Ulearngo

Center of Mass and Moments

Let’s begin by looking at the center of mass in a one-dimensional context. Consider a long, thin wire or rod of negligible mass resting on a fulcrum, as shown in Figure 6.62(a). Now suppose we place objects having masses \(m_{1}\) and \(m_{2}\) at distances \(d_{1}\) and \(d_{2}\) from the fulcrum, respectively, as shown in Figure 6.62(b).

This figure has two images. The first image is a horizontal line on top of an equilateral triangle. It represents a rod on a fulcrum. The second image is the same as the first with two squares on the line. They are labeled msub1 and msub2. The distance from msub1 to the fulcrum is dsub1. The distance from msub2 to the fulcrum is dsub2.
Figure 6.62 (a) A thin rod rests on a fulcrum. (b) Masses are placed on the rod.

The most common real-life example of a system like this is a playground seesaw, or teeter-totter, with children of different weights sitting at different distances from the center. On a seesaw, if one child sits at each end, the heavier child sinks down and the lighter child is lifted into the air. If the heavier child slides in toward the center, though, the seesaw balances. Applying this concept to the masses on the rod, we note that the masses balance each other if and only if \(m_{1} d_{1} = m_{2} d_{2} .\)

In the seesaw example, we balanced the system by moving the masses (children) with respect to the fulcrum. However, we are really interested in systems in which the masses are not allowed to move, and instead we balance the system by moving the fulcrum. Suppose we have two point masses, \(m_{1}\) and \(m_{2} ,\) located on a number line at points \(x_{1}\) and \(x_{2} ,\) respectively (Figure 6.63). The center of mass, \(\overset{–}{x} ,\) is the point where the fulcrum should be placed to make the system balance.

This figure is an image of the x-axis. On the axis there is a point labeled x bar. Also on the axis there is a point xsub1 with a square above it. Inside of the square is the label msub1. There is also a point xsub2 on the axis. Above this point there is a square. Inside of the square is the label msub2.
Figure 6.63 The center of mass \(\overset{–}{x}\) is the balance point of the system.

Thus, we have

\[\begin{aligned} m_{1} \left|\right. \overset{–}{x} - x_{1} \left|\right. & = & m_{2} \left|\right. x_{2} - \overset{–}{x} \left|\right. \\ m_{1} \left(\right. \overset{–}{x} - x_{1} \left.\right) & = & m_{2} \left(\right. x_{2} - \overset{–}{x} \left.\right) \\ m_{1} \overset{–}{x} - m_{1} x_{1} & = & m_{2} x_{2} - m_{2} \overset{–}{x} \\ \overset{–}{x} \left(\right. m_{1} + m_{2} \left.\right) & = & m_{1} x_{1} + m_{2} x_{2} \\ \overset{–}{x} & = & \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} . \end{aligned}\]

The expression in the numerator, \(m_{1} x_{1} + m_{2} x_{2} ,\) is called the first moment of the system with respect to the origin. If the context is clear, we often drop the word first and just refer to this expression as the moment of the system. The expression in the denominator, \(m_{1} + m_{2} ,\) is the total mass of the system. Thus, the center of mass of the system is the point at which the total mass of the system could be concentrated without changing the moment.

This idea is not limited just to two point masses. In general, if n masses, \(m_{1} , m_{2} ,\ldots , m_{n} ,\) are placed on a number line at points \(x_{1} , x_{2} ,\ldots , x_{n} ,\) respectively, then the center of mass of the system is given by

\[\overset{–}{x} = \frac{\sum_{i = 1}^{n} m_{i} x_{i}}{\sum_{i = 1}^{n} m_{i}} .\]

Theorem 6.9

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 \(m = \sum_{i = 1}^{n} m_{i}\) denote the total mass of the system. Then, the moment of the system with respect to the origin is given by

\[M = \sum_{i = 1}^{n} m_{i} x_{i}\]

and the center of mass of the system is given by

\[\overset{–}{x} = \frac{M}{m} .\]

We apply this theorem in the following example.

Example 6.29

Finding the Center of Mass of Objects along a Line

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

\[\begin{aligned} m_{1} = 30 \text{kg}, \text{placed at} x_{1} = −2 \text{m} & & & m_{2} = 5 \text{kg}, \text{placed at} x_{2} = 3 \text{m} \\ m_{3} = 10 \text{kg}, \text{placed at} x_{3} = 6 \text{m} & & & m_{4} = 15 \text{kg}, \text{placed at} x_{4} = −3 \text{m} . \end{aligned}\]

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

Solution

First, we need to calculate the moment of the system:

\[\begin{aligned} M & = \sum_{i = 1}^{4} m_{i} x_{i} \\ & = −60 + 15 + 60 - 45 = −30. \end{aligned}\]

Now, to find the center of mass, we need the total mass of the system:

\[\begin{aligned} m & = \sum_{i = 1}^{4} m_{i} \\ & = 30 + 5 + 10 + 15 = 60 \text{kg} . \end{aligned}\]

Then we have

\[\overset{–}{x} = \frac{M}{m} = \frac{−30}{60} = - \frac{1}{2} .\]

The center of mass is located 1/2 m to the left of the origin.

