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

1.4.10 · D2 · Physics › Momentum & Collisions › Centre of mass — derivation for common shapes (rod, triangle

Step 1 — See-saw: "balance point" ka asli matlab

KYA HAI. Ek plank par do bacche bithaao. Ek halka baccha mass $m_{1}$ ke saath position $x_{1}$ par, aur ek bhaari baccha mass $m_{2}$ ke saath position $x_{2}$ par. Pivot kahan rakhoge taaki plank tip na kare?

KYUN. Yahi sab kuch ka beej hai. Ek plank tab balance hoti hai jab dono sides pivot ke baare mein barabar turning effect exert karti hain. Ek bacche ka turning effect asliyat mein uska weight hota hai — uski mass $m$ times gravity $g$ (Zameen ka khichwav, sabke liye same number) — times pivot se kitni dur woh baith rahi hai. Woh product $m g d$ moment (ya torque) kehlata hai. "Tip nahi hoga" ka matlab hai left moment equals right moment. Aao uss sentence ko ek formula mein badlein aur dekhein $g$ kaise gayab ho jaata hai.

PICTURE. Figure mein, bhaari baccha pivot ko apni taraf kheench raha hai. Balance point beech mein nahi hoti — woh mass ki taraf jhukti hai.

Figure — Centre of mass — derivation for common shapes (rod, triangle, semicircle, hemisphere)

Pivot ki position ko $x_{c m}$ kahte hain. Baccha 1 pivot se $(x_{1} - x_{c m})$ door baitha hai; baccha 2 $(x_{2} - x_{c m})$ door. Balance = dono moments cancel ho jaate hain: $bacche 1 ka moment m_{1} g (x_{1} - x_{c m}) + bacche 2 ka moment m_{2} g (x_{2} - x_{c m}) = 0$ Har term mein same factor $g$ hai, isliye hum poori equation ko $g$ se divide kar sakte hain — gravity cancel ho jaati hai aur balance point par kabhi asar nahi dalti. (Yahi exact reason hai ki centre of mass aur centre of gravity uniform gravity mein coincide karte hain — dekho Centroid vs centre of mass vs centre of gravity.) Jo bachta hai woh pure mass hai: $m_{1} (x_{1} - x_{c m}) + m_{2} (x_{2} - x_{c m}) = 0.$ Ab $x_{c m}$ solve karo. Expand karo aur collect karo: $m_{1} x_{1} + m_{2} x_{2} - (m_{1} + m_{2}) x_{c m} = 0 ⟹ x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}} .$

Agar bacche barabar hote ( $m_{1} = m_{2}$ ) toh yeh collapse hokar $\frac{x _{1} + x _{2}}{2}$ ban jaata, plain middle — mass-weighting exactly unequal bacchon ke liye correction hai. Poori force picture ke liye dekho Centre of mass — definition and system of particles.

Step 2 — Kuch bacchon se bheed tak: $\sum$ sign

KYA HAI. Zyada particles add karo. Recipe nahi badlati — har particle apna $m_{i} x_{i}$ upar contribute karta hai, aur apna $m_{i}$ total mein.

KYUN. Hum bahut jald infinitely many particles (ek solid body atoms ki dense bheed hai) ki baat karne waale hain. Infinity par jaane se pehle, hum finite bheed ko clearly likhte hain. Wahi "moments pivot ke baare mein cancel hote hain" argument (jisme $g$ phir cancel ho jaata hai), $N$ particles ke liye kiya, neeche wali sum deta hai.

PICTURE. Ek line par paanch particles; har ek $x_{i}$ par $m_{i}$ height ki ek stack contribute karta hai. COM "stacks ka centre" hai.

Step 3 — Ek solid body ko slice karna: $d m$ aur $\int$ ka matlab

KYA HAI. Ek rod paanch bacche nahi hoti — yeh mass ka continuous felaav hai. Isliye hum ise itne patle slices mein kaatte hain ki har slice basically ek point particle ho. Ek slice ek tiny mass le jaati hai jise hum $d m$ kehte hain ("mass ka ek chhota sa tukda"), position $x$ par baitha.

