Question

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which would allow astronauts to experience a sort of artificial gravity when walking along the inner wall of the station's outer rim.1. Imagine one such station with a diameter of 110 m, where the apparent gravity is 2.70 m/s2 at the outer rim. How fast is the station rotating in revolutions per minute?2. How fast would the space station have to rotate, in revolutions per minute, for the artificial gravity that is produced to equal that at the surface of the Earth, 9.80 m/s2?

Answer 1

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is NOT INERTIAL, the centrifugal force is a fictitious force, the real force is the centripetal).

a.

The rotational speed  is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.02454 \frac{rad^2}{s^2} }$

$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

b.

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.2985 \frac{rev}{min} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 2.8502 \frac{rev}{min}$

Answer 2

Answer:

1)The station is rotating at 2.11 revolutions per minute

2) The space station will have to rotate at as peed of 4.03 revolutions per minute, for the artificial gravity to equal that at the surface of earth (9.8 m/s²)

Explanation:

1)

$a_{c}=$ω²r = ω²d/2

Here,

$a_{c}=$2.7 m/s²

d = 110 m

therefore,

ω² = (2.7 m/s²)(2)/(110 m)

ω = √0.049 rad/s²

ω = 0.2216 rad/s

To convert into rev/min

ω = (0.2216 rad/s)(1 rev/2π rad)(60 s/ 1min)

ω = 2.11 rev/min

2)

Here,

$a_{c}=g=$9.8 m/s²

d = 110 m

therefore,

ω² = (9.8 m/s²)(2)/(110 m)

ω = √0.178 rad/s²

ω = 0.4221 rad/s

To convert into rev/min

ω = (0.4221 rad/s)(1 rev/2π rad)(60 s/ 1min)

ω = 4.03 rev/min