James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 8.3 Applications to Physics and Engineering 561
of the system is located at
4
x −
o n
m i x i
i−1
o n
m i
i−1
−
o n
m i x i
i−1
m
where m − om i is the total mass of the system, and the sum of the individual moments
M − o n
i−1
m i x i
m£
y£
‹
y
⁄
m¡
›
0 fi x
x 2
m
is called the moment of the system about the origin. Then Equation 4 could be rewritten
as mx − M, which says that if the total mass were considered as being concentrated
at the center of mass x, then its moment would be the same as the moment of the
system.
Now we consider a system of n particles with masses m 1 , m 2 , . . . , m n located at the
points sx 1 , y 1 d, sx 2 , y 2 d, . . . , sx n , y n d in the xy-plane as shown in Figure 8. By analogy
with the one-dimensional case, we define the moment of the system about the y-axis
to be
M y − o n
5
m i x i
i−1
FIGURE 8
and the moment of the system about the x-axis as
6
M x − o n
i−1
m i y i
Then M y measures the tendency of the system to rotate about the y-axis and M x measures
the tendency to rotate about the x-axis.
As in the one-dimensional case, the coordinates sx, yd of the center of mass are given
in terms of the moments by the formulas
7
x − M y
m
y − M x
m
3
y
center of mass
8
0 4 x
where m − om i is the total mass. Since mx − M y and my − M x , the center of mass
sx, yd is the point where a single particle of mass m would have the same moments as
the system.
Example 3 Find the moments and center of mass of the system of objects that have
masses 3, 4, and 8 at the points s21, 1d, s2, 21d, and s3, 2d, respectively.
SOLUtion We use Equations 5 and 6 to compute the moments:
M y − 3s21d 1 4s2d 1 8s3d − 29
M x − 3s1d 1 4s21d 1 8s2d − 15
Since m − 3 1 4 1 8 − 15, we use Equations 7 to obtain
x − M y
m − 29
15
y − M x
m − 15
15 − 1
FIGURE 9
Thus the center of mass is s1 14
15 , 1d. (See Figure 9.) n
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