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2 years ago

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A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an "artificial gravity" of 9.80 m/s^2 at the rim?

1 answer:

Anestetic [448]2 years ago

5 0

Answer:

ω = 0.313 rad/s

Explanation:

D = 200 m ⇒ R = D / 2 = (200 m / 2) = 100 m

ac = 9.8 m/s²

The following equation can be used

ac = ω²*R ⇒ ω = √(ac / R)

then we insert the values in the last formula

ω = √(9.8 m/s² / 100 m) = 0.313 rad/s

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2 years ago

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2 years ago

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To solve this problem it is necessary to apply the kinematic equations of angular motion.

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$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

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Our values are given as

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Using the expression of angular acceleration we can find the to then find the torque, that is,

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$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

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Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

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