1328
Chapter 13
13.101.
IDENTIFY:
Eq.(13.39) says
2/
TI
m
g
d
π
=
.
SET UP:
Let the two distances from the center of mass be
12
and .
dd
There are then two relations of the form of
Eq. (13.39); with
22
1c
m
1
2c
m
2
and
II m
d
d
=+
.
EXECUTE:
These relations may be rewritten as
( )
2
m
1
4
mgd T
I
md
and
( )
2
m
2
4.
mgd T
I
md
Subtracting the expressions gives
()
( )
2
2
2
1 2
1212
44
.
mg d
d T
m d
d
m d
d
d
d
ππ
−=
−
=
−+
Dividing by the
common factor of
( )
md d
−
and letting
dd L
+
=
gives the desired result.
EVALUATE:
The procedure works in practice only if both pivot locations give rise to SHM for small oscillations.
13.102. IDENTIFY:
Apply
m
∑
F
=a
#
#
to the mass, with
2
rad
aa r
ω
==
.
SET UP:
The spring, when stretched, provides an inward force.
EXECUTE:
Using
2
l
′
for the magnitude of the inward radial acceleration,
2
0
mlk
l
l
′
=
−
, or
0
2
kl
l
km
=
′
−
.
(b)
The spring will tend to become unboundedly long.
EXECUTE:
As resonance is approached and
l
becomes very large, both the spring force and the radial acceleration
become large.
13.103. IDENTIFY:
For a small displacement
x
, the force constant
k
is defined by
x
Fk
x
= −
.
SET UP:
Let
00
,
so that
and
rR x
2
[]
.
bx
bx
FA
e
e
−−
=−
EXECUTE:
When
x
is small compared to
1
,
b
−
expanding the exponential function gives
[(1 2
)
(1
)]
,
F
A
bx
bx
Abx
≈−−
−
=
−
corresponding to a force constant of
579 N/m
Ab
=
.
EVALUATE:
Our result is very close to the value given in Exercise 13.40.
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View Full Document141
F
LUID
M
ECHANICS
14.1.
IDENTIFY:
Use Eq.(14.1) to calculate the mass and then use
wm
g
=
to calculate the weight.
SET UP:
/
mV
ρ
=
so
=
From Table 14.1,
33
7.8 10 kg/m .
=×
EXECUTE:
For a cylinder of length
L
and radius
R
,
22
4
3
(
)
(0.01425 m) (0.858 m)
5.474 10 m .
VR
L
ππ
−
==
=
×
Then
4
3
(7.8 10 kg/m )(5.474 10 m )
4.27 kg,
−
== ×
×
=
and
2
(4.27 kg)(9.80 m/s )
41.8 N
g
=
(about
9.4 lbs). A cart is not needed.
EVALUATE:
The rod is less than 1m long and less than 3 cm in diameter, so a weight of around 10 lbs seems
reasonable.
14.2.
IDENTIFY:
Convert gallons to kg. The mass
m
of a volume
V
of gasoline is
=
.
SET UP:
1 gal
3.788 L
3.788 10 m
−
×
.
3
1 m of gasoline has a mass of 737 kg.
EXECUTE:
3
1 gal
1 m
45.0 mi/gal
(45.0 mi/gal)
16.1 mi/kg
737 kg
−
⎛⎞
⎜⎟
×
⎝⎠
EVALUATE:
1 gallon of gasoline has a mass of 2.79 kg. The car goes fewer miles on 1 kg than on 1 gal, since
1 kg of gasoline is less gasoline than 1 gal of gasoline.
14.3.
IDENTIFY:
/
=
SET UP:
The density of gold is
19.3 10 kg/m
×
.
EXECUTE:
333
6
3
(5.0 10 m)(15.0 10 m)(30.0 10 m)
2.25 10 m
V
−−−
−
×
×
=
×
.
63
0.0158 kg
7.02 10 kg/m
m
V
−
= ×
×
. The metal is not pure gold.
EVALUATE:
The average density is only 36% that of gold, so at most 36% of the mass is gold.
14.4.
IDENTIFY:
Find the mass of gold that has a value of
6
$1.00 10
×
. Then use the density of gold to find the volume
of this mass of gold.
SET UP:
For gold,
. The volume
V
of a cube is related to the length
L
of one side by
3
VL
=
.
EXECUTE:
3
6
1 troy ounce
31.1035 10 kg
($1.00 10 )
72.9 kg
$426.60
1 troy ounce
m
−
×
=
.
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 Spring '06
 Buchler
 Physics, Center Of Mass, Mass, Pressure measurement, gauge pressure

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