Question

A space station, in the form of a wheel 124 m in diameter, rotates to provide an "artificial gravity" of 2.80 m/s2 for persons who walk around on the inner wall of the outer rim. Find the rate of the wheel's rotation in revolutions per minute that will produce this effect.

Answer 1

Answer:

0.103 rev/min

Explanation:

ω = Angular velocity

r = Radius = Diameter/2 = 124/2 = 62 m

n = Number of rotations

a = Acceleration

$a=\omega^2 r\\\Rightarrow \omega=\sqrt{\frac{a}{r}}\\\Rightarrow \omega=\sqrt{\frac{2.8}{62}}=0.213\ rad/s$

$\omega=2\pi n\\\Rightarrow n=\frac{\omega}{2r}\\\Rightarrow n=\frac{0.213}{2\times 62}=0.00171\ rev/s$

Converting to rev/min

0.00171×60 = 0.103 rev/min

The rate at which the wheel will spin is 0.103 rev/min