xaktly | Physics

Center of mass

Center of mass = balance point

Every physical object has a center of mass, or a single point at which the object would balance of suspended by that point.

It's not too difficult, and often very helpful, to calculate the center of mass, either of masses on a point, or arranged in a plane or in 3D space.

One-dimensional COM

The simplest system for which to calculate a center of mass is masses arranged on a line. It makes sense, for example, that the center of mass of two 1 Kg masses separated by 1 meter is halfway in between. It's the point at which, if the two masses were connected by a light rod (light because we want to be able to ingore its mass), the dumbell-looking object would balance.

The location of the center of mass of N masses arranged on a line can be calculated using the formula:

$$com = \frac{1}{M} \sum_{i = 1}^N \; m_i \cdot x_i$$

where mi is the mass of the ith object, xi is the location (say, on a ruler) of the ith object, and M is the sum of all N masses. The index of the sum runs from i = 1 to i = N.

Here's an example of how it works. Consider three masses arranged on a line like the ones below, a 5 Kg mass and two 2 Kg masses.

A "ruler" has been drawn in below. As we will see later, the location of the zero will have no effect; we'll always be able to get the location of the center of mass (com) just fine.

Now applying our formula, we get

$$ \begin{align} com &= \frac{1}{9 \; Kg} [(5 \; Kg)(-4 \; m) + 2 \cdot 1 + 2 \cdot 8] \\[5pt] &= -0.22 \; m \end{align}$$

Now that makes some sense. We expect the com to be located nearer to the larger mass. It's kind of like "where to you grab a shovel?" Of course, we know from experience that a shovel balances if we pick it up closer to the blade (heavier) end.

Here's our picture with the com sketched in:

What if we change the position of the ruler?

Now let's look at the same example, but this time we'll change our scale:

It's the same set of masses, separated by the same distances, but now the calculation looks like this:

$$ \begin{align} com &= \frac{1}{9 \; Kg} [(5 \; Kg)(0 \; m) + 2 \cdot 5 + 2 \cdot 12] \\[5pt] &= 3.78 \; m \end{align}$$

The label of our com has changed, but its real location has has not. Take a look:

It's in the same place; only it's numerical label has changed.

The relative scale of the measurement has no effect on the calculation of the center of mass.

Center of mass in two dimensions

We can also just as easily calculate the center of mass of a system of masses arranged in a two-dimensional lattice (grid). Here's an example:

The trick is to first mentally "collapse" all of the masses down to the x-axis, and calculate the x-coordinate of the center of mass. Here's the calculation:

$$ \begin{align} com_x &= \frac{1}{19}[5\cdot 2 + 2 \cdot 4 + 2 \cdot 7 + 5 \cdot 8 + 5 \cdot 10] \\[5pt] &= 6.42 \; m \end{align}$$

Likewise, we "collapse" all of the masses to the left (the y-axis) in order to calculate the y-coordinate of the com:

$$ \begin{align} com_x &= \frac{1}{19}[5\cdot 10 + 5 \cdot 9 + 2 \cdot 6 + 5 \cdot 5 + 2 \cdot 1] \\[5pt] &= 7.05 \; m \end{align}$$

So the com of this system is located (in this coordinate system) at (6.42, 7.05).

Here it is:

The centroid

Center of a mass distribuition

In our examples above, calculation of the center of mass was relatively easy because we used "point masses." That is, we assumed that all of the mass of each object was located at a single point on a line or on a plane. It is, however, possible to find the center of mass of objects that are more complicated. For example, think of a frisbee. We know that a Frisbee balances at the center – that's intuitive. There are a number of ways to calculate that, though. For one, we might approximate

The diagram below shows how we could represent a Frisbee with three, six and twenty fixed masses. It's easy to see that if we just kept adding masses and reducing their size (so that the sum of all masses would add to the mass of the Frisbee, we approximate the distribution of mass quite well. It's easy for any of those approximations to show that the center of mass of the Frisbee is in the center of the disk.

One interesting use for center of mass (really center of population) is in calculating the population centroid of the US. The idea is that we take the area of the 48 contiguous states, and approximate that each person has the same mass.

Knowing the locations of people from the census, we can then calculate the "balance point" of the country (minus Alaska, Hawaii and the territories).

Why is this useful?

Big companies that ship in the US often locate their main shipping hubs near the population centroid, often in Tennessee or Kentucky. FedEx is one of them. Shipping from there, reduces overall shipping costs, and the savings can be passed along to customers.

The center of mass of a Frisbee can be located by approximating the continuous disk as a distribution of discrete small masses.

Creative Commons License   optimized for firefox
xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.