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# Center of Mass – Definition, Formula, and Examples

Integral calculus has a wide range of applications in physics and engineering and that includes finding the**center of mass**through integration. In this article, weâ€™ll learn how to compute the center of mass of a given object through single and double integration.

**For us to learn to compute for the center of mass, it is essential that we know how to calculate for the area of the region enclosed by two curves. In case you need a quick refresher, head over to this link and check the article we wrote about this topic. Weâ€™ll also touch on the concept of moments briefly to better understand how we came up with the formula for the center of mass. For now, letâ€™s head over to the next section and understand the key components of the center of mass!**

*The center of mass represents the point within the object where the object is balanced horizontally when suspended from that given point. We can find the center of mass using definite integrals.***What is the center of mass?**

The **center of mass**of an object is simply the point where the mass of the given

**object is equally distributed**. When working with an

**object that has a uniform density**, normally represented by $\rho$, the objectâ€™s center of mass will also be its

**geometric center called the centroid**.

**balances the region horizontally**. Before we dive right into the mathematical definition for the center of mass, letâ€™s first understand what

**moments are and understand their significance when it comes to the objectâ€™s center of mass**. Imagine that thereâ€™s a rod with a negligible mass and attached to it are two masses, $m_1$ and $m_2$.

**center of mass be positioned at**$\boldsymbol{\overline{x}}$. Extending the equation above, we now have: \begin{aligned}m_1(\overline{x} â€“ x_1) &= m_2(x_2 – \overline{x})\\\overline{x} &= \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}\end{aligned} We call $m_1x_1$ and $m_2x_2$ as the

**moments of the two masses**, $m_1$ and $m_2$, respectively. We can then extend this concept to a system of $m_1, m_2, m_3,â€¦,m_n$ masses positioned at $x_1, x_2,x_3, â€¦,x_n$. The center of mass of the system can be expressed as shown below. \begin{aligned}\overline{x} &= \dfrac{\sum_{i =1}^{n} m_ix_i}{m}\end{aligned} Keep in mind that the denominator, $m$, represents the total mass or $\sum_{i=1}^{n}m_i$. We call the sum of the moments

**the total moment about the origin**. When working with a coordinate system, we can express the center of mass as an ordered pair,$(\overline{x} \overline{y})$.

\begin{aligned}\overline{x} &= \dfrac{M_y}{m}\end{aligned} | \begin{aligned}\overline{y} &= \dfrac{M_x}{m}\end{aligned} |

**Center of mass definition in terms of definite integral**

Weâ€™ve covered the center of mass of discrete points, so letâ€™s take a look at the center of mass of systems that are distributed continuously throughout a thin sheet of material. Weâ€™ll begin by observing that the **sheet is thin enough that we can consider it as a two-dimensional figure we call the lamina**. When we assume that the laminaâ€™s density is uniform, the laminaâ€™s center of mass is solely dependent on its shape. This means that weâ€™ll disregard its density first. Weâ€™ll apply a similar process to find the total mass and center of mass of the lamina. By symmetry principle, recall that when the region is symmetric about a line, the

**centroid is set to be to lie on the region as well**. For laminas that are rectangular in shape, we can guarantee that the

**center of mass (or centroid) is located at the center of the rectangle or where the diagonals intersect.**Now, letâ€™s say we have a general lamina, $R$, where the region is bounded by two vertical lines, $x =a$ and $x =b$, as shown below.

**as the**

**center of mass of the rectangle**and $\boldsymbol{f(\overline{x_i}) \Delta x}$

**as the area of the region**. Â This means that the

**lamina has a mass of**$\boldsymbol{\rho f(\overline{x_i}) \Delta x}$.

**Center of mass formula using definite integrals**

Finding the area between two curves is an **extension of finding the area under the functionâ€™s curve**. We can now derive the formula for the center of mass of the lamina using definite integrals. Using these expressions, we can now estimate the mass systems with respect to the $x$ and $y$-axis.

