The Center of Mass

We define the center of mass (com) of a system of particles (such as a person) in order to predict the possible motion of the system.

 The center of mass of a system of particles is the point that moves as though (1) all of the system’s mass were concentrated there and (2) all external forces were applied there.

In this section we discuss how to determine where the center of mass of a system of particles is located. We start with a system of only a few particles, and then we consider a system of a great many particles (a solid body, such as a baseball bat). Later in the chapter, we discuss how the center of mass of a system moves when external forces act on the system.

Systems of Particles

Figure 9-2a shows two particles of masses m₁ and m₂ separated by distance d. We have arbitrarily chosen the origin of an x axis to coincide with the particle of mass m₁. We define the position of the center of mass (com) of this two-particle system to be

Suppose, as an example, that m₂ = 0. Then there is only one particle, of mass m₁, and the center of mass must lie at the position of that particle; Eq. 9-1 dutifully reduces to x_com = 0. If m₁ = 0, there is again only one particle (of mass m₂), and we have, as we expect, x_com = d. If m₁ = m₂, the center of mass should be halfway between the two particles; Eq. 9-1 reduces to , again as we expect. Finally, Eq. 9-1 tells us that if neither m₁ nor m₂ is zero, x_com can have only values that lie between zero and d; that is, the center of mass must lie somewhere between the two particles.

Figure 9-2b shows a more generalized situation, in which the coordinate system has been shifted leftward. The position of the center of mass is now defined

Note that if we put x₁ = 0, then x₂ becomes d and Eq. 9-2 reduces to Eq. 9-1, as it must. Note also that in spite of the shift of the coordinate system, the center of mass is still the same distance from each particle.

We can rewrite Eq. 9-2 as

in which M is the total mass of the system. (Here, M = m₁ + m₂.) We can extend this equation to a more general situation in which n particles are strung out along the x axis. Then the total mass is M = m₁ + m₂ + … + m_n, and the location of the center of mass is

The subscript i is a running number, or index, that takes on all integer values from 1 to n. It identifies the various particles, their masses, and their x coordinates.

If the particles are distributed in three dimensions, the center of mass must be identified by three coordinates. By extension of Eq. 9-4, they are

We can also define the center of mass with the language of vectors. First recall that the position of a particle at coordinates x_i, y_i, and z_i is given by a position vector:

Here the index identifies the particle, and , , and  are unit vectors pointing, respectively, in the positive direction of the x, y, and z axes. Similarly, the position of the center of mass of a system of particles is given by a position vector:

The three scalar equations of Eq. 9-5 can now be replaced by a single vector equation,

where again M is the total mass of the system. You can check that this equation is correct by substituting Eqs. 9-6 and 9-7 into it, and then separating out the x, y, and z components. The scalar relations of Eq. 9-5 result.

Solid Bodies

An ordinary object, such as a baseball bat, contains so many particles (atoms) that we can best treat it as a continuous distribution of matter. The “particles” then become differential mass elements dm, the sums of Eq. 9-5 become integrals, and the coordinates of the center of mass are defined as

where M is now the mass of the object.

Evaluating these integrals for most common objects (such as a television set or a moose) would be difficult, so here we consider only uniform objects. Such objects have uniform density, or mass per unit volume; that is, the density ρ (Greek letter rho) is the same for any given element of an object as for the whole object:

Fig. 9-2 (a) Two particles of masses m₁ and m₂ are separated by distance d. The dot labeled com shows the position of the center of mass, calculated from Eq. 9-1. (b) The same as (a) except that the origin is located farther from the particles. The position of the center of mass is calculated from Eq. 9-2. The location of the center of mass with respect to the particles is the same in both cases.

where dV is the volume occupied by a mass element dm, and V is the total volume of the object. Substituting dm = (M/V) dV from Eq. 9-10 into Eq. 9-9 gives

You can bypass one or more of these integrals if an object has a point, a line, or a plane of symmetry. The center of mass of such an object then lies at that point, on that line, or in that plane. For example, the center of mass of a uniform sphere (which has a point of symmetry) is at the center of the sphere (which is the point of symmetry). The center of mass of a uniform cone (whose axis is a line of symmetry) lies on the axis of the cone. The center of mass of a banana (which has a plane of symmetry that splits it into two equal parts) lies somewhere in that plane.

The center of mass of an object need not lie within the object. There is no dough at the com of a doughnut, and no iron at the com of a horseshoe.

