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Linear Momentum: Conservation of Momentum

Center of Mass

Terms and Formulae

Center of Mass, page 2

page 1 of 2

Up to this point in our study of classical mechanics, we have studied primarily the motion of a single particle or body. To further our comprehension of mechanics we must begin to examine the interactions of many particles at once. To begin this study, we define and examine a new concept, the center of mass, which will allow us to make mechanical calculations for a system of particles.

The Center of Mass of Two Particles

We start by defining and explaining the concept of the center of mass for the simplest possible system of particles, one containing only two particles. From our work in this section we will generalize for systems containing many particles.

Before quantifying our idea of a center of mass, we must explain it conceptually. The concept of the center of mass allows us to describe the movement of a system of particles by the movement of a single point. We will use the center of mass to calculate the kinematics and dynamics of the system as a whole, regardless of the motion of the individual particles.

Center of Mass for Two Particles in One Dimension

If a particle with mass m 1 has a position of x 1 and a particle with mass m 2 has a position of x 2 , then the position of the center of mass of the two particles is given by:

x cm =    

Thus the position of the center of mass is a point in space that is not necessarily part of either particle. This phenomenon makes intuitive sense: connect the two objects with a light but rigid pole. If you hold the pole at the position of the center of mass of the objects, they will balance. That balancing point will often not exist within either object.

Center of Mass for Two Particles beyond One Dimension

Now that we have the position, we extend the concept of the center of mass to velocity and acceleration, and thus give ourselves the tools to describe the motion of a system of particles. Taking a simple time derivative of our expression for x cm we see that:

v cm =    

Thus we have a very similar expression for the velocity of the center of mass. Differentiating again, we can generate an expression for acceleration:

a cm =    

With this set of three equations we have generated the necessary elements of the kinematics of a system of particles.

From our last equation, however, we can also extend to the dynamics of the center of mass. Consider two mutually interacting particles in a system with no external forces. Let the force exerted on m 2 by m 1 be F 21 , and the force exerted on m 1 by m 2 by F 12 . By applying Newton's Second Law we can state that F 12 = m 1 a 1 and F 21 = m 2 a 2 . We can now substitute this into our expression for the acceleration of the center of mass:

a cm =

However, by Newton's Third Law F 12 and F 21 are reactive forces, and F 12 = - F 21 . Thus a cm = 0 . Thus, if a system of particles experiences no net external force, the center of mass of the system will move at a constant velocity.

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