KYUN yeh tool — integral? $\sum$ ek countable pile of particles ke liye kaam karta hai. Lekin hamare slices infinitely thin aur infinitely many hain. Integral sign $\int$ exactly woh tool hai jo "infinitely many infinitely small pieces add karo" ke liye invent kiya gaya tha — dekho Integration as continuous summation. Yeh us sawaal ka jawab deta hai jo plain $\sum$ nahi de sakta: ek smooth continuum ko total kaise karte ho?

PICTURE. Rod vertical slivers mein kati hui; ek sliver highlighted, width $d x$ , left end se $x$ distance par baitha.

Ek slice ka $d m$ kitna bada hai? Yeh is par depend karta hai ki mass kaise spread hai. Line of mass ke liye hum linear density $λ$ (Greek "lambda") use karte hain = mass per unit length, $λ = M / L$ , isliye $d x$ length ki ek slice mein $d m = λ d x$ mass hai. Do cousins baad mein aate hain:

Flat sheet ("lamina") ke liye hum surface density $σ$ (Greek "sigma") use karte hain = mass per unit area, isliye $d A$ area ki ek slice mein $d m = σ d A$ mass hai.
Solid 3-D body ke liye hum volumetric density $ρ$ (Greek "rho") use karte hain = mass per unit volume, isliye $d V$ volume ki ek slab mein $d m = ρ d V$ mass hai.

Har case mein density hai "ek unit cheez mein kitni mass packed hai", aur — uniform body ke liye top aur bottom dono mein same hone ki wajah se — yeh cancel ho jaayegi. Is page par hum $λ$ (line) aur $σ$ (sheet) use karte hain; $ρ$ (solid) yahan define kiya gaya hai taaki Step 7 mein "density" shabd ka matlab ho.

$d m = mass per length λ slice ki length d x .$

Step 4 — Rod nikal aata hai (aur density kyun vanish ho jaati hai)

KYA HAI. Rod ko master formula mein daalo. Origin left end par, rod $x = 0$ se $x = L$ tak — isliye limits $0$ se $L$ hain.

KYUN. Yeh sabse simple possible shape hai, isliye yahan hum machine check karte hain ki kaam karti hai isse pehle ki hum mushkil shapes par trust karein.

PICTURE. Wahi sliced rod, ab running position $x$ marked ke saath aur answer $L /2$ bilkul centre par flagged.

$x_{c m} = \frac{\int _{0}^{L} x ( λ d x )}{\int _{0}^{L} λ d x} = \frac{λ \int _{0}^{L} x d x}{λ \int _{0}^{L} d x} = \frac{λ \cdot \frac{L ^{2}}{2}}{λ \cdot L} = \frac{L}{2}$

$\int_{0}^{L} x d x = \frac{L ^{2}}{2}$ — mass-weighted total, line $y = x$ ke neeche area.
$\int_{0}^{L} d x = L$ — kul length, jise $λ$ se multiply karo toh total mass $M$ milti hai.
$λ$ top aur bottom cancel ho jaata hai. Yeh bahut badi baat hai: ek uniform body ke liye answer purely shape ke baare mein hai, na ki woh kitni bhaari hai. Woh shape-only balance point centroid hai — dekho Centroid vs centre of mass vs centre of gravity.

Step 5 — Triangle: base par mass kyun pile hoti hai

KYA HAI. Ek flat triangle (ek "lamina"), base $b$ , height $h$ . Origin ko base ke mid-point par fix karo, jisme $x$ -axis base ke saath aur $y$ -axis seedha apex ki taraf point kare. Toh base $x = - \frac{b}{2}$ (left corner) se $x = + \frac{b}{2}$ (right corner) tak $y = 0$ par chalti hai, aur apex $(x, y) = (0, h)$ par baitha hai. Khaas taur par $y = 0$ base hai aur $y = h$ apex hai — har height neeche se base se upar measure ki jaati hai. Triangle ko thin horizontal strips mein base ke parallel kaato; height $y$ par ek strip ek patla rectangle hai.

