Exercicios Resolvidos do Halliday sobre Rotação- Cap11- Exercicio 17

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17. The wheel has angular velocity ω0 = +1.5 rad/s = +0.239 rev/s2 at t = 0, and has constant value of
angular acceleration α < 0, which indicates our choice for positive sense of rotation. At t1 its angular
displacement (relative to its orientation at t = 0) is θ1 = +20 rev, and at t2 its angular displacement is
θ2 = +40 rev and its angular velocity is ω2 = 0.
(a) We obtain t2 using Eq. 11-15:
θ2 =
1
2
(ω0 + ω2) t2 =⇒ t2 = 2(40)0.239
which yields t2 = 335 s which we round off to t2 ≈ 340 s.
(b) Any equation in Table 11-1 involving α can be used to find the angular acceleration; we select
Eq. 11-16.
θ2 = ω2t2 − 12αt
2
2 =⇒ α = −
2(40)
3352
which yields α = −7.12× 10−4 rev/s2 which we convert to α = −4.5× 10−3 rad/s2.
(c) Using θ1 = ω0t1 + 12αt
2
1 (Eq. 11-13) and the quadratic formula, we have
t1 =
−ω0 ±
√
ω20 + 2θ1α
α
=
−0.239±√0.2392 + 2(20) (−7.12× 10−4)
−7.12× 10−4
which yields two positive roots: 98 s and 572 s. Since the question makes sense only if t1 < t2 we
conclude the correct result is t1 = 98 s.
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