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74. (a) We use Table 102(e) and the parallelaxis theorem to obtain the rod’s rotational
inertia about an axis through one end:
II
M
h
M
L M
L
ML
=+
=
+
F
H
G
I
K
J
=
com
22
2
2
1
12
2
1
3
where
L
= 6.00 m and
M
= 10.0/9.8 = 1.02 kg. Thus, the inertia is
2
12.2 kg m
I
=⋅
.
(b) Using
ω
= (240)(2
π
/60) = 25.1 rad/s, Eq. 1131 gives the magnitude of the angular
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Inertia

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