- Find $M_y$ by adding all the moments $(m_i \overline{x_i})$.
- We apply a similar process for $M_x$ but use $\dfrac{1}{2}f(\overline{x_i})$ as the distance.

\begin{aligned}\boldsymbol{M_y}\end{aligned} | \begin{aligned}\boldsymbol{M_x}\end{aligned} |

\begin{aligned} M_y &= \lim_{n \rightarrow \infty} \rho \overline{x_i} f(\overline{x_i}) \Delta x\\&= \rho \int_{a}^{b}Â xf(x) \phantom{x}dx Â \end{aligned} | \begin{aligned} M_x &= \lim_{n \rightarrow \infty} \rho \overline{x_i}\frac{1}{2} [f(\overline{x_i})]^2 \Delta x\\&= \rho \int_{a}^{b}Â \frac{1}{2}[f(x)]^2 \phantom{x}dxÂ \end{aligned} |

Centroidâ€™s $\boldsymbol{x}$-coordinate |
Centroidâ€™s $\boldsymbol{y}$-coordinate |

\begin{aligned} \overline{x} &= \dfrac{M_y}{m}\\&= \dfrac{\rho \int_{a}^{b} xf(x) \phantom{x}dx}{\rho \int_{a}^{b}f(x) \phantom{x}dx}\\&= \dfrac{\int_{a}^{b}x f(x)\phantom{x}dx}{\int_{a}^{b} f(x)\phantom{x}dx}\end{aligned} | \begin{aligned} \overline{y} &= \dfrac{M_x}{m}\\&= \dfrac{\rho \int_{a}^{b} \frac{1}{2}[f(x) ]^2\phantom{x}dx}{\rho \int_{a}^{b}f(x) \phantom{x}dx}\\&= \dfrac{\int_{a}^{b}\frac{1}{2}[f(x) ]^2 f(x)\phantom{x}dx}{\int_{a}^{b} f(x)\phantom{x}dx}\end{aligned} |

**laminaâ€™s center mass will not be affected by its density**, $\rho$. Weâ€™ve learned that $\text{Area} = A = \int_{a}^{b} f(x)\phantom{x}dx$, so we can further

**rewrite the plateâ€™s center of mass (and centroid) as shown below**. \begin{aligned} \boldsymbol{\overline{x}} &= \boldsymbol{\dfrac{1}{A}}\int_{\boldsymbol{a}}^{\boldsymbol{b}} \boldsymbol{x f(x)dx}\\ \boldsymbol{\overline{y}} &= \boldsymbol{\dfrac{1}{A}}\int_{\boldsymbol{a}}^{\boldsymbol{b}} \boldsymbol{\dfrac{1}{2}[f(x)]^2dx}\end{aligned} We can use these two equations to determine the center of mass of the a given plate or lamina.

**How to find the center of mass?**

In the earlier sections, weâ€™ve discussed how the center of mass is defined and how we came up with the formulas. Letâ€™s now break down the steps we need to apply when finding the coordinates of the plateâ€™s center of mass.
**Find the area of the region, $A = \int_{a}^{b} f(x) \phantom{x}dx$.**

__Step 1:__**Â Use the formula to find the center of massâ€™s $x$-coordinate, $\overline{x}$.**

__Step 2:__**Â Use the formula to find the center of massâ€™s $y$-coordinate, $\overline{y}$.**

__Step 3:__**Â Return the center of massâ€™ location as an ordered pair, $(\overline{x}, \overline{y})$. Keep in mind that**

__Step 4:__**when the region is symmetric along the**$y$-axis, there is no need for us to use the formula. Instead, we simply find the

**midpoint between the endpoints of**$\boldsymbol{x}$. Weâ€™ve summarized the helpful formulas youâ€™ll need when calculating for the $x$ and $y$-coordinates of the plateâ€™s center of mass.

