Applications — mass, centre of mass, moments of inertia: Visual walkthrough

4.4.22 · D2 · Maths › Multivariable Calculus › Applications — mass, centre of mass, moments of inertia

Yeh page parent note se zyada detail mein jaata hai. Wahi recipe use karta hai lekin har line ko drawing tak slow kar deta hai.

Step 1 — "Flat plate" kya hoti hai, aur yahan "density" ka matlab kya hai?

KYA. Socho ek patli metal ki sheet table par flat padi hai. Hum seedha upar se dekhte hain — woh kuch region fill karti hai jise hum $R$ kehte hain (page par bas ek shape, jaise triangle ya disc).

KYUN. Balance ki baat karne se pehle, humein ek tarika chahiye yeh kehne ka ki "is jagah par itna stuff hai." Poori plate ke liye ek akela number kaafi nahi hai kyunki sheet ek corner mein thick aur heavy ho sakti hai aur doosri jagah thin aur light.

PICTURE. Figure dekho. Gray outline region $R$ hai. Shading density dikhati hai — dark matlab heavy, light matlab light.

Step 2 — Plate ko tiny tiles mein chop karo, har ek ka apna tiny mass

KYA. Hum $R$ ko ek grid of tiny rectangles mein kaat dete hain. Ek tile chuno. Woh position $(x, y)$ par baith hai, uski width $d x$ aur height $d y$ hai, toh uska area ek tiny number hai jise hum $d A = d x d y$ kehte hain.

KYUN. Poori plate mein density jagah-jagah badalti hai, isliye hum ek density use nahi kar sakte. Lekin ek tile itni choti ki density uske across barely change kare, hum density ko ek single number treat kar sakte hain. Yeh har tile ko ek aasaan point mass bana deta hai — woh cheez jo hum pehle se handle karna jaante hain.

PICTURE. Figure mein orange tile hamara chosen piece hai. Uska tiny mass hai (wahan ki density) × (uska area).

Step 3 — Saare tiny masses add karo: yahi total mass hai

KYA. $R$ mein har tile par $d m$ add karo. Infinitely many infinitely small pieces add karna exactly wahi hai jo symbol $\iint_{R}$ mean karta hai — region par ek double integral.

KYUN. Do integral signs kyun? Kyunki hamare tiles do directions mein phele hain — across ( $x$ ) aur upar ( $y$ ). Ek sum ek row sweep karta hai, doosra sum rows ko stack karta hai. Yeh machinery Double Integrals over General Regions se hai.

PICTURE. Green arrows sweep dikhate hain: pehle ek vertical strip fill karo ( $y$ mein sum), phir us strip ko across slide karo ( $x$ mein sum).

Step 4 — "Balance" ka matlab: leverage, sirf mass nahi

KYA. Ek seesaw lagao. Pivot ko vertical line $x = \overset{x}{ˉ}$ par rakho (ek position jo hum abhi nahi jaante — $ˉ$ bar matlab "balance value"). Position $x$ par ek tile pivot se $(x - \overset{x}{ˉ})$ doori par baith hai. Uska turning effect (torque) uska mass times woh doori hai.

KYUN. Pivot se door ek chota mass paas ke ek bade mass ko balance kar sakta hai — socho seesaw par do bachche. Toh sirf position ya sirf mass balance point nahi dhundh sakta; humein mass times pivot-se-doori chahiye. Woh product lever arm effect hai.

PICTURE. Seesaw: ek light tile door left par aur ek heavy tile right ke paas red pivot line ke baare mein balance karte hain.

Step 5 — Balance equation ko $\overset{x}{ˉ}$ ke liye solve karo

KYA. Step 4 ka sum do pieces mein split karo, kyunki $\overset{x}{ˉ}$ ek constant hai (har tile ke liye same) aur integral ke bahar pull kiya ja sakta hai.

