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Unformatted text preview: 17. The wheel has angular velocity ω0 = +1.5 rad/s = +0.239 rev/s 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. 1015: θ2 = 1 ( ω0 + ω 2 ) t 2 2 t2 = 2(40 rev) = 335 s 0.239 rev/s which we round off to t2 ≈ 3.4 ×102 s . (b) Any equation in Table 101 involving α can be used to find the angular acceleration; we select Eq. 1016.
2 θ 2 = ω2t2 − α t2 1 2 α =− 2(40 rev) = −7.12 × 10−4 rev/s 2 2 (335 s) which we convert to α = – 4.5 × 10–3 rad/s2.
1 (c) Using θ 1 = ω 0t1 + 2 αt12 (Eq. 1013) and the quadratic formula, we have t1 = −ω0 ± ω02 + 2θ1α α = −(0.239 rev/s) ± (0.239 rev/s) 2 + 2(20 rev)( −7.12 ×10 −4 rev/s 2 ) −7.12 × 10 −4 rev/s 2 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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 Spring '08
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 Physics, Acceleration

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