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

WHAT tool: parent note ka rod result $x_{cm}=L/2$. WHY: uniform linear density $\lambda =M/L$ constant hai, isliye $\frac{\int x\lambda dx}{\int \lambda dx}$ mein yeh cancel ho jaata hai — COM purely geometric hai, yaani centroid, seedha middle. $x_{cm}=\frac{L}{2}=\frac{1.2}{2}=\boxed{0.6\text{ m}}$

WHAT tool: wire $\to \frac{2R}{\pi }$, disc $\to \frac{4R}{3\pi }$. WHY alag hain: wire apna sara mass rim par rakhta hai; disc ka mass centre ke paas gather hota hai, jo average ko neeche kheenchta hai. Silhouette same hai, mass distribution alag hai → answer alag hai. (Yeh exactly parent note ka Error 1 wala trap hai.) $y_{\text{wire}}=\frac{2R}{\pi }=\frac{2(7)}{\pi }\approx 4.456\text{ cm}$ $y_{\text{disc}}=\frac{4R}{3\pi }=\frac{4(7)}{3\pi }\approx 2.971\text{ cm}$

WHAT slice: position $x$ par width $dx$ ka ek tukda $dm=\lambda (x)dx={\lambda }_0\frac{x}{L}dx$ mass carry karta hai. WHY sirf $L/2$ nahi: density constant nahi hai, isliye $\lambda$ cancel nahi hota — heavier (right) end average ko apni taraf kheenchta hai, isliye hum expect karte hain $x_{cm}>L/2$. $x_{cm}=\frac{{\int }_0^{L}x{\lambda }_0\frac{x}{L}dx}{{\int }_0^L{\lambda }_0\frac{x}{L}dx}=\frac{{\int }_0^L{x}^{2}dx}{{\int }_0^{L}xdx}=\frac{L^3/3}{L^2/2}=\boxed{\frac{2L}{3}}$ Jaise predict kiya tha, $\frac{2L}{3}>\frac{L}{2}$: average heavy end ki taraf jhukta hai.

WHAT slice: height $y$ par ek horizontal strip, jiska width $w(y)=b\left(1-\frac{y}{h}\right)$ similar triangles se (red strip dekho — apex ke paas yeh zero ho jaata hai). Phir $dm=\sigma w(y)dy=\sigma b\left(1-\frac{y}{h}\right)dy$. WHY horizontal strips: hum $y_{cm}$ chahte hain, isliye hum same height $y$ wale saare mass ko ek $dm$ mein group karte hain — ek slice ke andar $y$ vary karne ki zaroorat nahi. Step 1 — full integrals likho, har factor ko carry karte hue: $y_{cm}=\frac{{\int }_0^{h}ydm}{{\int }_0^{h}dm}=\frac{{\int }_0^{h}y\sigma b\left(1-\frac{y}{h}\right)dy}{{\int }_0^h\sigma b\left(1-\frac{y}{h}\right)dy}$ Step 2 — dono integrals se constants bahar nikalo: $\sigma$ aur $b$ $y$ par depend nahi karte, isliye har integral $\sigma b\times (\text{ek }y\text{-integral})$ ban jaata hai: $y_{cm}=\frac{\sigma b{\int }_0^{h}y\left(1-\frac{y}{h}\right)dy}{\sigma b{\int }_0^h\left(1-\frac{y}{h}\right)dy}$ Step 3 — WHY yeh vanish ho jaate hain: same factor $\sigma b$ numerator aur denominator dono mein hai, isliye exactly cancel ho jaata hai. Isliye answer purely geometric hai — yeh depend nahi kar sakta ki triangle kitna heavy ($\sigma$) ya kitna wide ($b$) hai. $y_{cm}=\frac{{\int }_0^{h}y\left(1-\frac{y}{h}\right)dy}{{\int }_0^h\left(1-\frac{y}{h}\right)dy}=\frac{h^2/2-h^2/3}{h-h/2}=\frac{h^2/6}{h/2}=\frac{h}{3}$ $y_{cm}=\frac{9}{3}=\boxed{3\text{ cm}}$ Note: yeh $b$ se independent $h/3$ hai — exactly isliye kyunki $\sigma$ aur $b$ Step 3 mein cancel ho gaye.

WHAT tool — subtraction (negative-mass) trick: punched plate ko (full disc) minus (removed small disc) ki tarah treat karo. Superposition ke liye, $x_{cm}=\frac{M_{\text{full}}{x}_{\text{full}}-M_{\text{hole}}{x}_{\text{hole}}}{M_{\text{full}}-M_{\text{hole}}}$ WHY yeh kaam karta hai: COM formula mass mein linear hai, isliye material remove karna = negative mass add karna. Dekho Centre of mass — definition and system of particles.

