the 20-g centrifuge at NASAs research center in Mt. view Ca, is a horizontal cyl

Question

the 20-g centrifuge at NASAs research center in Mt. view Ca, is a horizontal cylindrical tube 58 ft long and is represented in the figure. assume an astronaut in training sits in a seat at one end facing the axis of rotation 29 feet away. determine the rotation rate in revolutions per second required to give the astronaut a centripetal acceleration of 20.0g.

I think i need help through this, 29ft is like the radius of the circle (or one rotation) then one revolution will be 58pi.

Then what? What is g in 20.0g? How can you change that to m/s since acceleration is 9.8m/s

Explanation / Answer

Given that r = 29 ft = 8.8392 m a = 20 g 1 rev = 58 pi g = acceleration due to gravity = 9.8 m/s^2 We know that centripetal acceleration a = v^2 / r ==> v^2 = a * r = 20 * g * 8.8392 = 20 * 9.8 * 8.8392 = 1732.4 ==> v = 41.62 m/s We know that v = r w ==> w = v / r = 41.62 / 8.8392 = 4.708 rad/s [ 1 rev = 2 * pi rad ] = 4.708 * 1 rev / 2*3.14 s = 0.7498 rev/s

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