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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
(12 cm)(0.14 kg)
1.31kg
1
1
1
1
2.8 cm.
1.31 kg
0.14 kg
HM
m
h
mM
§·
§
·
=+
−
=
+
−
=
¨¸
¨
¸
¨
¸
©¹
©
¹
8. (a) Since the can is uniform, its center of mass is at its geometrical center, a distance
H
/2 above its base. The center of mass of the soda alone is at its geometrical center, a
distance
x
/2 above the 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
MH
mH
Mm
H
=
+
+
=
//
22
2
bg
above the base, on the cylinder axis. With
H
= 12 cm, we obtain
h
= 6.0 cm.
(b) We now consider the can alone. The center of mass is
H
/2 = 6.0 cm 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
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Center Of Mass, Mass

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