Surface density $\sigma$ lo. Mass $\propto$ area:

• Full disc: $M_{\text{full}}=\sigma \pi R^2$, centred at $x_{\text{full}}=0$.
• Hole: radius $R/2$, isliye $M_{\text{hole}}=\sigma \pi (R/2{)}^2=\frac{1}{4}\sigma \pi R^2$, centred at $x_{\text{hole}}=R/2$.

$x_{cm}=\frac{\sigma \pi R^2(0)-\frac{1}{4}\sigma \pi R^2\cdot \frac{R}{2}}{\sigma \pi R^2-\frac{1}{4}\sigma \pi R^2}=\frac{-\frac{1}{8}R}{\frac{3}{4}}=-\frac{R}{6}$ $x_{cm}=-\frac{4}{6}=\boxed{-0.667\text{ cm}}$ COM hole se door shift hota hai (solid side mein) — sensible hai, kyunki humne $+x$ side se mass remove kiya. Symmetry se $y_{cm}=0$.

WHAT tool: har uniform square ko apne centre (centroid) par ek point mass maano, phir mass-weighted average lo. Equal tiles → equal mass $m$; masses cancel ho jaate hain aur hum centres ka average lete hain. WHY: poore body ka COM nikalne ke liye har square ka mass uske COM par ek point se replace kiya ja sakta hai — yeh Centre of mass — definition and system of particles ka ek theorem hai.

Centres: $A=(a/2,a/2)$, $B=(a/2,3a/2)$, $C=(3a/2,a/2)$. $x_{cm}=\frac{a/2+a/2+3a/2}{3}=\frac{5a/2}{3}=\frac{5a}{6},y_{cm}=\frac{a/2+3a/2+a/2}{3}=\frac{5a}{6}$ $a=2$ ke saath: $x_{cm}=y_{cm}=\frac{5(2)}{6}=\boxed{1.667\text{ cm}}$. COM corner tile se thoda bahar hai, dono outer squares ne diagonally kheencha.

WHAT tool: do known centroids ko mass-weighted average se combine karo. Rectangle: area $A_1=2R\cdot R=2R^2$, mass $M_1=2\sigma R^2$, centroid apne middle $y_1=\frac{R}{2}$ par. Semicircle: area $A_2=\frac{\pi R^2}{2}$, mass $M_2=\frac{\sigma \pi R^2}{2}$. Uska apna COM $\frac{4R}{3\pi }$ apne flat edge ke upar hai, aur woh flat edge $y=R$ par hai. Isliye uski centroid height hai $y_2=R+\frac{4R}{3\pi }$. WHY $R$ add karo: disc ka $\frac{4R}{3\pi }$ apne base se measure hota hai, jo humne $y=R$ tak lift kiya hai. Sub-COMs ko hamesha common origin par re-anchor karo.

$y_{cm}=\frac{M_1{y}_1+M_2{y}_2}{M_1+M_2}=\frac{2\sigma R^2\cdot \frac{R}{2}+\frac{\sigma \pi R^2}{2}\left(R+\frac{4R}{3\pi }\right)}{2\sigma R^2+\frac{\sigma \pi R^2}{2}}$ $\sigma R^2$ cancel karo. Numerator $=R+\frac{\pi }{2}\left(R+\frac{4R}{3\pi }\right)=R+\frac{\pi R}{2}+\frac{2R}{3}=\frac{5R}{3}+\frac{\pi R}{2}$. Denominator $=2+\frac{\pi }{2}$. $y_{cm}=\frac{\frac{5}{3}+\frac{\pi }{2}}{2+\frac{\pi }{2}}R\approx \frac{3.238}{3.571}R\approx \boxed{0.907R}$ Glue line ($y=R$) se thoda neeche — heavier rectangle balance point ko low rakhta hai, lekin disc usse seam ke paas upar kheenchti hai.

WHAT tool: solid-hemisphere result $z=\frac{3R}{8}$ ke saath negative-mass subtraction. Hemisphere ka volume $\propto R^3$; use karo $V=\frac{2}{3}\pi R^3$.

• Big: $M_1=\rho \cdot \frac{2}{3}\pi R^3$, $z_1=\frac{3R}{8}$.
• Removed small (radius $R/2$): $M_2=\rho \cdot \frac{2}{3}\pi (R/2{)}^3=\rho \cdot \frac{2}{3}\pi \frac{R^3}{8}=\frac{M_1}{8}$, $z_2=\frac{3(R/2)}{8}=\frac{3R}{16}$.

