How Fast Must an Astronaut Spin for 2.89g Acceleration?

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SUMMARY

The discussion focuses on calculating the rotation rate required for an astronaut to experience a centripetal acceleration of 2.89g in a circular motion with a radius of 9.46 meters. The calculations involve using the formula for radial acceleration, where the acceleration is determined to be 28.322 m/s². The linear velocity (v) is calculated to be 16.368 m/s, leading to a period (T) of 3.631 seconds for one complete revolution. Additionally, the discussion highlights the alternative method of finding angular velocity (ω) using the relationship between linear velocity and radius.

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Homework Statement



As their booster rockets separate, Space Shuttle astronauts typically feel accelerations up to 3g, where g = 9.80 m/s2. In their training, astronauts ride in a device where they experience such an acceleration as a centripetal acceleration. Specifically, the astronaut is fastened securely at the end of a mechanical arm that then turns at constant speed in a horizontal circle. Determine the rotation rate, in revolutions per second, required to give an astronaut a centripetal acceleration of 2.89g while in circular motion with radius 9.46 m.

Homework Equations



radial(centripetal) acceleration= v^(2) / r

v= 2(pi)(r) / T

The Attempt at a Solution



radial acc.= (9.8)(2.89)= 28.322 m/s^(s)

28.322 m/s^(2) = v^(2) / 9.46m

v= 16.368

T= 2(pi)(9.46m) / (16.368) = 3.631

Is that right?
 
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africanmasks said:

The Attempt at a Solution



radial acc.= (9.8)(2.89)= 28.322 m/s^(s)

28.322 m/s^(2) = v^(2) / 9.46m

v= 16.368

T= 2(pi)(9.46m) / (16.368) = 3.631

Is that right?

you want to find the angular velocity ω. So use v=rω and find ω.

Alternatively you could have used another expression for centripetal acceleration

a=ω2r
 

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