Version 095 – EX4 – ditmire – (58216)
1
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printout
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20
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before answering.
001
10.0 points
A siphon consists of a flexible tube with the
same cross section throughout the tube. It is
used to drain a special liquid from a tank. Let
the density of the liquid be 502 kg
/
m
3
and
the diameter of the tube be 2
.
6 cm
.
Let the
siphon discharge a distance 5
.
8 m below the
surface of the liquid.
h
5
.
8 m
2
.
6 cm
What is the maximum height
h
;
i.e.
, the
vertical distance between the surface of the
liquid to the point at the top of the siphon, for
this siphon to work? The acceleration of grav
ity is 9
.
8 m
/
s
2
and
P
atm
= 1
.
013
×
10
5
N
/
m
2
.
1. 11.95
2. 20.5911
3. 17.6094
4. 19.0014
5. 12.9533
6. 11.6667
7. 13.4418
8. 19.285
9. 12.3203
10. 14.7879
Correct answer: 20
.
5911 m.
Explanation:
Let :
P
atm
= 1
.
013
×
10
5
N
/
m
2
,
g
= 9
.
8 m
/
s
2
,
= 5
.
8 m
,
d
= 2
.
6 cm
,
and
ρ
= 502 kg
/
m
3
.
Applying Bernoulli’s equation at the sur
face of the liquid pool and at the top of the
siphon gives
P
atm
+
1
2
ρ
v
2
A
=
P
B
+
1
2
ρ
v
2
B
+
ρ
g h .
The limiting case occurs when velocity in the
tube and the pressure at the top of the tube
are essentially zero. Hence
h
=
P
atm
ρ
g
=
1
.
013
×
10
5
N
/
m
2
(502 kg
/
m
3
) (9
.
8 m
/
s
2
)
=
20
.
5911 m
.
002
10.0 points
A uniform plank of mass 1
.
48 kg and length
36
.
3 cm is pivoted at one end, and the op
posite end is attached to a spring of force
constant 474 N
/
m.
The height of the pivot
has been adjusted so that the plank will be in
equilibrium when it is horizontally oriented as
in the figure.
36
.
3 cm
1
.
48 kg
θ
474 N
/
m
Find the period of small oscillation about
the equilibrium point.
1. 0.202703
2. 0.290968
3. 0.182044
4. 0.325383
5. 0.3196
6. 0.279113
7. 0.258874
8. 0.379447
9. 0.228038
10. 0.285553
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Version 095 – EX4 – ditmire – (58216)
2
Correct answer: 0
.
202703 s.
Explanation:
Let :
L
= 36
.
3 cm
,
m
= 1
.
48 kg
,
and
k
= 474 N
/
m
.
τ
=
I
α
Using the parallel axis theorem, the rotational
inertia of the plank about the pivot point is
I
=
1
12
m L
2
+
m
L
2
2
=
1
3
m L
2
.
(1)
From the freebody diagram, after displace
ment by a small angle
θ
, we have
τ
=

m g
L
2
+
k y L
cos
θ
=

m g
L
2
+
k L
sin
θ
L
cos
θ
,
(2)
where
y
=
L
sin
θ
is the displacement from
equilibrium.
Using
I
from Eq. (1), and for
small
θ
, cos
θ
= 1 and sin
θ
=
θ
, we have
m g
L
2

k L
2
θ
=
1
3
m L
2
d
2
θ
dt
2
.
(3)
Substitute
θ
=
θ

m g
2
k L
and
d
2
θ
dt
2
=
d
2
θ
dt
2
, we
have
d
2
θ
dt
2
+
3
k
m
θ
= 0
,
(4)
where the coe
ffi
cient of
θ
is
ω
2
=
3
k
m
. There
fore
T
≡
2
π
ω
= 2
π
m
3
k
(5)
= 2
π
(1
.
48 kg)
3 (474 N
/
m)
= 0
.
202703 s
.
Alternate Solution:
Use conservation of
energy.
E
=
m g
y
2
+
1
2
I
ω
2
+
1
2
k y
2
=
m g
y
2
+
1
2
1
3
m L
2
ω
2
+
1
2
k y
2
=
1
2
m g y
+
1
6
m v
2
+
1
2
k y
2
Di
ff
erentiate with respect to
t
.
Note:
Total energy is a constant, thus
dE
dt
=
0
and
use
the
chain
rule
for
[
v
(
t
)]
2
and [
y
(
t
)]
2
.
d E
dt
=
d
dt
1
2
m g y
+
1
6
m v
2
+
1
2
k y
2
0 =
1
2
m g v
+
1
6
m a v
+
1
2
k y v
0 =
m g
+
1
3
m a
+
k y .
By substitution we can write,
d
2
y
dt
2
+
3
k
m
y
= 0
.
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 Spring '10
 SHIH
 Mass, Speed of sound, Wavelength, Correct Answer

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