CHECKPOINT 1 The figure shows a uniform square plate from which four identical squares at the corners will be removed. (a) Where is the center of mass of the plate originally? Where is it after the removal of (b) square 1; (c) squares 1 and 2; (d) squares 1 and 3; (e) squares 1, 2, and 3; (f) all four squares? Answer in terms of quadrants, axes, or points (without calculation, of course).

Sample Problem 9-1

Three particles of masses m₁ = 1.2 kg, m₂ = 2.5 kg, and m₃ = 3.4 kg form an equilateral triangle of edge length a = 140 cm. Where is the center of mass of this system?

Solution: A Key Idea to get us started is that we are dealing with particles instead of an extended solid body, so we can use Eq. 9-5 to locate their center of mass. The particles are in the plane of the equilateral triangle, so we need only the first two equations. A second Key Idea is that we can simplify the calculations by choosing the x and y axes so that one of the particles is located at the origin and the x axis coincides with one of the triangle’s sides (Fig. 9-3). The three particles then have the following coordinates:

The total mass M of the system is 7.1 kg.

From Eq. 9-5, the coordinates of the center of mass are

Fig. 9-3 Three particles form an equilateral triangle of edge length a. The center of mass is located by the position vector .

In Fig. 9-3, the center of mass is located by the position vector , which has components x_com and y_com.

Sample Problem 9-2

Figure 9-4a shows a uniform metal plate P of radius 2R from which a disk of radius R has been stamped out (removed) in an assembly line. Using the xy coordinate system shown, locate the center of mass com_P of the plate.

Solution: First, let us roughly locate the center of plate P by using the Key Idea of symmetry. We note that the plate is symmetric about the x axis (we get the portion below that axis by rotating the upper portion about the axis). Thus, com_P must be on the x axis. The plate (with the disk removed) is not symmetric about the y axis. However, because there is somewhat more mass on the right of the y axis, com_P must be somewhat to the right of that axis. Thus, the location of com_P should be roughly as indicated in Fig. 9-4a.

Another Key Idea here is that plate P is an extended solid body, so we can use Eqs. 9-11 to find the actual coordinates of com_P. However, that procedure is difficult. A much easier way is to use this Key Idea: In working with centers of mass, we can assume that the mass of a uniform object is concentrated in a particle at the object’s center of mass. Here is how we do so:

First, put the stamped-out disk (call it disk S) back into place (Fig. 9-4b) to form the original composite plate (call it plate C). Because of its circular symmetry, the center of mass com_S for disk S is at the center of S, at x = −R (as shown). Similarly, the center of mass com_C for composite plate C is at the center of C, at the origin (as shown). We then have the following:

Now we use the Key Idea of concentrated mass: Assume that mass m_S of disk S is concentrated in a particle at x_S = −R, and mass m_P is concentrated in a particle at x_P (Fig. 9-4c). Next treat these two particles as a two-particle system, using Eq. 9-2 to find their center of mass x_S+P. We get

Next note that the combination of disk S and plate P is composite plate C. Thus, the position x_S+P of com_S+P must coincide with the position x_C of com_C, which is at the origin; so x_S+P = x_C = 0. Substituting this into Eq. 9-12 and solving for x_P, we get

Now we seem to have a problem, because we do not know the masses in Eq. 9-13. However, we can relate the masses to the face areas of S and P by noting that

Fig. 9-4 (a) Plate P is a metal plate of radius 2R, with a circular hole of radius R. The center of mass of P is at point com_P. (b) Disk S has been put back into place to form a composite plate C. The center of mass com_S of disk S and the center of mass com_C of plate C are shown. (c) The center of mass com_S+P of the combination of S and P coincides with com_C, which is at x = 0.

Because the plate is uniform, the densities and thicknesses are equal; we are left with

Substituting this and x_S = −R into Eq. 9-13, we have

PROBLEM-SOLVING TACTICS

TACTIC 1 : Center-of-Mass Problems

Sample Problems 9-1 and 9-2 provide three strategies for simplifying center-of-mass problems. (1) Make full use of the symmetry of the object, be it about a point, a line, or a plane. (2) If the object can be divided into several parts, treat each of these parts as a particle, located at its own center of mass. (3) Choose your axes wisely: If your system is a group of particles, choose one of the particles as your origin. If your system is a body with a line of symmetry, let that be your x or y axis. The choice of origin is completely arbitrary; the location of the center of mass is the same regardless of the origin from which it is measured.