$z_{cm}=\frac{M_1{z}_1-M_2{z}_2}{M_1-M_2}=\frac{M_1\frac{3R}{8}-\frac{M_1}{8}\frac{3R}{16}}{M_1-\frac{M_1}{8}}$ $M_1$ cancel karo: numerator $=\frac{3R}{8}-\frac{3R}{128}=\frac{48R-3R}{128}=\frac{45R}{128}$. Denominator $=\frac{7}{8}$. $z_{cm}=\frac{45R/128}{7/8}=\frac{45R}{128}\cdot \frac{8}{7}=\frac{45R}{112}\approx \boxed{0.402R}$ Plain $0.375R$ se zyada: humne neeche se mass scoop kiya (chota hemisphere flat face ke paas rehta hai), isliye average upar uth jaata hai.

WHAT parts: arc aur diameter, dono uniform linear density $\lambda$ ke saath.

• Arc: length $\pi R$, mass $M_a=\lambda \pi R$, uska COM $y_a=\frac{2R}{\pi }$ par (parent result #3).
• Diameter: length $2R$, mass $M_d=2\lambda R$, axis par rakhi hai isliye $y_d=0$.

$y_{cm}=\frac{M_a{y}_a+M_d{y}_d}{M_a+M_d}=\frac{\lambda \pi R\cdot \frac{2R}{\pi }+0}{\lambda \pi R+2\lambda R}=\frac{2\lambda R^2}{\lambda R(\pi +2)}=\frac{2R}{\pi +2}$ $y_{cm}=\frac{2R}{\pi +2}\approx \boxed{0.389R}$ Limiting discussion: diameter $y=0$ par baitha hai, isliye use add karne se COM sirf neeche ja sakta hai. Isliye $0<\underset{\approx 0.389R}{\underbrace{\frac{2R}{\pi +2}}}<\underset{\approx 0.637R}{\underbrace{\frac{2R}{\pi }}}.$ Pure arc ($\frac{2R}{\pi }$) upper bound hai (axis par koi dead weight nahi); lower bound $0$ hoga agar diameter par infinite mass hoti. "D" beech mein land karta hai — exactly jaise ek mass-weighted average apni extremes ke beech mein hona chahiye. Yeh Symmetry arguments in physics principle aur Centre of mass — definition and system of particles averaging law dono ek saath hain.

The argument: solid hemisphere mein material har depth par hai — flat face se pole tak, including flat face ke paas bahut zyada mass (bade base discs). Shell apna sara mass curved surface par rakhta hai, jo average par flat plane se zyada door baitha hai. Zyada door material ka mass-weighted average upar land karta hai. Isliye solid $<$ shell. Koi numbers nahi chahiye — bas "mass kahan rehta hai?"

Quick verification (shell) — WHAT slice: top se polar angle $\theta$ par ek ring ki height $z=R\cos \theta$ hai aur mass $dm=\sigma (2\pi R\sin \theta )(Rd\theta )$ hai (circumference × slant width × density). Step 1 — $z_{cm}$ likho har factor carry karte hue: $z_{cm}=\frac{{\int }_0^{\pi /2}(R\cos \theta )\sigma (2\pi R\sin \theta )(Rd\theta )}{{\int }_0^{\pi /2}\sigma (2\pi R\sin \theta )(Rd\theta )}$ Step 2 — constants bahar nikalo: $\cos \theta ,\sin \theta$ ke siwa har factor $\theta$ mein constant hai. Numerator mein $\sigma \cdot 2\pi R^2\cdot R=2\pi \sigma R^3$ hai; denominator mein $\sigma \cdot 2\pi R^2=2\pi \sigma R^2$ hai: $z_{cm}=\frac{2\pi \sigma R^3{\int }_0^{\pi /2}\cos \theta \sin \theta d\theta }{2\pi \sigma R^2{\int }_0^{\pi /2}\sin \theta d\theta }$ Step 3 — WHY constants ek single $R$ mein collapse hote hain: shared factor $2\pi \sigma R^2$ top aur bottom cancel ho jaata hai; sirf numerator ke $z=R\cos \theta$ wala extra $R$ bachta hai. Isliye $z_{cm}=R\times (\text{dono }\theta \text{-integrals ka ratio})$ — answer ek pure number times $R$ hona chahiye, jaise koi bhi length honi chahiye. $z_{cm}=R\frac{{\int }_0^{\pi /2}\cos \theta \sin \theta d\theta }{{\int }_0^{\pi /2}\sin \theta d\theta }=R\cdot \frac{1/2}{1}=\frac{R}{2}$ Wakai $0.5R>0.375R$. ✓