Problem 4-72 Centripetal acceleration - Part 3 - C

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

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


Correct.

\(\text{So} \; a_c = 9.8\; m/s^2 = (v2)/0.5\times10^3\; m \\ v^2 = 4900 \\ v = 70\; m/s\)

The time to make one revolution (the period \(T\)), where \(D\) is the diameter, is:

(A)   \(\pi D/v\)

(B)    \(D/v\)

(C)   \(\pi Dv\)