If the space station is 190 m in diameter, what angular velocity would produce an "artificial gravity" of 9.80 m/s2 at the rim? Give your answer in rad/s.

Respuesta :

Answer:

The angular velocity produced is 0.321 rad/s.

Explanation:

Given :

Diameter of space station , D = 190 m.

Therefore, radius , [tex]R=\dfrac{D}{2}=\dfrac{190}{2}=95\ m.[/tex]

Also, acceleration , [tex]a=9.8\ m/s^2.[/tex]

We know, angular velocity , [tex]\omega=\sqrt \dfrac{a}{R}[/tex].

Putting value of g and R in above equation.

We get ,

[tex]\omega=\sqrt \dfrac{9.8\ m/s^2}{95\ m}[/tex]

[tex]\omega=0.321\ rad/s.[/tex]

Hence, this is the required solution.