\subsection{Center of Mass and Translational Motion of Systems}

This subsection extends Newton's second law from a single particle to a system of particles. The key idea is that the overall translational motion of the system is described by its center of mass, even when the particles exert complicated internal forces on one another.

\dfn{Center of mass for a system}{Consider $N$ particles labeled by an index $i=1,2,\dots,N$. Let $m_i$ denote the mass of particle $i$, let $\vec{r}_i$ denote the position vector of particle $i$, and let
\[
M=\sum_{i=1}^N m_i
\]
denote the total mass of the system. The \emph{center of mass} is the vector
\[
\vec{r}_{\mathrm{cm}}=\frac{1}{M}\sum_{i=1}^N m_i\vec{r}_i.
\]
Its velocity and acceleration are
\[
\vec{v}_{\mathrm{cm}}=\frac{d\vec{r}_{\mathrm{cm}}}{dt},
\qquad
\vec{a}_{\mathrm{cm}}=\frac{d\vec{v}_{\mathrm{cm}}}{dt}.
\]
For a continuous mass distribution, let $dm$ denote a differential mass element located at position vector $\vec{r}$. Then the continuous analog is
\[
\vec{r}_{\mathrm{cm}}=\frac{1}{M}\int \vec{r}\,dm.
\]}

\thm{Newton's second law for a system}{For the system above, let $\vec{F}_{\mathrm{ext},i}$ denote the net external force on particle $i$. Let
\[
\sum \vec{F}_{\mathrm{ext}}=\sum_{i=1}^N \vec{F}_{\mathrm{ext},i}
\]
denote the net external force on the whole system. Then
\[
\sum \vec{F}_{\mathrm{ext}}=M\vec{a}_{\mathrm{cm}}.
\]
Therefore the center of mass moves as if all the system mass $M$ were concentrated at the center of mass and acted on by the net external force.}

\nt{Internal forces can change the separations, shape, or rotation of the parts of a system, but they do not change the motion of the center of mass. When the internal forces are added over the whole system, they cancel in equal-and-opposite pairs. As a result, only external forces determine $\vec{a}_{\mathrm{cm}}$. In particular, if $\sum \vec{F}_{\mathrm{ext}}=\vec{0}$, then $\vec{a}_{\mathrm{cm}}=\vec{0}$, so the center of mass moves with constant velocity even if an explosion or collision makes the individual parts fly apart.}

\pf{Derivation by summing Newton's second law over the particles}{For each particle $i$, let $\vec{a}_i$ denote its acceleration, and let $\vec{F}_{\mathrm{int},i}$ denote the net internal force on that particle from the other particles in the system. Newton's second law for particle $i$ gives
\[
m_i\vec{a}_i=\vec{F}_{\mathrm{ext},i}+\vec{F}_{\mathrm{int},i}.
\]
Now sum over all particles:
\[
\sum_{i=1}^N m_i\vec{a}_i=\sum_{i=1}^N \vec{F}_{\mathrm{ext},i}+\sum_{i=1}^N \vec{F}_{\mathrm{int},i}.
\]
By Newton's third law, the internal forces cancel in pairs, so
\[
\sum_{i=1}^N \vec{F}_{\mathrm{int},i}=\vec{0}.
\]
Thus,
\[
\sum_{i=1}^N m_i\vec{a}_i=\sum \vec{F}_{\mathrm{ext}}.
\]
From the definition
\[
\vec{r}_{\mathrm{cm}}=\frac{1}{M}\sum_{i=1}^N m_i\vec{r}_i,
\]
differentiate twice with respect to time:
\[
\vec{a}_{\mathrm{cm}}=\frac{1}{M}\sum_{i=1}^N m_i\vec{a}_i.
\]
Multiplying by $M$ gives
\[
M\vec{a}_{\mathrm{cm}}=\sum_{i=1}^N m_i\vec{a}_i.
\]
Substitute this into the previous result to obtain
\[
\sum \vec{F}_{\mathrm{ext}}=M\vec{a}_{\mathrm{cm}}.
\]}

\qs{Worked example}{Two carts move on a horizontal frictionless track. Cart 1 has mass $m_1=2.0\,\mathrm{kg}$ and cart 2 has mass $m_2=3.0\,\mathrm{kg}$. At the instant of interest, cart 1 is at position coordinate $x_1=0.40\,\mathrm{m}$ and cart 2 is at position coordinate $x_2=1.60\,\mathrm{m}$. The carts interact with each other through a light compressed spring between them, so the spring forces are internal to the two-cart system. A student pulls on cart 1 so that the net external force on the two-cart system is a constant horizontal force of magnitude $10.0\,\mathrm{N}$ to the right. At that instant the center of mass is moving to the right with speed $v_{\mathrm{cm},0}=1.2\,\mathrm{m/s}$.

Find the $x$-coordinate of the center of mass, the acceleration of the center of mass, and the speed of the center of mass $t=2.0\,\mathrm{s}$ later.}

\sol For one-dimensional motion along the $x$-axis, let $x_{\mathrm{cm}}$ denote the $x$-coordinate of $\vec{r}_{\mathrm{cm}}$. Also let
\[
M=m_1+m_2=2.0\,\mathrm{kg}+3.0\,\mathrm{kg}=5.0\,\mathrm{kg}.
\]
Then the center-of-mass coordinate is
\[
x_{\mathrm{cm}}=\frac{m_1x_1+m_2x_2}{M},
\]
Substitute the given values:
\[
x_{\mathrm{cm}}=\frac{(2.0\,\mathrm{kg})(0.40\,\mathrm{m})+(3.0\,\mathrm{kg})(1.60\,\mathrm{m})}{5.0\,\mathrm{kg}}.
\]
Therefore,
\[
x_{\mathrm{cm}}=\frac{0.80+4.80}{5.0}\,\mathrm{m}=1.12\,\mathrm{m}.
\]

Now use Newton's second law for the system:
\[
\sum \vec{F}_{\mathrm{ext}}=M\vec{a}_{\mathrm{cm}}.
\]
The net external force has magnitude $10.0\,\mathrm{N}$ to the right, so the center-of-mass acceleration has magnitude
\[
a_{\mathrm{cm}}=\frac{10.0\,\mathrm{N}}{5.0\,\mathrm{kg}}=2.0\,\mathrm{m/s^2}
\]
to the right. Equivalently, $\vec{a}_{\mathrm{cm}}$ points in the positive $x$-direction with magnitude $2.0\,\mathrm{m/s^2}$.

Because the external force is constant, the center of mass has constant acceleration during the $2.0\,\mathrm{s}$ interval. Let $v_{\mathrm{cm}}$ denote the center-of-mass speed at the later time. Then
\[
v_{\mathrm{cm}}=v_{\mathrm{cm},0}+a_{\mathrm{cm}}t.
\]
Substituting gives
\[
v_{\mathrm{cm}}=1.2\,\mathrm{m/s}+(2.0\,\mathrm{m/s^2})(2.0\,\mathrm{s})=5.2\,\mathrm{m/s}.
\]

So the center of mass is at
\[
x_{\mathrm{cm}}=1.12\,\mathrm{m},
\]
its acceleration has magnitude
\[
a_{\mathrm{cm}}=2.0\,\mathrm{m/s^2},
\]
with $\vec{a}_{\mathrm{cm}}$ pointing in the positive $x$-direction, and its speed $2.0\,\mathrm{s}$ later is
\[
v_{\mathrm{cm}}=5.2\,\mathrm{m/s}.
\]
Even if the spring pushes the carts apart strongly, that spring force is internal to the chosen system. It can change the individual motions of the carts, but it does not change these center-of-mass results.