Question

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 $9.80 \textrm{ m/s}^2$ at the rim?

$0.313 \textrm{ rad/s}$

# OpenStax College Physics, Chapter 6, Problem 19 (Problems & Exercises)

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This is College Physics Answers with Shaun Dychko. This space station has some simulated gravity by rotating the space station around and we're told the diameter of it is 200 meters. We'll find the radius by dividing that by two so we get 100 meter radius. We're meant to have an acceleration that is experienced by this astronaut here equal to what the acceleration due to gravity would be at the surface of the earth. So, centripetal acceleration is g in other words. Now, centripetal acceleration is the radius multiplied by the angular velocity squared and we're asked to find out what angular velocity would be needed in order to have acceleration due to gravity be the same as the centripetal acceleration. So we divide both sides by r here, and then take the square root of both sides and switch the sides around and that solves for omega. So the angular velocity is the square root of centripetal acceleration over r, and so that's the square root of 9.8 meters per second squared divided by 100 meters, which gives 0.313 radians per second.

Hello, we take the square root of both sides of the equation for $\omega^2$ in order to solve for $\omega$. The line that I skipped showing in the solution is $\omega^2 = \dfrac{a_\textrm{c}}{r}$