p09_009.pdf
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79 KB
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Pobierz
Chapter 9 - 9.9
9. (a) Since the can is uniform, its center of mass is at its geometrical center, a distance
H/
2aboveits
base. The center of mass of the soda alone is at its geometrical center, a distance
x/
2abovethe
base of the can. When the can is full this is
H/
2. Thus the center of mass of the can and the soda
it contains is a distance
h
=
M
(
H/
2)+
m
(
H/
2)
M
+
m
=
H
2
above the base, on the cylinder axis.
(b) We nowconsider the can alone. The center of mass is
H/
2 above the base, on the cylinder axis.
(c) As
x
decreases the center of mass of the soda in the can at first drops, then rises to
H/
2 again.
(d) When the top surface of the soda is a distance
x
above the base of the can, the mass of the soda in
the can is
m
p
=
m
(
x/H
), where
m
is the mass when the can is full (
x
=
H
). The center of mass
of the soda alone is a distance
x/
2 above the base of the can. Hence
h
=
M
(
H/
2)+
m
p
(
x/
2)
M
+
m
p
=
M
(
H/
2)+
m
(
x/H
)(
x/
2)
M
+(
mx/H
)
=
MH
2
+
mx
2
We find the lowest position of the center of mass of the can and soda by setting the derivative of
h
with respect to
x
equal to 0 and solving for
x
. The derivative is
dh
dx
=
2
mx
2(
MH
+
mx
)
−
(
MH
2
+
mx
2
)
m
2(
MH
+
mx
)
2
=
m
2
x
2
+2
MmHx
MmH
2
2(
MH
+
mx
)
2
−
.
The solution to
m
2
x
2
+2
MmHx
−
MmH
2
=0is
x
=
MH
m
1+
1+
m
M
.
The positive root is used since
x
must be positive. Next, we substitute the expression found for
x
into
h
=(
MH
2
+
mx
2
)
/
2(
MH
+
mx
). After some algebraic manipulation we obtain
h
=
HM
m
1+
m
M
−
1
.
2(
MH
+
mx
)
.
−
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