GPS Problem 8.20 (2
nd
ed. 8.10) Modi
fi
ed Planar Double Pendulum
This problem is a variation of the planar double pendulum problem, illustra
trated in Fig. 1.4.
The solution to that problem, GPS 1.22 or 2
nd
ed. 1.20, will
be useful here. There are two major modi
fi
cations that are required. First, we
put
m
1
= 0
and
m
2
=
m
. Second, the angle
θ
1
is now a driven coordinate with
a uniform rotation frequency, that will be denoted as
ω
.
So we have
·
θ
1
=
ω
and
θ
1
(
t
) =
ωt
, with
θ
1
(0) = 0
for simplicity.
The remaining changes are minor: we
let
l
1
=
a
,
l
2
=
l
, and
θ
2
=
θ
.
Because the motion of the mass point is con
fi
ned
to the vertical plane, the system has only one degree of freedom, described by
the angle
θ
,
θ
2
in Fig. 1.4, as the generalized coordinate. The angles will be
measured from a
y
axis that points downwards so that
θ
1
and
θ
2
are positive as
shown in Fig. 1.4.
The kinetic energy of the pendulum is
T
=
m
2
(
·
x
2
+
·
y
2
)
.
(1)
In the polar coordinate system we have
x
=
a
sin
ωt
−
l
sin
θ,
(2)
y
=
a
cos
ωt
+
l
cos
θ,
(3)
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 Fall '10
 Wilemski
 mechanics, Energy, Kinetic Energy, Cos, Polar coordinate system

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