Examples

CoM of a parabolic plate

Find the CoM of the region depicted:

Solution

(1) Compute the total mass:

Area under the curve with density factor $\rho$:

$$m={\int }_0^2\rho x^{2}dx\gg \gg \rho {\frac{x^3}{3}|}_0^2\gg \gg \frac{8\rho }{3}$$

---

(2) Compute $M_y$:

Formula:

$$M_y={\int }_a^b\rho xdA$$

Interpret and calculate:

$$\begin{array}{c}{M}_y={\int }_0^2\rho xf(x)dx\gg \gg \rho {\int }_0^2{x}^{3}dx\\ \\ \gg \gg 4\rho =M_y\end{array}$$

---

(3) Compute $M_x$:

Formula:

$$M_x={\int }_c^d\rho ydA$$

Width of horizontal strips between the curves:

$$w(y)=2-\sqrt{y}$$

Interpret $dA$:

$$dA=(2-\sqrt{y})dy$$

Calculate integral:

$$\begin{array}{c}{M}_x={\int }_c^d\rho ydA\gg \gg {\int }_0^4\rho y(2-\sqrt{y})dy\\ \\ \gg \gg {\int }_0^4\rho y(2-\sqrt{y})dy\gg \gg {\int }_0^4\rho 2ydy-{\int }_0^4\rho y^{3/2}dy\\ \\ \gg \gg \frac{16\rho }{5}=M_x\end{array}$$

---

(4) Compute CoM coordinates from moments:

CoM formulas:

$$\bar{x}=\frac{M_y}{m},\bar{y}=\frac{M_x}{m}$$

Insert data:

$$\begin{array}{c}\bar{x}=\frac{4\rho }{8\rho /3}\gg \gg \frac{3}{2},\bar{y}=\frac{16\rho /5}{8\rho /3}\gg \gg \frac{6}{5}\\ \\ \text{CoM}=(\bar{x},\bar{y})=(\frac{3}{2},\frac{6}{5})\end{array}$$

Computing CoM using only vertical strips

Find the CoM of the region:

Solution

(1) Compute the total mass $m$:

Area under the curve times density $\rho$:

$${\int }_0^{\pi /2}\rho \cos xdx=\rho \sin x{|}_0^{\pi /2}=\rho$$

---

(2) Compute $M_x$, using vertical strips and the ‘midpoints’ method:

Plug $f_2(x)=f(x)=\cos x$ and $f_1(x)=0$ into formula:

$$M_x={\int }_0^{\pi /2}\rho \frac{1}{2}{f}_2^{2}dx\gg \gg {\int }_0^{\pi /2}\rho \frac{1}{2}{\cos }^{2}xdx$$

Integration by ‘power to frequency conversion’:

$$\begin{array}{c}{\int }_0^{\pi /2}\rho \frac{1}{2}{\cos }^{2}xdx\gg \gg \frac{\rho }{4}{\int }_0^{\pi /2}(1+\cos 2x)dx\\ \\ \gg \gg \frac{\rho }{4}x{|}_0^{\pi /2}+{\frac{\rho \sin 2x}{8}|}_0^{\pi /2}=\frac{\pi \rho }{8}\end{array}$$

---

(3) Compute $M_y$ using vertical strips and the regular method:

Plug $f(x)=\cos x$ into formula:

$$M_y={\int }_a^b\rho xf(x)dx\gg \gg {\int }_0^{\pi /2}\rho x\cos xdx$$

Integration by parts. Set $u=x$, $v^{\prime }=\cos x$ and so $u^{\prime }=1$, $v=\sin x$:

$$\begin{array}{c}{\int }_0^{\pi /2}\rho x\cos xdx\gg \gg \rho x\sin x{|}_0^{\pi /2}-\rho {\int }_0^{\pi /2}\sin xdx\\ \\ \gg \gg \frac{\pi \rho }{2}\cdot 1-\rho (-\cos \frac{\pi }{2}--\cos 0)=\rho (\frac{\pi }{2}-1)\end{array}$$

---

(4) Compute CoM:

CoM formulas:

$$\bar{x}=\frac{M_y}{m},\bar{y}=\frac{M_x}{m}$$

Plug in data:

$$\begin{array}{c}\bar{x}=\frac{\rho (\pi /2-1)}{\rho }\gg \gg \frac{\pi }{2}-1\\ \\ \bar{y}=\frac{\pi \rho /8}{\rho }\gg \gg \frac{\pi }{8}\\ \\ \text{CoM}=(\bar{x},\bar{y})=(\frac{\pi }{2}-1,\frac{\pi }{8})\end{array}$$

CoM of region between line and parabola

Compute the CoM of the region below $y=x$ and above $y=x^2$ with $x\in [\mathrm{0,1}]$.

Solution

(1) Compute total mass $m$:

Name the functions: $f_1(x)=x^2$ (lower) and $f_2(x)=x$ (upper) over $x\in [\mathrm{0,1}]$.

Mass is area (between curves) times density:

$$M={\int }_0^1\rho (f_2-f_1)dx\gg \gg \rho {\int }_0^{1}x-x^{2}dx\gg \gg \frac{\rho }{6}$$

---

(2) Compute $M_y$ using vertical strips:

$$\begin{array}{c}{M}_y={\int }_0^1\rho x(f_2-f_1)dx\\ \\ \gg \gg \rho {\int }_0^{1}x(x-x^2)dx\gg \gg \frac{\rho }{12}\end{array}$$

---

(3) Compute $M_x$ also using vertical strips:

$$\begin{array}{c}{M}_x={\int }_0^1\rho \frac{1}{2}(f_2^2-f_1^2)dx\\ \\ \gg \gg \rho {\int }_0^1\frac{1}{2}(x^2-x^4)dx\gg \gg \frac{2\rho }{30}\end{array}$$

---

(4) Compute CoM using moment formulas:

$$\begin{array}{c}\bar{x}=\frac{\rho /12}{\rho /6}\gg \gg \frac{1}{2}\\ \\ \bar{y}=\frac{2\rho /30}{\rho /6}\gg \gg \frac{2}{5}\\ \\ \text{CoM}=(\bar{x},\bar{y})=(\frac{1}{2},\frac{2}{5})\end{array}$$

Center of mass using moments and symmetry

Find the center of mass of the two-part region:

Solution

(1) Symmetry: CoM on $y$-axis

Because the region is symmetric in the $y$-axis, the CoM must lie on that axis. Therefore $\bar{x}=0$.

---

(2) Additivity of moments:

Write $M_x$ for the total $x$-moment (distance measured to the $x$-axis from above).

Write $M_x^{\mathrm{tri}}$ and $M_x^{\mathrm{circ}}$ for the $x$-moments of the triangle and circle.

Additivity of moments equation:

$$M_x=M_x^{\mathrm{tri}}+M_x^{\mathrm{circ}}$$

---

(3) Find moment of the circle $M_x^{\mathrm{circ}}$ using Symmetry:

By symmetry we know ${\bar{x}}^{\mathrm{circ}}=0$.

By symmetry we know ${\bar{y}}^{\mathrm{circ}}=5$.

Area of circle with $r=2$ is $A=4\pi$, so total mass is $M=4\pi \rho$.

Centroid-from-moments equation:

$${\bar{y}}^{\mathrm{circ}}=\frac{M_x^{\mathrm{circ}}}{m}$$

Solve the equation for $M_x^{\mathrm{circ}}$:

$$\begin{array}{c}{\bar{y}}^{\mathrm{circ}}=\frac{M_x^{\mathrm{circ}}}{m}\gg \gg 5=\frac{M_x^{\mathrm{circ}}}{4\pi \rho }\\ \\ \gg \gg M_x^{\mathrm{circ}}=20\pi \rho \end{array}$$

---

(4) Find moment of the triangle $M_x^{\mathrm{tri}}$ using Symmetry and the “$h/3$ trick”:

By Symmetry, the CoM of the triangle must lie on the vertical line $x=0$.

By the “$h/3$ trick,” the CoM of the triangle must lie on the horizontal line at $y=h/3$. In this case $h=3$, so it lies on the line $y=1$.