KYUN horizontal strips? Kyunki har aisi strip ek single clean height $y$ par hoti hai, isliye woh ek single particle ki tarah behave karti hai us height par. Yeh 2D shape ko strips ka 1D stack mein convert karta hai — wapas Step 3 ki machinery par. Yeh ek slice-cleverly move hai.

PICTURE. Base ke paas ek wide strip, apex ke paas ek chhoti strip — tum dekh sakte ho ki base strips zyada area (zyada mass) carry karti hain. Strip ki width height ke saath linearly shrink hoti hai.

Pehle, horizontal coordinate — free mein. Kyunki origin mid-base par hai aur apex seedhe uske upar hai, triangle $y$ -axis ( $x = 0$ ) ke baare mein left↔right mirror image hai. Har strip us axis par centred hai, isliye har strip ka apna balance point uske middle par, $x = 0$ par baitha hai. Is symmetry ki wajah se poore triangle ka COM us axis par hai: $x_{c m} = 0 (the mid-base line).$ Toh hamen sirf height $y_{c m}$ nikalni hai, base se measure ki gayi.

Similar triangles se, height $y$ par ek strip ki width hai $w (y) = b (1 - \frac{y}{h})$

$y = 0$ par (base): $w = b$ (poori width, sabse bhaari).
$y = h$ par (apex): $w = 0$ (vanish ho jaata hai).

Iski mass hai $d m = σ w (y) d y$ , jahan $σ$ (surface density, Step 3 mein define kiya) wahi role play karta hai jo $λ$ ne kiya tha. Kyunki $y = 0$ base hai aur $y = h$ apex hai, strips poore triangle ko cover karti hain jab $y$ $0 \to h$ se chalti hai — yahi limits hain:

$y_{c m} = \frac{\int _{0}^{h} y σ b ( 1 - \frac{y}{h} ) d y}{\int _{0}^{h} σ b ( 1 - \frac{y}{h} ) d y} = \frac{\int _{0}^{h} ( y - \frac{y ^{2}}{h} ) d y}{\int _{0}^{h} ( 1 - \frac{y}{h} ) d y}$

$σ$ aur $b$ cancel ho jaate hain (woh top aur bottom dono ko equally multiply karte hain). Ab chaar chhote integrals ek ek karke karo, single rule $\int_{0}^{h} y^{n} d y = \frac{h ^{n + 1}}{n + 1}$ use karke ( $y^{n}$ ke neeche area):

$\int_{0}^{h} y d y = \frac{h ^{2}}{2}$ — straight line $y$ ke neeche area.
$\int_{0}^{h} y^{2} d y = \frac{h ^{3}}{3}$ , isliye $\int_{0}^{h} \frac{y ^{2}}{h} d y = \frac{1}{h} \cdot \frac{h ^{3}}{3} = \frac{h ^{2}}{3}$ .
$\int_{0}^{h} 1 d y = h$ — plain length.
$\int_{0}^{h} \frac{y}{h} d y = \frac{1}{h} \cdot \frac{h ^{2}}{2} = \frac{h}{2}$ .

Yeh pieces substitute karo: $y_{c m} = \frac{\frac{h ^{2}}{2} - \frac{h ^{2}}{3}}{h - \frac{h}{2}} = \frac{\frac{h ^{2}}{6}}{\frac{h}{2}} = \frac{h}{3}$

Numerator $\frac{h ^{2}}{2} - \frac{h ^{2}}{3} = \frac{3 h ^{2} - 2 h ^{2}}{6} = \frac{h ^{2}}{6}$ : heights ko strip mass se weight kiya gaya.
Denominator $h - \frac{h}{2} = \frac{h}{2}$ : total (mass $σ b$ ki units mein).
Result $h /3$ : base se ek-tihaayi upar, exactly jaisa "bottom-heavy" picture ne promise kiya tha. $h /2$ nahi — upar ka narrowing top ko kill karta hai.

