\chapter{Center of Mass}

{\bf \index{seesaw principle} The seesaw principle.} Suppose that a light-weight board 5 meters long
is laid out between a point 10 meters from a reference point and a point 15
meters from the reference point.  Suppose that someone weighing 50 kg is
sitting at one end of the board (at 10 meters from the reference point),
and someone weighing 75 kg is sitting at the other end.  We want to know
where to put the fulcrum so as to balance the board with the people on it.
Here we will assume that the weight of the board is negligible, and that
the people are \index{point-masses} point-masses concentrated at 10 and 15 on our axis:

\begin{center}
%-figure FA -----------------------------
\begin{psfigure}{4in}{1in}{figFA.ps}
%\putcrosshair{0,0}
\puttext{-170., -15.}{\Small$0$}
\puttext{-22., -15.}{\Small$10$}
\puttext{-22., 10.}{\Small$50$ kg}
\puttext{52.5, -15.}{\Small$15$}
\puttext{52.5, 10.}{\Small$75$ kg}
\puttext{20., -20.}{\Small$\bar{x}$}
\end{psfigure}\\
%------------------------------
\end{center}

The location of the \index{balancing point} balancing point $\overline{x}$ is obtained by taking
the ratio of the total {\bf \index{moment} moment} $M$ to the total {\it mass} $m$.  Here
$m$ is simply $50+75=125$ kg.  The moment (in the $x$-direction, relative
to the reference point 0) is the sum of (mass$\times x$-coodinate), i.e.,
$50\cdot 10+75\cdot 15=1625$ kg-m.  (The units are kilogram-meters, i.e.,
mass times distance.)  Then $\overline{x}=M/m=1625/125=13$ meters.  That
is, the balancing point is 3/5 the way from the lighter person to the
heavier person.  (We would have gotten the same result if we had chosen our
reference point differently, for example, if we had chosen one end of the
board to be 0.)

\smallskip

If there are a number of point-masses located at various places on the
$x$-axis, the procedure is the same.  Suppose, for instance, that we have
five masses: a mass $m_1$ located at position $x_1$, a mass $m_2$ located
at $x_2$, and so on.  Then 
$$
\overline{x}=\frac{M}{
m}=\frac{m_1x_1+m_2x_2+m_3x_3+m_4x_4+m_5x_5}{ m_1+m_2+m_3+m_4+m_5}.
$$

\smallskip

We shall call this the ``seesaw principle:'' {\it the center of mass is
equal to total moment divided by total mass, where moment is computed by
adding up each piece of mass with the position of the piece of mass
multiplied in.} In reality, mass is usually not all concentrated in a few
points, but rather is distributed continuously over some region.  That is
where integration (rather than taking a finite sum) comes in.

\smallskip

The \index{center of mass} center of mass is important for understanding the motion of large
objects.  This is because the vector sum of all external forces on the
object is equal to the mass times the acceleration of the center of mass.
For example, when a baton is twirled and thrown into the air (Example
11.1 in the Math 124 notes), its center travels along a simple parabolic
falling body trajectory (but the path of the tip of the baton is more
complicated).

\smallskip

There are many types of center of mass problems, involving three-dimensional
shapes with varying densities.  However, we shall deal only with problems of
a particularly straightforward type: the center of mass of a flat uniform
shape (also called its ``\index{centroid} centroid'').

\smallskip

So suppose we have some shape in the $xy$-plane, for example, the region
contained between two curves (a ``top curve'' and a ``bottom curve'')
between $x=a$ and $x=b$.  Suppose that this shape is made up of a uniform
substance having a certain constant density $D$ measured in mass per unit
area --- say, gm/cm${}^2$.  (If the substance has a known density in the
sense of mass per volume --- say, gm/cm${}^3$ --- then $D$ is equal to this
volume density times the thickness of the plate.)  The mass of the uniform
plate is simply $D$ times its area, i.e., 
\[
m=D\int_a^b(\text{top~function~}-\text{~bottom~function})dx\,.
\]  
Let $f(x)$ denote the distance across the plate from top to bottom measured
at position $x$ on the $x$-axis, i.e., $f(x)$ is obtained by subtracting
the bottom function from the top function.  (In order to keep the notation
simple, in this section we will not introduce separate symbols for the top
function and the bottom function.)  Then 
\[
    m=D\int_a^bf(x)dx \,.
\]

