A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m) and spun at a constant angular velocity of 13.2 rpm. Answer the following, show all work: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 experienced by the astronaut? Give magnitude and direction.F) Explain how and when the astronaut would experience a torque in this situation.

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boffeemadrid boffeemadrid · Answer 1 · 2020-01-10T05:51:40+01:00

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

[tex]a=2.92164g[/tex]

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

[tex]\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s[/tex]

Angular velocity is 1.3823 rad/s

Linear velocity is given by

[tex]v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s[/tex]

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

[tex]a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2[/tex]

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

[tex]\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g[/tex]

[tex]a=2.92164g[/tex]

Centripetal force is given by

[tex]F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N[/tex]

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.