ch10-p028 - 28. (a) The angular acceleration is = 0 150 rev...

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α ω == =− Δ Δ t 0 150 22 114 rev / min h)(60 min /1h) rev / min 2 (. .. (b) Using Eq. 10-13 with t = (2.2) (60) = 132 min, the number of revolutions is () 2 3 0 11 (150 rev/min)(132 min) 1.14 rev/min 132min 9.9 10 rev. tt θω =+ = + (c) With r = 500 mm, the tangential acceleration is ar t F H G I K J F H G I K J 114 2 .r e v / m i n 2r a d 1rev 1m in 60 s (500 mm) 2 ch π which yields a t = –0.99 mm/s 2 . (d) The angular speed of the flywheel is (75 rev/min)(2 rad/rev)(1 min/ 60 s)
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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.

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