\smallskip

\begin{centering}
%-figure FB -----------------------------
\begin{psfigure}{2.5in}{2.2in}{figFB.ps}
%\putcrosshair{0,0}
\puttext{5, -45.}{\Small$\bar{x}$}
\puttext{-55., 10.}{\Small$\bar{y}$}
\puttext{-24., -45.}{\Small$x$}
%\puttext{-12.0, 10}{\Small$f(x)$}
\end{psfigure}
%------------------------------
\hspace{1cm}
%-figure FC -----------------------------
\begin{psfigure}{2.5in}{2.2in}{figFC.ps}
%\putcrosshair{0,0}
\puttext{-24., 45.}{\tiny$\Delta x$}
\puttext{-24., -45.}{\Small$x$}
\puttext{-5., 10.}{\Small$f(x)$}
\end{psfigure}\\
%------------------------------
\end{centering}

The uniform plate will balance at some point with coordinates
$(\overline{x},\overline{y})$. The practical meaning of $\overline{x}$
and $\overline{y}$ is as follows. If you take a razor-edge parallel to
the $y$-axis and put it under the flat plate at a position $x$ which is
to the left of $\overline{x}$, then the plate will overbalance to the
right; if you put the razor-edge under the plate at a position
$x>\overline{x}$, then the plate will overbalance to the left. If you
put the razor-edge exactly at the position $\overline{x}$, then the
plate will balance. Similarly, if you take a razor-edge parallel to the
$x$-axis and put it under the plate at a position $y$ which is less than
or greater than $\overline{y}$, then the plate will overbalance on one
side or the other; but if you put it parallel to the $x$-axis exactly at
$\overline{y}$, then the plate will balance. The intersection of the
line parallel to the $y$-axis where the plate balances and the line
parallel to the $x$-axis where it balances is the point
$(\overline{x},\overline{y})$. If you stuck a pencil under the plate at
this point, the plate would spin around on the pencil point without
falling to any side.

\smallskip
 
We first find $\overline{x}$, which is the ratio of total moment {\bf in
the $x$-direction} to the total mass.  We already know how to find the
total mass: $m=D\int_a^bf(x)dx$.  To find the moment, we use the same type
of procedure as when we derived the integral formula for the area under a
curve or the volume of a solid of revolution.  Namely, we divide the
interval from $a$ to $b$ into a large number of small subintervals of
length $\Delta x=(b-a)/n$.  The left endpoints of these subintervals
are at $a$, $a+\Delta x$, $a+2\Delta x,\ldots$ $a+(n-1)
\Delta x$.  If we assume that the strip above each of the subintervals is
a rectangle (this is approximately the case if $\Delta x$ is small), then
the total mass of all of the strips is obtained by summing
$$D\,f(a)\Delta x+D\,f(a+\Delta x)\Delta x+D\,f(a+2\Delta x)\Delta x+\cdots+
D\,f(a+(n-1)\Delta x)\Delta x.$$
However, we want the {\it moment}, which is equal to mass times location
in the $x$-direction.  We make the simplifying assumption that the
entire strip above a subinterval has the same $x$-location as the left
endpoint of the interval (this is
approximately the case if $\Delta x$ is small); so to get the moment we
multiply each term in the above sum by the $x$-coordinate of the left
endpoint.  In other words, the moment is approximately
\begin{eqnarray*}
D\,a\,f(a)\Delta x&+&D(a+\Delta x)f(a+\Delta x)\Delta x+
D(a+2\Delta x)f(a+2\Delta x)\Delta x+\\
\cdots&+&D(a+(n-1)\Delta x) f(a+(n-1)\Delta x)\Delta x,
\end{eqnarray*}
\noindent
which is the Riemann sum for the function $D\,x\,f(x)$.  Thus, in the
limit we get the definite integral from $a$ to $b$ of $D\,x\,f(x)$:
$${\rm moment}=D\int_a^bxf(x)dx.$$

