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What frequency of rotation, in rpm, is required to give an acceleration of 1.4 g to an astronaut's feet, if her feet are 1.6 m from the platform's rotational axis

User Eknumbat
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1 Answer

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Answer:

The frequency of rotation of the platform is approximately 27.970 revolutions per minute.

Step-by-step explanation:

In this case, the astronaut experiments a centrifugal acceleration as a reaction to the centripetal acceleration due to the rotation of the platform. (Second and Third Newton's Laws). Centripetal acceleration (
a_(r)), measured in meters per second, is expressed as:


a_(r) = \omega^(2)\cdot r (Eq. 1)

Where:


\omega - Angular velocity, measured in radians per second.


r - Distance of the astronaut's feet from the rotational axis of the platform, measured in meters.

Now we clear the angular velocity:


\omega = \sqrt{(a_(r))/(r) }

If we know that
a_(r) = 13.730\,(m)/(s^(2)) and
r = 1.6\,m, the angular velocity of the platform is:


\omega = \sqrt{(13.730\,(m)/(s^(2)) )/(1.6\,m) }


\omega \approx 2.929\,(rad)/(s)

And this outcome is converted into revolutions per minute:


\dot n = 2.929\,(rad)/(s)* (60\,s)/(1\,min)* (1\,rev)/(2\pi\,rad)


\dot n \approx 27.970\,(rev)/(min)

The frequency of rotation of the platform is approximately 27.970 revolutions per minute.

User Shia
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