9.6 Center of mass  (Page 2/13)

As a result of this fractional force, the momentum of each particle gets changed:

The net force $\overrightarrow{F}$ on the object is the vector sum of these forces:

This net force changes the momentum of the object as a whole, and the net change of momentum of the object must be the vector sum of all the individual changes of momentum of all of the particles:

Combining [link] and [link] gives

Let’s now think about these summations. First consider the internal forces term; remember that each ${\overrightarrow{f}}_j^{\text{int}}$ is the force on the j th particle from the other particles in the object. But by Newton’s third law, for every one of these forces, there must be another force that has the same magnitude, but the opposite sign (points in the opposite direction). These forces do not cancel; however, that’s not what we’re doing in the summation. Rather, we’re simply mathematically adding up all the internal force vectors. That is, in general, the internal forces for any individual part of the object won’t cancel, but when all the internal forces are added up, the internal forces must cancel in pairs. It follows, therefore, that the sum of all the internal forces must be zero:

(This argument is subtle, but crucial; take plenty of time to completely understand it.)

For the external forces, this summation is simply the total external force that was applied to the whole object:

As a result,

This is an important result. [link] tells us that the total change of momentum of the entire object (all N particles) is due only to the external forces; the internal forces do not change the momentum of the object as a whole. This is why you can’t lift yourself in the air by standing in a basket and pulling up on the handles: For the system of you + basket, your upward pulling force is an internal force.

Force and momentum

Remember that our actual goal is to determine the equation of motion for the entire object (the entire system of particles). To that end, let’s define:

${\overrightarrow{p}}_{\text{CM}}=$ the total momentum of the system of N particles (the reason for the subscript will become clear shortly)

Then we have

and therefore [link] can be written simply as

Since this change of momentum is caused by only the net external force, we have dropped the “ext” subscript.

This is Newton’s second law, but now for the entire extended object. If this feels a bit anticlimactic, remember what is hiding inside it: ${\overrightarrow{p}}_{\text{CM}}$ is the vector sum of the momentum of (in principle) hundreds of thousands of billions of billions of particles $(6.02\times {10}^{23})$ , all caused by one simple net external force—a force that you can calculate.

Center of mass

Our next task is to determine what part of the extended object, if any, is obeying [link] .

It’s tempting to take the next step; does the following equation mean anything?

If it does mean something (acceleration of what, exactly?), then we could write

and thus