Suppose that there are three point masses arranged as shown in the figure at right. Where is the center of mass of this 3-object system?
So, the center of mass of the system is at the point (2.0 m, 1.7 m).
Note: It would have been quicker and easier to notice that the masses in the diagram at left are symmetric about x = 2 m, so the x-coordinate of the center of mass has to be 2.0 m.
Note: Given this arrangement of masses, you would reduce the amount of calculation by placing the origin of the coordinate system at the location of one of the points. I didn't do that for reasons you will soon see.
Suppose that a 2 m by 3 m rectangle is cut from a square piece of plywood which originally had sides of length 4 m (as shown at right). What is the center of mass of the resulting "U-shaped" piece of plywood?
Well, the center of mass of the of a homogeneous rectangle is in the geometric center of the rectangle (by symmetry). Why not divide the U-shaped piece into three rectangles as shown in the diagram at right - two 2m x 3 m rectangles and one 4 m x 1 m rectangle. (This is not the only way to divide the U-shape into rectangles, and any of the other ways will work just as well.)
The centers of mass of the three rectangles are indicated in the figure at right. (Does it look familiar?) We can now replace the rectangular sheets by their centers of mass, and using the area of the rectangle as a "stand-in" for its mass, you get:
Note: Compare this to the example above.
Moral: If you can "chop up" a continuous object into pieces whose centers of mass are easy to find, you can reduce the problem to finding the center of mass of a system of discrete points.
Perhaps chopping perfectly-good objects into little pieces does not appeal to you. Here's an alternative, based on the idea that the center of mass of an object is in the same place no matter how you calculate it.
The center of mass of the originall 4 m x 4 m piece of plywood is at its geometric center (middle dot at right), so the y-coordinate of the center of mass of the original square is 2 m.
On the other hand, you could find the center of mass of the original square by finding the center of mass of the 2-point system consisting of the centers of mass of the U-shape and the center of mass of the plywood that originally fit in the "hole."
where Y is the y-coordinate of the original square, au is the area of the U-shape, yu is the y-coordinate of the center of mass of the U-shape, ah = area of the "hole", yh is the y-coordinate of the plywood that originally filled the hole, and A is the area of the original square (which equals au + ah). A little algebra gives:
which matches the results above.
last update February 1, 2010 by JL Stanbrough