Exercises — Applications — mass, centre of mass, moments of inertia

Recipe "$1,r,r^2$" se seedha padho:

• (a) mass → $f=1$.
• (b) $M_x$ = $x$-axis ke baare mein moment → $x$-axis tak lever arm height $y$ hai, first power → $f=y$.
• (c) $M_y$ = $y$-axis ke baare mein moment → lever arm $x$ hai, first power → $f=x$.
• (d) $I_x$ = $x$-axis ke baare mein inertia → wahi lever arm $y$ lekin squared → $f=y^2$.
• (e) $I_y$ → $f=x^2$.
• (f) $I_0$ = origin ke baare mein inertia → lever arm $r$, squared → $f=r^2=x^2+y^2$.

Notice karo $I_0=\iint (x^2+y^2)\rho dA=I_x+I_y$ — flat plate ke liye perpendicular-axis theorem (L4.2 mein picture ke saath explain hua hai).

Tiles mein kaato; har tile ka mass $\rho dA=(xy)dxdy$ hai. Rectangle par add karo. Limits constant hain kyunki region ek box hai: $m={\int }_0^2{\int }_0^{3}xydydx.$ Bas itna hi L1 maangta hai — yeh pehchanna ki kaunsa factor $f(=1)$ aur kaunsi limits.

Inner integral ($x$ ko constant maano, $y$ mein integrate karo): ${\int }_0^{3}xydy=x\cdot \frac{y^2}{2}{|}_0^3=x\cdot \frac{9}{2}=\frac{9x}{2}.$ Inner pehle kyun? Ek fixed vertical strip mein $x$ chalti hai, $y$ upar se neeche sweep karta hai; hum us strip ko ek number mein collapse karte hain, phir $x$ sweep karte hain. Outer integral: ${\int }_0^2\frac{9x}{2}dx=\frac{9}{2}\cdot \frac{x^2}{2}{|}_0^2=\frac{9}{2}\cdot 2=9.$ $\boxed{m=9.}$

Region round hai, isliye polar coordinates natural tool hai (circle ki boundary flat line $r=a$ ban jaati hai). Substitute karo: $x=r\cos \theta ,y=r\sin \theta ,dA=rdrd\theta ,y^2=r^2{\sin }^2\theta .$ Upper half ⇒ $\theta$, $0$ se $\pi$ tak; $r$, $0$ se $2$ tak. $I_x={\int }_0^{\pi }{\int }_0^2(r^2{\sin }^2\theta )rdrd\theta ={\int }_0^{\pi }{\sin }^2\theta d\theta {\int }_0^2{r}^{3}dr.$ Separate kyun hota hai: integrand ek $\theta$-only piece aur ek $r$-only piece ka product hai rectangular $r\theta$-box par, isliye integrals split ho jaate hain. ${\int }_0^2{r}^{3}dr=\frac{r^4}{4}{|}_0^2=\frac{16}{4}=4,{\int }_0^{\pi }{\sin }^2\theta d\theta =\frac{\pi }{2}.$ $\boxed{I_x=4\cdot \frac{\pi }{2}=2\pi .}$

Region: $0\le x\le 2,0\le y\le 2-x$ (slanted line $x+y=2$ ke neeche, figure mein cyan edge).

Mass ($f=1$, times $\rho =y$): $m={\int }_0^2{\int }_0^{2-x}ydydx.$ Inner: ${\int }_0^{2-x}ydy=\frac{(2-x{)}^2}{2}$. Outer: $m={\int }_0^2\frac{(2-x{)}^2}{2}dx$. Substitution, explained: $u=2-x$ rakh lo. Tab $du=-dx$, yaani $dx=-du$. Jab $x=0$, $u=2$; jab $x=2$, $u=0$ — limits neeche $2\to 0$ run karti hain. $dx=-du$ ka minus sign aur downward limits milkar: ek definite integral ki limits palat do to sign palat jaata hai, aur woh flip cancel ho jaata hai, clean upward $0\to 2$ restore karta hai: ${\int }_{x=0}^{x=2}\frac{(2-x{)}^2}{2}dx={\int }_{u=2}^{u=0}\frac{u^2}{2}(-du)={\int }_{u=0}^{u=2}\frac{u^2}{2}du=\frac{1}{2}\cdot \frac{u^3}{3}{|}_0^2=\frac{1}{2}\cdot \frac{8}{3}=\frac{4}{3}.$

$M_x=\iint y\cdot \rho dA=\iint y\cdot ydA=\iint y^{2}dA$ (lever arm $y$ times density $y$): $M_x={\int }_0^2{\int }_0^{2-x}{y}^{2}dydx.$ Inner: ${\int }_0^{2-x}{y}^{2}dy=\frac{(2-x{)}^3}{3}$. Outer wahi $u=2-x$ ke saath (phir se $-du$ downward limits ko cancel karta hai): $M_x={\int }_0^2\frac{(2-x{)}^3}{3}dx={\int }_0^2\frac{u^3}{3}du=\frac{1}{3}\cdot \frac{u^4}{4}{|}_0^2=\frac{1}{3}\cdot \frac{16}{4}=\frac{4}{3}.$

