How do I find the rev/s to give a centripetal acceleration of 19.6g given the following
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How do I find the rev/s to give a centripetal acceleration of 19.6g given the following

[From: ] [author: ] [Date: 11-09-30] [Hit: ]
0 ft away. Determine the rotation rate, in revolutions per second, required to give the astronaut a centripetal acceleration of 19.6g.How do I find this?......
The 20-g centrifuge at NASA's Ames Research Center in Mountain View, California, is a horizontal, cylindrical tube 58 ft long and is represented in the figure below. Assume an astronaut in training sits in a seat at one end, facing the axis of rotation 29.0 ft away. Determine the rotation rate, in revolutions per second, required to give the astronaut a centripetal acceleration of 19.6g.

_____________rev/s

How do I find this? If you would provide equations that are used that would be great. Thanks in advance for the help.

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All you need is the centripital acceleration needed

Fc = V^2/r

Where:

V= Tangential velocity
Fc = 19.6 g = 19.6 * 32.17 ft/s^2

Cercumference = Pi * d = 3.14... * 2*29 = 182.21 ft

19.6 * 32.17 = V^2/r
630.53 = V^2/29
V^2 = 18,285.43 ft^2/s^2
V = 135.22 ft/s

What is that in revolutions/s?

Each revolution = 182.21 ft

135.22/182.21 = 0.75 rev/s

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Depends what units your are supposed to work in. Easiest in feet and seconds.
g (= 9.8m/s^2) = 32ft/s^2.

19.6g = 19.6x32 = 627.2ft/s^2

Centripetal acceleration =ω^2.r where ω is angular speed in radians/second

ω^2.r = 627.2
ω =sqrt(627.2/29.0) = 4.651rad/s

Since 1 revolution = 2π radians

ω =4.651/(2π) revs/s
= 0.74rev/s
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