Features Extended:

  1. We can give a more rigorous proof that 2-dimensional systems of uniform density with at least 2 axes of reflection symmetry must have their centers of mass at the points where their reflection axes intersect.

    1. First, note a system with at least 2 axes of reflection symmetry CANNOT necessarily be rotated into itself by a rotation from one axis to the next. This statement is proven by a single counterexample to the opposite statement. A square is a good counterexample. A line that divides a square into equal rectangles and a diagonal are both reflection axes, but the square is NOT rotated into itself by rotating it from one to the other.

    2. Proof: Give a reflection axis the coordinate y and give a perpendicular axis the coordinate x. At every point y by reflection symmetry, there are 2 mass-weighted displacements specified by (-x)ρdxdy and xρdxdy, where ρ is the uniform density. The sum of these 2 mass-weighted displacements is 0. The entire x line at y consists of such canceling pairs, and thus that line's center of mass is at x = 0. If you sum on all the x line centers of mass, the center of mass of the system is found to be located on the y axis: i.e., on the reflection axis.

      Now consider a distinct second reflection axis. By the same argument as above, the center of mass must lie on this second reflection axis.

      Since there is only one center of mass, the center of mass must be at the intersection of the 2 distinct reflection axes. QED.

    3. Corollary: Since there was nothing special about the 2 reflection axes discussed in the above proof, all reflection axes of a system must pass through the center of mass.

    4. Corollary: Since members of some subsets (some/all of which may be the whole set) of reflection axes can be rotated around the center of mass into each other with the whole system being rotated into itself, the center of mass is the highest symmetry point of the system by a reasonable definition of highest symmetry point.

  2. Note that in most of the illustrated cases in the diagram, you can see obviously that the contributions to the center of mass from differential mass elements cancel pairwise about the point of highest symmetry, and so that point must be the center of mass. This mental trick does NOT work for the regular pentagon and pentagram. However, if you think of symmetrical sets of 5 points ...

File: Mechanics file: center_of_mass_2d_1bb.html.