Centre of mass height (recipe box se $\bar{y}=M_x/m$ use karo): $\bar{y}=\frac{M_x}{m}=\frac{4/3}{4/3}=1.$ $\boxed{m=\frac{4}{3},M_x=\frac{4}{3},\bar{y}=1.}$ Sanity check: triangle $y\in [0,2]$ par hai aur density $y$ ke saath badhti hai, isliye mass upar jhukti hai — $\bar{y}=1$ geometric centroid height ($2/3$) se upar baitha hai. Bilkul sahi.

$I_x\propto b^2$ aur $I_y\propto a^2$. Inertia badhta hai jitna mass axis se door hota hai. Mass $x=a=3$ tak phela hai ($y$-axis se door) lekin sirf $y=b=1$ tak ($x$-axis ke paas). Isliye $y$-axis ke baare mein spin karna mushkil hai: $I_y>I_x$. $\frac{I_y}{I_x}=\frac{a^2}{b^2}=\frac{9}{1}=9.$ $\boxed{I_y\text{ is }9\times \text{larger}.}$ Lesson: inertia us direction se dominate hoti hai jis direction mein mass door tak jaata hai — kyunki distance squared hoti hai.

Region: $0\le x\le 1,0\le y\le 1-x$. Origin ke baare mein lever arm squared $x^2+y^2$ hai: $I_0=\iint (x^2+y^2)\rho dA={\int }_0^1{\int }_0^{1-x}(x^2+y^2)(x+y)dydx.$ Integrand expand karo: $(x^2+y^2)(x+y)=x^3+x^{2}y+x{y}^2+y^3$. Inner integral $y$ mein $[0,1-x]$ par, $L=1-x$ likhke (antiderivative phir $y=L$ plug karo): ${\int }_0^L(x^3+x^{2}y+x{y}^2+y^3)dy=x^{3}L+\frac{x^2{L}^2}{2}+\frac{x{L}^3}{3}+\frac{L^4}{4}.$ Outer integral $x$ mein $[0,1]$ par. $L=1-x$ substitute karo aur har term integrate karo. Cleanest way: har term ko $L$ ki power mein convert karo swap $u=1-x$ use karke (so $x=1-u$, $dx=-du$, limits $x:0\to 1\Rightarrow u:1\to 0$, aur $-du$ phir se flipped limits cancel karta hai, ${\int }_0^1(\cdots )du$ deta hai): ${\int }_0^1{x}^{3}Ldx={\int }_0^1(1-u{)}^{3}udu,{\int }_0^1\frac{x^2{L}^2}{2}dx={\int }_0^1\frac{(1-u{)}^2{u}^2}{2}du,$ ${\int }_0^1\frac{x{L}^3}{3}dx={\int }_0^1\frac{(1-u)u^3}{3}du,{\int }_0^1\frac{L^4}{4}dx={\int }_0^1\frac{u^4}{4}du.$ Har ek Beta-type integral hai ${\int }_0^1{u}^p(1-u{)}^{q}du=\frac{p!q!}{(p+q+1)!}$. Evaluate karo: ${\int }_0^1(1-u{)}^{3}udu=\frac{1!3!}{5!}=\frac{6}{120}=\frac{1}{20},$ $\frac{1}{2}{\int }_0^1(1-u{)}^2{u}^{2}du=\frac{1}{2}\cdot \frac{2!2!}{5!}=\frac{1}{2}\cdot \frac{4}{120}=\frac{1}{60},$ $\frac{1}{3}{\int }_0^1(1-u)u^{3}du=\frac{1}{3}\cdot \frac{3!1!}{5!}=\frac{1}{3}\cdot \frac{6}{120}=\frac{1}{60},$ $\frac{1}{4}{\int }_0^1{u}^{4}du=\frac{1}{4}\cdot \frac{1}{5}=\frac{1}{20}.$ Add karo: $I_0=\frac{1}{20}+\frac{1}{60}+\frac{1}{60}+\frac{1}{20}=\frac{3}{60}+\frac{1}{60}+\frac{1}{60}+\frac{3}{60}=\frac{8}{60}=\frac{2}{15}.$ $\boxed{I_0=\frac{2}{15}.}$ Radius of gyration: $R_g=\sqrt{\frac{I_0}{m}}=\sqrt{\frac{2/15}{1/3}}=\sqrt{\frac{2}{15}\cdot 3}=\sqrt{\frac{2}{5}}\approx \mathrm{0.632.}$ $\boxed{R_g=\sqrt{2/5}\approx \mathrm{0.632.}}$ Interpretation: agar saari mass scrape karke origin ke around $R_g\approx 0.63$ radius ki ring mein chipka do, to woh real plate jitna hi spin resist karegi.

