For any realistic object, carrying out the summations of Equations 12.4 over all the atoms in the object is not practical. Instead, as FIGURE 12.6 shows, we can divide an extended object into many small cells or boxes, each with the same very small mass $\text{Δ}m$. We will number the cells 1, 2, 3, …, just as we did the particles. Cell i has coordinates $\left(x_i\text{,}{y}_i\right)$ and mass $m_i=\text{Δ}m$. The center-of-mass coordinates are then

FIGURE 12.6 Calculating the center of mass of an extended object.

Now, as you might expect, we’ll let the cells become smaller and smaller, with the total number increasing. As each cell becomes infinitesimally small, we can replace $\text{Δ}m$ with $dm$ and the sum by an integral. Then

Equations 12.5 are a formal definition of the center of mass, but they are not ready to integrate in this form. First, integrals are carried out over coordinates, not over masses. Before we can integrate, we must replace $dm$ by an equivalent expression involving a coordinate differential such as $dx$ or $dy$. Second, no limits of integration have been specified. The procedure for using Equations 12.5 is best shown with an example.

EXAMPLE 12.2 The center of mass of a rod

Find the center of mass of a thin, uniform rod of length L and mass M. Use this result to find the tangential acceleration of one tip of a 1.60-m-long rod that rotates about its center of mass with an angular acceleration of $6.0\text{rad}\text{/}{\text{s}}^2$.

VISUALIZE FIGURE 12.7 shows the rod. We’ve chosen a coordinate system such that the rod lies along the x-axis from 0 to L. Because the rod is “thin,” we’ll assume that $y_{\text{cm}}=0$.

FIGURE 12.7 Finding the center of mass of a long, thin rod.

FIGURE 12.7 Full Alternative Text

SOLVE Our first task is to find $x_{\text{cm}}$, which lies somewhere on the x-axis. To do this, we divide the rod into many small cells of mass $dm$. One such cell, at position x, is shown. The cell’s width is $dx$. Because the rod is uniform, the mass of this little cell is the same fraction of the total mass M that $dx$ is of the total length L. That is,

$${\displaystyle \frac{dm}{M}}={\displaystyle \frac{dx}{L}}$$

Consequently, we can express $dm$ in terms of the coordinate differential $dx$ as

$$dm={\displaystyle \frac{M}{L}}dx$$

NOTE   The change of variables from $dm$ to the differential of a coordinate is the key step in calculating the center of mass.

With this expression for $dm$, Equation 12.5 for $x_{\text{cm}}$ becomes

$$x_{\text{cm}}={\displaystyle \frac{1}{M}}\left({\displaystyle \frac{M}{L}}{\displaystyle {\int }_{}^{}}xdx\right)={\displaystyle \frac{1}{L}}{\displaystyle {\int }_0^L}xdx$$

where in the last step we’ve noted that summing “all the mass in the rod” means integrating from $x=0$ to $x=L$. This is a straightforward integral to carry out, giving

$$x_{\text{cm}}=\frac{1}{L}{\left[\frac{x^2}{2}\right]}_0^L=\frac{1}{L}\left[\frac{L^2}{2}-0\right]=\frac{1}{2}L$$

The center of mass is at the center of the rod, as you probably guessed. For a 1.60-m-long rod, each tip of the rod rotates in a circle with $r=\frac{1}{2}L=0.80\text{m}$. The tangential acceleration, the rate at which the tip is speeding up, is

$$a_t=r\alpha =(0.80\text{ m})\left(6.0{\text{ rad/s}}^2\right)=4.8{\text{ m/s}}^2$$

To see where the center-of-mass equations come from, FIGURE 12.8 shows an object rotating about its center of mass. Particle i is moving in a circle, so it must have a centripetal acceleration. Acceleration requires a force, and this force is due to tension in the molecular bonds that hold the object together. Force $\vec{T}{}_{\text{i}}$ on particle i has magnitude

where we used Equation 12.3 for $a_r$. All points in a rigid rotating object have the same angular velocity, so $\omega$ doesn’t need a subscript.

FIGURE 12.8 Finding the center of mass.

At every instant of time, the internal tension forces are all paired as action/reaction forces, equal in magnitude but opposite in direction, so the sum of all the tension forces must be zero. That is, $\sum \vec{T}{}_{\text{i}}=\vec{0}$. The x-component of this sum is

You can see from Figure 12.8 that $\text{cos}{\theta }_i=(x_{\text{cm}}-x_i)\text{/}{r}_i$. Thus

This equation will be true if the term in parentheses is zero. $x_{\text{cm}}$ is a constant, so we can bring it outside the summation to write

where we used the fact that $\sum m_i$ is simply the object’s total mass M. Solving for $x_{\text{cm}}$, we find the x-coordinate of the object’s center of mass to be

This was Equation 12.4. The y-equation is found similarly.