The Navy operated a "man-rated" centrifuge in Warminster, PA. It was used for testing with the Mercury astronauts and the X15 project. The gondola pictured above, which holds the person, is on the end of a 50 ft boom. Without special equipment and training most people will black-out at an acceleration of about 6 g (6×9.8 m/s2 ).
(a) At what speed would a person in the gondola have a centripetal acceleration of 8.3 g?
(b)What is the period of the gondola at that speed?
I have no idea how to do this, AT ALL. Can someone show me how? Will rate best.
(a) At what speed would a person in the gondola have a centripetal acceleration of 8.3 g?
(b)What is the period of the gondola at that speed?
I have no idea how to do this, AT ALL. Can someone show me how? Will rate best.
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(a) given r = 50 feet = 15.24 m
=>By a = v^2/r
=>8.3 x 9.8 = v^2/15.24
=>v = √1239.62
=>v = 35.21 m/s
(b) By ω = 2π/T
=>T = 2π/ω = [2π/(v/r)] {as v = rω}
=>T = 2πr/v
=>T = (2 x 3.14 x 15.24)/35.21
=>T = 2.72 sec
=>By a = v^2/r
=>8.3 x 9.8 = v^2/15.24
=>v = √1239.62
=>v = 35.21 m/s
(b) By ω = 2π/T
=>T = 2π/ω = [2π/(v/r)] {as v = rω}
=>T = 2πr/v
=>T = (2 x 3.14 x 15.24)/35.21
=>T = 2.72 sec
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T= 1/2pi the formula is on the web. try jiskha homework it helps. try it out. good luck.