Centers of Mass

We will now begin discussing the idea of weighted averages which arises in several interesting ways. As we will see, this will give us a means of condensing a vast amount of data which, say, comes out of an experiment into a few pieces of information which somehow characterize the data. Before we do that, however, let's think about a simpler kind of problem: determining how to balance a beam with various weights placed on it. This is the problem of the teeter-totter.


The Law of the Lever

First, we will consider what happens when we have some masses $ m_i $ placed along a board at positions $ x_i $ . The situation is demonstrated below. Notice at first, the two masses are equal to one another and the beam is balanced by placing the fulcrum halfway in between the two. This means that the balance point is at the average of the two x coordinates.

Now vary one of the masses and notice that the balance point moves towards that mass: it is no longer at the average of the two x coordinates. You can also change the position of the masses, add new ones and remove old ones.

When there are two masses, the condition for balance is this: if $ l_1 $ is the distance from the first mass to the fulcrum and $ l_2 $ the distance from the second to the fulcrum, then :

\begin{eqnarray*} &\frac{l_1}{l_2} = \frac{m_2}{m_1} \\ \mbox{or }&m_1 l_1 = m_2 l_2 \end{eqnarray*}

This is known as the law of the lever . We can express is slightly differently in terms of the x coordinates. Namely, if $ \bar{x} $ is the coordinate of the fulcrum, then $ l_1 = \bar{x} - x_1 $ and $ l_2 = x_2 - \bar{x} $ . The law of the lever can then be rewritten as

\begin{eqnarray*} &m_1(\bar{x} - x_1) = m_2(x_2 - \bar{x}) \\ \mbox{or }&m_1(\bar{x} - x_1) + m_2(\bar{x} - x_2) = 0 \end{eqnarray*}

In the case that there are more masses, this condition becomes

\[ m_1(\bar{x} - x_1) + \ldots + m_i(\bar{x} - x_i) + \ldots + m_n(\bar{x} - x_n) = 0 \]

We can use this to find an expression for the balance point:

\begin{eqnarray*} m_1\bar{x} - m_1 x_1 +\ldots + m_i\bar{x} - m_ix_i + \ldots + m_n\bar{x} - m_nx_n & = & 0 \\ (m_1 + \ldots + m_n)\bar{x} - (m_1x_1 + \ldots + m_nx_n) & = & 0 \\ \bar{x} & = & \frac{m_1x_1 + \ldots + m_nx_n}{m_1 + \ldots + m_n} \end{eqnarray*}

It is typical to call the balance point $ \bar{x} $ the center of mass.

Observations:

  1. The term in the denominator is the total mass.

  2. If all the masses are equal to one another, say $ m_i = m $ , then

    \[ \bar{x} = \frac{m(x_1 + \ldots + x_n)}{mn} = \frac{x_1 + \ldots + x_n}{n} \]

    In other words, when all the masses are equal, the balance point is simply the average of the x coordinates.

More generally, we say that $ \bar{x} $ is the weighted average of the x coordinates. That is, points that have greater mass have their corresponding x coordinate count more in the determination of $ \bar{x} $ .


The Continuous Case

Now instead of the case when a set of discrete masses are laid out on a board, we will consider the case of a beam whose density varies continuously. By $ \rho(x) $ , we will denote the linear density whose units are say, kilograms per meter. Basically, it tells us the mass of a certain width of the beam.

To understand the center of mass in this situation, we will break up the beam into many little pieces. The width of each piece will be denoted by $ dx $ and the mass by $ dm $ . Now the density relates the mass of a little piece to its width: $ dm = \rho dx $ and the expression for the center of mass becomes

\[ \bar{x} = \frac{\int x~dm}{\int dm} = \frac{\int_0^L x\rho ~dx}{\int_0^L \rho ~dx} \]

Examples:

  1. Let's begin with the case where $ \rho $ is constant meaning that the mass of each little piece is the same. Then

    \[ \bar{x} = \frac{\int_0^L x\rho~dx}{\int_0^L \rho~dx} = \frac{\frac 12 \rho L^2}{\rho L} = \frac 12 L \]

    In other words, the center of mass is in the middle of the beam which is again the average of all the x coordinates. Notice again that the term in the denominator is the total mass of the beam.

  2. $ \rho(x) = x $

    \[ \bar{x} = \frac{\int_0^L x^2~dx}{\int_0^L x~dx} = \frac{\frac 13L^3}{\frac 12L^2} = \frac 23 L \]

    Here, the center of mass is drawn over to the right side of the beam. This is not too surprising if we consider the density function:

    Here we see that the beam is heavier on the right side and lighter on the left so the result is not too surprising.