Region aur $\rho =x+y$ ki $x\leftrightarrow y$ symmetry se, $I_x=I_y$ hai. Isliye sirf ek chahiye. $I_x=\iint y^2(x+y)dA={\int }_0^1{\int }_0^{1-x}(x{y}^2+y^3)dydx.$ Inner ($L=1-x$): ${\int }_0^L(x{y}^2+y^3)dy=x\frac{L^3}{3}+\frac{L^4}{4}.$ Outer $x\in [0,1]$ par, wahi swap $u=1-x$ (so $x=1-u$): $I_x={\int }_0^1(x\frac{L^3}{3}+\frac{L^4}{4})dx=\frac{1}{3}{\int }_0^1(1-u)u^{3}du+\frac{1}{4}{\int }_0^1{u}^{4}du.$ First term: $\frac{1}{3}\cdot \frac{3!1!}{5!}=\frac{1}{3}\cdot \frac{6}{120}=\frac{1}{60}$. Second term: $\frac{1}{4}\cdot \frac{1}{5}=\frac{1}{20}=\frac{3}{60}$. $I_x=\frac{1}{60}+\frac{3}{60}=\frac{4}{60}=\frac{1}{15},I_y=\frac{1}{15}.$ Tab $I_x+I_y=\frac{1}{15}+\frac{1}{15}=\frac{2}{15}=I_0.$ ✓ Perpendicular-axis theorem hold karta hai — aur karna hi chahiye, kyunki yeh plate flat hai ($xy$-plane mein ek lamina). In identities ke geometric cousin ke liye Centroids and the Pappus Theorems dekho.

Round region + radial density ⇒ polar coordinates decisively win karti hain. $r$ se $1$ to $2$, $\theta$ se $0$ to $2\pi$, $dA=rdrd\theta$. Mass: $m={\int }_0^{2\pi }{\int }_1^2(\underset{\rho }{\underbrace{r}})\underset{dA}{\underbrace{rdrd\theta }}={\int }_0^{2\pi }{\int }_1^2{r}^{2}drd\theta .$ Inner: ${\int }_1^2{r}^{2}dr=\frac{r^3}{3}{|}_1^2=\frac{8-1}{3}=\frac{7}{3}$. Outer $\times 2\pi$: $m=\frac{7}{3}\cdot 2\pi =\frac{14\pi }{3}.$ Moment of inertia ($f=r^2$): $I_0={\int }_0^{2\pi }{\int }_1^2{r}^2\cdot r\cdot rdrd\theta ={\int }_0^{2\pi }{\int }_1^2{r}^{4}drd\theta .$ $r$'s count karo: ek $\rho =r$ se, ek $r^2$ (lever arm squared) se, ek Jacobian $dA=rdrd\theta$ se → $r^4$. Inner: ${\int }_1^2{r}^{4}dr=\frac{r^5}{5}{|}_1^2=\frac{32-1}{5}=\frac{31}{5}$. Outer $\times 2\pi$: $I_0=\frac{31}{5}\cdot 2\pi =\frac{62\pi }{5}.$ $\boxed{m=\frac{14\pi }{3},I_0=\frac{62\pi }{5}.}$

Ek rigid body ki rotational kinetic energy $E=\frac{1}{2}{I}_0{\omega }^2$ hai — exactly isliye humne $I_0$ mein $r^2$ define kiya: har tile ki speed $v=\omega r$ hai, aur energy $\frac{1}{2}(dm)v^2=\frac{1}{2}{\omega }^2(r^{2}dm)$ sum hokar $\frac{1}{2}{\omega }^2{I}_0$ banti hai. (Poori kahani Rotational Kinetic Energy and Angular Momentum mein.) $E=\frac{1}{2}\cdot \frac{62\pi }{5}\cdot 3^2=\frac{1}{2}\cdot \frac{62\pi }{5}\cdot 9=\frac{558\pi }{10}=\frac{279\pi }{5}\approx \mathrm{175.3.}$ $\boxed{E=\frac{279\pi }{5}\approx 175.3\text{J}.}$

Mass factor ::: $f=1$ ($\rho dA$ integrate karo). Moment factor ::: lever arm first power mein ($x$ ya $y$), sign rakho. Inertia factor ::: lever arm squared ($x^2$, $y^2$, ya $x^2+y^2$), hamesha positive. Centre of mass ::: $\bar{x}=M_y/m,\bar{y}=M_x/m$. Polar area element ::: $dA=rdrd\theta$ — kabhi nahi $drd\theta$. Perpendicular-axis theorem ::: $I_0=I_x+I_y$, sirf flat plates ke liye (saari mass $z=0$ par). Rotational energy ::: $E=\frac{1}{2}{I}_0{\omega }^2$.