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48. (a) The rotational inertia of a uniform rod with pivot point at its end is
I
=
mL
2
/12 +
mL
2
= 1/3
ML
2
. Therefore, Eq. 1529 leads to
()
2
2
1
3
2
3
2
28
ML
gT
T
Mg L
π
=
¡
so that
L
= 0.84 m.
(b) By energy conservation
bottom of swing
end of swing
mm
E
EK
U
=
¡
=
where
U
Mg
=−
A
(c
o
s
)
1
θ
with
A
being the distance from the axis of rotation to the center
of mass. If we use the small angle approximation (
cos
≈−
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 Spring '08
 Any
 Physics, Energy, Inertia

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