Moments and Center of Mass
Center of mass of a Wire
Suppose we have a wire
feet long whose density is
pounds per foot at the point
feet from the left hand end of the wire. What is the total mass of the wire and where is its
center of mass
, i.e., the point cm about which the total moment of the wire is 0?

Mass
Chop the wire into n small pieces each
feet long and pick an arbitrary point
in each piece. An approximation to the mass of the ith piece of wire is
, so an approximation to the total mass is
. This approximate mass is a
Riemann sum
approximating the integral
, and so the mass of the wire is defined as the value of this integral.

Center of mass
: Chopping as above, the
approximate moment
of the ith piece about the center of mass cm is
and so the total approximate moment is
. This is seen to be a Riemann sum approximating the integral
. But the center of mass is defined as the point about which the total moment is zero so the integral satisfies the equation
. Using properties of integrals, we can solve this equation for cm, to get the ratio of integrals
. Note the top integral represents the total moment of the wire about its left end (x=0) and the bottom integral is the total mass of the wire.
Exercise: Find the center of mass of a wire 1 foot long whose density at a point x inches from the left end is 10 + x + sin(x) lbs/inch.
Ex. 1
Locate the centroid of the plane area enclosed between the curve
, and between the y axis and the line x=3.
Warning, the name changecoords has been redefined
> 
C[x]:=int(x*f(x),x=0..3)/A;

> 
C[y]:=int(f(x)^2,x=0..3.0)/(2*A);

> 
p2:=pointplot([[C[x],C[y]]],symbol=circle):

The centroid lies on the line of symmetry of the surface area as could be anticipated. Now we add a slight complication.
Ex. 2
We plot sin(x) and a circle centered at (
) with radius 1 and locate the centroid of the resulting figure.
Warning, the name changecoords has been redefined
> 
Eq1:=(xPi/2)^2+(y1)^2=1;

> 
plot({f(x),g(x)},x=0..3,scaling=constrained);

Now we will find the centroid of the area enclosed by the two plots. First we note that the area we are interested in is described by f(x)g(x). The center of each strip is
above the x axis. The length of each strip is f(x)g(x) .
The formulas for the centroid therefore require modification to:
=
First we find the limits of integration by finding the points of intersection of the two curves.
> 
x1:=fsolve(f(x)=g(x),x=0..1);

> 
x2:= fsolve(f(x)=g(x),x=2..3);

> 
A:=int(f(x)g(x),x=x1..x2);

> 
C[x]:=int(x*(f(x)g(x)),x=x1..x2)/A;

> 
C[y]:=int((f(x)^2g(x)^2),x=x1..x2)/(2*A);

> 
p1:=plot({f(x),g(x)},x=0..3,scaling=constrained, labels=[x,y]):

> 
p2:=pointplot([[C[x],C[y]]],symbol=circle):

This certainly looks about right.
Pappus' Theorem
Ex. 3
Given the ellipse:
. Find the surface area of the solid of revolution about the x axis. We rotate the ellipse around the x axis.
Warning, the name changecoords has been redefined
> 
Eq1:=(x5)^2/9+(y7)^2/4=1;

plot3d
> 
plot3d({[x,E[1]*cos(theta),E[1]*sin(theta)],[x,E[2]*cos(theta),E[2]*sin(theta)]} ,x=2..8,theta=0..2*Pi,scaling=constrained,axes=boxed,color=theta);

Since, by symmetry, we know that the centroid of the ellipse is at (5,7) and the area of an ellipse is
(semimajor axis)
x
(semiminor axis), we immediately have, for the volume of the resulting solid;
Center of mass of a solid of revolution
If
for
, then let S be the
solid of revolution
obtained by rotating the region under the graph of f around the x axis. We know how to express the volume of S as an integral: Just integrate from a to b the crossectional area
of the solid S to get
Now how would we find the
center of mass
of the solid, assuming it's made of a homogeneous material? Well, it's clear that the center of mass will be somewhere along the xaxis between a and b. Let CM be the center of mass. Partition
into n subintervals
and using planes perpendicular to
approximate the solid S with the n disks where the ith one has volume
Now the signed moment of the ith disk about the point CM is
and the sum of these moments will be approximately 0, since CM is the center of mass. If we let
go to zero this approximate equation becomes an equation for the center of mass:
> 
CMequation := int((xCM)*Pi*f(x)^2,x=a..b) =0;

Useing properties of integrals, we can solve this equation for CM.
> 
sol := solve(CMequation,{CM} ) ;

Notice that the center of mass of the solid of revolution is the same as the center of mass of a wire whose density at
is the area of the crosssection.
We can define a word cenmass which takes a function f, an interval [a,b], and locates the center of the solid of revolution.
> 
cenmass := proc(f,a,b)
int(x*f(x)^2,x=a..b)/int(f(x)^2,x=a..b) end:

For example, the center of the solid obtained by rotating the region R under the graph of
for x between 0 and
is
Now we can define a word to draw the solid and locate the center of mass.
> 
drawit := proc(f,a,b)
local cm, solid;
### WARNING: the definition of the type `symbol` has changed'; see help page for details
cm := plots[pointplot3d]([evalf(cenmass(f,a,b)),0,0],color=red,symbol=box,thickness=3):
solid := plots[tubeplot]([x,0,0],x=a..b,radius=f(x),numpoints=20,tubepoints=30, style=wireframe);
plots[display]([cm,solid],scaling=constrained); end:

Test this out.
We can animate the motion of the center of mass as the solid changes.
> 
plots[display]( [seq(drawit(2+cos,0,(i/numFrames)*b), i=1..numFrames)],
insequence=true);

Practice
Practice
1. Find the centroid of each of the following figures.
a. The triangle formed by the x axis, the y axis and the line
b. The area enclosed by the x axis, the y axis and the curve
c. The area enclose by the curves:
and
2. An equilateral triangle, 2 units on each side, is rotated around a line parallel to, and 2 units from, one side. Find the surface area and the volume of the resulting solid.
3. Find the center mass of a homogeneous hemispherical solid.
4. A homogeneous solid is in the shape of a parabolic solid of revolution obtained by rotating the graph of y=x^2, x in [0,a] around the y axis, for some positive number a. If the center of mass is at y=10, what's a?