Problem 4-62 Centripetal acceleration - Part 3 - C

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

The expression for centripetal acceleration is: \(a_c = v^2/r\)

\(\text{The speed} \; v = d/t \; \text{where}\; d = 47.1 \;m \)

Correct!

\(d = 2\pi r = 2\pi(7.5 \;m) = 47.1 \;m\)

The time for one revolution is:

(A)   \(1 \; \text{min}\)

(B)   \(30 \;s\)

(C)   \(1/30\; \text{min}\)

(D)   \(1/30\; s\)