# Problem 4-62 Centripetal acceleration - 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**

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

Correct!

\(a_c = v^2/r\)

Text Eq. 4-15.

In fact, the other choice \(a_c = 4\pi ^2r/T^2\) is also correct. We will use \(a_c = v^2/r\) but at the end you should go back and follow the alternate solution as well.

For this solution we need \(v,\) the speed of the astronaut. This is given by:

(A) \(v = v_0 + at\)

(B) \(v^2 = v_0{^2} + 2ad\)

(C) \(v = d/t\)