KYUN. Hum $\overset{x}{ˉ}$ akele ek side par chahte hain. Integrals par algebra allowed hai: ek constant factor sum se bahar slip ho jaata hai, bilkul jaise $2 a + 2 b = 2 (a + b)$ .

$ise M_{y} kehte hain \iint_{R} x ρ d A - \overset{x}{ˉ} = m \iint_{R} ρ d A = 0$

Pehla blob ek moment hai — mass weighted by uski $x$ -position. Hum ise $M_{y}$ kehte hain (moment about the $y$ -axis, kyunki $x$ $y$ -axis se doori hai).
Doosra blob Step 3 ka mass $m$ hi hai, $\overset{x}{ˉ}$ bahar pull hoke.

PICTURE. Figure do blobs literally dikhata hai: left par $x \cdot d m$ bars ka stack, right par total mass $m$ .

$M_{y} - \overset{x}{ˉ} m = 0$ rearrange karne par:

Bilkul same argument se ek horizontal line $y = \overset{y}{ˉ}$ ke baare mein balance karte hue: $\overset{y}{ˉ} = \frac{M _{x}}{m} = \frac{\iint _{R} y ρ d A}{\iint _{R} ρ d A}, M_{x} = \iint_{R} y ρ d A .$

Step 6 — Poora worked case: triangle with $ρ = x + y$

KYA. Region $R$ : triangle with corners $(0, 0), (1, 0), (0, 1)$ , density $ρ (x, y) = x + y$ . Hum poori recipe run karte hain: $m$ dhundho, phir $M_{y}$ , phir $\overset{x}{ˉ}$ .

KYUN. Variable density wahan hai jahan density sach mein matter karti hai — balance point plain geometric centre se shift ho jaata hai. Yeh dikhata hai ki machinery kaam aa rahi hai.

PICTURE. Density shading hypotenuse ki taraf dark hoti hai (jahan $x + y$ sabse bada hai). Purple dot woh balance point hai jo hum compute karenge.

Ek fixed $x$ ke liye, tile column bottom edge $y = 0$ se slanted line $y = 1 - x$ tak jaati hai. Toh inner sum $y$ mein hai.

Mass. $m = \int_{0}^{1} \int_{0}^{1 - x} (x + y) d y d x .$ Inner: $\int_{0}^{1 - x} (x + y) d y = x (1 - x) + \frac{( 1 - x ) ^{2}}{2} = \frac{1 - x ^{2}}{2} .$ Outer: $\int_{0}^{1} \frac{1 - x ^{2}}{2} d x = \frac{1}{2} (1 - \frac{1}{3}) = \frac{1}{3} .$ Toh $m = \frac{1}{3}$ .

Moment $M_{y} = \iint x ρ d A$ . Inner: $\int_{0}^{1 - x} x (x + y) d y = x \cdot \frac{1 - x ^{2}}{2} .$ Outer: $\int_{0}^{1} \frac{x - x ^{3}}{2} d x = \frac{1}{2} (\frac{1}{2} - \frac{1}{4}) = \frac{1}{8} .$

Balance point. $\overset{x}{ˉ} = \frac{M _{y}}{m} = \frac{1/8}{1/3} = \frac{3}{8} .$ Kyunki $ρ = x + y$ $x$ aur $y$ mein symmetric hai, same computation deta hai $\overset{y}{ˉ} = \frac{3}{8}$ . Balance point hai $(\frac{3}{8}, \frac{3}{8})$ — heavy hypotenuse ki taraf push hua, plain centroid $(\frac{1}{3}, \frac{1}{3})$ par nahi.

Step 7 — Edge aur degenerate cases (kabhi surprise mat lo)

KYA / KYUN / PICTURE — teen failure-proofing scenarios.

