Calculus 2nd Edition Rogawski

SECTION 9.3 Center of Mass 491
9.3 Center of Mass
Every object has a balance point called the center of mass (Figure 1). When a rigid object
such as a hammer is tossed in the air, it may rotate in a complicated fashion, but its center
of mass follows the same simple parabolic trajectory as a stone tossed in the air. In this
section we use integration to compute the center of mass of a thin plate (also called a
lamina) of constant mass density ρ.
The center of mass (COM) is expressed in terms of quantities called moments. The
moment of a single particle of mass m with respect to a line L is the product of the particle’s
mass m and its directed distance (positive or negative) to the line:
Moment with respect to line L = m × directed distance to L
The particular moments with respect to the x- and y-axes are denoted M x and M y . For a
particle located at the point (x, y) (Figure 2),
FIGURE 1 This acrobat with Cirque du
Soleil must distribute his weight so that his
arm provides support directly below his
center of mass.
y
M x = my
M y = mx
x
Mass m
located at (x, y)
y
(mass times directed distance to x-axis)
(mass times directed distance to y-axis)
y
m 1 x 1
x 2
m 2
m 3
COM
x 3
m 4
x
x 4
x
FIGURE 2
FIGURE 3
By definition, moments are additive: the moment of a system of n particles with
coordinates (x j ,y j ) and mass m j (Figure 3) is the sum
CAUTION The notation is potentially
confusing: M x is defined in terms of
y-coordinates and M y in terms of
x-coordinates.
M x = m 1 y 1 + m 2 y 2 +···+m n y n
M y = m 1 x 1 + m 2 x 2 +···+m n x n
The center of mass (COM) is the point P = (x CM ,y CM ) with coordinates
4
3
2
1
y
2
4
COM
1 2 3 4 5 6
FIGURE 4 Centers of mass for Example 1.
8
7
x
x CM = M y
M ,
y CM = M x
M
where M = m 1 + m 2 +···+m n is the total mass of the system.
EXAMPLE 1 Find the COM of the system of three particles in Figure 4, having masses
2, 4, and 8 at locations (0, 2), (3, 1), and (6, 4).
Solution The total mass is M = 2 + 4 + 8 = 14 and the moments are
M x = m 1 y 1 + m 2 y 2 + m 3 y 3 = 2 · 2 + 4 · 1 + 8 · 4 = 40
M y = m 1 x 1 + m 2 x 2 + m 3 x 3 = 2 · 0 + 4 · 3 + 8 · 6 = 60
Therefore, x CM = 60
14 = 30 7 and y CM = 40
14 = 20 7 . The COM is ( 30
7
, 20 )
7 .