek solid hemisphere ka COM bilkul uski height ke aadhe par hota hai.

False — yeh

\frac{3 R}{8} = 0.375 R

par hota hai, midpoint se neeche, kyunki flat face ke paas wale chaude slabs zyada mass carry karte hain aur average ko neeche kheenchte hain.

Hemispherical shell ka COM (

R /2

) solid hemisphere ke COM (

3 R /8

) se zyada uuncha hai.

True — shell apna saara mass curved surface par bahar rakhti hai, flat face se average par zyada door, isliye uska balance point mass-filled solid se zyada uuncha hota hai.

Ek uniform rod ka mass double karne se uska COM shift ho jaata hai.

False — COM position ek mass-weighted average hai; har

d m

ko same factor se scale karna numerator aur denominator mein cancel ho jaata hai, isliye location unchanged rehti hai.

Agar kisi body mein symmetry ka ek plane ho, toh uska COM us plane par hi hona chahiye.

True — ek taraf ke har mass element ke liye doosri taraf ek mirror-image element hota hai; unke off-plane coordinate par contributions pairs mein cancel ho jaate hain, average ko plane par force karte hue (Symmetry arguments in physics).

Triangle ka COM depend karta hai ki aap kaunsi side ko "base" keh rahe hain.

False — woh point teen medians ka intersection hai (vertex average

\frac{1}{3} (r_{1} + r_{2} + r_{3})

); "

h /3

from base" description usi fixed point ko ek chosen edge se describe karti hai.

Spot the error

Har item ek galat line of reasoning batata hai; reveal mein flaw aur fix dono hain.

"Semicircular disc ke liye maine shells par

d m = σ d r

use kiya."

Galat — yeh 1D linear pattern copy karta hai. Ek half-ring shell ka area

d A = π r d r

(length

\times

thickness) hota hai, isliye

d m = σ π r d r

; aapko slice ki area geometry include karni hi hogi.

"Triangle ka COM

h /2

par hai, jaise rod

L /2

par tha."

Galat — rod ki constant width hai isliye mass evenly spread hai, lekin triangle apex ki taraf sankra hota hai, use bottom-heavy banata hai. Sahi height

h /3

hai.

"Maine semicircular wire ke liye safe rehne ko

x

aur

y

dono integrate kiye."

Wasteful hai, galat nahi — flat side

x

-axis par hai isliye shape

x = 0

ke baare mein symmetric hai, to

x_{c m} = 0

symmetry se turant mil jaata hai; sirf

y_{c m}

ko integration chahiye.

"Solid hemisphere ke liye maine slice mass

d m = ρ d z

set kiya."

Galat — height

z

par ek slab radius

R^{2} - z^{2}

ki full disc hai, isliye

d V = π (R^{2} - z^{2}) d z

aur

d m = ρ π (R^{2} - z^{2}) d z

. Aapne disc ka area factor drop kar diya.

"Semicircular arc element ki length

d l = d θ

hai."

Galat — arc length radius times angle hoti hai,

d l = R d θ

R

bhoolne se density

λ = M / (π R)

aur final answer

\frac{2 R}{π}

dono break ho jaate hain.

"Disc COM nikalne ke liye maine har half-ring ko uski apni radius

r

se weight kiya."

Galat — har half-ring ka COM

r

par nahi balki uski apni height

\frac{2 r}{π}

par hai (wire case ka result). Har shell ke mass ko

r

se nahi

\frac{2 r}{π}

se weight karo.

"Density

σ

ek uniform lamina ke COM ko affect karti hai, isliye maine use final answer mein rakha."

Galat — uniform body ke liye

σ

constant hai aur

\int x d m

aur

\int d m

ke beech cancel ho jaati hai; result purely geometric hai, usme koi

σ

nahi hota.

"Triangle mein height

y

par ek strip ki width

b

har jagah hai."

Galat — similar triangles ke zariye width linearly shrink hoti hai:

w (y) = b (1 - \frac{y}{h})

, apex par zero. Constant

b

use karne par galti se

h /2

milega.

Why questions

Conceptual "why" prompts — reveal mein mechanism hai, sirf fact nahi.

Ek continuous body ke liye sum

\sum m_{i} x_{i} / \sum m_{i}

ko integral se replace karna kyun kaam karta hai?

Ek continuous body infinitely many infinitesimal particles

d m

ka limit hai; discrete average exactly

\frac{\int x d m}{\int d m}

ban jaata hai jaise summation integration ban jaati hai (Integration as continuous summation).

Integrate karne se pehle origin aur axes clever tarike se choose karne ki takleef kyun uthate hain?

Ek accha choice symmetry ko ek ya do coordinates immediately zero kar deta hai aur density function ko simple banata hai, isliye ek messy multi-variable integral ki jagah ek clean 1D integral integrate karte hain.