Step 6 — Semicircular wire: angles aur $sin$ laana

KYA HAI. Ek bent wire, half-ring radius $R$ ki, flat side $x$ -axis par, upar arch karti hui. Hum iski balance height chahte hain.

KYUN naया coordinate? Wire curved hai, isliye $x$ ya $y$ se slice karna ugly hai — ek tiny piece ko sabse naturally us angle $θ$ se locate kiya jaata hai jo woh $x$ -axis ke saath banata hai. Hum position $sin$ aur $cos$ se measure karte hain kyunki yahi exactly woh tools hain jo ek angle ko circle par height aur width mein turn karte hain.

PICTURE. Arc jisme ek tiny piece $d l$ angle $θ$ par marked hai; iski height $y = R sin θ$ ek vertical drop ki tarah piece se axis tak drawn hai.

$d θ$ angle ki tiny arc ki length $d l = R d θ$ hai (arc = radius × angle). Total length $π R$ hai, isliye $λ = \frac{M}{π R}$ aur $d m = λ d l = \frac{M}{π R} R d θ = \frac{M}{π} d θ .$ Piece ki height $y = R sin θ$ hai — $sin θ$ $R$ ka woh fraction hai jo height mein count hota hai. Angle $0$ se $π$ tak sweep karta hai poori half-ring cover karne ke liye, isliye yahi limits hain.

$y_{c m} = \frac{1}{M} \int_{0}^{π} (R sin θ) \frac{M}{π} d θ = \frac{R}{π} \int_{0}^{π} sin θ d θ = \frac{R}{π} [- cos θ]_{0}^{π} = \frac{R}{π} (1 + 1) = \frac{2 R}{π}$

$\int_{0}^{π} sin θ d θ = 2$ : saare arc pieces ka total "height-pull".
$\frac{2 R}{π} \approx 0.64 R$ — kaafi upar, kyunki wire ki saari mass rim par apex ke paas baithti hai.

Step 7 — Edge aur degenerate cases: kya machine toot jaati hai?

KYA HAI. Formula ko limits par test karo, jahan formulas explode karna pasand karte hain.

KYUN. Ek derivation jis par tum trust karte ho use corners survive karne chahiye: zero size, ek point par saari mass, ek symmetry axis. Parent note inhe state karta hai; yahan hum dekhte hain ki yeh kyun hold karte hain.

PICTURE. Teen panels: (a) ek degenerate zero-height triangle rod mein collapse hota hua, (b) ek symmetric shape apna COM axis par pinned kiye, (c) ek point mass jahan COM = the point.

Worked example Teen checks

(a) $h \to 0$ wala triangle. Yeh ek line segment mein flatten ho jaata hai. $y_{c m} = h /3 \to 0$ — COM line par sink ho jaata hai, jaisa hona chahiye. Division-by-zero nahi: bottom $h /2$ top ke saath saath shrink karta hai. (b) Symmetry. Wire ke liye, $+ x$ par har piece ka $- x$ par ek mirror twin hota hai. Unki $x$ -contributions pairs mein cancel hoti hain, $x_{c m} = 0$ kisi bhi integration se pehle force karte hain. Free symmetry hamesha pehle spend karo. (c) Ek akela point mass. Yahan poora "body" $x_{0}$ par single particle hai. Formula deta hai $\frac{\int x d m}{\int d m} = \frac{x _{0} M}{M} = x_{0}$ — COM exactly point par baitha hai, jaisa common sense maangta hai. (Alag se, notice karo ki koi bhi uniform density — $λ$ , $σ$ , ya Step 3 ki volumetric $ρ$ — hamesha top aur bottom par aati hai, isliye cancel ho jaati hai: COM value kabhi is par depend nahi karti ki kitni mass hai, sirf kahan hai. Yahan "cancels" density ko refer karta hai, COM coordinate ke zero jaane ko nahi.)

