N a laboratory test of tolerance for high acceleration, a pilot is swung in a circle 11.5 m in diameter. it is found that the pilot blacks out when he is spun at 30.6 rpm (rev/min). at what acceleration (in si units) does the pilot black out?

Question

Physics

Scott Casey

15 February, 01:35

N a laboratory test of tolerance for high acceleration, a pilot is swung in a circle 11.5 m in diameter. it is found that the pilot blacks out when he is spun at 30.6 rpm (rev/min). at what acceleration (in si units) does the pilot black out?

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Joaquin Tran · Answer 1

59.027m/s^2 Centripetal force is give by F = mv^2/r Since F = ma The acceleation accoiated with this centripetal force is a=v^2/r the radius is found by r=d/2=5.75m velocity is found by v=d/t distance is the circumference of the circle c=pi*d=11.5*3.14=36.11m time=60s/30.6rpm=1.96s v=36.11/1.96=18.423m/s plugging this back in to a=v^2/r = (18.423m/s) ^2/5.75m=59.027m/s^2