Center of Mass, Momentum Principle and Kinetic Energy for a
multi-particle system
Objective: To state the momentum principle for a multi-particle system; to describe the point called the
“center of mass”; to write the total kinetic energy of a multi-particle system that has translational motion,
rotational motion about the center of mass, and vibrational motion about the center of mass.
Center of Mass
The momentum principle for a multi-particle system is
~
Fnet =d~
Ptot
dt (1)
where the total momentum of the system is equal to the sum of the momenta of the particles in the system,
~
Ptot =~p1+~p2+... (2)
For speeds much less than the speed of light, we can write the momentum of a particle as mass times
velocity. Thus,
~
Ptot =m1~v1+m2~v2+... v << c (3)
But how can we treat a system of particles as a single particle? There is a certain point called the center
of mass, and that point has the same momentum as the total momentum of the system. Therefore, the
motion of that point is described by the momentum principle as if all of the external forces on the system
acted at that one point. Just consider it this way, if Superman crushes the system so that all of the mass is
concentrated at the center of mass, then this fictitious particle’s motion will be the same as the motion of
the real center of mass of the system. If Mis the total mass of the system, then
~
Ptot =M~vcm v << c (4)
To find the center of mass, we calculate a weighted average. An example of a weighted average is the
calculation of your course grade. Since your homework is weighted 20%, final exam 15%, mid-term exam
15%, quizzes 25%, and lab 25%, then your grade is the weighted average:
grade =0.2H W + 0.25LAB + 0.25Q+ 0.15F E + 0.15M T E
1(5)
But what if you haven’t completed the final exam yet, and you want to know “where you stand in the
class”? Well, just calculate the weighted average without the final exam. It would be
grade =0.2H W + 0.25LAB + 0.25Q+ 0.15M T E
0.85 (6)
Note that the denominator is the sum of the weights of homework, lab, quizzes, and the mid-term exam.
How does this related to center of mass? To calculate center of mass, you calculate a weighted average.
Suppose you have object A of mass 2.0 kg at x = 1.0 m and object B of mass 1.0 kg at the x=0.5 m. Where
is the center of mass? You might be tempted to average their x-coordinates to get 0.75 m. But this doesn’t
take into account that object A is more massive than object B. Thus, to get the center of mass, do a weighted
average:
xcm =m1x1+m2x2+...
m1+m2+... (7)
Therefore, in this example, the center of mass would be at 0.83 m, definitely not midway between the
objects.
What if you have something like a solid wheel, half of which is made of lead and half of which is made
of aluminum? You could always break it into two pieces, find the center of mass of each piece and use these
calculations to find the center of mass of the system. However, there’s another way to think about it. Break
the wheel into lots of very small pieces, each of mass dm. Then, add up the product of xdm for each little
piece and divide by the total mass to get
xcm =Σxdm
Σdm (8)
