homework 01 – HILL, WILLIAM – Due: Sep 7 2007, 1:00 am
1
Question 1, chap 1, sect 6.
part 1 of 1
10 points
A piece of pipe has an outer radius, an
inner radius, and length as shown in the figure
below.
36 cm
4
.
5 cm
2
.
9 cm
The density
is 8
.
9 g
/
cm
3
.
What is the mass of this pipe?
Correct answer:
11
.
9177
kg (tolerance
±
1
%).
Explanation:
Let :
r
1
= 4
.
5 cm
,
r
2
= 2
.
9 cm
,
ℓ
= 36 cm
,
and
ρ
= 8
.
9 g
/
cm
3
.
Basic Concepts:
The volume of the pipe will be the cross
sectional area times the length.
Solution:
V
= (
π r
2
1
−
π r
2
2
)
ℓ
=
π
[
r
2
1
−
r
2
2
]
ℓ
=
π
[(4
.
5 cm)
2
−
(2
.
9 cm)
2
] (36 cm)
= 1339
.
07 cm
3
.
Thus the density is
ρ
=
m
V
so
m
=
ρ V
=
ρ π
[
r
2
1
−
r
2
2
]
ℓ
= (8
.
9 g
/
cm
3
)
π
[(4
.
5 cm)
2
−
(2
.
9 cm)
2
] (36 cm)
= 11917
.
7 g = 11
.
9177 kg
.
Question 2, chap 1, sect 6.
part 1 of 2
10 points
Consider the dimension content of the cen
tripetal force. The force should depend on the
mass
m
, the tangential speed
v
and the radius
r
. We write it in the form
F
=
m
x
v
y
r
z
.
Based on dimensional analysis, determine the
powers
x
,
y
and
z
.
1.
x
=
−
1,
y
=
−
2,
z
= 1.
2.
x
=
−
1,
y
=
−
2,
z
=
−
1.
3.
x
= 1,
y
= 2,
z
= 1.
4.
x
=
−
1,
y
= 2,
z
= 1.
5.
x
=
−
1,
y
= 2,
z
=
−
1.
6.
x
= 1,
y
=
−
2,
z
= 1.
7.
x
= 1,
y
= 2,
z
=
−
1.
correct
8.
x
= 1,
y
=
−
2,
z
=
−
1.
Explanation:
In order that
F
=
m
x
v
y
r
z
to be dimen
sionally correct:
[
F
] =
M
1
L
1
T
−
2
[
m
x
v
y
r
z
] =
M
x
parenleftBig
L
T
parenrightBig
y
L
z
=
M
x
L
y
+
z
T
−
y
.
By equating the powers of
M
,
L
, and
T
x
= 1,
y
+
z
= 1, and
−
y
=
−
2.
Therefore,
x
= 1,
y
= 2, and
z
=
−
1.
Question 3, chap 1, sect 6.
part 2 of 2
10 points
Consider the following set of equations,
where
s
,
s
0
,
x
and
r
have units of length,
t
has units of time,
v
has units of velocity,
g
and
a
have units of acceleration, and
k
is
dimensionless.
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homework 01 – HILL, WILLIAM – Due: Sep 7 2007, 1:00 am
2
Which one is dimensionally
incorrect
?
1.
v
2
= 2
a s
+
k s v
t
2.
t
=
v
a
+
x
v
3.
s
=
s
0
+
v t
+
v
2
a
4.
a
=
g
+
k v
t
+
v
2
s
0
5.
t
=
k
radicalbigg
s
g
+
a
v
correct
Explanation:
For an equation to be dimensionally cor
rect, all its terms must have the same units.
(1):
t
=
v
a
+
x
v
[
t
] =
T
bracketleftBig
v
a
bracketrightBig
+
bracketleftBig
x
v
bracketrightBig
=
L T
−
1
L T
−
2
+
L
L T
−
1
=
T
+
T
=
T
is consistent.
(2):
a
=
g
+
k v
t
+
v
2
s
0
[
a
] =
L T
−
2
bracketleftbigg
g
+
k v
t
+
v
2
s
0
bracketrightbigg
=
L T
−
2
+
L T
−
1
T
+
L
2
T
−
2
L
=
L T
−
2
is also consistent.
(3):
t
=
k
radicalbigg
s
g
+
a
v
[
t
] =
T
bracketleftbigg
k
radicalbigg
s
g
+
a
v
bracketrightbigg
=
radicalbigg
L
L T
−
2
+
L T
−
2
L T
−
1
=
T
+
T
−
1
is
not
dimensionally consistent.
(4):
v
2
= 2
a s
+
k s v
t
[
v
2
] =
L
2
T
−
2
bracketleftbigg
2
a s
+
k s v
t
bracketrightbigg
=
L T
−
2
L
+
L L T
−
1
T
−
1
=
L
2
T
−
2
is also consistent.
(5):
s
=
s
0
+
v t
+
v
2
a
[
s
] =
L
bracketleftbigg
s
0
+
v t
+
v
2
a
bracketrightbigg
=
L
+
L T
−
1
T
+
L
2
T
−
2
L T
−
2
=
L
is also consistent.
So only
t
=
k
radicalbigg
s
g
+
a
v
is dimensionally
incorrect
.
Question 4, chap 1, sect 6.
part 1 of 2
10 points
s
is a distance with unit [L],
t
is a time with
unit [T] and
θ
is an angle in radians.
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 Spring '08
 Turner
 Physics, Vectors, Dot Product, Work, Correct Answer, Orders of magnitude, vy

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