Exercicios Resolvidos do Halliday sobre Rotação- Cap11- Exercicio 71

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71. 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 further level of detail in case that hint is not
obvious or – simply – in case one wishes to see how the calculus supports our intuition.
(a) The (centripetal) force exerted on an infinitesimal portion of the blade with mass dm located a
distance r from the rotational axis is (Newton’s second law) dF = (dm)ω2r, where dm can be
written as (M/L)dr and the angular speed is ω = (320)(2π/60) = 33.5 rad/s. Thus for the entire
blade of mass M and length L the total force is given by
F =
∫
dF =
∫
ω2 r dm
=
M
L
∫ L
0
ω2 r dr
=
Mω2r2
2L
∣∣∣∣
L
0
=
Mω2L
2
=
(110 kg)(33.5 rad/s)2(7.80m)
2
= 4.8× 105 N .
(b) About its center of mass, the blade has I = ML2/12 according to Table 11-2(e), and using the
parallel-axis theorem to “move” the axis of rotation to its end-point, we find the rotational inertia
becomes I =ML2/3. Using Eq. 11-37, the torque (assumed constant) is
τ = Iα
=
(
1
3
ML2
)(
∆ω
∆t
)
=
1
3
(110 kg)(7.8m)2
(
33.5 rad/s
6.7 s
)
= 1.1× 104 N·m .
(c) Using Eq. 11-44, the work done is
W = ∆K =
1
2
Iω2 − 0
=
1
2
(
1
3
ML2
)
ω2
=
1
6
(110 kg)(7.80m)2(33.5 rad/s)2
= 1.3× 106 J .
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