Center of Mass

This is a useful thing to know about.

  • The center of mass (CM) or center of gravity is the balance point of an object, (but see the note at the bottom of the page if you want extreme precision- technically there can be a tiny difference between CM and CG).
  • When the momentum of a system is constant, the CM has zero acceleration. A discussion of this type of application is near the bottom of this page.

How to find the CM:

  1. If it has enough symmetry, the CM is the geometric center. A golf ball, a board that is straight and free of knots, a straight pipe are a few examples.
    CM almost at the geometric center: a sheet of paper with holes punched in it, a book, a bowling ball, a door, a planet.
  2. Balance the thing if possible. If you can hang it by a string, record the line extending straight down below where the string is attached; CM is on that line. Then hang it again from another point, and again record the line. The intersection of the two lines is the CM.
  3. If the object or group of objects is made of parts whose CMs are known, then choose an origin for your x, y and z axes, and calculate
    xcm = (S mi xi)/mtotal. Similarly for ycm and zcm. Note that the mass units cancel and you have distance.
  4. In general xcm = (ò xdm)/mtotal , so you need to find dm, the mass of an infinitesimal volume, in terms of dx. This is useful for shapes that can be expressed by equation(s). But in real life that is rare.

Here is an example of a calculation of the #3 type above. The picture below shows a table. The table top is 60.0 inches long, 1.00 inch thick and weighs 35.0 pounds. Each of the 4 legs is uniform, centered 4 inches from the table top edges, is 36.0 inches long and weighs 6.00 pounds. So the total weight is 59.0 pounds and the center of each leg is 19.0 inches below the top surface. The center of the table top is 1/2 inch below the top surface.

Using the upper left edge of the table top as the origin, we find xcm by symmetry is 30 inches. Taking y positive down,
y cm = (30× 0.5 + 24× 19)/59 = 7.98 inches below the top.

 

Note 1: Any units can be used, even weight units in place of mass units, but see note 2.

Note 2: If you want to be super fussy, weight varies with altitude, so the center mass is not exactly the same as the center of gravity. As far as I know, this fact is of no practical use whatsoever. A 1.00000 Newton weight at sea level weighs 0.99997 N at a 100 meter elevation, so a vertical 100 m rod has CG a gnat’s eyelash below the CM, but who cares?  So use them interchangeably except around an instructor who wants you to know the difference.

APPLICATION: No external forces- then acm = 0.
For example, a 200 lb fisherman is in a 300 lb boat, at rest. He moves 4 feet with respect to the boat toward shore. How much closer to the shore is he now? Neglect drag force of the water. The problem is, the boat moved away from the shore a distance x. So he is 4 - x closer to shore. Find x. Don't peek.

Solution 1: Call the two masses M and m. Choose the initial location of the boat's center of mass as the origin, and let's say the man is initially a distance xm from there, away from shore. Then
xcm = [M(0) + mxm]/(m+M). After the man moves in the negative direction a distance 4-x and the boat moves in the positive direction a distance x, we have
xcm = [Mx + m(xm - 4 + x)]/(m+M). Equate the two and note that the denominators cancel, then do some more algebra and get
x = 4m/(M+m) ft. Weight is proportional to mass, and the mass units cancel, so plug in 200 and 300 and get x = 1.6 ft, so he is 2.4 feet closer to shore. (4-1.6=2.4)

Solution 2: Think about this:
If S mx/(m+M) = constant, then S mx = constant, then
S mD x=0. So if the boat moved x in the positive direction and the man moved
4-x in the negative direction then
Mx + m(-4+x)=0, and solve this for x.

 

My main pages:
Mechanics
Fluids, heat, electricity and magnetism
Vibrations and waves
Quantum
brief definitions & principles in my index.  Comments, questions: fredrick.gram @ tri-c.edu  (remove spaces)