Problem 4-72 Centripetal acceleration - Part 5 - 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
\(a_c = v^2/r \\ a = 9.8 m/s^2 \\ r = 0.5\times10^3 \;m \\ v = 70 \;m/s \\ T = 45 \;s\)
No. The period is the time for 1 revolution, so the frequency is the number of revolutions in \(1\; s\) and so is the reciprocal of the period. Check the units
\([f] = s^{-1}\) \([1/T] = 1/s = s^{-1}\)