## Fun Fact - Center of Mass

The center of area of a shape S in the plane is a point c, such that any line l drawn through c cuts S into two pieces, and each piece has half the area of S. If S is a circle then c is the center. Any line l drawn through c becomes a diameter, and half the circle is on one side and half is on the other. If S is an ellipse, the center still works. Draw a line l through c, even a slanted line, and the region on one side of l can be spun around to coincide with the region on the other side of l. The same holds for a rectangle or a parallelogram. Whenever S has rotational symmetry, i.e. it looks the same when spun around 180 degrees, then its center is in fact the center of area.

Most shapes do not have a center of area. There is no such point c. I'll illustrate by pasting six squares together to make some stairs. Put three squares on the bottom row, two squares on the next row, and one square at the top, all left justified. S looks like a set of stairs going up and to the left. For convenience, color the left two squares on the bottom and middle rows red, and color the rightmost and upper most squares blue. Thus two blue squares hang off a block of four red squares.

Let l be a vertical line and sweep l from left to right. At the start, all of S is on the right side of l. As l moves from left to right, more of S moves to the left of l, until finally all of S is on the left. At some point in this process,l cuts S in half. This happens when l runs along the right edge of the first column of squares. There are three squares to the left of l and three squares to the right. If c exists it must lie on this vertical line.

Let l be a horizontal line and sweep l from bottom to top. l cuts S in half when l separates the bottom row of squares from the top two rows. If c exists it must lie on this horizontal line. Therefore c is at the center of the four red squares.

If l runs up and to the right at a 45 degree angle, passing through c, there is no trouble. By symmetry, half of S lies on one side of l and half on the other. However, let l run down and to the right at a 45 degree angle. The region below l consists of one whole red square and two half red squares, giving an area of 2. The region above l has an area of 4. This staircase shape has no center of area. Indeed, most shapes do not have a center of area.

There is another concept called center of mass which is well defined for every shape. If c is the center of mass of S, also called the centroid of S, or the center of gravity, then S can be supported at the point c and it will perfectly balance. Again, the center of a circle will work. Take a disk made of metal or wood and place it on the tip of an upturned pencil, or the Eifel Tower, or any such structure, and the disk balances.

Balance brings in the concept of torque. Imagine a long steel rod, thin and light, supported at the center c. Place a weight on each end of the rod, like a dumbbell. Your life experience tells you the rod is in balance. It will not tip left or right. Now move the left weight in towards the center. There is just as much weight on either side of c as there was before, but now the rod tips over to the right. The distance from c matters. A weight that is close to c doesn't do much, but a weight far away from c tips the rod over. In fact, torque is linear with distance. Move a weight twice as far from c and it exerts twice the torque. This inspires the following definition of center of mass.

The point c is the centroid of a shape S if the following holds for every line l passing through c. The aggregate sum, or integral if you prefer, of area times distance from l, to the right of l, equals the integral of area times distance from l, to the left of l. The torque is the same on either side, and S balances upon l.

Look at a square whose sides have length 2, with c at the center. Let c be the origin in the plane. Draw a vertical line l through c, cutting the square in half. Thus l is the y axis. As x marches from 0 to 1, it defines a vertical stripe of height 2. This is multiplied by the distance to l, which is x. Integrate 2x from 0 to 1 and get 1. Moving to the left of l, let x run from -1 to 0. The distance to l is now -x, and the integral of -2x from -1 to 0 is 1, so we're good. Both sides of l come out the same. Of course we knew they wood by symmetry. For each x there is a stripe of height 2 at a distance x from l on either side of l. It has to come out the same.

A shortcut in the above is to assume distance to the right of l is positive, and distance to the left of l is negative; then the entire integral of distance times area comes out zero. In our square example we have the integral from -1 to 1 of 2x. Evaluate x2 at 1 and -1, and 1-1 = 0.

The square is rather trivial, because of its symmetry, so let's go back to the stair shape shown above. Draw a 45 degree line from lower left to upper right, passing through the red squares, and S is symmetric about this line. For each distance d from l, the amount of S cut by a line parallel to l, at a distance d to the right of l, is the same as the amount of S cut by a line parallel to l at a distance d to the left of l. The integral comes out 0, and c must lie on this line. The center of mass is somewhere on the diagonal through the red squares from lower left to upper right. This is rather intuitive. Picture the shape as a rigid structure and balance it on a point; that point has to be on the diagonal line, and it probably lies just inside the upper right red square.

