Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 s to speed up from rest to its top speed of 1 rotation every 1.40 s. The astronaut is strapped into a seat 4.30 m from the axis. What is the astronaut's tangential acceleration during the first 40.0 s? How many g's of acceleration does the astronaut experience when the device is rotating at top speed? Each 9.80 m/s^2 of acceleration is 1 g.
2007-09-26 15:04:54 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Angular vekicity at the end of 40 s is ω = [2π / 1.4] = 4.49 radian /s
Linear velocity at the end of 40 second is r ω = 4.3 x 4.49 = 19.3 m/s
Tangential acceleration is change in velocity / time = 19.3 / 40 = 0.485 m/s^2.
The centripetal acceleration is r ω^2 = r ω* ω = 19.3 x 4.49 = 86.657 m/s^2
= 86.657/ 9.8 = 8.84 g force.
2007-09-26 16:37:37 · answer #1 · answered by Pearlsawme 7 · 5⤊ 0⤋
average tangential acceleration ,a = (final velocity - initial velocity)/time
a = 2x3.14x4.30/1.40 = 19.29 m/s^2
Centripetal acceleration of austronaut = (2x3.14/1.40)x(2x3.14/1.40)x4.30 = 86.52 = 8.8 g
2007-09-26 16:34:31 · answer #2 · answered by Let'slearntothink 7 · 2⤊ 0⤋