AE 353: HW 3 Solutions
3.50
HintProvided
Say
y
(0)
=
0 and ˙
y
(0)
=
0. Will the conductor move at all? Note that when ˙
y
(0)
=
0
then
α
=
0, so
L
(
α
)
=
0.
Hence ¨
y
=
0.
Conductor stays at rest.
Only reason
why conductor moves is, if there is an initial velocity or nonzero initial position.
Assume for simplicity
y
(0)
=
0. and ˙
y
(0)
=
ǫ
(very small since oscillations are of
low frequency).
Since
ǫ
is small, we can do linearization about ˙
y
=
0. Do linearization term by
term and then do laplace transform.
Though
ǫ
is small if poles of
Y
(
s
) are in RHP, then
y
(
t
) increases with
t
. We
want poles of
Y
(
s
) to be in LHP i.e. roots of denominator of
Y
(
s
) should be in LHP
and Routh criterion gives the su
ffi
cient conditions for this.
solution
If ˙
y
is small compared to
ν
, then
α
=
−
tan
−
1
(
˙
y
ν
)
≈−
˙
y
ν
. Now,
L
(
α
)
≈
L
(0)
+
∂
L
∂α
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
α
=
0
α
≈
L
(0)
−
∂
L
∂α
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
α
=
0
˙
y
ν
Since
L
(0)
=
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 Spring '08
 Voulgaris,P
 K1 K1, K1 K2

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