12.68. Model:
The bar is a solid body rotating through its center.
Visualize:
Solve:
(a)
The two forces form a couple. The net torque on the bar about its center is
net
I
LF
I
F
L
α
τα
==
⇒
=
where
F
is the force produced by one of the air jets. We can find
I
and
as follows:
22
2
2
10
1
0
11
(0.50 kg)(0.60 m)
0.015 kg m
12
12
(
)
150 rpm
5.0 rad/s
0 rad
(10 s
0 s)
0.50 rad/s
(0.015 kg m )(0.5 rad/s )
0.0393 N
(0.60 m)
IM
L
tt
F
ωωα
π
=
=+ −⇒
=
=
+
− ⇒=
⇒=
=
The force
39 mN.
F
=
(b)
The torque of a couple is the same about any point. It is still
net
.
LF
τ
=
However, the moment of inertia has
changed.
2
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 Fall '09
 geller
 Force, kg, 0.6 m, 0.015 kg, 0 rad, 0.060 kg

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