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9.1
SYSTEMS OF PARTICLES
EXERCISES
Section 9.1 Center of Mass
12.
Take the
x
axis along the seesaw in the direction of the father, with origin at the center. The center of mass of the
child and her father is at the origin, so
xm
x
m
x
cm
c
c
f
f
==
+
0,
where the masses are given, and
x
c
=−
(.
)
/
35
2
m
(half the length of the seesaw in the negative
x
direction). Thus,
x
m
fc
c
f
=
=
/(
/
)
(
.).
28 65 1 75
75 4
mc
m
from
the center.
13.
INTERPRET
This problem is about center of mass. Our system consists of three masses located at the vertices of
an equilateral triangle. If two masses are known and the location of the center of mass is also known, then the third
mass can be calculated.
DEVELOP
The center of mass of a system of particles is given by Equation 9.2:
r
rr
r
mr
m
M
ii
i
i
i
i
cm
∑
∑
∑
We shall choose
xy
coordinates with origin (0,0) at the midpoint of the base. With this arrangement, the center of
the mass is located at
x
cm
=
0 and
yy
cm
=
3
2
/, where
y
3
is the position of the third mass (and of course,
12
0
for the equal masses
mmm
on the base).
EVALUATE
Using Equation 9.2, the
y
coordinate of the center of mass is
y
my
y
m
cm
=
++
=
+
11
2 2
33
123
00
()
mm
+
=
+
3
3
3
2
2
cm
Solving for
m
3
, we have
22
m
+=
, or
3
2
=
.
ASSESS
From symmetry consideration, it is apparent that
x
cm
=
0. On the other hand, we have
mm m
2
at the
bottom two vertices of the triangle. Since
cm
=
3
2
/, i.e.,
y
cm
is halfway to the top vertex, we expect the mass there
to be 2
m
(See Example 9.2).
14.
With origin at the barbell’s center,
x
1
75
cm, and
x
2
75
=
cm. For part (a):
x
cm
=
−+
+
(
)
(
)
60
75
60
75
60
60
kg
cm
kg
cm
kg
kg
=
0
For part
(b)
:
x
cm
=
+
(
)
(
)
50
75
80
75
50
80
kg
cm
kg
cm
kg
kg
cm
=
17 3
.
(toward the heavier mass). The addition of 75 cm to the above coordinates transforms these answers to the frame of
Example 9.1.
15.
INTERPRET
This problem is about locating the center of mass. Our system consists of three equal masses located
at the vertices of an equilateral triangle of side
L
.
9
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View Full Document 9.2
Chapter 9
DEVELOP
We take
xy
coordinates with origin at the center of one side as shown. The center of mass of a system
of particles is given by Equation 9.2:
r
rr
r
mr
m
M
ii
i
i
i
i
cm
==
∑
∑
∑
EVALUATE
From the symmetry (for every mass at
x
, there is an equal mass at
−
x
), we have
x
cm
=
0. As for
y
cm
, since
y
=
0 for the two masses on the
x
axis, and
yL
L
3
3
2
60
=°
=
sin
for the third mass, Equation 9.2 gives
y
my
mmm
L
m
cm
/
=
++
=
11
2 2
33
123
00
3
2
()
+
+
mm
LL
3
6
0 289
.
ASSESS
From symmetry consideration, it is apparent that
x
cm
=
0. On the other hand, we have
mm m
+=
2
at the
bottom two vertices of the triangle, and
m
at the top of the vertex. Therefore, we should expect
y
cm
to be one third
of
y
3
. This indeed is the case, as
y
cm
can be rewritten as
yy
cm
=
3
3
/.
16.
With origin at the center of the Earth,
x
MM
r
cm
Em
=
+
+
=
×
+
(. )
(.
)
(
0
7 35 3 85 10
597
5
km
735
4680
.)
=
km,
or about 1690 km below the Earth’s surface.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.
 Fall '10
 Pheong

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