7.5

Center of Mass

In previous sections, we have encountered situations in which objects interact with one another, such as the two skaters pushing off in Example 6. In these situations, the mass of the system is located in several places, and the various objects move relative to each other before, after, and even during the interaction. It is possible, however, to speak of a kind of average location for the total mass by introducing a concept known as the center of mass (abbreviated as “cm”). With the aid of this concept, we will be able to gain additional insight into the principle of conservation of linear momentum.
The center of mass is a point that represents the average location for the total mass of a system. Figure 7-16, for example, shows two particles of mass m1 and m2 that are located on the x axis at the positions x1 and x2, respectively. The position xcm of the center-of-mass point from the origin is defined to be

| (7.10)

Each term in the numerator of this equation is the product of a particle’s mass and position, while the denominator is the total mass of the system. If the two masses are equal, we expect the average location of the total mass to be midway between the particles. With , Equation 7.10 becomes , which indeed corresponds to the point midway between the particles. Alternatively, suppose that and , while and . Then we expect the average location of the total mass to be located closer to particle 2, since it is more massive. Equation 7.10 is also consistent with this expectation, for it gives

If a system contains more than two particles, the center-of-mass point can be determined by generalizing Equation 7.10. For three particles, for instance, the numerator would contain a third term m3x3, and the total mass in the denominator would be . For a macroscopic object, which contains many, many particles, the center-of-mass point is located at the geometric center of the object, provided that the mass is distributed symmetrically about the center. Such would be the case for a billiard ball. For objects such as a golf club, the mass is not distributed symmetrically and the center-of-mass point is not located at the geometric center of the club. The driver used to launch a golf ball from the tee, for instance, has more mass in the club head than in the handle, so the center-of-mass point is closer to the head than to the handle.

Figure 7-16 The center of mass cm of the two particles is located on a line between them and lies closer to the more massive particle.

Equation 7.10 (and its generalization to more than two particles) deals with particles that lie along a straight line. A system of particles, however, may include particles that do not all lie along a single straight line. For particles lying in a plane, an equation like Equation 7.10 uses the x coordinates of each particle to give the x coordinate of the center of mass. A similar equation uses the y coordinates of each particle to give the y coordinate of the center of mass. If a system contains rigid objects, each of them may be treated as if it were a particle with all of the mass located at the object’s own center of mass. In this way, a collection of rigid objects becomes a collection of particles, and the x and y coordinates of the center of mass of the collection can be determined as described earlier in this paragraph.
To see how the center-of-mass concept is related to momentum conservation, suppose that the two particles in a system are moving, as they would be during a collision. With the aid of Equation 7.10, we can determine the velocity vcm of the center-of-mass point. During a time interval , the particles experience displacements of and , as Figure 7-17 shows. They have different displacements during this time because they have different velocities. Equation 7.10 can be used to find the displacement of the center of mass by replacing xcm by , x1 by , and x2 by :

Now we divide both sides of this equation by the time interval . In the limit as becomes infinitesimally small, the ratio becomes equal to the instantaneous velocity vcm of the center of mass. (See Section 2.2 for a review of instantaneous velocity.) Likewise, the ratios and become equal to the instantaneous velocities v1 and v2, respectively. Thus, we have

| (7.11)

Figure 7-17 During a time interval , the displacements of the particles are and , while the displacement of the center of mass is .

The numerator () on the right-hand side in Equation 7.11 is the momentum of particle 1 () plus the momentum of particle 2 (), which is the total linear momentum of the system. In an isolated system, the total linear momentum does not change because of an interaction such as a collision. Therefore, Equation 7.11 indicates that the velocity vcm of the center of mass does not change either. To emphasize this important point, consider the collision discussed in Example 7. With the data from that example, we can apply Equation 7.11 to determine the velocity of the center of mass before and after the collision:

Thus, the velocity of the center of mass is the same before and after the objects interact during a collision in which the total linear momentum is conserved.
In a two-dimensional collision such as that discussed in Example 9, the velocity of the center of mass is also the same before and after the collision, provided that the total linear momentum is conserved. Figure 7-18 illustrates this fact. The drawing reproduces Figure 7-15a, except that the initial and final center-of-mass locations are indicated, along with the velocity vector for the center of mass at these two places (see the vectors labeled vcm). As the drawing shows, this velocity vector is the same before and after the balls collide. Thus, the center of mass moves along a straight line as the two balls approach the collision point and continues along the same straight line following the collision.

Figure 7-18 This drawing shows the two balls in Figure 7-15a and the path followed by the center-of-mass point as the balls approach the collision point and then move away from it. Because of momentum conservation, the velocity of the center of mass of the balls is the same before and after the collision (see the vectors labeled vcm). As a result, the center of mass moves along the same straight-line path before and after the collision.

Check Your Understanding 4

Water, dripping at a constant rate from a faucet, falls to the ground. At any instant, there are many drops in the air between the faucet and the ground. Where does the center of mass of the drops lie relative to the halfway point between the faucet and the ground: above it, below it, or exactly at the halfway point? (The answer is given at the end of the book.)
Background: The solution to this question involves an understanding of the concepts of accelerated motion (the falling drops) and the center of mass.
For similar questions (including calculational counterparts), consult Self-Assessment Test 7.2. The test is described next.

Several important ideas are discussed in Sections 7.3, 7.4 and 7.5:

· | Collisions in One and Two Dimensions
· | The Center of Mass of a System of Particles

Use Self-Assessment Test 7.2 to check your understanding of these concepts.

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.