\smallskip

One way to think of this formula is to look at the contribution to
the moment that comes from the part of the region that
lies between $x$ and $x+\Delta x$, where we suppose that $\Delta x$ is
very small.  This is a thin strip $\Delta x$ across whose height at the
left is $f(x)$.  Since $\Delta x$ is small, the strip is very close to
being a rectangle, and so its area is very close to $f(x)\Delta x$, and its
mass is very close to $D\,f(x)\Delta x$.  Its contribution to moment is then
very close to $x$ times $D\,f(x)\Delta x$.  Now to find the moment of the
entire flat plate, we have to sum up the contributions as $x$ goes from
$a$ to $b$.  In the limit as $\Delta x$ goes to zero this sum becomes
an integral, and we write $dx$ instead of $\Delta x$.  That is, we get
$\int_a^b\bigl(Dx\,f(x)\bigr)dx$, as before.

\smallskip

Thus, {\it the \index{total moment} total moment of a uniform flat plate is equal to the same
type of formula as the total mass, except with $x$ multiplied by $f(x)$
inside the integral}.  Finally, to find $\overline{x}$ we divide
$M=D\int_a^bxf(x)\,dx$ by $m=D\int_a^bf(x)\,dx$.  The $D$'s cancel (that
is, the location of the balancing point does not depend on whether the flat
plate is made of a heavy or light substance), and we conclude that 
\[
   \overline{x}=\frac{\int_a^b x f(x)\,dx}{\int_a^bf(x)\,dx}.
\]

\smallskip

It remains to find a formula for $\overline{y}$.  We shift perspectives,
and look at the same shape from left to right as we move up the $y$-axis.
Let $g(y)$ denote the distance across the shape from left to right at the
position $y$.  To find $g(y)$, we express the right boundary of the shape
in the form $x=g_1(y)$ and the left boundary in the form $x=g_2(y)$; then
$g(y)=g_1(y)-g_2(y)$.  We suppose that in the $y$-direction the shape is
contained between $y=c$ and $y=d$, which will be our limits of integration.
(This is what we did when finding the area between two curves in cases when
we integrated with respect to $y$ rather than $x$.)  The derivation of a
formula for $\overline{y}$ is exactly the same as for $\overline{x}$.  The
result is: 
$$
\overline{y}=\frac{\int_c^dyg(y)\,dy}{\int_c^dg(y)\,dy}.
$$

\smallskip

Actually, the denominator in this formula has the same value as in the
formula for $\overline{x}$ --- in both cases it is the area of the plate.
The numerators, however, are different.

\begin{example}
%{\bf Example 1.} 
A uniform flat shape has the form of a rectangular ruler
of width $w$ and length 2 feet, with the last foot tapered off to a point,
i.e., it is in the shape of a rectangle of length 1 ft surmounted by an
isosceles triangle of altitude 1 ft.  Find the balancing point.

Our first task is to put this shape in the $xy$-plane in a convenient way.
Let us take the $x$-axis as the axis of symmetry, and put the pointed end
at the origin.  (That way the slanted line has the simplest possible
equation.)  Thus, the shape is bounded by the two lines joining the origin
to the points $(1,\pm w/2)$, the two horizontal lines joining $(1,\pm w/2)$
to $(2,\pm w/2)$, and the vertical line joining $(2,-w/2)$ to $(2,+w/2)$.
See the drawing below, where the equations of these 5 lines are indicated:

\begin{centering}
%-figure FD -----------------------------
\begin{psfigure}{4in}{1in}{figFD.ps}
%\putcrosshair{0,0}
\puttext{0, -7.}{\tiny$1$}
\puttext{50., 25.}{\Small$ y = w/2$}
\puttext{50., -25.}{\Small$ y = -w/2$}
\puttext{-50., 25.}{\Small$ y = \frac{w}{2}x$}
\puttext{-50., -25.}{\Small$ y = -\frac{w}{2}x$}
\puttext{200., 30}{\Small$x=2$}
\put(190,20){\vector(-2,-1){20}}
\end{psfigure}\\
%------------------------------
\end{centering}