Type of Region |
$\boldsymbol{x}$-coordinate |
$\boldsymbol{y}$-coordinate |

Area of region bounded by $f(x)$ and the $x$-axis: \begin{aligned}A&=\int_{a}^{b} f(x)\phantom{x}dx\end{aligned} | \begin{aligned}\overline{x} &= \dfrac{1}{A} \int_{a}^{b} xf(x)\phantom{x} dx\end{aligned} | \begin{aligned}\overline{y} &= \dfrac{1}{A} \int_{a}^{b} \dfrac{1}{2}[f(x)]^2\phantom{x} dx\end{aligned} |

Area of region bounded by two curves $(f(x) \geq g(x))$: \begin{aligned}A&=\int_{a}^{b} [f(x) â€“ g(x)]\phantom{x}dx\end{aligned} | \begin{aligned}\overline{x} &= \dfrac{1}{A} \int_{a}^{b} x\left[ f(x)- g(x) \right]\phantom{x} dx\end{aligned} | \begin{aligned}\overline{y} &= \dfrac{1}{A} \int_{a}^{b} \dfrac{1}{2}\left\{[f(x)]^2 – [g(x)]^2\right\}\phantom{x} dx\end{aligned} |

**What is the center of mass of the plate that is bounded by the region shown below?**

*Example 1*__Solution__Although we can find the area under the curve of $y = \sqrt{16 â€“ x^2}$ using definite integrals, itâ€™s much easier and faster when we simply use our previous knowledge of semicircles. The area of a semicircle is equal to $\dfrac{1}{2}\pi r^2$, where $r$ is the semicircleâ€™s radius. \begin{aligned}A &= \dfrac{1}{2} \pi 4^2 \\&= 8\pi \end{aligned} We can see that the semicircular plate is symmetric with respect to the $y$-axis, so there is no need for us to use the formula weâ€™ve derived to find the $x$-coordinate of the center of mass. By the

**symmetry principle**, we simply find the midpoint between $x= -4$ and $x = 4$, so $\overline{x} =0$. Letâ€™s now focus on calculating the $y$-coordinate of the center of mass using the formula, $\overline{y} = \dfrac{1}{A} \int_{a}^{b} \dfrac{1}{2}[f(x)]^2 \phantom{x}dx$. \begin{aligned} A&= 8\pi\\ a&= -4\\ b&= 4 \end{aligned} \begin{aligned} \overline{y} &= \dfrac{1}{8\pi} \int_{-4}^{4} \dfrac{1}{2}[\sqrt{16 – x^2}]^2 \phantom{x}dx\\ &= \dfrac{1}{8\pi} \int_{-4}^{4} \dfrac{1}{2}(16 – x^2 )\phantom{x}dx\\&= \dfrac{1}{8\pi} \cdot \dfrac{1}{2} \int_{-4}^{4} (16 – x^2 )\phantom{x}dx \\&= \dfrac{1}{16 \pi} \left[\int_{-4}^{4} 16\phantom{x}dx – \int_{-4}^{4} x^2\phantom{x}dx \right ]\\&= \dfrac{1}{16 \pi}\left\{\left[16x \right ]_{-4}^{4} – \left[\dfrac{x^3}{3} \right ]_{-4}^{4}\right\}\\&= \dfrac{1}{16 \pi}\left\{\left[16(4) -16(-4) \right ]_{-4}^{4} – \left[\dfrac{(4)^3}{3} -\dfrac{(-4)^3}{3} \right ]\right\}\\&= \dfrac{1}{16 \pi} \cdot \dfrac{128}{3}\\&= \dfrac{16}{3 \pi}\end{aligned} We have $\overline{x} = 0$ and $\overline{y} = \dfrac{16}{3\pi}$, so the semicircular plateâ€™s center of the mass is located at $\left(0, \dfrac{16}{3\pi}\right)$.