Triangle ka COM teen vertices ka average kyun hota hai?

Kisi bhi side ke parallel strips mein area ko split karke integrate karne par us side se

h /3

milta hai; teen sides ke liye aisa karne par point medians ke intersection par pin ho jaata hai, jo

\frac{1}{3} (r_{1} + r_{2} + r_{3})

ke barabar hai.

Filled disc ka COM (

\frac{4 R}{3 π}

) wire ke COM (

\frac{2 R}{π}

) se neeche kyun aata hai?

Disc ka flat diameter ke paas kaafi zyada area hai (chhota

y

), aur woh neeche wale shells mass-weighted average ko neeche kheenchte hain, unlike wire jiska mass poora bahar wale uunche rim par hai.

Momentum conservation sirf COM ko kyun track karta hai, internal motion ko ignore karta hai?

Internal forces action–reaction pairs mein aate hain jo cancel ho jaate hain, isliye sirf external forces COM ko move karti hain; system ka total momentum

M v_{c m}

hai chahe parts kaise bhi hilte rahein (Momentum conservation).

Hum semicircular disc ko un half-rings se build kyun kar sakte hain jinke COM hum pehle se jaante hain?

Whole apne shells ka sum hai; har shell ko uski apni COM height

\frac{2 r}{π}

par point mass maankar aur un points ko average karke exact disc COM milta hai — re-derive karne ki jagah reuse karo.

Solid-hemisphere calculation mein

π

cancel kyun ho jaata hai lekin semicircle answers mein appear karta hai?

Hemisphere mein numerator aur denominator dono disc area se

π

carry karte hain, isliye cancel ho jaata hai; semicircle mein

π

sirf total length/area

π R

ke through aata hai, isliye

\frac{2 R}{π}

aur

\frac{4 R}{3 π}

mein survive karta hai.

Ek heavier particle COM ko apni taraf kyun "pull" karta hai?

Weighted average

\frac{\sum m _{i} x _{i}}{\sum m _{i}}

mein uska bada

m_{i}

uski position ko zyada strongly multiply karta hai, isliye balance point us direction mein shift ho jaata hai — exactly see-saw effect.

Edge cases

Boundary aur degenerate inputs — ensure karo ki koi bhi scenario surprise na kare.

Ek rod ka COM kahan hoga jiska density ek end se linearly badhti hai,

λ (x) = k x

L /2

par nahi — mass dense end ki taraf bottom-heavy hai, isliye

x_{c m} = \frac{\int _{0}^{L} x k x d x}{\int _{0}^{L} k x d x} = \frac{2 L}{3}

, heavier end ki taraf shift.

Jab triangle degenerate hokar thin sliver ban jaata hai (apex angle

\to 0

), triangle COM formula

h /3

ka kya hota hai?

Yeh hold karta rehta hai —

h /3

ek triangle ke liye scale- aur shape-independent hai; needle-thin triangle ka bhi COM base se height ke ek-tehai upar hota hai.

Jab semicircle "open out" hokar full circle ban jaata hai, COM

y

kya ho jaana chahiye?

Zero — ek full ring ya disc upar-neeche symmetric hai, isliye woh upward pull jo

\frac{2 R}{π}

deta tha ek equal downward half se cancel ho jaata hai,

y_{c m} \to 0

Hemisphere

R \to 0

ke limit mein, COM kahan hai?

Origin par —

\frac{3 R}{8} \to 0

; ek vanishingly small body apna COM apne hi point par collapse kar leta hai, jaise koi bhi average karta hai.

Agar aap ek symmetric plate ke geometric centre par exactly ek chhota hole kaadte hain, kya COM shift hota hai?

Nahi — removed piece COM par hi baitha hai, isliye use hatane se mass-weighted average shift nahi hoti (COM ke baare mein uski "leverage" zero hai).

Kya do alag balls ke system ka COM kisi bhi ball ke andar hota hai?

Zaroor nahi — yeh unhe milane wali line par mass-weighted point par hota hai, jo generally unke beech empty space mein hota hai, heavier ball ke zyada paas.

Kya ek bahut tall object ke liye non-uniform gravitational field mein "centre of gravity" aur COM ek hi hote hain?

Nahi — COM purely ek mass average hai, jabki centre of gravity local

g

se weight karta hai; yeh sirf uniform field mein coincide karte hain, aur jab

g

body across vary karta hai toh alag ho jaate hain (Centroid vs centre of mass vs centre of gravity).

Ek uniform full circular disc vs ek uniform full circular ring ka COM kahan hota hai?

Dono ka centre par — full circular symmetry COM ko geometric centre par force karti hai chahe mass rim par ho ya area par spread ho.