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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

x = i = 1 n m i x i 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 = 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 = i = 1 n m i x i

and the center of mass of the system is given by

x = 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:

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 .

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:

M = i = 1 4 m i x i = −60 + 15 + 60 45 = −30.

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

m = i = 1 4 m i = 30 + 5 + 10 + 15 = 60 kg .

Then we have

x = M m = −30 60 = 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:

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 .

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

M = 24 , x = 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 = 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 = i = 1 n m i y i and M y = i = 1 n m i x i .

Also, the coordinates of the center of mass ( x , y ) of the system are

x = M y m and y = 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):

m 1 = 2 kg, placed at ( −1 , 3 ) , m 2 = 6 kg, placed at ( 1 , 1 ) , m 3 = 4 kg, placed at ( 2 , −2 ) .

Find the center of mass of the system.

First we calculate the total mass of the system:

m = i = 1 3 m i = 2 + 6 + 4 = 12 kg .

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

M y = i = 1 3 m i x i = −2 + 6 + 8 = 12 , M x = i = 1 3 m i y i = 6 + 6 8 = 4.

Then we have

x = M y m = 12 12 = 1 and y = M x m = 4 12 = 1 3 .

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

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Practice Key Terms 6

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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