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. ...
View
Full
Document
This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

Click to edit the document details