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1
x
2
x
d
1
m
2
m
x
0
Chapter 6 System of Particles
6.1
System of Particles
6.2 Newton’s Second Law for a System of Particles
6.3 The Momentum of a System of Particles
6.4
Appendix – Centers of Mass of Uniform Bodies
6.1
System of Particles
The center of mass (CM) is one point of an extended object whose motion under the
influence of external forces can be analyzed just like that of a simple point particle. It is a
point represents the average location for the total mass of a system.
(1) Two particles in 1D
Total mass:
M
m
m
=
+
1
2
Center of mass (CM):
M
x
m
x
m
m
m
x
m
x
m
x
2
2
1
1
2
1
2
2
1
1
CM
+
=
+
+
≡
If
m
m
m
1
2
=
=
,
then
2
2
)
(
2
1
2
1
CM
x
x
m
x
x
m
x
+
=
+
=
(2)
N
particles in 1D
M
x
m
m
m
m
m
x
m
x
m
x
m
x
N
i
i
i
N
N
N
∑
=
=
+
⋅
⋅
⋅
⋅
⋅
⋅
+
+
+
+
⋅
⋅
⋅
⋅
⋅
⋅
+
+
≡
1
3
2
1
2
2
1
1
CM
(3)
N
particles in 3D
)
,
,
(
1
1
1
1
z
y
x
r
=
)
,
,
(
N
N
N
N
z
y
x
r
=
=
=
=
∑
∑
∑
=
=
=
N
i
i
i
N
i
i
i
N
i
i
i
z
m
M
z
y
m
M
y
x
m
M
x
1
CM
1
CM
1
CM
1
1
1
1
1
m
2
m
3
m
1
r
2
r
3
r
O
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∑
=
=
N
i
i
i
r
m
M
r
1
CM
1
For example, for a uniform rectangular block,
∆
=
∆
=
∑
∑
i
i
i
i
i
i
m
y
M
y
m
x
M
x
1
1
CM
CM
When
∆
m
i
→
0 ,
=
=
∫
∫
dm
y
M
y
xdm
M
x
1
1
CM
CM
In general,
∫
=
dm
r
M
r
1
CM
Remark
:
The center of gravity (CG) is a concept similar to center of mass. The force of gravity
actually acts on all different parts of an object, but for purposes of determining the motion of
an object as a whole, we can assume that the entire weight of the object acts at a point called
the center of gravity. For nearly all practical purposes, the CM and CG of an object are at the
same point. There would be difference between them only in the unusual case of an object so
large that the acceleration due to gravity,
g
, was different at different parts of the object.
Example 61
Two masses
m
1
= 0.260 kg and
m
2
= 0.170 kg are separated by
0.500 m as shown in the figure. What is the distance from
m
1
to the center of mass of the system?
Answer:
Method 1
We know that the center of mass lies on the line joining
m
1
and
m
2
. Now, let
O
be the center of mass and
x
1
be the
distance between
m
1
and
O
. We can write
2
1
1
1
)
500
.
0
(
m
x
m
x

=
That is,
)
170
.
0
)(
500
.
0
(
)
260
.
0
(
1
1
x
x

=
, and we obtain
198
.
0
1
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