Checkpoint 6.29

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

\[\begin{aligned} m_{1} = 12 \text{kg}, \text{placed at} x_{1} = −4 \text{m} & & & m_{2} = 12 \text{kg}, \text{placed at} x_{2} = 4 \text{m} \\ m_{3} = 30 \text{kg}, \text{placed at} x_{3} = 2 \text{m} & & & m_{4} = 6 \text{kg}, \text{placed at} x_{4} = −6 \text{m} . \end{aligned}\]

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

We can generalize this concept to find the center of mass of a system of point masses in a plane. Let \(m_{1}\) be a point mass located at point \(\left(\right. x_{1} , y_{1} \left.\right)\) in the plane. Then the moment \(M_{x}\) of the mass with respect to the x-axis is given by \(M_{x} = m_{1} y_{1} .\) Similarly, the moment \(M_{y}\) with respect to the y-axis is given by \(M_{y} = m_{1} x_{1} .\) Notice that the x-coordinate of the point is used to calculate the moment with respect to the y-axis, and vice versa. The reason is that the x-coordinate gives the distance from the point mass to the y-axis, and the y-coordinate gives the distance to the x-axis (see the following figure).

This figure has the x and y axes labeled. There is a point in the first quadrant at (xsub1, ysub1). This point is labeled msub1.
Figure 6.64 Point mass \(m_{1}\) is located at point \(\left(\right. x_{1} , y_{1} \left.\right)\) in the plane.

If we have several point masses in the xy-plane, we can use the moments with respect to the x- and y-axes to calculate the x- and y-coordinates of the center of mass of the system.

Theorem 6.10

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 \(\left(\right. x_{1} , y_{1} \left.\right) , \left(\right. x_{2} , y_{2} \left.\right) ,\ldots , \left(\right. x_{n} , y_{n} \left.\right) ,\) respectively, and let \(m = \sum_{i = 1}^{n} m_{i}\) 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

\[M_{x} = \sum_{i = 1}^{n} m_{i} y_{i} \text{and} M_{y} = \sum_{i = 1}^{n} m_{i} x_{i} .\]

Also, the coordinates of the center of mass \(\left(\right. \overset{–}{x} , \overset{–}{y} \left.\right)\) of the system are

\[\overset{–}{x} = \frac{M_{y}}{m} \text{and} \overset{–}{y} = \frac{M_{x}}{m} .\]

The next example demonstrates how to apply this theorem.

Example 6.30

Finding the Center of Mass of Objects in a Plane

Suppose three point masses are placed in the xy-plane as follows (assume coordinates are given in meters):

\[\begin{aligned} m_{1} = 2 \text{kg},\text{ placed at} \left(\right. −1 , 3 \left.\right) , \\ m_{2} = 6 \text{kg},\text{ placed at} \left(\right. 1 , 1 \left.\right) , \\ m_{3} = 4 \text{kg},\text{ placed at} \left(\right. 2 , −2 \left.\right) . \end{aligned}\]

Find the center of mass of the system.

Solution

First we calculate the total mass of the system:

\[m = \sum_{i = 1}^{3} m_{i} = 2 + 6 + 4 = 12 \text{kg} .\]

Next we find the moments with respect to the x- and y-axes:

\[\begin{aligned} \\ \\ M_{y} = \sum_{i = 1}^{3} m_{i} x_{i} = −2 + 6 + 8 = 12 , \\ M_{x} = \sum_{i = 1}^{3} m_{i} y_{i} = 6 + 6 - 8 = 4. \end{aligned}\]

Then we have

\[\overset{–}{x} = \frac{M_{y}}{m} = \frac{12}{12} = 1 \text{and} \overset{–}{y} = \frac{M_{x}}{m} = \frac{4}{12} = \frac{1}{3} .\]

The center of mass of the system is \(\left(\right. 1 , 1 / 3 \left.\right) ,\) in meters.

Checkpoint 6.30

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

\[\begin{aligned} m_{1} = 5 \text{kg},\text{ placed at} \left(\right. −2 , −3 \left.\right) , \\ m_{2} = 3 \text{kg},\text{ placed at} \left(\right. 2 , 3 \left.\right) , \\ m_{3} = 2 \text{kg},\text{ placed at} \left(\right. −3 , −2 \left.\right) . \end{aligned}\]

Find the center of mass of the system.

This lesson is part of:

Applications of Integration

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