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Unformatted text preview: 14. (a) Eq. 1013 gives θ − θo = ωo t + 12 2 αt = 0 + 2 (1.5 rad/s²) t12 1 where θ − θo = (2 rev)(2π rad/rev). Therefore, t1 = 4.09 s. (b) We can find the time to go through a full 4 rev (using the same equation to solve for a new time t2) and then subtract the result of part (a) for t1 in order to find this answer. (4 rev)(2π rad/rev) = 0 + 2 (1.5 rad/s²) t22 Thus, the answer is 5.789 s – 4.093 s ≈ 1.70 s.
1 t2 = 5.789 s. ...
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 Spring '08
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 Physics

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