Nguyen, Phan – Homework 1 – Due: Jan 27 2004, 4:00 am – Inst: V Kaplunovsky
2
is also consistent.
(5):
s
=
s
0
+
vt
+
v
2
a
[
s
] =
L
•
s
0
+
vt
+
v
2
a
‚
=
L
+
LT

1
T
+
L
2
T

2
LT

2
=
L
is also consistent.
So only
t
=
k
r
s
g
+
a
v
is dimensionally
in
correct
.
003
(part 1 of 2) 10 points
Stokes law says
F
= 6
πrηv.
F
is a force,
r
the radius and
v
the velocity.
The parameter
η
has the dimension of
1.
[
η
] =
M
T L
correct
2.
[
η
] =
T
2
L
2
M
3.
[
η
] =
M
T L
2
4.
[
η
] =
T
M
5.
[
η
] =
T
2
L
M
6.
[
η
] =
M
T
7.
[
η
] =
M
T
2
L
2
8.
[
η
] =
M
T
2
L
9.
[
η
] =
T L
M
10.
[
η
] =
T L
2
M
Explanation:
Recall that the dimension of force is
[
F
] = [
m a
] =
M
L
T
2
.
The number 6
π
is dimensionless. The dimen
sions of the radius and velocity are
[
r
] =
L
;
[
v
] =
L
T
so
[
F
] = [6
π r η v
]
M
L
T
2
=
L
L
T
[
η
]
Solving for [
η
] yields
[
η
] =
M
LT
.
004
(part 2 of 2) 10 points
Consider a simple pendulum which consists of
a string with a bob attached to its end.
Its
period, (
i.e.
, the time interval taken for the
bob to complete one cycle of motion), may be
written in the form
T
=
k m
x
g
y
b
z
,
where
k
is a dimensionless constant,
g
the
magnitude of the gravitational acceleration,
m
the mass of the bob, and
b
the length of the
string.
The appropriate
x
,
y
, and
z
values are given
respectively by
1.
(
x, y, z
) =
1
,
1
2
,

1
2
¶
2.
(
x, y, z
) =
0
,

1
2
,

1
2
¶
3.
(
x, y, z
) =
1
,

1
2
,
1
2
¶
4.
(
x, y, z
) =
0
,

1
2
,
1
2
¶
correct
5.
(
x, y, z
) =
0
,
1
2
,

1
2
¶
6.
(
x, y, z
) =
1
,

1
2
,

1
2
¶
7.
(
x, y, z
) = (1
,

1
,
1)
8.
(
x, y, z
) =
0
,
1
2
,
1
2
¶
9.
(
x, y, z
) = (0
,

1
,
1)