1.7: Moments and Centers of Mass

Mass and Linear Density

Definition: Linear Density

The linear density of a one-dimensional object (a rod of negligible thickness) is mass per unit length. That is,

$$\rho = \dfrac{m}{L}. \nonumber$$

Theorem: Mass–Linear Density and Weight-Linear Weight Density Formulas of a One-Dimensional Object

Given a thin rod oriented along the $x$-axis over the interval $[a,b]$, let $\rho(x)$ denote a linear density function giving the density of the rod at a point $x$ in the interval. Then the mass of the rod is given by

$$m=\int ^b_a \rho(x) \, dx. \nonumber$$ If given the linear weight density function, $\gamma(x)$, then the weight of the rod is given by $$F = \int_a^b \gamma(x) \, dx. \nonumber$$

Lecture Example $\PageIndex{1}$

Consider a thin rod oriented on the $x$-axis over the interval $\left[ 0, \frac{\pi}{2}\right]$. If the linear density of the rod is given by $\rho(x)=\cos x$, what is the mass of the rod?

Mass and Radial Density

Definition: Area Density

The area density of a two-dimensional object (or a three-dimensional object of negligible thickness) is mass per unit area. That is,

$$\rho = \dfrac{m}{A}. \nonumber$$ If the density varies in terms of the radius from a central point, we call the area density function a radial density.

Theorem: Mass–Radial Density and Weight-Radial Weight Density Formulas of a Circular Object

Let $\rho(x)$ be an integrable function representing the radial density of a disk of radius $r$. Then the disk's mass is given by

$$m=\int ^r_0 2 \pi x\rho(x) \, dx. \label{massEq1}$$ If given the radial weight density function, $\gamma(x)$, the weight of the disk is given by $$F = \int^r_0 2 \pi x \gamma(x) \, dx. \label{weightEq1}$$

Lecture Example $\PageIndex{2}$

Let $\rho(x)=\cosh (x^2 + 1)$ represent the radial density of a disk. Calculate the mass of a disk of radius 4.

Center of Mass and Moments: The Finite 1-D Case

This entire subsection and the next one (The Finite 2-D Case) are covered in the following video:

Definition: Center of Mass

The point, $P$, on a thin plate of any shape where the plate will balance is called the center of mass.

Theorem: Archimedes' Law of the Lever

Given two masses, $m_1$ and $m_2$, on opposite sides of a fulcrum, each a distance of $d_1$ and $d_2$, respectively, from the fulcrum,

$$m_1 d_1 = m_2 d_2.\nonumber$$

Theorem: Center of Mass of Two Objects on a Line

If $\bar{x}$ is the center of mass on the $x$-axis and $x_1$ and $x_2$ are the placements along the $x$-axis of the masses, then

$$\bar{x} = \dfrac{m_1 x_1+m_2 x_2}{m_1+m_2}.\nonumber$$

Definition: Moment

$m_i x_i$ is called the moment of mass $i$ with respect to the origin.

Note

The moment represents the ability for a mass to rotate about the origin.

Theorem: Center of Mass of Objects on a Line

Let $m_1,m_2, \ldots ,m_n$ be point masses placed on a number line at points $x_1,x_2, \ldots ,x_n$, respectively, and let $m= \displaystyle \sum_{i=1}^n m_i$ denote the total mass of the system. Then, the moment of the system with respect to the origin is given by

$$M=\sum_{i=1}^n m_i x_i \nonumber$$ and the center of mass of the system is given by $$\bar{x}=\dfrac{M}{m}. \nonumber$$

Center of Mass and Moments: The Finite 2-D Case

Theorem: Center of Mass of Objects in a Plane

Let $m_1, m_2, \ldots , m_n$ be point masses located in the $xy$-plane at points $(x_1,y_1),(x_2,y_2), \ldots ,(x_n,y_n)$, respectively, and let $m = \displaystyle \sum_{i=1}^n m_i$ denote the total mass of the system. Then the moments $M_x$ and $M_y$ of the system with respect to the $x$- and $y$-axes, respectively, are given by

$$M_x = \sum_{i=1}^n m_i y_i \nonumber$$ and $$M_y = \sum_{i=1}^n m_i x_i. \nonumber$$ Also, the coordinates of the center of mass $(\bar{x},\bar{y})$ of the system are $$\bar{x} = \dfrac{M_y}{m} \nonumber$$ and $$\bar{y}=\dfrac{M_x}{m}. \nonumber$$

Note

Another way to write that last theorem is $m \bar{x} = M_y$. This states that, if all the mass was gathered at $\left(\bar{x},\bar{y}\right)$, its "tendency to rotate" about the $y$-axis would be $M_y$ and about the $x$-axis is $M_x$.