Ek-picture summary

Upar sab kuch ek sentence hai chaar frames mein: har position ko uski mass se weight karo, sum karo, divide karo. Rod, triangle, aur wire sirf is mein differ karte hain ki ek slice $d m$ kaisi shaped hai.

Ab tak har shape ne ek coordinate ( $x$ , ya $y$ , ya ek height) use kiya kyunki symmetry ne baaki ko kill kar diya. Generally ek point in space ko apne saare coordinates ek saath chahiye. Hum unhe ek symbol mein bundle karte hain $r = (x, y, z)$ — position vector, origin se point tak ka ek arrow. Master formula ko har coordinate par separately apply karna, $x_{c m} = \frac{\int x d m}{M}$ , $y_{c m} = \frac{\int y d m}{M}$ , $z_{c m} = \frac{\int z d m}{M}$ , exactly wahi hai jo neeche wali single vector line package karti hai:

Recall Feynman: poora walk plain words mein

Socho ek stick ko apni ungli par balance karna. Tumhari ungli wahan honi chahiye jahan left wali cheez exactly utni hi strongly kheenche jitni right wali — bhaari cheez zyada kheenchti hai, isliye ungli uski taraf slide karti hai. Woh "fair balance" spot centre of mass hai. Gravity har bit par equally kheenchti hai, isliye woh sum se bilkul nikal jaati hai — sirf mass kahan baithe hai maayene rakhta hai. Do bacchon ke see-saw ke liye yeh ek simple weighted average hai, aur tum ise prove kar sakte ho yeh demand karke ki ungli ke baare mein do turning-effects cancel ho jaayein. Ek solid object sirf bahut bade crowd of tiny bacche hain, isliye kuch numbers add karne ki jagah hum integral use karte hain — infinitely many tiny pieces add karne ki machine. Object ko slivers mein kaato, note karo ki har sliver ki kitni mass hai aur kahan hai, multiply karo, saare add karo, aur total mass se divide karo. Ek plain stick ke liye fair spot middle hai. Triangle ke liye wide bottom bhaari hota hai, isliye spot ek-tihaayi upar hota hai (aur symmetry se left-right bilkul centre mein). Ek half-circle mein bent wire ke liye saara weight rim par ride karta hai, isliye spot upar float karta hai, radius ka kaafi do-tihaayi. Har baar same recipe — sirf ek sliver ki shape change hoti hai.

Recall Self-test

Continuous body ke liye $\sum m_{i} x_{i}$ ki jagah kya aaya? ::: $\int x d m$
COM formula mein gravity $g$ kyun nahi aata? ::: Yeh har moment ko equally multiply karta hai, isliye balance equation ko $g$ se divide karne par yeh remove ho jaata hai — sirf mass aur position bachte hain.
Master formula apply karne se pehle, kya do cheezein fix karni hain? ::: Measure karne ke liye ek origin, aur integration limits jo poore body ko cover karein.
Uniform rod ke liye $λ$ kyun cancel hota hai? ::: Yeh top aur bottom par barabar baitha hai; uniform bodies ek shape-only (centroid) answer deti hain.
Triangle ke liye $h /3$ kyun na ki $h /2$ ? ::: Base ke paas strips wider (zyada mass) hoti hain, isliye average height neeche khiich jaati hai.
Triangle COM horizontally kahan hai, aur kyun? ::: Mid-base line $x = 0$ par — left-right symmetry ise wahan pin karti hai.
Kaun sa tool arc angle ko height mein turn karta hai, aur kyun? ::: $sin θ$ , kyunki $R sin θ$ circle par ek point ka vertical part hota hai.
Symmetry kaam kaise bachata hai? ::: Ek mirror axis off-axis contributions ko pairs mein cancel karta hai, isliye woh coordinate instantly $0$ ho jaata hai.

Connected ideas: 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 · Moment of inertia — derivation for common shapes · Momentum conservation