Midterm 03 - Bautista Aldo Midterm 3 Due 10:00 pm Inst...

Bautista, Aldo – Midterm 3 – Due: Nov 16 2005, 10:00 pm – Inst: Maxim Tsoi 1
This print-out should have 17 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering. 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 ,

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
.