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Problem 1.13
[Difficulty: 5]
mg
kV
t
Given:
Data on sphere and terminal speed from Problem 1.12.
Find:
Distance traveled to reach 99% of terminal speed; plot of distance versus time.
Solution:
Use given data; integrate equation of motion by separating variables.
The data provided are:
M11
0
13
−
×
slug
⋅
=
V
t
0.2
ft
s
⋅
=
Newton's 2nd law for the general motion is (ignoring buoyancy effects)
M
dV
dt
⋅
Mg
⋅
kV
⋅
−
=
(1)
Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects)
⋅
t
⋅
=
so
k
⋅
V
t
=
k11
0
13
−
×
slug
⋅
32.2
×
ft
s
2
⋅
s
0.2 ft
⋅
×
lbf s
2
⋅
slug ft
⋅
×
=
k
1.61
10
11
−
×
lbf s
⋅
ft
⋅
=
To find the distance to reach 99% of
V
t
, we need
V
(
y
).
From 1:
M
dV
dt
⋅
M
dy
dt
⋅
dV
dy
⋅
=
MV
⋅
dV
dy
⋅
=
⋅
⋅
−
=
VdV
⋅
g
k
M
V
⋅
−
dy
=
Separating variables
Integrating and using limits
y
M
2
g
⋅
k
2
−
ln 1
k
⋅
V
⋅
−
⎛
⎝
⎞
⎠
⋅
M
k
V
⋅
−
=
We must evaluate this when
V
0.99 V
t
⋅
=
V
0.198
ft
s
⋅
=
y1
1
0
13
−
⋅
slug
⋅
()
2
32.2 ft
⋅
s
2
⋅
ft
1.61 10
11
−
⋅
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 Fall '07
 Lear
 Fluid Mechanics

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