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

Read straight off the recipe "$1,r,r^2$":

• (a) mass → $f=1$.
• (b) $M_x$ = moment about the $x$-axis → lever arm to the $x$-axis is the height $y$, first power → $f=y$.
• (c) $M_y$ = moment about the $y$-axis → lever arm is $x$, first power → $f=x$.
• (d) $I_x$ = inertia about the $x$-axis → same lever arm $y$ but squared → $f=y^2$.
• (e) $I_y$ → $f=x^2$.
• (f) $I_0$ = inertia about the origin → lever arm $r$, squared → $f=r^2=x^2+y^2$.

Notice $I_0=\iint (x^2+y^2)\rho dA=I_x+I_y$ — the perpendicular-axis theorem for a flat plate (unpacked with a picture in L4.2).

Chop into tiles; each tile has mass $\rho dA=(xy)dxdy$. Add over the rectangle. The limits are constant because the region is a box: $m={\int }_0^2{\int }_0^{3}xydydx.$ That's all L1 asks — recognising which factor $f(=1)$ and which limits.

Inner integral (treat $x$ as a constant, integrate in $y$): ${\int }_0^{3}xydy=x\cdot \frac{y^2}{2}{|}_0^3=x\cdot \frac{9}{2}=\frac{9x}{2}.$ Why inner first? For a fixed vertical strip $x$ runs along, $y$ sweeps top to bottom; we collapse that strip to a number, then sweep $x$. Outer integral: ${\int }_0^2\frac{9x}{2}dx=\frac{9}{2}\cdot \frac{x^2}{2}{|}_0^2=\frac{9}{2}\cdot 2=9.$ $\boxed{m=9.}$

The region is round, so polar coordinates are the natural tool (a circle's boundary becomes the flat line $r=a$). Substitute $x=r\cos \theta ,y=r\sin \theta ,dA=rdrd\theta ,y^2=r^2{\sin }^2\theta .$ Upper half ⇒ $\theta$ from $0$ to $\pi$; $r$ from $0$ to $2$. $I_x={\int }_0^{\pi }{\int }_0^2(r^2{\sin }^2\theta )rdrd\theta ={\int }_0^{\pi }{\sin }^2\theta d\theta {\int }_0^2{r}^{3}dr.$ Why it separates: the integrand is a product of a $\theta$-only piece and an $r$-only piece over a rectangular $r\theta$-box, so the integrals split. ${\int }_0^2{r}^{3}dr=\frac{r^4}{4}{|}_0^2=\frac{16}{4}=4,{\int }_0^{\pi }{\sin }^2\theta d\theta =\frac{\pi }{2}.$ $\boxed{I_x=4\cdot \frac{\pi }{2}=2\pi .}$

Region: $0\le x\le 2,0\le y\le 2-x$ (below the slanted line $x+y=2$, the cyan edge in the figure).

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: put $u=2-x$. Then $du=-dx$, i.e. $dx=-du$. When $x=0$, $u=2$; when $x=2$, $u=0$ — the limits run downward $2\to 0$. The minus sign from $dx=-du$ and the downward limits combine: flipping a definite integral's limits flips its sign, and that flip cancels the minus, restoring the clean upward $0\to 2$: ${\int }_{x=0}^{x=2}\frac{(2-x{)}^2}{2}dx={\int }_{u=2}^{u=0}\frac{u^2}{2}(-du)={\int }_{u=0}^{u=2}\frac{u^2}{2}du=\frac{1}{2}\cdot \frac{u^3}{3}{|}_0^2=\frac{1}{2}\cdot \frac{8}{3}=\frac{4}{3}.$

$M_x=\iint y\cdot \rho dA=\iint y\cdot ydA=\iint y^{2}dA$ (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 with the same $u=2-x$ (again the $-du$ cancels the downward limits): $M_x={\int }_0^2\frac{(2-x{)}^3}{3}dx={\int }_0^2\frac{u^3}{3}du=\frac{1}{3}\cdot \frac{u^4}{4}{|}_0^2=\frac{1}{3}\cdot \frac{16}{4}=\frac{4}{3}.$

Centre of mass height (using $\bar{y}=M_x/m$ from the recipe box): $\bar{y}=\frac{M_x}{m}=\frac{4/3}{4/3}=1.$ $\boxed{m=\frac{4}{3},M_x=\frac{4}{3},\bar{y}=1.}$ Sanity: the triangle spans $y\in [0,2]$ and the density grows with $y$, so mass leans upward — $\bar{y}=1$ sits above the geometric centroid height ($2/3$). Good.

$I_x\propto b^2$ and $I_y\propto a^2$. Inertia grows with how far mass sits from the axis. Mass is spread out to $x=a=3$ (far from the $y$-axis) but only to $y=b=1$ (close to the $x$-axis). So spinning about the $y$-axis is harder: $I_y>I_x$. $\frac{I_y}{I_x}=\frac{a^2}{b^2}=\frac{9}{1}=9.$ $\boxed{I_y\text{ is }9\times \text{larger}.}$ The lesson: inertia is dominated by the direction the mass reaches far — because distance is squared.

