Problem 6-19: The angular velocity of an “artificial gravity”


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/s2 at the rim?


Solution:

We are given the following quantities:

\text{radius} = \frac{\text{diameter}}{2} = \frac{200\ \text{m}}{2} = 100 \ \text{m}
\text{centripetal acceleration}, a_c = 9.80 \ \text{m/s}^2

Centripetal acceleration is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. The formula for centripetal acceleration is

a_{c} = r \omega ^2

If we solve for the angular velocity in terms of the other quantities, we have

\omega = \sqrt{\frac{a_c}{r}}

Substituting the given quantities,

\begin{align*}
\omega & = \sqrt{\frac{a_c}{r}}  \\ \\
\omega & = \sqrt{\frac{9.80 \ \text{m/s}^2}{100\ \text{m}}} \\ \\
\omega & = 0.313 \ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}


Advertisements
Advertisements