Center Of Mass Troubles

Question

I understand the concept of Center Of Mass(com), but I am having a difficult time interpreting the equation of the simplified case of one-dimension.

The book I am reading defines the position of the com of a two-particle system to be $x_{com}= \Large\frac{m_1x_1+m_2x_2}{m_1+m_2}$ I'm sorry if this seems like a trivial question, but could someone explain to me the interpretation of this definition? Perhaps even why they defined it to be this way.

Here is an excerpt from my textbook:

"An ordinary object, such as a baseball bat, contains so many particles (atoms)that we can best treat it as a continuous distribution of matter. The “particles” then become differential mass elements $dm$, the sums of Eq. 9-5 become integrals, and the coordinates of the center of mass are defined as (9-9)...Evaluating these integrals for most common objects (such as a television set or a moose) would be difficult, so here we consider only uniform objects. Such objects have uniform density,or mass per unit volume; that is, the density $\rho$ (Greek letter rho) is the same for any given element of an object as for the whole object. From Eq. 1-8, we can write $\rho = \large\frac {dm}{dV} = \frac {m}{V}$"

What does the author mean by, "the 'particles' then become differential mass elements $dm$? Is $\rho = \Large\frac{dm}{dV}$ the derivative of the density function? If so, how would I interpret that? Furthermore, if it is indeed the derivative of the density function, why is it also equal to the original function of density?

Answer 1

The book I am reading defines the position of the com of a two-particle system to be $x_{com}= \Large\frac{m_1x_1+m_2x_2}{m_1+m_2}$ I'm sorry if this seems like a trivial question, but could someone explain to me the interpretation of this definition? Perhaps even why they defined it to be this way.

It's a weighted average of the position of the particles where the weighting is the mass.

To see this, consider the two-particle, discrete case where the masses are the same, then it reduces to $x_{com} = \frac{x_1 + x_2}{2}$ which is clearly an average position.

It is useful because it turns out that you can often (but not always) factor out CoM motion from motion relative to the CoM and simplify your life. Physicists like to simplify their own lives.

What does the author mean by, "the 'particles' then become differential mass elements $dm$?

That is just the usual continuum limit. You imagine breaking a continuous distribution into little boxes and treating them with the discrete equation then let the size of the boxes get arbitrarily small.

Answer 2

$\rho = \large\frac {dm}{dV} = \frac {m}{V}$

Is just the definition of density. For a very small element of mass $dm$ in a very small volume $dv$ we can divide mass/volume and get local density. If we assume that all the mass is the same we can sum all the $dm$ and $dv$ to get $m$ and $v$ and an overall average density.

It's not a differential equation as such - but it is the fundemental principle of calculus.

Answer 3

If there is no external force (but only an internal force $F_{12}$), then the equation of motion for $x_{com}$ is "free". It describes a free motion of the system as a whole. It is obtained by adding two Newton equations where the "internal forces" cancel out.