With this in mind, let's find c. Place the origin at the lower left of S, and let each square be one unit on a side. Thus S runs from 0 to 3 along the x axis and from 0 to 3 up the y axis. Place c somewhere in the upper right red square and draw a vertical line l through c. Compute the integral in three parts, for the first column of squares, the second column of squares, and the last blue square on the right. In the first integral, x runs from 0 to 1, the height is 3, and the distance to l is x-c. (In fact the distance to l will always be x-c.) The integrand is 3(x-c), the integral is 3x2/2 - 3cx, and when you plug in 0 and 1 for x, the result is 3/2 - 3c.

Move to the second column of squares and the integrand is 2(x-c), giving an integral of x2 - 2cx, and a result of 3 - 2c.

Let x run from 2 to 3 to cover the last blue square. The integrand is x-c, the integral is x2/2 - cx, and the result is 5/2 - c. Add these three pieces together and get 7 - 6c. This has to equal 0, hence c = 7/6. The center of mass has the coordinates [7/6,7/6]. c is just a bit up and to the right of the center of the four red squares, which is what you would expect from intuition. S will balance precisely at this point.

We haven't yet proved that the centroid is well defined. If you find c by an x coordinate calculation as above, and a y coordinate calculation, then every line l passing through c balances - every line l passing through c admits a distance * area integral of 0. S tips neither left nor right upon l, and S is perfectly balanced at c. The proof isn't hard, but it entails some algebra, so I'll just give an overview.

Put the centroid at the origin. Area is the double integral of 1 over the region S. (Every time I say double integral in this paragraph, I mean the double integral over the region S.) To balance on the y axis, multiply by the distance from the y axis, which is x. Thus the double integral of x equals 0. Similarly, the double integral of y equals 0. Now draw another line l through the origin at an angle θ. this line is the u axis, and the perpendicular is the v axis. Here is the map from the xy plane to the uv plane.

u = cos(θ) x + sin(θ) y
v = -sin(θ) x + cos(θ) y

We want the double integral of v. Integrate by substitution and the integrand becomes -sin(θ)x + cos(θ)y. (The jacobian comes out 1.) change this to a linear combination of double integrals over x and y. Those are both 0, hence the integral of v is 0, hence c is the centroid for any line l.

### Centroid of a Triangle

There is a wonderful procedure to find the centroid of any triangle. Draw a line segment from each corner to the midpoint of the opposite side. these are the three medians of the triangle, and they intersect in the centroid. The proof is simple and elegant. Consider the triangle ABC, and let M be the midpoint of BC. Orient ABC so that AM runs along the x axis. Find the double integral over the triangle ABC of y, the distance to AM. Remember that M is midway between B and C. Since BC is a line, the y coordinate of B is minus the y coordinate of C. Call this y coordinate h, thus y runs from -h to h. The horizontal stripes parallel to AM have the length of AM at y = 0, and shrink linearly to 0 as y runs from 0 up to h and as y runs from 0 down to -h. The integral is symmetric about the x axis, and comes out 0. The centroid lies somewhere on the line AM. This holds for each of the three medians, hence the medians intersect, and that intersection is the center of gravity. The triangle will balance on this point.

To illustrate, let S be the right triangle with corners [0,0], [1,0], and [0,1]. Draw the medians from [1,0] to [0,½], and from [0,1] to [½,0]. These intersect in [⅓,⅓]. This is the centroid of the triangle. Each median divides the triangle into two equal areas, but if you pass a vertical line through c, the two pieces have 4/9 and 5/9 the area of S. As mentioned earlier, few shapes have a true center of area.

### Puzzle

Find a shape that has a well defined center of area that is different from its center of mass.

Answer: Let the origin be the center of area of S. Draw a circle of radius 1 about the origin, and a circle of radius 1.2. The ring between these circles is called an annulus. Let S contain a section of this annulus about -1, having a span of 10 degrees.

Next imagine circles of radius 2 and approximately 2.1. Let S contain a section of this annulus about 2, having a span of 10 degrees. As the line l pivots about the origin, it sweeps through equal areas of S on the left and on the right. The arc on the right is thinner, about half as thin as the arc on the left, but l passes through its length twice as fast. The origin is the center of area, but the center of mass is about halfway between the two pieces of S, at x = (approximately) ½. If you are bothered by S being disconnected then let a very thin, carefully crafted rod join the two annular arcs.