A space station of diameter 80 m is turning about its axis at a constant rate. If the acceleration of the outer rim of the station is 2.5 m/s², what is the period of revolution of the space station? A) 22 s B) 19 s C) 25 s D) 28 s E) 40 s

Question

Quizplus · Accepted Answer

The acceleration of the outer rim of the station is equal to the tangential acceleration, which can be calculated using the formula:
a = rω^2
where r is the radius (half the diameter) of the station and ω is the angular velocity.
Rearranging the formula gives:
ω = sqrt(a/r) = sqrt(2.5/40) = 0.25 rad/s
The period of revolution can be calculated using the formula:
T = 2π/ω = 2π/0.25 ≈ 25 s
Therefore, the answer is choice C.

A Space Station of Diameter 80 M Is Turning About