CHAPTER
8
Systems of Particles and Conservation of Momentum
1* ·
Give an example of a threedimensional object that has no mass at its center of mass.
A hollow sphere.
2
·
Three point masses of 2 kg each are located on the
x
axis at the origin,
x
= 0.20 m, and
x
= 0.50 m. Find the center
of mass of the system.
Use Equ. 84; note that
y
cm
= 0
x
cm
= [(2
×
0 + 2
×
0.2 + 2
×
0.5)/6] m = 0.233 m
3
·
A 24kg child is 20 m from an 86kg adult. Where is the center of mass of this system?
Take the origin at the position of the child.
Use Equ. 84
x
cm
= (86
×
20/110) m = 15.6 m
4
·
Three objects of 2 kg each are located in the
xy
plane at points (10 cm, 0), (0, 10 cm), and (10 cm, 10 cm). Find
the location of the center of mass.
Use Equ. 84
x
cm
= [(10
×
2 + 10
×
2)/6] cm = 6.67 cm;
y
cm
= [(10
×
2 + 10
×
2)/6] cm = 6.67 cm
5* ·
Find the center of mass
x
cm
of the three masses in Figure 846.
Use Equ. 84
x
cm
= [(1
×
1 + 2
×
2 + 8
×
4)/11] m = 3.36 m
6
·
Alley Oop’s clubax consists of a symmetrical 8kg stone attached to the end of a uniform 2.5kg stick that is 98
cm long. The dimensions of the clubax are shown in Figure 847. How far is the center of mass from the handle end
of the clubax?
1. Locate CM of stick and of stone
2. Use Equ. 84
By symmetry,
x
cm
(stick) = 0.49 m;
x
cm
(stone) = 0.89 m
x
cm
= [(2.5
×
0.49 + 8
×
0.89)/10.5] m = 0.795 m
7
·
Three balls
A, B,
and
C
, with masses of 3 kg, 1 kg, and 1 kg, respectively, are connected by massless rods. The
balls are located as in Figure 848. What are the coordinates of the center of mass?
Use Equ. 84
x
cm
= [(3
×
2 + 1
×
1 + 1
×
3)/5] m = 2 m
y
cm
= [(3
×
2 + 1
×
1 + 1
×
0)/5] m = 1.4 m
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Chapter 8
Systems of Particles and Conservation of Momentum
8
·
By symmetry, locate the center of mass of an equilateral triangle of side length
a
with one vertex on the
y
axis
and the others at (
a
/2, 0) and (+
a
/2, 0).
1. Draw the triangle; assume vertex at
y
> 0
2. Locate the intersection of the bisectors; see sketch
3. Give the coordinates of CM
x
cm
= 0;
y
cm
=
1/2
a
tan 30
o
= 0.289
a
9* ··
The uniform sheet of plywood in Figure 849 has a mass of 20 kg. Find its center of mass.
We shall consider this
as two sheets, a square sheet of 3 m side length and mass
m
1
and a rectangular sheet 1m
×
2m with a mass of 
m
2
. Let
coordinate origin be at lower left hand corner of the sheet. Let
σ
be the surface density of the sheet.
1. Find
x
cm
(
m
1
),
y
cm
(
m
1
) and
x
cm
(
m
2
),
y
cm
(
m
2
)
2. Determine
m
1
and
m
2
3. Use Equ. 84
By symmetry,
x
cm
(
m
1
) = 1.5 m,
y
cm
(
m
1
) = 1.5 m
and
x
cm
(
m
2
) = 1.5 m,
y
cm
(
m
2
) = 2.0 m
m
1
= 9
σ
kg,
m
2
= 2
σ
kg
x
cm
= (9
σ
×
1.5  2
σ
×
1.5)/7
σ
= 1.5 m
y
cm
= (9
σ
×
1.5  2
σ
×
2.0)/7
σ
= 1.36 m
10 ··
Show that the center of mass of a uniform semicircular disk of radius
R
is at a point (4/3
π
)
R
from the center of
the circle.
1. The semicircular disk is shown; we also show here the surface element
dA
2. Use Equ. 85 to find
y
cm
;
x
cm
= 0 by symmetry.
y
cm
= (1/
M
)
∫
y
σ
dA
3.
y
=
r
sin
θ
,
dA
=
r d
θ
dr
, and
M
=
π
R
2
σ
/2; make the appropriate
substitutions
4. y
cm
=
∫
R
M
0
σ
∫
π
0
r
2
sin
θ
d
θ
dr
=
∫
R
M
0
2
σ
r
2
dr
=
M
3
2
σ
R
3
=
π
3
4
R
0
11 ··
A baseball bat of length
L
has a peculiar linear density (mass per unit length) given by
λ
=
λ
0
(1 +
x
2
/
L
2
). Find the
x
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 Spring '09
 Eduardo
 Physics, Center Of Mass, Energy, Kinetic Energy, Mass, Momentum, Velocity

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