Question

2. An 85.5 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 12.0 m) and spun at a constant angular velocity of 18.4 rpm. Answer the following:
(a) What is the angular velocity of the centrifuge in ?
(b) What is the linear velocity of the astronaut at the outer edge of the centrifuge?
(c) What is the centripetal acceleration of the astronaut at the end of the centrifuge?
(d) How many g’s does the astronaut experience?
(e) What is the centripetal force experienced by the astronaut?

Answer 1

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Since one revolution around a circle is 2 pi radian; hence 1 rpm equals 2 pi radians per minute. And because a minute has 60 seconds, 1 rpm equals 1/60 revolution per second. Therefore, we have 2 pie/60 * 15.3 = 0.2513 rps. The linear velocity v = wr where w is the angular velocity in rad/s and r is the distance. So we have 0.2513 * 10.0 = 2.513 rad/s The centripetal acceleration is given by a = w^2 r = (2.513)^2 * 10 = 63.15 rad/s2 The centripetal force F = mass * centripetal acc = 75 * 63.15 = 4736.25 N The torque = centripetal force * distance = 4736 * 63.15 = 299078.4 Nm Two forces acts on the astronaut. The normal force and acceleration due to gravity.

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