Center of Mass

Consider a system of particles 1,2,...,n. Their total mass is:

M = $\sum$i=1,...,n mi

Their total momentum is:

= $\sum$i=1,...,n i = $\sum$i=1,...,n mi i

Define the center-of-mass coordinate vector:

cm = (1/M) $\sum$i=1,...,n mi i

The x-coordinate of the center of mass is, for example:

\[ \rm \mathbf{x_{cm} = \frac{m_{1}x_{1} + m_{2}x_{2} + ... + m_{n}x_{n}}{M}} \]

Why is this useful?

We can find the equation of motion for the center of mass by taking two time derivatives and multiplying with M => Newton's Second Law for a system of particles:

Note that only the net external force enters here. The internal forces between the particles pairwise add up to 0 because of Newton's Third Law.

Consequences:

  1. No matter how complicated the motion of the individual particles, the motion of the center of mass is usually very simple.
  2. If we calculate in a coordinate system that is moving with the center of mass, then the total momentum is zero in this coordinate system, and all momenta of all particles must add up to zero.

Since there is some calculus on this page, we have provided the same page in calculus-free condition for all of you calc-avoiders.

© MultiMedia Physics, 1999