Dropped Particle with Rebound Problem
(a)
If we neglect the very small “centrifugal force” term in GPS Eq.(4.91),
the equation of motion for a particle of mass
m
in the rotating frame of the
earth is
m
d
v
dt
=
m
g
+ 2
m
v
×
ω
,
(1)
where
v
is the particle velocity,
ω
is the earth’s angular velocity vector, and
g
is the acceleration of gravity at the earth’s surface,
g
=
−
g
e
z
.
All the vectors
and derivatives are measured in a body
fi
xed frame in which the
z
axis is normal
to the earth’s surface and the
x

y
plane is tangent to the earth’s surface and is
located at
z
= 0
. Assume that we are in the Northern hemisphere, and let the
y
axis point north and the
x
axis point east. As we saw in class, the solutions
to Eq.(1) for
v
and
r
to
fi
rst order in
ω
are
v
=
v
0
+
g
t
+ 2(
v
0
×
ω
)
t
+ (
g
×
ω
)
t
2
,
(2)
r
=
r
0
+
v
0
t
+
1
2
g
t
2
+ (
v
0
×
ω
)
t
2
+
1
3
(
g
×
ω
)
t
3
,
(3)
where the subscript
0
denotes an initial value.
Falling particle:
This problem is solved in the textbook and in the class
notes, and the answer is that the de
fl
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 Fall '10
 Wilemski
 mechanics, Centrifugal Force, Force, Mass, René Descartes, Euclidean geometry, gth ez

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