Therefore the triangle CoM is:

$$({\bar{x}}^{\text{tri}},{\bar{y}}^{\text{tri}})=(\mathrm{0,1})$$

To get moments, multiply by mass $m=\rho \cdot \frac{1}{2}(4)(3)=6\rho$:

$$M_y^{\text{tri}}=0,M_x^{\text{tri}}=(6\rho )(1)=6\rho$$

---

(5) Apply additivity:

$$M_x=M_x^{\mathrm{tri}}+M_x^{\mathrm{circ}}\gg \gg \rho (20\pi +6)$$

---

(6) Total mass of region:

Area of circle is $4\pi$. Area of triangle is $\frac{1}{2}\cdot 4\cdot 3=6$. Thus $m=\rho A=\rho (4\pi +6)$.

---

(7) Compute center of mass $\bar{y}$ from total $M_x$ and total $M$:

We have $M_x=\rho (20\pi +6)$ and $M=\rho (4\pi +6)$. Plug into formula:

$$\begin{array}{c}\bar{y}=\frac{M_x}{m}\gg \gg \frac{\rho (20\pi +6)}{\rho (4\pi +6)}\approx 3.71\\ \\ \text{CoM}=(\bar{x},\bar{y})=(\mathrm{0,3.71})\end{array}$$

Center of mass - two part region

Find the center of mass of the region which combines a rectangle and triangle (as in the figure) by computing separate moments. What are those separate moments? Assume the mass density is $\rho =1$.

Solution

(1) Apply symmetry to rectangle:

By symmetry, the center of mass of the rectangle is located at $(-\mathrm{1,2})$.

Thus ${\bar{x}}^{\text{rect}}=-1$ and ${\bar{y}}^{\text{rect}}=2$.

---

(2) Find moments of the rectangle:

Total mass of rectangle $=m^{\text{rect}}=\rho \times \text{area}=1\cdot 8=8$. Thus:

$$\begin{array}{cc}{\bar{x}}^{\text{rect}}=\frac{M_y^{\text{rect}}}{m^{\text{rect}}}& \gg \gg M_y^{\text{rect}}=-8\\ & \\ {\bar{y}}^{\text{rect}}=\frac{M_x^{\text{rect}}}{m^{\text{rect}}}& \gg \gg M_x^{\text{rect}}=16\end{array}$$

---

(3) Find moments of the triangle:

CoM of triangle using “$h/3$ trick:”

From the base on the $x$-axis, we have $h=4$ so the CoM lies on the line $y=4/3$.

From the base on the $y$-axis, we have $h=4$ so the CoM lies on the line $x=4/3$.

Multiply by the triangle’s mass to get its moment:

$$m^{\text{tri}}=\rho \cdot \frac{1}{2}(4)(4)\gg \gg 8$$

Therefore:

$$\begin{array}{c}{M}_x^{\text{tri}}=(8\rho )(4/3)\gg \gg \frac{32}{3}\\ \\ M_y^{\text{tri}}=(8\rho )(4/3)\gg \gg \frac{32}{3}\end{array}$$

---

(4) Add up total moments:

General formulas: $M_x=M_x^{\text{tri}}+M_x^{\text{rect}}$ and $M_y=M_y^{\text{tri}}+M_y^{\text{rect}}$

Plug in data: $M_x=\frac{32}{3}+16=\frac{80}{3}$ and $M_y=\frac{32}{3}-8=\frac{8}{3}$

---

(5) Find center of mass from moments:

Total mass of triangle $=M^{\text{tri}}=\rho \times \text{area}=1\cdot \frac{1}{2}\cdot 4\cdot 4=8$.

Total combined mass $=m=m^{\text{tri}}+m^{\text{rect}}=8+8=16$.

Apply moment relation:

$$\begin{array}{c}\bar{x}=\frac{M_y}{m}\gg \gg \frac{8/3}{16}\gg \gg \frac{1}{6}\\ \\ \bar{y}=\frac{M_x}{m}\gg \gg \frac{80/3}{16}\gg \gg \frac{5}{3}\\ \\ \text{CoM}=(\bar{x},\bar{y})=(\frac{1}{6},\frac{5}{3})\end{array}$$