# Physics with Calculus/Mechanics/Center of Mass

When you throw a ball, which is composed of many, many interacting atoms, it is by no means obvious that it will follow a parabolic path, even though a single particle does. What do we even mean that a ball follows a parabolic path? Is there some special point to follow? We will show that indeed there is a special point to follow called the center of mass and we will show that Newton's laws fantastically scale from the very small to the very large.

Let $r_{i}$ be the vector from the origin to the ith particle. Let $m_{i},v_{i},f_{i}$ be the mass, velocity, and force of the ith particle respectively. In order to see if there are any particularly special points, define an arbitrary vector R. Now, every $r_{i}$ can be written as $r_{i}'+R$ . Applying Newton's second law to every particle,

$f_{i}=m_{i}{\frac {dv_{i}}{dt}}$ .

But

$v_{i}={\frac {dr_{i}'}{dt}}+{\frac {dR}{dt}}=v_{i}'+V$ .

(the last equation is just definition). So

$f_{i}=m_{i}{\frac {dv_{i}'}{dt}}+m_{i}{\frac {dV}{dt}}$ .

Summing,

$F=\sum _{i}f_{i}={\frac {d}{dt}}\sum _{i}m_{i}v_{i}'+M{\frac {dV}{dt}}.$ This takes on the simplest form, and indeed the form of Newton's second law for a particle, when the derivative of the sum is zero. Now, the derivative is zero when the sum is constant. That is,

$\sum _{i}m_{i}{\frac {dr_{i}'}{dt}}=P$ .

Where P is a constant. Thus,

$\sum _{i}m_{i}r_{i}'=Pt+C$ where C is a constant. Substituting the definition of R,

$\sum _{i}m_{i}R-\sum m_{i}r_{i}=Pt+C$ .

This is equivalent to

$R={\frac {1}{\sum _{i}m_{i}}}\sum _{i}m_{i}r_{i}+{\frac {1}{\sum _{i}m_{i}}}Pt+{\frac {1}{\sum _{i}m_{i}}}C$ .

We don't want R to depend on time, so take P = 0 (this is allowed because P was a constant of integration). Since C is arbitrary, and we want the simplest possible form, take C = 0. Although there is no great harm in taking it to not be zero, it will just be very cumbersome.

Thus, the form of the equations takes on a particularly simple form when

$R={\frac {1}{\sum _{i}m_{i}}}\sum _{i}m_{i}r_{i}$ .

We define this to be the center of mass vector. It has the very important property that the entire system follows

$F=M{\frac {d^{2}R}{dt^{2}}}$ where

$F=\sum _{i}f_{i}$ and

$M=\sum _{i}m_{i}.$ That is, the entire system, no matter how complicated, behaves as a single particle of mass M located at the center of mass!

Another interesting property comes from the fact that we can split up the force on each particle

$\sum f_{i}=\sum _{i}f_{i}^{external}+\sum _{i,j}f_{ij}$ where $f_{ij}$ is the force on the ith particle caused by the jth particle. By Newton's third law, the entire second sum on the right hand side cancels out. Thus, F is the sum of the external forces -- a body at rest cannot accelerate itself. The motion of the center of mass is completely determined by the external forces. This has some very interesting, but intuitive consequences. It means that if you are in deep space, no matter what you do, you can never move. Similarly if you are on a very slippery ice rink, there is nothing you can do except by throwing things to get off. So how do rockets leave the earth? They expel exhaust out the back -- it is leaving behind a great deal of mass so the center of mass remains the same.

Now, we have just showed that Newton's laws miraculously scale from a single particle to many particles. If you think about it though, Newton would not have discovered his laws if they didn't scale -- he discovered them for great groups of particles, such as a marble or anything in ordinary life. It is, in fact, almost a requirement that the law scale! However, although Newton's law scales up, it does not have to scale down. Although it would fit many of our everyday experiences if it did, it does not -- the very small is the realm of quantum mechanics. The reason it took so long to discover quantum mechanics is that it does not scale up; the effects get smaller and smaller as the things involved get bigger and bigger.