We first observe that, because of the symmetry of the shape, the balancing
point is obviously on the $x$-axis, i.e., $\overline{y}=0$.  Thus, we need
only find $\overline{x}$.  In general, when finding the center of mass of a
symmetrical shape you can reduce the amount of computation required by
taking into account that the balancing point must be on the axis of
symmetry.

\smallskip

To compute $\overline{x}$, we find the following formulas for the distance
across from top to bottom (i.e., top function minus bottom function):
$$
f(x)=\begin{cases}wx,&\text{if $0\le x\le 1$;}\\ 
       w,& \text{if $1\le x\le 2$.}
    \end{cases}
$$ 
Thus, the numerator (the moment integral) in the formula
for $\overline{x}$ is 
\[
     \int_0^1wx^2dx+\int_1^2wx\,dx=w/3+3w/2=11w/6 \,.
\]
The
denominator can be found either by integration or simply by using the
formulas for the area of a rectangle and triangle: this gives us $3w/2$.
Thus, $\overline{x}=(11w/6)/(3w/2)=11/9=1.222$.  In other words, the
balancing point is in the rectangular part of the shape a distance 2/9 the
way from the base of the triangle to the opposite side of the rectangle.

\smallskip

Two observations should be made about this answer.  (1) The width $w$
canceled out; this means that the balancing point does not depend on
whether the shape is wide or narrow (just as it does not depend on whether
it is made of a heavy or light substance).  (2) A rough check on the
correctness of the answer is that we would expect an answer between 1 and
1.5, since a razor-edge placed at the base of the triangle would cause the
shape to overbalance to the right, whereas a razor-edge place in the middle
of the rectangular part would case the shape to overbalance to the left.

\end{example}

\begin{example}
%{\bf Example 2.} 
Find the center of mass of a uniform flat plate in the
shape of a quarter circle of radius 1 (say, the quarter of the unit circle
that is in the first quadrant of the $xy$-plane) on top of a square of side
1 (i.e., the unit square in the corner of the fourth quadrant).

\smallskip

Here we do not have symmetry to help us, so we must compute both
$\overline{x}$ and $\overline{y}$.  The denominator in both cases is the
area of our shape, which is 
\[
      \bigl(\frac{1}{4}\pi\cdot 1^2\bigr)+\bigl(1^2 \bigr)=(\pi/4)+1\,.
\]
The distance from top to bottom is $f(x)=1+\sqrt{1-x^2}$.
Thus, in computing $\overline{x}$, we find that the moment integral is
\[
   \int_0^1x\left(1+\sqrt{1-x^2}\right)dx=
\int_0^1x\,dx+\int_0^1x\sqrt{1-x^2}dx\,.
\]
The first of these two integrals is $0.5$.  The second integral can be
evaluated by the \index{method of substitution} method of substitution, since the derivative of what's
under the radical is essentially $x\,dx$.  That is, we substitute
$u=1-x^2$; then $du=-2xdx$, and so the integral becomes
$-0.5\int\sqrt{u}\,du$.  As for the limits of integration, $x=0$ implies
$u=1-0^2=1$, and $x=1$ implies $u=1-1^2=0$.  Thus, the \index{moment integral} moment integral is
\[
0.5-0.5\int_1^0\sqrt{u}\,du
=0.5-0.5(-2/3)=5/6,
\text{ so that }
\overline{x}=\frac{5/6}{(\pi/4)+1}=0.4667.
\]

