文档介绍：Chapter Thirteen
More Integration
Some Applications
Think now for a moment back to elementary school physics. Suppose we have a
system of point masses and forces acting on the masses. Specifically, suppose that for
each i = 12, ,K,n we have a point mass mi whose position in space at time t is given by
the vector ri .. Assume moreover that there is a force fi acting on this mass. Thus
according to Sir Isaac Newton, we have
d 2 r
f = m i
i i dt 2
for each i. Now sum these equations to get
n n d 2 r
= = i
F å fi å mi 2 , or
i =1 i =1 dt
æ n ö
ç m r ÷
d 2 å i i
F = M ç i =1 ÷,
dt ç n ÷
ç å mi ÷
è i =1 ø
n
where M = å mi . Reflect for a moment on this equation. If we define R by
i =1
n
m r
å i i d 2 R
R = i =1 , then the equation es F = M . Thus the sum of the external
n dt 2
å mi
i =1
forces on the system of masses is the total mass times the acceleration of the mystical
point R. This point R is called the center of mass of the system.
In case the total mass is continuously distributed in space, the "sum" in the
equation for R es an integral. Let's look at what this means in two dimensions.
Suppose we have a plate and the mass density of the plate at (x,y) is given by r (x, y) .
To find the center of mass of the plate, we approximate its location by chopping it into a
bunch of small pieces and treating each of these pieces as a point mass.
* *
Now choose a point ri = xi i + yi j in each rectangle. The mass of this rectangle will be
* *
approximately r (xi , yi )DAi , where DAi is the area of the rectangle. The equation for the
center of mass of this system of rectangles is then
n n
* *
å mi ri å r (xi , yi )ri DAi
~ i =1 i =1
R = n = n
* *
å mi å r (xi , yi )DAi
i =1 i =1
1 ì é n ù é n ù ü
= r * * * D + r * * *D
n í êå (xi , yi )xi Ai úi êå ( xi , yi ) yi Ai ú jý
* * î ëi =1 û ëi =1 û þ
å r ( xi , yi )DAi
i =1
The three sums in the previous line are Rieman