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()
(
)
2
2
22
0
5
110kg
33.5 rad s
7.80m
4.81 10 N.
L
MM
L
Fd
F
r
d
m
r
d
r
L
ω
ωω
==
=
=
=
=×
³³
³
(b) About its center of mass, the blade has
I
ML
=
2
12
/
according to Table 102(e), and
using the parallelaxis theorem to “move” the axis of rotation to its endpoint, we find the
rotational inertia becomes
I
ML
=
2
/ 3. Using Eq. 1045, the torque (assumed constant) is
(
)
2
2
4
1
1
33.5rad/s
110kg 7.8 m
1.12 10 N m.
33
6
.
7
s
IM
L
t
τα
§·
Δ
§
·
=
=
×
⋅
¨¸
¨
¸
Δ
©¹
©
¹
(c) Using Eq. 1052, the work done is
(
)
2
6
11
1
1
0
110kg 7.80m
33.5rad/s
1.25 10 J.
3
6
WK
I
M
L
=Δ =
− =
=
=
×
75. The
Hint
given in the problem would make the computation in part (a) very
straightforward (without doing the integration as we show here), but we present this
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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

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