Question

Consider a cylindrical space station 500 m in diameter rotating about its central axis. Astronauts walk on the inside surface of the space station. What rotation period will provide earth gravitational acceleration of 9.8 m/s/s

Answer 1

Answer:

The rotation period is 44.88 seconds

Explanation:

The acceleration experienced by the rotating cylindrical space station is called the centripetal acceleration.

centripetal acceleration = $\frac{v^{2} }{r}$ = $9.8m/s^{2}$

where v = velocity of the rotating body

r = radius of rotation measured from its axis = 500m

$9.8=\frac{v^{2} }{500}$

$v^{2}= 4900$

$\\v =\sqrt{4900}=70$m/s

The velocity of the rotating body and its period are related with the following formula

$v=\frac{2\pi r}{T}$

from that, we have that

$T= \frac{2\pi r}{v}$

substituting in values, the period can be evaluated as

$T=\frac{2\pi \times 500}{70} = 44.88 seconds$

∴ The period of the rotating body is 44.88 seconds