Bautista, Aldo – Midterm 3 – Due: Nov 16 2005, 10:00 pm – Inst: Maxim Tsoi
1
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The due time is Central
time.
001
(part 1 of 1) 10 points
Assume:
The buoyant force of the air is neg
ligible.
Given:
The density of the oil is 822 kg
/
m
3
.
The density of the water salt mixture is
1030 kg
/
m
3
.
A rectangular parallelepiped block of uni
form density floats in a container which con
tains oil and salt water as shown.
The
block
sticks
up
above
the
oil
by
a
dis
tance
d
1
= 0
.
43 m.
The oil thickness is
d
2
= 0
.
38 m.
The block’s depth in the salt
water is
d
3
= 0
.
17 m. The horizontal area of
the block is 0
.
034 m
2
.
M
d
1
d
3
d
2
Calculate the mass of the block.
Correct answer: 16
.
5736 kg.
Explanation:
The mass of the salt water displaced is
m
3
=
ρ
w
d
3
a
= (1030 kg
/
m
3
) (0
.
17 m) (0
.
034 m
2
)
= 5
.
9534 kg
.
The mass of the oil displaced is
m
2
=
ρ
o
d
2
a
= (822 kg
/
m
3
) (0
.
38 m) (0
.
034 m
2
)
= 10
.
6202 kg
.
Since the mass of the block
m
b
is equal to the
liquid displace, we have
m
b
=
m
2
+
m
3
= (10
.
6202 kg) + (5
.
9534 kg)
= 16
.
5736 kg
.
002
(part 1 of 1) 10 points
A 24 kg mass and a 16 kg mass are suspended
by a pulley that has a radius of 4
.
7 cm and
a mass of 7 kg.
The cord has a negligible
mass and causes the pulley to rotate without
slipping. The pulley rotates without friction.
The masses start from rest 1
.
8 m apart. Treat
the pulley as a uniform disk.
The acceleration of gravity is 9
.
8 m
/
s
2
.
1
.
8 m
4
.
7 cm
7 kg
ω
24 kg
16 kg
Determine the speeds of the two masses as
they pass each other.
Correct answer: 1
.
80115 m
/
s.
Explanation:
Let :
M
= 7 kg
,
R
= 4
.
7 cm
,
m
1
= 24 kg
,
m
2
= 16 kg
,
h
= 1
.
8 m
,
v
=
ω R ,
I
=
1
2
M R
2
,
and
K
disk
=
1
2
I ω
2
=
1
4
M v
2
.
From conservation of energy
K
1
+
K
2
+
K
disk
=

Δ
U
or
m
1
v
2
2
+
m
2
v
2
2
+
M v
2
4
= (
m
1

m
2
)
g
h
2
m
1
+
m
2
+
M
2
¶
v
2
= (
m
1

m
2
)
g h ,
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Bautista, Aldo – Midterm 3 – Due: Nov 16 2005, 10:00 pm – Inst: Maxim Tsoi
2
where
h
2
is the height. Taking no slipping into
account, we can solve for
v
v
=
s
2 (
m
1

m
2
)
g h
2 (
m
1
+
m
2
) +
M
(1)
=
s
141
.
12 J
43
.
5 kg
=
1
.
80115 m
/
s
.
Alternative Solution:
The forces in the
vertical directions for
m
1
and
m
2
,
and the
torque on the pulley, give us
m
1
g

T
1
= +
m
1
a
m
2
g

T
2
=

m
2
a ,
so
T
1
=
m
1
(
g

a
)
T
2
=
m
2
(
g
+
a
)
,
and
I α
= [
T
1

T
2
]
R
1
2
M R
2
‡
a
R
·
=
h
m
1
(
g

a
)]
R

[
m
2
(
g
+
a
)
i
R
1
2
M a
=
m
1
g

m
1
a

m
2
a

m
2
g
a
=
2 (
m
1

m
2
)
M
+ 2 (
m
1
+
m
2
)
.
The velocity is
v
2
f
=
v
2
i
+
s a
Δ
y
v
f
=
r
2
a
h
2
=
s
2 (
m
1

m
2
)
g h
2 (
m
1
+
m
2
) +
M
,
which is the same as Eq. 1.
003
(part 1 of 1) 10 points
A solid sphere of radius 19 cm is positioned
at the top of an incline that makes 22
◦
angle
with the horizontal.
This initial position of
the sphere is a vertical distance 2
.
6 m above
its position when at the bottom of the incline.
The sphere is released and moves down the
incline.
The acceleration of gravity is 9
.
8 m
/
s
2
.
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 Angular Momentum, Energy, Kinetic Energy, Moment Of Inertia, Correct Answer

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