  3. $ \rho(x) = \sin(\frac{\pi}{2L}x) $

    The graph of the density is shown below. Again the beam is heavier on the right than on the left so we expect the center of mass to lie towards the right hand side.

    Now the total mass is given by

    \[ \int_0^L \sin(\frac{\pi}{2L}x)~dx = -\frac{2L}{\pi} \cos(\frac{\pi}{2L}x)|_0^L = \frac{2L}{\pi} \]

    The integral in the numerator can be handled using integration by parts.

    \begin{eqnarray*} \int_0^l x\sin(\frac{\pi}{2L}x)~dx & = & -\frac{2L}{\pi}x\cos(\frac{\pi}{2L}x)|_0^L + \frac{2L}{\pi}\int_0^L \cos(\frac{\pi}{2L}x)~dx \\ & = & (\frac{2L}{\pi})^2 \sin(\frac{\pi}{2L}x)|_0^L \\ & = & \frac{4L^2}{\pi^2} \end{eqnarray*}

    This means that the center of mass is located at

    \[ \bar{x} = \frac{\int_0^L x\rho ~dx}{\int_0^L \rho ~dx} =\frac 2\pi L \approx 23 L \]


Centroids

We can also use these ideas to find the center of mass of two dimensional homogeneous objects. In this case, the center of mass is called the centroid. Let's look at an example.

Suppose we have a homogenous plate of constant density $ \rho $ in the shape of a semi-circle of radius r.

It will be helpful to put this onto a coordinate grid as:

We will denote the centroid, the point at which the plate would balance, by $ (\bar{x},\bar{y}) $ . One thing is clear: because the plate is symmetrical about the x axis, the centroid should lie on that axis; that is, $ \bar{y} = 0 $ . Our work is now reduced to finding $ \bar{x} $ .

To accomplish this, let's take a different view of the plate. Rather than looking at it from above, let's turn it so that we see it from the bottom side; that is, view it from the negative y axis. If we do that, what we see resembles the beam we have been working with.

If we think of slicing this beam up into small pieces of width $ dx $ , then we can see that the mass of each of these pieces will vary since we are in fact taking different slices of the plate. In particular, $ dm = \rho ~dA $ where $ dA $ is the area of the slice. To understand this quantity, we can approximate it by a rectangle of width $ dx $ . The height is given by the difference in the y coordinates at the top and bottom of the circle.

Since $ x^2 + y^2 = r^2 $ describes the circle, we can see that $ y = \pm \sqrt{r^2 - x^2} $ where the top of the cirlce is given by choosing the plus sign. This means that the height of the rectangle is $ 2\sqrt{r^2 - x^2} $ and hence the area is $ dA = 2\sqrt{r^2 - x^2}~dx $ .

Putting everything together, we have that $ dm = 2\rho\sqrt{r^2 - x^2}~dx $ and so

\[ \bar{x} = \frac{\int_0^r 2\rho x\sqrt{r^2 - x^2}~dx}{\int_0^r 2\rho\sqrt{r^2 -x^2}~dx} \]

The integral in the denominator is actually pretty easy: it is the area under the curve defining a circle. Since we only integrate from x = 0 to r , we have

\[ \int_0^r 2\rho\sqrt{r^2-x^2}~dx = \frac 14\pi r^2 2\rho = \frac{\rho\pi r^2}{2} \]

The integral in the numerator can be evaluated by the substitution $ u = r^2 - x^2 $ . Then we find that

\begin{eqnarray*} \int_0^r 2\rho x\sqrt{r^2 - x^2}~dx & = & 2\rho \int_{r^2}^0 \sqrt{u} (-\frac 12)~du \\ & = & -\frac 23 \rho u^{3/2}|_{r^2}^0 \\ & = & \frac 23 \rho r^3 \end{eqnarray*}

This means that

\[ \bar{x} = \frac{\frac 23 \rho r^3}{\frac 12 \rho\pi r^2} = \frac{4}{3\pi}r \]

In other words, the centroid lies at $ (\frac{4}{3\pi} r,0) $ . We can approximate the x coordinate by $ \frac{4}{3\pi} r \approx \frac 49 r $ . This shows that the centroid lies slightly closer to the left side of the plate than the right. This makes sense since there is more of the plate on the left.