
1 
Physics 112  HW #14 Solutions
Spring 2005
1354
[Driven Oscillations]
If
k
=
m
ω
d
, then
A
1
=
F
max
(k  m
ω
d
2
)
2
+ b
1
2
ω
d
2
=
F
max
b
1
ω
d
(a)
A
=
F
max
(3b
1
)
ω
d
=
A
1
3
(b)
A
=
F
max
(b
1
/2)
ω
d
=
2A
1
1390
[Lrod Oscillator]
The CM of the swinging system (two rods combined) is located a distance from the pivot equal to:
d
=
L/2
2
=
L
2
2
.
The system's moment of inertia about the pivot point is:
I
=
1
3
mL
2
+
1
3
mL
2
=
2
3
mL
2
=
1
3
ML
2
[where
M = 2m is the total mass of the system]
So the freq. of its SHM oscillation is:
f
=
1
2
π
Mgd
I
f =
1
2
π
Mg
L
2
2
1
3
ML
2
=
1
2
π
3g
2
2L
14.11
[Diving Bell]
(a)
Gauge pressure p =
ρ
g h = (1.03 x 10
3
kg/m
3
)(9.81 m/s
2
)(250 m) =
2.52 x 10
6
Pa
(b)
Force on window F = p A = p
π
r
2
= (2.52 x 10
6
Pa)(3.142)(0.150 m)
2
=
1.78 x 10
5
N
CM
d
L/2
L/2
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2 
14.54
[Hot Air Balloon]
Total weight of balloon + gases inside is
W =
ρ
g
V g + 900 N + 1700 N + 3200 N =
ρ
g
V g + 5800 N
Bouyancy force is equal to the weight of air displaced by the balloon.
F
b
=
ρ
a
V g = (1.23 kg/m
3
)(2200 m
3
)(9.81 m/s
2
) = 2.655 x 10
4
N
For balloon to be just able to lift, F
b
= W, so
2.655 x 10
4
N =
ρ
g
V g + 5800 N
ρ
g
V g = 2.075 x 10
4
N
ρ
g
=
2.075 x 10
4
N
(2200 m
3
) (9.81 m/s
2
)
=
0.961 kg/m
3
14.64
[Block and Beaker on Scales]
(a)
The liquid in the beaker is pushing up on the block with force f
b
=
ρ
l
V
block
g, the bouyant
force.
So the block must be pushing DOWN on the liquid with a force of the same
magnitude (N’s 3
rd
law.)
Then the reading on balance E, times g, is the weight of the liquid plus the beaker, plus (the
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 Fall '07
 LECLAIR,A
 Physics, mechanics, Force, Mass, kg

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