Moments and Centers of Mass: Fresh Take

Lumen Learning

Moments and Centers of Mass: Fresh Take

Find the balance point (center of mass) of straight objects and flat surfaces
Utilize a shape’s symmetry to find the centroid, or geometric center, of flat objects
Use Pappus’s theorem to calculate the volume of an object

Center of Mass and Moments

The Main Idea

Center of Mass:
- The balancing point of an object or system
- Point where the total mass can be concentrated without changing the object’s behavior
Moments:
- - Measure of the rotational force applied to an object
  - First moment with respect to origin: $M = \sum_{i=1}^n m_i x_i$
One-Dimensional Systems:
- Center of mass: $\overline{x} = \frac{\sum_{i=1}^n m_i x_i}{\sum_{i=1}^n m_i}$
Two-Dimensional Systems:
- Moments: $M_x = \sum_{i=1}^n m_i y_i$ , $M_y = \sum_{i=1}^n m_i x_i$
- Center of mass: $\overline{x} = \frac{M_y}{m}$ , $\overline{y} = \frac{M_x}{m}$

Problem-Solving Strategy

Identify the system (point masses or continuous object)
Choose an appropriate coordinate system
Calculate the total mass of the system
Compute moments with respect to axes
Use formulas to find the center of mass coordinates

Find the center of mass for three point masses:

$m_1 = 2\text{ kg}$ at $(-1, 3)$
$m_2 = 6\text{ kg}$ at $(1, 1)$
$m_3 = 4\text{ kg}$ at $(2, -2)$

Total mass:

$m = 2 + 6 + 4 = 12\text{ kg}$

Moments:

$M_y = (-1 \cdot 2) + (1 \cdot 6) + (2 \cdot 4) = 12$
$M_x = (3 \cdot 2) + (1 \cdot 6) + (-2 \cdot 4) = 4$

Center of mass:

$\overline{x} = \frac{M_y}{m} = \frac{12}{12} = 1$
$\overline{y} = \frac{M_x}{m} = \frac{4}{12} = \frac{1}{3}$

Center of mass:

$(1, \frac{1}{3})$

Suppose four point masses are placed on a number line as follows:

\begin{array}{cccc} m_{1} = 12 kg, placed at x_{1} = - 4 m & m_{2} = 12 kg, placed at x_{2} = 4 m \\ m_{3} = 30 kg, placed at x_{3} = 2 m & m_{4} = 6 kg, placed at x_{4} = - 6 m . \end{array}

$\begin{array}{cccc}{m}_{1}=12\text{kg,}\text{ placed at }{x}_{1}=-4\text{m}\hfill & & & {m}_{2}=12\text{kg,}\text{ placed at }{x}_{2}=4\text{m}\hfill \\ {m}_{3}=30\text{kg,}\text{ placed at }{x}_{3}=2\text{m}\hfill & & & {m}_{4}=6\text{kg,}\text{ placed at }{x}_{4}=-6\text{m}.\hfill \end{array}$

Find the moment of the system with respect to the origin and find the center of mass of the system.

$M=24,\overline{x}=\frac{2}{5}\text{m}$

Suppose three point masses are placed on a number line as follows (assume coordinates are given in meters):

\begin{matrix} m_{1} = 5 kg, placed at (- 2, - 3), \\ m_{2} = 3 kg, placed at (2, 3), \\ m_{3} = 2 kg, placed at (- 3, - 2) . \end{matrix}

$\begin{array}{c}{m}_{1}=5\text{kg, placed at}(-2,-3),\hfill \\ {m}_{2}=3\text{kg, placed at}(2,3),\hfill \\ {m}_{3}=2\text{kg, placed at}(-3,-2).\hfill \end{array}$

Find the center of mass of the system.

(- 1, - 1)

$(-1,-1)$ m

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “2.6 Moments and Centers of Mass” here (opens in new window).

Center of Mass of Thin Plates

The Main Idea

Lamina:
- Thin sheet with uniform density
- Mass distributed continuously across a 2D region
Centroid:
- Geometric center of a 2D shape
- Coincides with center of mass for uniform density
Key Formulas:
- Mass: $m = \rho \int_a^b f(x) dx$
- $x$ -axis Moment: $M_x = \rho \int_a^b \frac{[f(x)]^2}{2} dx$
- $y$ -axis Moment: $M_y = \rho \int_a^b x f(x) dx$
- Center of Mass: $\overline{x} = \frac{M_y}{m}, \overline{y} = \frac{M_x}{m}$
Symmetry Principle:
- If a region is symmetric about a line, its centroid lies on that line

Problem-Solving Strategy

Identify the region $R$ and its bounding function $f(x)$
Set up integrals for mass and moments ( $M_x$ and $M_y$ )
Evaluate the integrals
Calculate center of mass coordinates using $M_x$ , $M_y$ , and $m$
Simplify and interpret the results

Let $R$ be the region bounded above by the graph of the function $f(x)={x}^{2}$ and below by the $x$ -axis over the interval $\left[0,2\right].$ Find the centroid of the region.

