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Goldstein 815 (3
rd
ed. 8.25)
This problem di
f
ers from 814 because the cylinder is now driven to rotate
at a constant angular speed
ω
, and its kinetic energy is no longer variable. Since
the cylinder’s angular motion is prescribed, we only need to consider the motion
of the particle of mass
m
. Its kinetic energy is
T
=
m
2
[
a
2
·
θ
2
+
·
z
2
]
,
(1)
where
θ
is measured from a reference axis
f
xed in space. The potential energy
of the system is simply
V
=
mgz .
(2)
The vertical displacement
z
of the particle is measured from its initial position
at the top of the cylinder,
z
=0
. Since the helical track that the particle moves
on is rigidly fastened to the cylinder it is convenient to express
z
in terms of the
particle’s angular displacement
α
on the cylinder,
z
=
−
b
α
,
(3)
where
b
is the pitch of the helix. It should be clear that
α
is related to the
angular displacements in the
f
xed reference frame by the simple equation,
α
=
θ
−
ω
t,
(4)
where
ω
>
0
if the cylinder is rotating counterclockwise as viewed from above.
Similar conventions apply to
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 Fall '10
 Wilemski
 mechanics, Energy, Kinetic Energy

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