Problem 4-62 Centripetal acceleration - Part B3 - 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:  \( a_c = 4\pi ^2r/T^2\)
The period \(T = 2 \;s\)
Centripetal acceleration \(a_c = 74\; m/s^2\)

Correct!

\(T = 1/30 \;min = 60/30 = 2\; s\)

Therefore \(a_c = 4\pi ^2r/T^2 = (2\pi /T)^2r =(2\pi /2)^27.5 = 74 \;m/s^2\)

Expressed in \(g’s\) this is:

(A)   \(74g\)

(B)   \(7.6g\)

(C)   \(725g\)