Answer:
The frequency of rotation of the platform is approximately 27.970 revolutions per minute.
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 ([tex]a_{r}[/tex]), measured in meters per second, is expressed as:
[tex]a_{r} = \omega^{2}\cdot r[/tex] (Eq. 1)
Where:
[tex]\omega[/tex] - Angular velocity, measured in radians per second.
[tex]r[/tex] - Distance of the astronaut's feet from the rotational axis of the platform, measured in meters.
Now we clear the angular velocity:
[tex]\omega = \sqrt{\frac{a_{r}}{r} }[/tex]
If we know that [tex]a_{r} = 13.730\,\frac{m}{s^{2}}[/tex] and [tex]r = 1.6\,m[/tex], the angular velocity of the platform is:
[tex]\omega = \sqrt{\frac{13.730\,\frac{m}{s^{2}} }{1.6\,m} }[/tex]
[tex]\omega \approx 2.929\,\frac{rad}{s}[/tex]
And this outcome is converted into revolutions per minute:
[tex]\dot n = 2.929\,\frac{rad}{s}\times \frac{60\,s}{1\,min}\times \frac{1\,rev}{2\pi\,rad}[/tex]
[tex]\dot n \approx 27.970\,\frac{rev}{min}[/tex]
The frequency of rotation of the platform is approximately 27.970 revolutions per minute.