The centroid of the region is $(\frac{3}{2},\frac{6}{5}).$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “2.6 Moments and Centers of Mass” here (opens in new window).

Let R be the region bounded above by the graph of the function $f(x)=6-{x}^{2}$ and below by the graph of the function $g(x)=3-2x.$ Find the centroid of the region.

The centroid of the region is $(1,\frac{13}{5}).$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “2.6 Moments and Centers of Mass” here (opens in new window).

Find the centroid of the region bounded by $y = \sqrt{x}$ and the $x$ -axis from $x = 0$ to $x = 4$ .

Mass:

$m = \int_0^4 \sqrt{x} dx = \frac{2}{3}x^{3/2}|_0^4 = \frac{16}{3}$

Moments:

$M_x = \int_0^4 \frac{x}{2} dx = \frac{1}{4}x^2|_0^4 = 4$
$M_y = \int_0^4 x\sqrt{x} dx = \frac{2}{5}x^{5/2}|_0^4 = \frac{64}{5}$

Center of Mass:

$\overline{x} = \frac{M_y}{m} = \frac{64/5}{16/3} = \frac{12}{5}$
$\overline{y} = \frac{M_x}{m} = \frac{4}{16/3} = \frac{3}{4}$

Centroid:

$(\frac{12}{5}, \frac{3}{4})$

Center of Mass of a Region Bounded by Two Functions

The Main Idea

Region Definition:
- Bounded above by function $f(x)$
- Bounded below by function $g(x)$
- Bounded on sides by $x = a$ and $x = b$
Key Formulas:
- Mass: $m = \rho \int_a^b [f(x) - g(x)] dx$
- $x$ -axis Moment: $M_x = \rho \int_a^b \frac{1}{2}([f(x)]^2 - [g(x)]^2) dx$
- $y$ -axis Moment: $M_y = \rho \int_a^b x[f(x) - g(x)] dx$
- Center of Mass: $\overline{x} = \frac{M_y}{m}, \overline{y} = \frac{M_x}{m}$
Centroid of Rectangle:
- Located at [latex](x_i^, \frac{f(x_i^) + g(x_i^*)}{2})[/latex]

Problem-Solving Strategy

Identify bounding functions $f(x)$ and $g(x)$ and integration limits
Set up integrals for mass and moments ( $M_x$ and $M_y$ )
Evaluate the integrals, often requiring integration by parts or substitution
Calculate center of mass coordinates using $M_x$ , $M_y$ , and $m$
Simplify and interpret the results

The Symmetry Principle

The Main Idea

Symmetry Principle:
- If a region is symmetric about a line, its centroid lies on that line
- Simplifies calculations for symmetric shapes
Theorem of Pappus for Volume:
- Volume = Area of region × Distance traveled by centroid
- Applies to solids of revolution

Problem-Solving Strategy

Identify symmetry in the given region
Use symmetry to simplify centroid calculations
For solids of revolution:
1. Calculate the area of the region
2. Determine the path of the centroid
3. Apply the Theorem of Pappus

Let R be the region bounded above by the graph of the function $f(x)=1-{x}^{2}$ and below by $x$ -axis. Find the centroid of the region.

The centroid of the region is $(0,2\text{/}5).$

Let $R$ be a circle of radius $1$ centered at $(3,0).$ Use the theorem of Pappus for volume to find the volume of the torus generated by revolving $R$ around the $y$ -axis.

$6{\pi }^{2}$ units³

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “2.6 Moments and Centers of Mass” here (opens in new window).

Find the volume of a washer-shaped solid formed by revolving the region between concentric circles of radii $3$ and $5$ centered at $(4,0)$ around the $y$ -axis.

Area of the region:

$A = \pi(5^2 - 3^2) = 16\pi$

Centroid location: Due to symmetry, centroid is at $(4,0)$

Distance traveled by centroid:

$d = 2\pi(4) = 8\pi$

Volume using Pappus’s Theorem:

$V = A \cdot d = 16\pi \cdot 8\pi = 128\pi^2$

The volume of the washer-shaped solid is $128\pi^2$ cubic units.