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 200m in diameter, what angular velocity (in rad/s ) would produce an “artificial gravity” of 9.80m/s at the rim?
To calculate the angular velocity (ω) that would produce an "artificial gravity" of 9.80 m/s at the rim of the rotating space station, we can use the centripetal acceleration formula:
centripetal acceleration = ω^2 * radius
First, let's convert the artificial gravity value to centripetal acceleration:
centripetal acceleration = 9.80 m/s^2
Next, let's convert the diameter of the space station to radius:
radius = diameter / 2
radius = 200 m / 2
radius = 100 m
Now, we can rearrange the centripetal acceleration formula and solve for angular velocity (ω):
centripetal acceleration = ω^2 * radius
9.80 m/s^2 = ω^2 * 100 m
Dividing both sides of the equation by 100 m:
0.098 m/s^2 = ω^2
Now, we can take the square root of both sides:
ω = √(0.098 m/s^2)
ω ≈ 0.313 rad/s
Therefore, an angular velocity of approximately 0.313 rad/s would produce an "artificial gravity" of 9.80 m/s at the rim of the rotating space station.