\smallskip

To compute $\overline{y}$, we see that the distance across our shape from
the $y$-perspective (i.e., from left to right) is 
\[
g(y)=\begin{cases}
     1,& \text{if $-1\le y\le 0$;}\\
    \sqrt{1-y^2},&\text{ if $0\le y\le 1$.}
\end{cases}
\] 
Thus, the numerator of the $\overline{y}$ formula is
$\int_{-1}^0y\,dy+\int_0^1y\sqrt{1-y^2}dy$.  The first of these integrals
is $-1/2$, while the second integral happens to be exactly the same
integral that we already computed in the last paragraph, except with a
different letter as the variable of integration.  That is,
$\int_0^1y\sqrt{1-y^2}dy$ has the same value as $\int_0^1x\sqrt{1-x^2}dx$,
namely, $1/3$.  We conclude that 

\[
\overline{y}=\frac{-\frac{1}{2}+\frac{1}{3} }{ \frac{\pi}{4}+1}=-0.09335.
\] 
The reader should sketch a picture of the shape, and check that
$(0.4667,-0.09335)$ is a reasonable location for the balancing point.

\end{example}

\begin{example}
%{\bf Example 3.} 
Find the \index{centroid of Nicaragua} centroid of Nicaragua (see the map on the next page).  As an approximation to
the actual shape of the country, suppose that it is a quadrilateral with
vertices at the points

$11^\circ$ northern latitude \& $86^\circ$ western longitude, 

$11^\circ$ northern latitude \& $83^\circ$ western longitude,

$15^\circ$ northern latitude \& $83^\circ$ western longitude, 

$13^\circ$ northern latitude \& $88^\circ$ western longitude.

\noindent
Also suppose that Nicaragua is a plane shape, i.e., disregard the
mountains, curvature of the earth, etc.

\smallskip

\begin{centering}
%-figure FE -----------------------------
\begin{psfigure}{3in}{2.5in}{figFE.ps}
%\putcrosshair{0,0}
\puttext{0,0}{Nicaragua}
\end{psfigure}\\
%------------------------------
\end{centering}


Here our first problem is to put the shape in the $xy$-plane in a
convenient way.  We could choose the conventional longitude/latitude
coordinates as our $xy$-coordinates.  But that would be a little
cumbersome, because the origin of that coordinate system is at the Equator
along the Greenwich Meridian, which is far from Nicaragua.  Also note that
``west longitude'' means the negative direction from the Greenwich
Meridian, so we would be using fairly large negative numbers for the
$x$-coordinates of the corners (namely, $-86$, $-83$, $-88$).  A better
idea is to choose the origin at the intersection of the two slanted sides,
which is ($88^\circ$ west, $13^\circ$ north).  Then the other three corners
have the following coordinates relative to this origin: $(2,-2)$, $(5,-2)$,
$(5,2)$.  The two slanted sides have equations $y=0.4x$ (or $x=2.5y$) and
$y=-x$ (or $x=-y$).  The horizontal side is $y=-2$, and the vertical side
is $x=5$.  Thus, the distance $f(x)$ from top to bottom and the distance
$g(y)$ from left to right are: 
$$
f(x)=\begin{cases}1.4x,& \text{if $0\le x\le 2$;}\\ 
    0.4x+2,& \text{ if $2\le x\le 5$;}
\end{cases}
\qquad\qquad
g(y)=\begin{cases}5+y,& \text{if $-2\le y\le 0$;}\\ 
     5-2.5y,&  \text{if $0\le y\le 2$.}
\end{cases}
$$ 
The rest of this example is routine, and the details will
be left to the reader.  The answer is $(3.10^\circ,-0.31^\circ)$.  This
must be added to the longitude $-88^\circ$ and the latitude $13^\circ$ of
the origin.  Thus, the centroid of Nicaragua is at $12.7^\circ$ northern
latitude \& $84.9^\circ$ western longitude, i.e., about 70 miles east of
Matagalpa, on the Rio Grande.


\end{example}


