Theory

Theory 1

Moment

The moment of a region to an axis is the total (integral) of mass times distance to that axis:

Moment to $x$-axis:

$$\begin{array}{cc}{M}_x& =\int \rho ydA\text{(general formula)}\\ & \\ M_x& ={\int }_c^d\rho y(g_2(y)-g_1(y))dy\text{(region between functions }{g}_2\text{ and }{g}_1\text{)}\end{array}$$

Moment to $y$-axis:

$$\begin{array}{cc}{M}_y& =\int \rho xdA\text{(general formula)}\\ & \\ M_y& ={\int }_a^b\rho x(f_2(x)-f_1(x))dx\text{(region between functions }{f}_2\text{ and }{f}_1\text{)}\end{array}$$

Notice the swap in letters

• $M_y$ integrand has $x$ factor
• $M_x$ integrand has $y$ factor

Notice the total mass

If you remove $x$ or $y$ factors from the integrands, the integrals give total mass $M$.

These formulas are obtained by slicing the region into rectangular strips that are parallel to the axis in question.

The area per strip is then:

• $f(x)dx$ — region under $y=f(x)$
• $(f_2-f_1)dx$ — region between $f_1$ and $f_2$
• $g(y)dy$ — region ‘under’ $x=g(y)$
• $(g_2-g_1)dy$ — region between $g_1$ and $g_2$

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The idea of moment is related to:

• Torque balance and angular inertia
• Center of mass

The center of mass (CoM) of a solid body is a single point with two important properties:

1. $\text{CoM}=$ “average position” of the body
   • The average position determines an effective center of dynamics. For example, gravity acting on every bit of mass of a rigid body acts the same as a force on the CoM alone.
2. $\text{CoM}=$ “balance point” of the body
   • The net torque (rotational force) about the CoM, generated by a force distributed over the body’s mass, equivalently a force on the CoM, is zero.

Centroid

When the body has uniform density, then the CoM is also called the centroid.

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Center of mass from moments

Coordinates of the CoM:

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

Here $M$ is the total mass of the body.

Center of mass from moments - explanation

Notice how these formulas work. The total mass is always $M=\int \rho dA$. The moment to $y$ (for example) is $M_y=\int \rho xdA$. Dividing these two values:

$$\bar{x}=\frac{M_y}{m}\gg \gg \frac{\int \rho xdA}{\int \rho dA}\gg \gg \frac{\int xdA}{\int dA}\gg \gg \frac{\int xdA}{A}$$

where $A=\text{total area}$.

In other words, through the formula $\bar{x}=\frac{M_y}{M}$, we find that $\bar{x}$ is the average value of $x$ over the region with area $A$.

Theory 2

A downside of the technique above is that to find $M_x$ we needed to convert the region into functions in $y$. This would be hard to do if the region was given as the area under a curve $y=f(x)$ but $f^{-1}(y)$ cannot be found analytically. An alternative formula can help in this situation.

Midpoint of strips for opposite variables

When the region lies between $f_1(x)$ and $f_2(x)$, we can find $M_x$ with an $x$-integral:

$$M_x={\int }_a^b\rho \frac{1}{2}(f_2^2-f_1^2)dx\text{(region between }{f}_1\text{ and }{f}_2\text{)}$$

When the region lies between $g_1(y)$ and $g_2(y)$, we can find $M_y$ with a $y$-integral:

$$M_y={\int }_c^d\rho \frac{1}{2}(g_2^2-g_1^2)dy\text{(region between }{g}_1\text{ and }{g}_2\text{)}$$

Region under a curve

For the region “under the curve” $y=f(x)$, just set:

$$f_1=0,f_2=f$$

For the region “under the curve” $x=g(x)$, set:

$$g_1=0,g_2=g$$

The idea for these formulas is to treat each vertical strip as a point concentrated at the CoM of the vertical strip itself.

The height to this midpoint is $\frac{1}{2}f(x)$, and the area of the strip is $f(x)dx$, so the integral becomes $\int \rho \frac{1}{2}f{(x)}^{2}dx$.

Midpoint of strips formula - full explanation

• If the strip is located at some $x$, with $y$ values from $0$ up to $f(x)$, then:

$$\text{CoM of strip}=(x,\frac{1}{2}f(x))$$

• The area of the strip is $dA=f(x)dx$. So the integral formula for $M_x$ can be recast:

$$M_x=\int ydA\gg \gg {\int }_a^b\frac{1}{2}f(x)\cdot f(x)dx\gg \gg {\int }_a^b\frac{1}{2}{f}^{2}dx$$

• If the vertical strips are between $f_1(x)$ and $f_2(x)$, then the midpoints of the strips are given by the ‘average’ function:

$$\frac{1}{2}(f_1(x)+f_2(x))$$

• The height of each strip is $f_2(x)-f_1(x)$, so $dA=(f_2-f_1)dx$.
• Putting this together:

$$\begin{array}{c}{M}_x=\int ydA\gg \gg {\int }_a^b\frac{1}{2}(f_1+f_2)\cdot (f_2-f_1)dx\\ \\ \gg \gg {\int }_a^b\frac{1}{2}(f_2^2-f_1^2)dx\end{array}$$

Theory 3

Two useful techniques for calculating moments and (thereby) CoMs:

• Additivity principle
• Symmetry

Additivity says that you can add moments of parts of a region to get the total moment of the region (to a given axis).

A symmetry principle is that if a region is mirror symmetric across some line, then the CoM must lie on that line.

Theory 4

CoM of a triangle: “ $h/3$ trick”

Given a triangle with edge $b$ and altitude $h$ (measured along a perpendicular to $b$), the CoM of the triangle lies on the line parallel to $b$ and shifted upwards by $h/3$.

For example, if the triangle is positioned with base $b$ on the $x$-axis, then its CoM will lie on the line $y=h/3$.

Proof of “ $h/3$ trick”

Quick Linear Interpolation Function:

$$\begin{array}{c}w(y)=b+\frac{0-b}{h}(y-0)\\ \\ \gg \gg w(y)=b-\frac{b}{h}y\end{array}$$

Thus:

$$\begin{array}{c}{M}_x=\rho {\int }_0^{h}yw(y)dy\\ \\ \gg \gg \rho {\int }_0^{h}y(b-\frac{b}{h}y)dy\\ \\ \gg \gg \rho {(\frac{b{y}^2}{2}-\frac{b{y}^3}{3h})|}_0^h\gg \gg \rho \frac{b{h}^2}{6}\end{array}$$

Total mass:

$$m=\rho \cdot (\text{area})=\rho \cdot \frac{1}{2}bh$$

Therefore:

$$\bar{y}=\frac{M_x}{m}\gg \gg \frac{\rho b{h}^2/6}{\rho bh/2}\gg \gg \frac{h}{3}$$