Question

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.

Answer 1

Answer:

The angular velocity produced is 0.321 rad/s.

Explanation:

Given :

Diameter of space station , D = 190 m.

Therefore, radius , $R=\dfrac{D}{2}=\dfrac{190}{2}=95\ m.$

Also, acceleration , $a=9.8\ m/s^2.$

We know, angular velocity , $\omega=\sqrt \dfrac{a}{R}$.

Putting value of g and R in above equation.

We get ,

$\omega=\sqrt \dfrac{9.8\ m/s^2}{95\ m}$

$\omega=0.321\ rad/s.$

Hence, this is the required solution.