Lecture Example $\PageIndex{1}$

Four people are placed on a flat, rigid plane of negligible mass itself. Abby, Bela, Cameron, and Delia have masses 62 kg, 56 kg, 78 kg, and 70 kg, respectively. Their positions with respect to the "origin" are (-1,2), (4,8), (2,-1), and (-3,-7), respectively. Where is their center of mass?

Center of Mass: The Infinite Case

Definition: Centroid

The center of mass of a continuous object is called the centroid of the object.

Theorem: Symmetry Principle

If a region $\mathbf{R}$ is symmetric about a line $\mathcal{l}$, then the centroid of $\mathbf{R}$ lies on $\mathcal{l}$.

Theorem: Center of Mass of a Lamina Bounded by Two Functions

Let $\mathbf{R}$ denote a region bounded above by the graph of a continuous function $f(x)$, below by the graph of the continuous function $g(x)$, and on the left and right by the lines $x=a$ and $x=b$, respectively. Let $\rho$ denote the area density of the associated lamina. Then we can make the following statements:

1. The mass of the lamina is $$m = \dfrac{m}{A} \cdot A = \rho \cdot A = \rho \int ^b_a[f(x)−g(x)] \, dx. \nonumber$$
2. The moments $M_x$ and $M_y$ of the lamina with respect to the $x$- and $y$-axes, respectively, are $$M_x= \rho \displaystyle \int ^b_a \frac{1}{2} \left([f(x)]^2−[g(x)]^2\right) \, dx \nonumber$$ and $$M_y= \rho \int ^b_a x [f(x)−g(x)] \, dx. \nonumber$$
3. The coordinates of the center of mass $(\bar{x},\bar{y})$ are $$\bar{x}=\dfrac{M_y}{m} \nonumber$$ and $$\bar{y}=\dfrac{M_x}{m}. \nonumber$$

Note

Unless the density varies with position (which it will not for us), we can ignore $\rho$ when computing $\bar{x}$ and $\bar{y}$.

We illustrate this theorem in the following example.

Lecture Example $\PageIndex{2}$

Determine the centroid for the region bounded by $y = 2 \sin⁡(2x)$ and $y=0$ on the interval $\left[ 0 , \frac{\pi}{2} \right]$.

Hint

The antiderivative of $x \sin\left( 2x \right)$ is $-\frac{1}{2} x \cos\left( 2x \right) + \frac{1}{4} \sin\left( 2x \right)$.

Lecture Example $\PageIndex{3}$

Determine the centroid for the region bounded by $y = x^4$ and $y = x^6$ on the $x$-interval $\left[ 0,1 \right]$.

Theorem of Pappus

Theorem of Pappus for Volume

Let $\mathbf{R}$ be a region in the plane and let $\mathcal{l}$ be a line in the plane that does not intersect $\mathbf{R}$. Then the volume of the solid of revolution formed by revolving $\mathbf{R}$ around $\mathcal{l}$ is equal to the area of $\mathbf{R}$ multiplied by the distance, $d$, traveled by the centroid of $\mathbf{R}$.

Proof

We can prove the case when the region is bounded above by the graph of a function $f(x)$ and below by the graph of a function $g(x)$ over an interval $[a,b]$, and for which the axis of revolution is the $y$-axis. In this case, the area of the region is $A= \displaystyle \int ^b_a[f(x)−g(x)]\,dx$. Since the axis of rotation is the $y$-axis, the distance traveled by the centroid of the region depends only on the $x$-coordinate of the centroid, $\bar{x}$, which is $$x=\dfrac{M_y}{m}, \nonumber$$ where $$m= \rho \int ^b_a[f(x)−g(x)] \, dx \nonumber$$ and $$M_y= \rho \int ^b_ax[f(x)−g(x)] \, dx. \nonumber$$ Then, $$d = 2 \pi \dfrac{ \rho \int ^b_ax[f(x)−g(x)] \, dx }{ \rho \int ^b_a[f(x)−g(x)] \, dx} \nonumber$$ and thus $$d \cdot A = 2 \pi \int ^b_ax[f(x)−g(x)] \, dx. \nonumber$$ However, using the Method of Cylindrical Shells, we have $$V = 2 \pi \int ^b_ax[f(x)−g(x)] \, dx. \nonumber$$ So, $$V=d \cdot A \nonumber$$ and the proof is complete.

Lecture Example $\PageIndex{4}$

Find the volume of the solid obtained by rotating the triangular plate with vertices (2,3), (2,5), and (5,4) and having a constant density about the $x$-axis.