University Physics I - Classical Mechanics, 2019

116 CHAPTER 6. INTERACTIONS, PART 2: FORCES
just equal to Mv cm (compare Eq. (3.11), in the “Momentum” chapter). So we have
F ext,all = dp sys
= d dt dt (Mv cm) (6.10)
This extends a previous result. We already knew that in the absence of external forces, the momentum
of a system remained constant. Now we see that the system’s momentum responds to the
net external force as if the whole system was a single particle of mass equal to the total mass M
and moving at the center of mass velocity v cm . In fact, assuming that M does not change we can
rewrite Eq. (6.10) intheform
F ext,all = Ma cm (6.11)
where a cm is the acceleration of the center of mass. This is the key result that allows us to treat
extended objects as if they were particles: as far as the motion of the center of mass is concerned,
all the internal forces cancel out (as we already saw in our study of collisions), and the point
representing the center of mass responds to the sum of the external forces as if it were just a
particle of mass M subject to Newton’s second law, F = ma. Theresult(6.11) applies equally well
to an extended solid object that we choose to mentally break up into a collection of particles, as
to an actual collection of separate particles, or even to a collection of separate extended objects;
in the latter case, we would just have each object’s motion represented by the motion of its own
center of mass.
Finally, note that all the results above generalize to more than one dimension. In fact, forces
are vectors (just like velocity, acceleration and momentum), and all of the above equations, in 3
dimensions, apply separately to each vector component. In one dimension, we just need to be aware
of the sign of the forces, whenever we add several of them together.
6.2 Forces and potential energy
In the last chapter I mentioned a special case that we encounter often, in which a lighter object is
interacting with a much more massive one, so that the massive one essentially does not move at all
as a result of the interaction. Note that this does not contradict Newton’s 3rd law, Eq. (6.5): the
forces the two objects exert on each other are the same in magnitude, but the acceleration of each
object is inversely proportional to its mass, so F 12 = −F 21 implies
m 2 a 2 = −m 1 a 1 (6.12)
and so if, for instance, m 2 ≫ m 1 ,weget|a 2 | = |a 1 |m 1 /m 2 ≪|a 1 |. In words, the more massive
object is less responsive than the less massive one to a force of the same magnitude. This is just
how we came up with the concept of inertial mass in the first place!
Anyway, you’ll recall that in this situation I could just write the potential energy function of the
whole system as a function of only the lighter object’s coordinate, U(x). I am going to use this