Problem 4-72 Centripetal acceleration - Part 4 - A

To produce artificial gravity on a space colony, it is proposed that the colony should rotate. Suppose that the acceleration required is equal in magnitude to the acceleration due to gravity on the earth. For a colony that is \(1.0 \;km\) in diameter, determine the frequency of rotation, the period of rotation, as well as the speed of a person at the edge of the colony (relative to the centre of the colony). [Ans. \(2.2 \times 102\; Hz;\) \(45\; s;\) \(7.0 \times 101 \;m/s\)]

Accumulated Solution

\(a_c = v^2/r \\ a = 9.8 m/s^2 \\ r = 0.5\times10^3 \;m \\ v = 70 \;m/s\)

Correct.

\(T = \pi D/v = (p \times 1.0\times10^3)/70 = 45\; s\)

The frequency of rotation is:

(A)    \(1/T\)

(B)    \(2T\)

(C)    \(\text{unrelated to the period.}\)