HRW Hints for exercise. Chapter 09.pdf

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C
HAPTER
9
H
INT FOR
P
ROBLEM
6
Symmetry indicates that the center of mass of the slab lies on the plane halfway through
the thickness of the slab and halfway from front to back. You need to calculate its position
along an axis that runs from left to right through the center of the sample. Place an
x
axis
1.40 cm above the lower face and 6.5 cm from the front edge, with its origin at the left edge
of the slab. Replace the iron with a particle of mass
M
i
=
ρ
i
V
i
, located at its center: on the
x
axis at
x
= 5.5 cm. Here
ρ
i
is the density of iron and
V
i
is the volume of iron in the slab.
Replace the aluminum with a particle of mass
M
a
=
ρ
a
V
a
, located at its center: on the
x
axis at
x
= 16.5 cm. Here
ρ
a
is the density of aluminum and
V
a
is the volume of aluminum
in the slab. Find the center of mass of the two replacement particles.
C
HAPTER
9
H
INT FOR
P
ROBLEM
7
First consider the hydrogen atoms alone. Their center of mass is at the center of the triangle.
To see this, replace two of the atoms with an atom of twice the mass at the midpoint of a
triangle side (at A). Now, draw the line from the opposite vertex to A. Since there is twice
as much mass at A as at the vertex, the center of mass of the three atoms lies on the dotted
line, one third the distance to the vertex (that is, at B). This is directly beneath the nitrogen
atom.
Replace the three atoms with one of three times the mass, at B. Find the coordinates of the
center of mass of this atom and the nitrogen atom. Place the origin of a coordinate system
at the nitrogen atom and take the
z
axis to be positive downward. If
d
is the distance from
the nitrogen to the plane of hydrogens below, the
z
coordinate of the center of mass is given
by 3m
H
d/(3m
H
+
m
N
). The distance
d
can be found by applying the Pythagorean theorem
to the triangle formed by the nitrogen, the center of the base triangle and a hydrogen atom.
ans:
6.75
×
10
−
12
m directly below the nitrogen atom
C
HAPTER
9
H
INT FOR
P
ROBLEM
8
Replace each of the five sides with a single particle with mass equal to the mass of the side
and positioned at the center of mass of the side. Let
m
be the mass of a side. The center of
mass of the side is at the geometrical center of the side. The center of mass of the box is at
the same position as the center of mass of the five particles. Use
x
com
=
y
com
=
and
z
com
=
1
M
1
M
mx
i
,
my
i
,
1
mz
i
M
to compute the coordinates of the center of mass of the box. Here
M
(= 5m) is the mass of
the box and
x
i
,
y
i
, and
z
i
are the coordinates of particle
i.
C
HAPTER
9
H
INT FOR
E
XERCISE
10
The net external force acting on the system consisting of the two skaters and the pole is
zero, so the velocity of the center of mass does not change. The forces of the skaters on the
pole and of the pole on the skaters are all internal forces and sum to zero. Since the center
of mass was at rest to begin with, it remains at the same position. The skaters meet at their
center of mass. To find how far the 40 kg skater moves, calculate the original distance from
that skater to the center of mass.
Place the origin of the coordinate system at the 40 kg skater. The 65 kg skater is then at
x
= 10 m. Calculate the coordinate of the center of mass.
C
HAPTER
9
H
INT FOR
E
XERCISE
13
(a) Use kinematics to find the coordinates of the stones, then calculate the coordinate of the
center of mass.
Take the
y
axis to be downward with the origin at the release point. The coordinate of the
first stone is then given by
1
y
1
=
gt
2
2
and the coordinate of the second stone is given by
1
y
2
=
g(t
−
t
0
)
2
,
2
where
t
0
= 200 ms. Evaluate these expressions for
t
= 300 ms, then use
y
com
=
with
m
2
= 2m
1
.
(b) The velocity of the first stone is given by
v
1
=
gt
and that of the second is given by
v
2
=
g(t
−
t
0
). Evaluate these expressions for
t
= 300 ms, then use
v
com
=
m
1
v
1
+
m
2
v
2
.
m
1
+
m
2
m
1
y
1
+
m
2
y
2
,
m
1
+
m
2
ans:
(a) 28 cm; (b) 2.3 m/s

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