Husain, Zeena – Homework 1 – Due: Sep 4 2003, 4:00 am – Inst: H L Berk
1
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The due time is Central time.
001
(part 1 of 1) 10 points
One cubic meter (1.0 m
3
) of aluminum has a
mass of 2700 kg, and a cubic meter of iron has
a mass of 7860 kg.
Find the radius of a solid aluminum sphere
that has the same mass as a solid iron sphere
of radius 3
.
82 cm.
Correct answer: 5
.
45443 cm.
Explanation:
Let :
m
Al
= 2700 kg
,
m
Fe
= 7860 kg
,
and
r
Fe
= 3
.
82 cm
.
Basic Concept
Density is
ρ
=
m
V
Solution:
Since the masses are the same,
ρ
Al
V
Al
=
ρ
Fe
V
Fe
ρ
Al
4
3
π r
3
Al
¶
=
ρ
Fe
4
3
π r
3
Fe
¶
r
Al
r
Fe
¶
3
=
ρ
Fe
ρ
Al
r
Al
=
ρ
Fe
ρ
Al
¶
1
3
(
r
Fe
)
=
7860 kg
2700 kg
¶
1
3
(3
.
82 cm)
= 5
.
45443 cm
002
(part 1 of 2) 10 points
This problem shows how dimensional analysis
helps us check our work and sometimes even
help us find a formula. A rope has a cross sec
tion
A
= 9 m
2
and density
ρ
= 1710 kg
/
m
3
.
The “linear” density of the rope
μ
, is de
fined to be the mass per unit length, in the
form
μ
=
ρ
x
A
y
.
Based on dimensional analysis,
find the
powers
x
and
y
.
1.
x
= 1
, y
=

1
2.
x
= 1
, y
= 2
3.
x
=

2
, y
= 2
4.
x
=

2
, y
= 1
5.
x
=

1
, y
=

1
6.
x
=

1
, y
= 1
7.
x
=

2
, y
=

1
8.
x
=

1
, y
= 2
9.
x
= 1
, y
= 1
correct
Explanation:
Basic Concepts:
Kilogram (kg): a unit of mass (M).
Meter (m): a unit of length (L).
[
x
] means ”the units of
x
”.
The units of both sides of any equation must
be the same for the equation to make sense.
Solution:
The units of the left hand side (LHS) are given
as
[
μ
] =
M
L
=
ML

1
and the right hand side has
[
ρ
x
A
y
] =
M
L
3
¶
x
×
(
L
2
)
y
=
M
x
L

3
x
L
2
y
=
M
x
L
2
y

3
x
The powers on the units of mass and length
need to be the same as for the LHS above, so
x
= 1
2
y

3
x
=

1
2
y
=

1 + 3
= 2
y
= 1
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Husain, Zeena – Homework 1 – Due: Sep 4 2003, 4:00 am – Inst: H L Berk
2
Thus the answer is (
x, y
) = (1
,
1).
003
(part 2 of 2) 10 points
A simple pendulum is made out of a string
with length
L
and a mass
m
attached to one
end of the string. Its period of oscillation
T
may depend on the gravitational acceleration
g
, and also depend on
L
and
m
.
Based on dimensional analysis, which of the
following expressions is dimensionally accept
able, where
k
is a dimensionless constant?
1.
T
=
k
g
L
2.
T
=
k
s
L
g
correct
3.
T
=
k
m g
L
4.
T
=
k
r
g
L
5.
T
=
k
s
L
m g
6.
T
=
k
m L
g
7.
T
=
k
L
g
8.
T
=
k
r
m g
L
Explanation:
Here we proceed in the same way:
a pe
riod is a measure of time, thus the correct
expression must have units of time.
"
k
s
L
g
#
= 1
·
s
L
L/T
2
=
T
is the correct one. As for the others,
h
k
mg
L
i
= 1
·
ML/T
2
L
=
MT

2
•
k
mL
g
‚
= 1
·
ML
L/T
2
=
MT
2
•
k
r
mg
L
‚
= 1
·
r
ML/T
2
L
=
M
1
/
2
T

1
"
k
s
L
mg
#
= 1
·
s
L
ML/T
2
=
M

1
/
2
T
•
k
L
g
‚
= 1
·
L
L/T
2
=
T
2
h
k
g
L
i
= 1
·
L/T
2
L
=
T

2
•
k
r
g
L
‚
= 1
·
r
L/T
2
L
=
T

1
so these are all incorrect.
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 Fall '08
 Turner
 Acceleration, Mass, Work, Velocity, Correct Answer, Husain

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