134
Chapter 22.
Solutions: Case Study: Robot Control: Swinging Like a Pendulum
so the solution to the diferential equation is
y
(
t
)=
α
1
·
1
λ
1
¸
e
λ
1
t
+
α
2
·
1
λ
2
¸
e
λ
2
t
,
where
α
1
and
α
2
are constants determined by two additional conditions. IF the
discriminant satis±es
c
2
/
(4
m
2
±
2
)
−
g/± >
0, then the solution decays; otherwise it
can have an oscillatory component in addition to a decaying one.
CHALLENGE 22.2.
We note that
v
(0
,
0) = 0, and it is easy to see that
v>
0
For all other values oF its arguments.
We diferentiate:
d
d
t
v
(
y
(
t
)) =
g
sin
θ
(
t
)
±
d
θ
(
t
)
d
t
+
d
θ
(
t
)
d
t
d
2
θ
(
t
)
d
t
2
=
g
sin
θ
(
t
)
±
d
θ
(
t
)
d
t
−
d
θ
(
t
)
d
t
1
m±
(
c
d
θ
(
t
)
d
t
+
mg
sin(
θ
(
t
)))
=
−
c
m±
µ
d
θ
(
t
)
d
t
¶
2
≤
0
.
ThereFore, we can now conclude that the point
θ
=0
,
d
θ/
d
t
= 0 is stable For both the
damped (
c>
0) and undamped (
c
= 0) cases. ²or the undamped case, d
v
(
y
(
t
))
/
d
t
is identically zero, and we cannot conclude that we have asymptotic stability. ²or
the damped case, we note that the set de±ned by d
v
(
y
(
t
))
/
d
t
= 0 contains all points
(
θ,
d
d
t
= 0), and the only invariant set is the one containing the single point (0
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
 Fall '11
 Dr.Robin

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