Question

A 75.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 8.0 m) and spun at a constant angular velocity of 25.3 rpm. Answer the following:
a. What is the angular velocity of the centrifuge in rad/s?
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 and net torque experienced by the astronaut? Give magnitudes and directions.

Answer 1

a. angular velocity \(\omega = \dfrac{25.3 revolution}{minute}*\dfrac{2\pi}{revolution}*\dfrac{minute}{60seconds}= 2.65~~ rad/second\)

Answer 2

b. linear velocity \(v = \omega * r = 2.65~~rad/s *8.0m =21.196~~ m/s\)

Answer 3

c. centripetal acceleration \(\alpha = \omega^2 *r = (2.65~~rad/s)^2*8.0m= 56.18 \)

Answer 4

d. \(\alpha *\dfrac{g}{g} = \dfrac{56.18g}{9.8} = 5.73g \)
e. centripetal force F = \(\dfrac{m\omega^2}{r}= \dfrac{75.0~kg *2.65^2}{8.0}=65.84N\)

Answer 5

e. net Torque \(\sum \tau\) =\( I ~ \alpha = m r^2\alpha = 75*8^2*56.81 = 2.7*10^5Nm\)

Answer 6

apply right hand rule, this torque has the direction upward if the reentry rotates counter clock wise and downward if it rotates in clock wise direction