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

0.103 rev/min

Step-by-step 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{(a)/(r)}\\\Rightarrow \omega=\sqrt{(2.8)/(62)}=0.213\ rad/s`

`\omega=2\pi n\\\Rightarrow n=(\omega)/(2r)\\\Rightarrow n=(0.213)/(2* 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