Problem 4-62 Centripetal acceleration - Part 5 - A

An astronaut in training is spinning around in a device that rotates at 30 revolutions per minute. (a) If the astronaut is \(7.5\; m\) from the centre of the device, what is the magnitude of the centripetal acceleration? (b) Express the answer in terms of "\(g,\)" the magnitude of the acceleration due to gravity.
[Ans. (a) \(74\; m/s^2\)   (b) \(7.6\; g\)]

Accumulated Solution

\(\text{The expression for centripetal acceleration is:} \; ac = v^2/r \\ \text {The speed} \; v = d/t\; \text{where}\; d = 47.1\; m\; \text {and} \; t = 2 \;s \\ \text{Therefore} \; v =23.6\; m/s \\ \text{and} \; a = v^2/r = 74\; m/s^2\)

Correct!

\(\text{The time}\; t = (1/30) min \times 60 (s/min) = 2\;s \\ \text{Therefore} \; v = 47.1\; m/2 \;s = 23.6\; m/s \\ \text{and} \; a = v^2/r = (23.6 \;m/s) / 7.5\; m = 74 \;m/s^2\)

Expressed in \(g’s\) this is

(A)   \(74\; g\)

(B)   \(7.6\; g\)

(C)   \(725\; g\)