# Example: Center of mass

#### Prerequisites

## Understanding the situation

In our discussion of the center of mass, we did everything with abstract equations. Let's do some examples that show how they play out in explicit cases. These illustrate two important general ideas:

- In applying general (powerful) equations to specific cases, putting things in terms of a specific well-chosen coordinate system helps a lot in carrying out the calculation.
- Considering limiting cases can help a lot in making sense of what an equation is telling you.

## Presenting a sample problem

Let's begin by considering the two-mass case that we considered in the introduction to the idea of center of mass (CM), explicitly put in a coordinate system, and carry out the calculation to find the position of the CM.

In the window below, we have created a geogebra window that allows you to change each of two masses connected by a massless rod and to see what happens to the position of the CM as a result.

(a) For an object consisting of two masses, show that the CM of the combined object is along the line connecting them and find an expression showing how far along that line the CM is.

(b) For the three cases m_{A} = m_{B} , m_{A }>> m_{B} , and m_{A} << m_{B} , show that the result makes physical sense.

## Solving this problem

(a) The equation for the position of the CM of two objects is

$$ \overrightarrow{r}_{CM} = (\frac{m_A}{m_A + m_B}) \overrightarrow{r}_A + (\frac{m_B}{m_A + m_B}) \overrightarrow{r}_B $$

Let's choose our coordinate system so that mass A is at the origin. This makes our calculation a lot easier. Since we will not be changing x and y directions during this problem, we can take i and j as fixed and therefore use (x,y) notation. Mass A is at the origin so its position vector is (0,0). Mass B is on the x axis a distance *L* from the origin so its position vector is (*L*,0). Our result for the CM vector is therefore

$$ \overrightarrow{r}_{CM} = (\frac{m_A}{m_A + m_B}) (0,0)+ (\frac{m_B}{m_A + m_B}) (L,0) $$

$$ \overrightarrow{r}_{CM} = (\frac{m_B}{m_A+m_B}L,0)$$

Since there is no y component, the answer lies along the x-axis. Since $m_B$ is always less than $m_A+m_B$, the fraction in front of the $L$ must be between 0 (if $m_B = 0$) and 1 (if $m_B$ gets very large so $m_A$ can be ignored). Therefore, the CM of the two masses lies along the line connecting them a distance $\frac{m_B }{m_A + m_B}L$ from mass A.

(b) If the two masses are both equal (to, say, m), we get the result is mL/2m = L/2, halfway between the two masses. This agrees with what symmetry demands.

If mass A gets very large compared to B, the denominator of the fraction *m*_{B}/(*m*_{A}+*m*_{B}) gets large while the numerator stays fixed, so the fraction gets small (approaches 0). So if the mass A is much bigger, the CM is near 0 -- almost at the same position as mass A.

If mass B gets very large compared to A, you can ignore the *m*_{A} in the denominator of the fraction *m*_{B}/(*m*_{A}+*m*_{B}), so the fraction gets approaches 1. So if the mass B is much bigger, the CM is almost at the same position as mass B.

These both make sense and the symmetry of the result is important. We could have chosen either mass to be at the origin. Notice that since we have described where the CM is with respect to the masses rather than in terms of its coordinates, the result does NOT depend on the coordinate system chosen -- which was arbitrary!

This calculation illustrates a number of useful general ideas: (1) the choice of your coordinate system can simplify your calculations; (2) taking limiting cases (a parameter gets large or goes to 0) can help you both check your work and see that it makes sense; and (3) expressing your results in terms of physical things rather than coordinates gives you a result that doesn't depend on an arbitrary choice that you are making -- and that you might make differently the next time!

Joe Redish and Mark Eichenlaub 8/17/15

Last Modified: February 24, 2019