Problem 1.3
This is easiest to do by straightforward substitution
[]
[]
()
Time
Time
Time
Length
Length
g
l
g
l
T
=
=
⎟
⎠
⎞
⎜
⎝
⎛
=
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎦
⎤
⎢
⎣
⎡
=
2
2
2
2
π
Problem 1.4
(a)
[ ]
T
L
M
T
ML
T
ML
T
ML
L
T
L
M
T
L
M
T
ML
mgh
mv
mv
2
1
2
2
2
2
2
2
2
2
2
2
2
0
2
2
1
2
1
+
=
+
=
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
=
+
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎦
⎤
⎢
⎣
⎡
The second term doesn’t have the same dimensions due to the presence of the
square root operator.
(b)
[] [ ]
[]
()
LT
T
L
T
T
L
T
L
T
L
at
v
v
at
v
v
+
=
⎟
⎠
⎞
⎜
⎝
⎛
+
=
+
=
+
=
2
2
0
2
0
The problem with this equation is that the second term had an extra factor of
t
.
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[][]
2
2
2
2
2
T
L
T
ML
v
ma
v
ma
=
=
=
(T
he lefthand side has a spare factor of m and needs another factor with
dimensions of length to be dimensionally correct.
Problem 1.7
(a) Three significant digits
(b) Four significant digits
(c) Three significant digits
d) Two significant digits
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 Summer '08
 MEITZLER
 mechanics, Cartesian Coordinate System, Polar Coordinates, Heat, Polar coordinate system, Table tennis

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