Physics, 22.06.2019 03:00 bwheele6791

Arotating 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?

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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.

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Answer from: Geo777

0.313 rad/s

Explanation:

The equation that relates the velocity 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 :

$V=\sqrt{a_{c}.r}$ (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 $9.8m/s^{2}$ .

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Answer from: Quest

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