αω==−=−ΔΔt0 15022114rev / minh)(60 min /1h)rev / min2(...(b) Using Eq. 10-13 with t= (2.2) (60) = 132 min, the number of revolutions is ()23011(150 rev/min)(132 min)1.14 rev/min132min9.9 10 rev.ttθω=+=+−=×(c) With r= 500 mm, the tangential acceleration is art−FHGIKJFHGIKJ1142.rev/min2rad1rev1min60 s(500 mm)2chπwhich yields at= –0.99 mm/s2.(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.