Imagine yourself floating in a space capsule deep in space. Suppose you take an object like that shown in FIGURE 12.4a and spin it so that it simply rotates but has no translational motion as it floats beside you. About what point does it rotate? That is the question we need to answer.

FIGURE 12.4 Rotation about the center of mass.

Full Alternative Text

An unconstrained object (i.e., one not on an axle or a pivot) on which there is no net force rotates about a point called the center of mass. The center of mass remains motionless while every other point in the object undergoes circular motion around it. You need not go deep into space to demonstrate rotation about the center of mass. If you have an air table, a flat object rotating on the air table rotates about its center of mass.

To locate the center of mass, FIGURE 12.4b models the object as a set of particles numbered $i=1\text{,}2\text{,}3\text{,}\dots$. Particle i has mass $m_i$ and is located at position $\left(x_i\text{,}{y}_i\right)$. We’ll prove later in this section that the center of mass is located at position

where $M=m_1+m_2+m_3+\cdots$ is the object’s total mass.

Let’s see if Equations 12.4 make sense. Suppose you have an object consisting of N particles, all with the same mass m. That is, $m_1=m_2=\cdots =m_N=m$. We can factor the m out of the numerator, and the denominator becomes simply Nm. The m cancels, and the x-coordinate of the center of mass is

In this case, $x_{\text{cm}}$ is simply the average x-coordinate of all the particles. Likewise, $y_{\text{cm}}$ will be the average of all the y-coordinates.

This does make sense! If the particle masses are all the same, the center of mass should be at the center of the object. And the “center of the object” is the average position of all the particles. To allow for unequal masses, Equations 12.4 are called weighted averages. Particles of higher mass count more than particles of lower mass, but the basic idea remains the same. The center of mass is the mass-weighted center of the object.

EXAMPLE 12.1 The center of mass of a barbell

A barbell consists of a 500 g ball and a 2.0 kg ball connected by a massless 50-cm-long rod.

1. Where is the center of mass?

2. What is the speed of each ball if they rotate about the center of mass at 40 rpm?

MODEL Model the barbell as a rigid body.

VISUALIZE FIGURE 12.5 shows the two masses. We’ve chosen a coordinate system in which the masses are on the x-axis with the 2.0 kg mass at the origin.

FIGURE 12.5 Finding the center of mass.

FIGURE 12.5 Full Alternative Text

SOLVE

1. We can use Equations 12.4 to calculate that the center of mass is

   $$\begin{array}{lll}{x}_{\text{cm}}\hfill & =\hfill & {\displaystyle \frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}}\hfill \\ \hfill & =\hfill & {\displaystyle \frac{(2.0\text{ kg})\left(0.0\text{ m}\right)+\left(0.50\text{ kg}\right)\left(0.50\text{ m}\right)}{2.0\text{ kg}+0.50\text{ kg}}}=0.10\text{ m}\hfill \end{array}$$

   $y_{\text{cm}}=0$ because both masses are on the x-axis. The center of mass is 20% of the way from the 2.0 kg ball to the 0.50 kg ball.

2. Each ball rotates about the center of mass. The radii of the circles are $r_1=0.10\text{m}$ and $r_2=0.40\text{m}$. The tangential velocities are ${\left(v_i\right)}_t=r_i\omega$, but this equation requires $\omega$ to be in $\text{rad}\text{/}\text{s}$. The conversion is

   $$\omega =40{\displaystyle \frac{\text{rev}}{\text{min}}}\times {\displaystyle \frac{1\text{min}}{60\text{s}}}\times {\displaystyle \frac{2\pi \text{rad}}{1\text{rev}}}=4.19\text{rad}\text{/}\text{s}$$

   Consequently,

   $$\begin{array}{l}{\left(v_1\right)}_t=r_1\omega =(0.10\text{ m})\left(4.19\text{ rad/s}\right)=0.42\text{ m/s}\hfill \\ {\left(v_2\right)}_t=r_2\omega =(0.40\text{ m})\left(4.19\text{ rad/s}\right)=1.68\text{ m/s}\hfill \end{array}$$

REVIEW The center of mass is closer to the heavier ball than to the lighter ball. We expected this because $x_{\text{cm}}$ is a mass-weighted average of the positions. But the lighter mass moves faster because it is farther from the rotation axis.