The diameter of the space station is 1000m, and the outer edge of the space stat

Question

The diameter of the space station is 1000m, and the outer edge of the space station moves to give an astronaut the experience of a reduced “gravity” of 5m/s^2. You may assume that the astronaut is standing on the inner wall at a distance of nearly 500m from the axis of the space station and that the space station has a circumference of approximately 3000m.
What would be the resulting period of rotation of the space station?

Explanation / Answer

w^2*d = 9.8-5= 4.8 =>w= square root(4.8/500) =0.098 rad/sec to cover 1 rotation 2pi rads time = 2*pi/.098 = 64.11sec

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