Question

A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of 9.80 m/s^{2} 2 at the rim?

Answer 1

Answer: 0.313 rad/s

Explanation:

The equation that relates the velocity $V$ and the angular velocity $\omega$ in the uniform circular motion is:

$V=\omega.r$   (1)

Where $r=d/2=100m$ is the radius of the space station (with a diaeter of 200m) that describes the uniform circular motion.

Isolating $\omega$ from (1):

$\omega=\frac{V}{r}$  (2)

On the other hand, we are told the “artificial gravity” produced by the cetripetal acceleration $a_{c}$ is $9.8m/s^{2}$, and is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$   (3)

Isolating $V$:

$V=\sqrt{a_{c}.r}$   (4)

$V=31.3049m/s$   (5)

Substitutinng (5) in (2):

$\omega=\frac{31.3049m/s}{100m}$  (6)

$\omega=0.313rad/s$ This is the angular velocity that would produce an “artificial gravity” of 9 $9.8m/s^{2}$.

Answer 2

Centripetal acceleration is towards the center. The angular velocity of the rim that will produce the artificial gravity is 9.80 m/s².

What is centripetal acceleration?

The centripetal acceleration is caused due to change in direction of the body which is in a circular motion, the acceleration is towards the center of the circle. It is calculated using the formula,

$a_c = \dfrac{V^2}{r}$

We know that the acceleration produced by centripetal acceleration can be given as,

$a_c = \dfrac{V^2}{r}$,

We also know that the velocity in the form of angular velocity can be written as,

$V = \omega r$

Substituting the value of the velocity in the acceleration, we will get,

$a_c = \dfrac{(\omega r)^2}{r}\\\\a_c = \dfrac{\omega^2 r^2}{r}\\\\a_c = \omega^2 r\\\\\omega = \sqrt{\dfrac{a_c}{r}}$

As it is given to us that the acceleration of the artificial gravity is 9.80 m/s², while the radius is 200 m, and we know that the radius is half the diameter, therefore, the angular velocity can be written as,

$\omega = \sqrt{\dfrac{a_c}{r}}\\\\\omega = \sqrt{\dfrac{9.80}{100}}\\\\\omega = 0.313\rm\ rad/s$

Hence, the angular velocity of the rim that will produce the artificial gravity is 9.80 m/s².

Learn more about Centripetal Force:

https://brainly.com/question/10596517