Region: $0\le x\le 1,0\le y\le 1-x$. About the origin the lever arm squared is $x^2+y^2$: $I_0=\iint (x^2+y^2)\rho dA={\int }_0^1{\int }_0^{1-x}(x^2+y^2)(x+y)dydx.$ Expand the integrand: $(x^2+y^2)(x+y)=x^3+x^{2}y+x{y}^2+y^3$. Inner integral in $y$ over $[0,1-x]$, writing $L=1-x$ (antiderivative then plug $y=L$): ${\int }_0^L(x^3+x^{2}y+x{y}^2+y^3)dy=x^{3}L+\frac{x^2{L}^2}{2}+\frac{x{L}^3}{3}+\frac{L^4}{4}.$ Outer integral in $x$ over $[0,1]$. Substitute $L=1-x$ and integrate each term. It is cleanest to convert every term to a power of $L$ using the swap $u=1-x$ (so $x=1-u$, $dx=-du$, limits $x:0\to 1\Rightarrow u:1\to 0$, and the $-du$ again cancels the flipped limits, giving ${\int }_0^1(\cdots )du$): ${\int }_0^1{x}^{3}Ldx={\int }_0^1(1-u{)}^{3}udu,{\int }_0^1\frac{x^2{L}^2}{2}dx={\int }_0^1\frac{(1-u{)}^2{u}^2}{2}du,$ ${\int }_0^1\frac{x{L}^3}{3}dx={\int }_0^1\frac{(1-u)u^3}{3}du,{\int }_0^1\frac{L^4}{4}dx={\int }_0^1\frac{u^4}{4}du.$ Each is a Beta-type integral ${\int }_0^1{u}^p(1-u{)}^{q}du=\frac{p!q!}{(p+q+1)!}$. Evaluate: ${\int }_0^1(1-u{)}^{3}udu=\frac{1!3!}{5!}=\frac{6}{120}=\frac{1}{20},$ $\frac{1}{2}{\int }_0^1(1-u{)}^2{u}^{2}du=\frac{1}{2}\cdot \frac{2!2!}{5!}=\frac{1}{2}\cdot \frac{4}{120}=\frac{1}{60},$ $\frac{1}{3}{\int }_0^1(1-u)u^{3}du=\frac{1}{3}\cdot \frac{3!1!}{5!}=\frac{1}{3}\cdot \frac{6}{120}=\frac{1}{60},$ $\frac{1}{4}{\int }_0^1{u}^{4}du=\frac{1}{4}\cdot \frac{1}{5}=\frac{1}{20}.$ Add: $I_0=\frac{1}{20}+\frac{1}{60}+\frac{1}{60}+\frac{1}{20}=\frac{3}{60}+\frac{1}{60}+\frac{1}{60}+\frac{3}{60}=\frac{8}{60}=\frac{2}{15}.$ $\boxed{I_0=\frac{2}{15}.}$ 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: if you scraped all the mass and glued it as a ring of radius $R_g\approx 0.63$ around the origin, it would resist spinning exactly as much as the real plate.

By the $x\leftrightarrow y$ symmetry of both the region and $\rho =x+y$, we have $I_x=I_y$. So we need only one. $I_x=\iint y^2(x+y)dA={\int }_0^1{\int }_0^{1-x}(x{y}^2+y^3)dydx.$ Inner ($L=1-x$): ${\int }_0^L(x{y}^2+y^3)dy=x\frac{L^3}{3}+\frac{L^4}{4}.$ Outer over $x\in [0,1]$, same 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: $\frac{1}{4}\cdot \frac{1}{5}=\frac{1}{20}=\frac{3}{60}$. $I_x=\frac{1}{60}+\frac{3}{60}=\frac{4}{60}=\frac{1}{15},I_y=\frac{1}{15}.$ Then $I_x+I_y=\frac{1}{15}+\frac{1}{15}=\frac{2}{15}=I_0.$ ✓ The perpendicular-axis theorem holds — as it must, because this plate is flat (a lamina in the $xy$-plane). See Centroids and the Pappus Theorems for the geometric cousin of these identities.

Round region + radial density ⇒ polar coordinates win decisively. $r$ from $1$ to $2$, $\theta$ from $0$ 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 ($f=r^2$): $I_0={\int }_0^{2\pi }{\int }_1^2{r}^2\cdot r\cdot rdrd\theta ={\int }_0^{2\pi }{\int }_1^2{r}^{4}drd\theta .$ Count the $r$'s: one from $\rho =r$, one from $r^2$ (lever arm squared), one from the Jacobian $dA=rdrd\theta$ → $r^4$. Inner: ${\int }_1^2{r}^{4}dr=\frac{r^5}{5}{|}_1^2=\frac{32-1}{5}=\frac{31}{5}$. Outer $\times 2\pi$: $I_0=\frac{31}{5}\cdot 2\pi =\frac{62\pi }{5}.$ $\boxed{m=\frac{14\pi }{3},I_0=\frac{62\pi }{5}.}$

The rotational kinetic energy of a rigid body is $E=\frac{1}{2}{I}_0{\omega }^2$ — this is exactly why we defined $I_0$ with an $r^2$: each tile's speed is $v=\omega r$, and energy $\frac{1}{2}(dm)v^2=\frac{1}{2}{\omega }^2(r^{2}dm)$ sums to $\frac{1}{2}{\omega }^2{I}_0$. (Full story in Rotational Kinetic Energy and Angular Momentum.) $E=\frac{1}{2}\cdot \frac{62\pi }{5}\cdot 3^2=\frac{1}{2}\cdot \frac{62\pi }{5}\cdot 9=\frac{558\pi }{10}=\frac{279\pi }{5}\approx \mathrm{175.3.}$ $\boxed{E=\frac{279\pi }{5}\approx 175.3\text{J}.}$

Mass factor ::: $f=1$ (integrate $\rho dA$). Moment factor ::: lever arm to first power ($x$ or $y$), sign kept. Inertia factor ::: lever arm squared ($x^2$, $y^2$, or $x^2+y^2$), always positive. Centre of mass ::: $\bar{x}=M_y/m,\bar{y}=M_x/m$. Polar area element ::: $dA=rdrd\theta$ — never $drd\theta$. Perpendicular-axis theorem ::: $I_0=I_x+I_y$, flat plates only (all mass at $z=0$). Rotational energy ::: $E=\frac{1}{2}{I}_0{\omega }^2$.