Moments and Center of Mass

Center of mass of a Wire

Suppose  we  have a wire   l   feet long  whose  density is rho(x)   pounds per foot at the point x  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?

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.

 Center of mass of a solid of revolution

If   0 <= y*`=`*f(x)  for   a <= x*`<=`*b , 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 Pi*f(x)^2  of the solid S to get   Volume = Int(Pi*f(x)^2,x = a .. b)

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 x-axis between a and b.    Let CM be the center of mass.  Partition   [a, b]  into n subintervals [x[i], x[i+1]]  and using planes  perpendicular to [a, b]  approximate the solid S with the n disks where the ith one has volume    Delta*V[i] = Pi*f(x[i])^2*Delta*x[i]

Now the signed moment of the ith disk about the point CM is    M[i] = (x-CM)*Delta*V[i]   and the sum of these moments will be approximately 0, since CM is the center of mass.  If we let Delta*x[i]  go to zero  this  approximate equation becomes an equation for the center of mass:

>    CMequation :=  int((x-CM)*Pi*f(x)^2,x=a..b) =0;

CMequation := int((x-CM)*Pi*f(x)^2,x = a .. b) = 0

Useing properties of integrals, we can solve this equation for CM.

>    sol := solve(CMequation,{CM} )   ;

sol := {CM = int(f(x)^2*x,x = a .. b)/int(f(x)^2,x = a .. b)}

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 x  is the area of the cross-section.

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   y = cos(x)  for x between 0 and pi/2  is

>    cenmass(cos,0,Pi/2);

4*(1/16*Pi^2-1/4)/Pi

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,style=wireframe);
plots[display]([cm,solid],scaling=constrained); end:

Test this out.

>     drawit(cos+2,0,7);

[Maple Plot]

We can animate the motion of the center of mass as the solid changes.

>    plots[display]([seq(drawit(2+cos,0,t),t=1..10)],
insequence=true);

Exercise:   Find the center of mass of a homogeneous hemispherical solid.

Exercise:   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 the y axis, for some positive number a.  If the center of mass is at  y= 10, what's a?