**Find the centroid of the region enclosed by the curve of $y = \sin x$ over the interval, $\left[0, \dfrac{\pi}{2}\right]$.**

*Example 2*__Solution__Hereâ€™s the sketch of the enclosed region and you use the graph as a guide. Weâ€™re looking for the actual coordinates of the center of mass (or centroid). This is marked as the black point on the $xy$-plane.

\begin{aligned}u &= x\end{aligned} | \begin{aligned}dv &= \sin x \phantom{x}dx\end{aligned} |

\begin{aligned}du &= dx\end{aligned} | \begin{aligned}v &= -\cos x\end{aligned} |

**Find the center of mass of the region enclosed by the graphs of $y = 2x$ and $y = x^2$.**

*Example 3*__Solution__Graph the two curves to check which of the two is lying above the other. The graph can also show us the intersection between two points. We can also find the points of intersection algebraically by equating $2x$ and $x^2$.

$\boldsymbol{x}$-coordinate |
$\boldsymbol{y}$-coordinate |

\begin{aligned}\overline{x} &= \dfrac{1}{A} \int_{a}^{b} x\left[ f(x)- g(x) \right]\phantom{x} dx\end{aligned} | \begin{aligned}\overline{y} &= \dfrac{1}{A} \int_{a}^{b} \dfrac{1}{2}\left\{[f(x)]^2 – [g(x)]^2\right\}\phantom{x} dx\end{aligned} |

\begin{aligned}\boldsymbol{\overline{x}}\end{aligned} | \begin{aligned}\boldsymbol{\overline{y}}\end{aligned} |

\begin{aligned}\overline{x} &= \dfrac{1}{4/3} \int_{0}^{2} x(2x- x^2)\phantom{x} dx\\&= \dfrac{3}{4}\int_{0}^{2} (2x^2 -x^3)\phantom{x}dx\\&= \dfrac{3}{4}\left[2\int_{0}^{2} x^2\phantom{x}dx -\int_{0}^{2} x^3\phantom{x}dx\right]\\&= \dfrac{3}{4}\left\{2\left[\dfrac{x^3}{3} \right]_{0}^{2} -\left[\dfrac{x^4}{4} \right]_{0}^{2}\right\}\\&= \dfrac{3}{4}\left[2\left(\dfrac{2^3}{3} -0 \right )-\left(\dfrac{2^4}{4} -0 \right ) \right ]\\&= 1\end{aligned} | \begin{aligned}\overline{y} &= \dfrac{1}{4/3} \int_{0}^{2} \dfrac{1}{2}\left[ (2x)^2- (x^2)^2\right ]\phantom{x} dx\\&= \dfrac{3}{4}\cdot\dfrac{1}{2}\int_{0}^{2} (4x^2 -x^4)\phantom{x}dx\\&= \dfrac{3}{8}\left[4\int_{0}^{2} x^2\phantom{x}dx -\int_{0}^{2} x^4\phantom{x}dx\right]\\&= \dfrac{3}{8}\left\{4\left[\dfrac{x^3}{3} \right]_{0}^{2} -\left[\dfrac{x^5}{5} \right]_{0}^{2}\right\}\\&= \dfrac{3}{8}\left[4\left(\dfrac{2^3}{3} -0 \right )-\left(\dfrac{2^5}{5} -0 \right ) \right ]\\&= \dfrac{8}{5}\end{aligned} |

**Practice Questions**

1. What is the center of mass of the plate that is bounded by the region shown below?
**Answer Key**

1. $\left( 0, \dfrac{20}{3\pi}\right)$
2. $\left( 0, \dfrac{4r}{3\pi}\right)$
3. $\left(\dfrac{\pi}{2} – 1, \dfrac{\pi}{8}\right)$
4. $\left(-1, -\dfrac{16}{5}\right)$
5. $\left( \dfrac{3}{5}, \dfrac{12}{35}\right)$
*Images/mathematical drawings are created with GeoGebra.*

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