Uniform density. Agar $ρ$ ek constant $c$ hai, toh woh top aur bottom se factor out ho jaata hai aur cancel ho jaata hai: $\overset{x}{ˉ} = \frac{c \iint x d A}{c \iint d A} = \frac{\iint x d A}{area} .$ Ab balance point pure geometry hai — centroid. Dekho Centroids and the Pappus Theorems.
Hole wali / concave region. Balance point material ke bahar bhi ja sakta hai (jaise crescent). Theek hai — $\overset{x}{ˉ}, \overset{y}{ˉ}$ phir bhi correct averages hain; koi nahi kehta pivot plate par hi baith e.
Everywhere zero density ( $ρ \equiv 0$ ). Tab $m = 0$ aur $\overset{x}{ˉ} = 0/0$ undefined hai — sahi baat hai: ek empty plate ka koi balance point nahi hota. Formula zero se divide karke warn karta hai.
Ek akela dominant heavy point (density spike). Jaise ek tiny region bahut zyada heavy ho jaata hai, $\overset{x}{ˉ}, \overset{y}{ˉ}$ us point ki taraf slide karte hain — limit mein balance point wahi point ban jaata hai. Mass-weighted average gracefully point-mass answer tak degrade ho jaata hai.

Step 8 — Polar version: same recipe, ek extra factor

KYA. Round regions ke liye hum polar coordinates mein switch karte hain: $x = r cos θ$ , $y = r sin θ$ . Ek polar tile curved wedge hai, rectangle nahi.

KYUN. Wedge ki do sides hain — ek radial length $d r$ aur ek arc length $r d θ$ (arc = radius × angle). Toh uska area hai $d A = radial d r \times arc r d θ = r d r d θ,$ nahi $d r d θ$ . Woh extra $r$ Jacobian factor hai Polar Coordinates and the Jacobian se.

PICTURE. Wedge apni short inner arc aur long outer arc ke saath — clearly centre se door zyada wide hai.

Wahi $r^{2}$ idea moments of inertia aur Rotational Kinetic Energy and Angular Momentum ko power karta hai; poora apparatus solids tak lift ho jaata hai Triple Integrals aur Change of Variables and Jacobians ke zariye.

Ek-picture summary

Ek canvas par poori derivation: $R$ ko tiles mein chop karo → har tile ek point mass hai $d m = ρ d A$ → $m$ ke liye unhe add karo → position se weight karo aur $M_{y}, M_{x}$ ke liye add karo → balance point $(\overset{x}{ˉ}, \overset{y}{ˉ})$ paane ke liye divide karo.

Recall Feynman retelling — plain words mein poora walkthrough

Socho ek unevenly cheesy pizza table par padi hai. Pehle main ise hundreds of tiny squares mein kaatta hoon. Har chota square itna chota hai ki uski cheese basically even hai, toh main keh sakta hoon "is square ka weight itna hai" — density times uska chota area. Saare squares add karo aur mujhe pizza ka total weight pata hai; woh mass hai. Ab mujhe woh fingertip spot chahiye jahan woh balance kare. Main ek pivot line imagine karta hoon aur har square se poochta hoon ki woh pivot se kitna door hai times apna weight ke barabar turning force se neeche dhako. Jab left pushes exactly right pushes cancel karen, woh pivot line balance line hai. Algebra karne par, balance $x$ nikalta hai $x$ ka weighted average — har square ki position counted according to woh kitna heavy hai. Left-right aur up-down dono karo aur mujhe balance point mil jaata hai. Agar cheese perfectly even hai toh weights matter nahi karte aur mujhe plain geometric centre milta hai. Agar pizza mein ek bite liya gaya hai, balance point empty air ke upar bhi float kar sakta hai — aur agar pizza hi nahi hai, toh simply koi balance point nahi hai, jo exactly wahi hai jo zero se divide karna mujhe bata raha hai. Round pizzas ke liye main pie-slice tiles use karta hoon, aur kyunki outer arc inner arc se longer hai, har slice-tile secretly ek extra factor of $r$ carry karta hai. Har baar same recipe: chop, weight, add, divide.