6.6 Moments and centers of mass (Page 2/14)

Calculus volume 1 Page 2 / 14

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

\bar{x} = \frac{\sum_{i = 1}^{n} m_{i} x_{i}}{\sum_{i = 1}^{n} m_{i}} .

Center of mass of objects on a line

Let $m_{1}, m_{2},…, m_{n}$ be point masses placed on a number line at points $x_{1}, x_{2},…, x_{n},$ 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

\bar{x} = \frac{M}{m} .

We apply this theorem in the following example.

Finding the center of mass of objects along a line

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

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

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

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

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

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

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

Then we have

\bar{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.

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

\begin{matrix} 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{matrix}

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

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

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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 $(x_{1}, y_{1})$ 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. — Point mass $m_{1}$ is located at point $(x_{1}, y_{1})$ 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.

Center of mass of objects in a plane

Let $m_{1}, m_{2},…, m_{n}$ be point masses located in the xy -plane at points $(x_{1}, y_{1}), (x_{2}, y_{2}),…, (x_{n}, y_{n}),$ 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} and M_{y} = \sum_{i = 1}^{n} m_{i} x_{i} .

Also, the coordinates of the center of mass $(\bar{x}, \bar{y})$ of the system are

\bar{x} = \frac{M_{y}}{m} and \bar{y} = \frac{M_{x}}{m} .

The next example demonstrates how to apply this theorem.

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{matrix} m_{1} = 2 kg, placed at (−1, 3), \\ m_{2} = 6 kg, placed at (1, 1), \\ m_{3} = 4 kg, placed at (2, −2) . \end{matrix}

Find the center of mass of the system.

First we calculate the total mass of the system:

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

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

\begin{matrix} 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{matrix}

Then we have

\bar{x} = \frac{M_{y}}{m} = \frac{12}{12} = 1 and \bar{y} = \frac{M_{x}}{m} = \frac{4}{12} = \frac{1}{3} .

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

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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