How Fast Must an Astronaut Spin for 2.89g Acceleration?

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To achieve a centripetal acceleration of 2.89g (28.322 m/s²) for an astronaut in a circular motion with a radius of 9.46 m, the required tangential velocity is calculated to be 16.368 m/s. Using the formula T = 2πr/v, the period of rotation is determined to be approximately 3.631 seconds. Additionally, the angular velocity (ω) can be derived using the relationship v = rω, or by applying the alternative centripetal acceleration formula a = ω²r. The calculations confirm the necessary parameters for maintaining the specified acceleration. The discussion emphasizes the importance of understanding both linear and angular motion in astronaut training scenarios.
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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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