Question

At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spaceship's diameter is 35 mm , and give your answer as the time needed for one revolut

Answer 1

Answer:

Time needed for one revolution is 0.38 s

Explanation:

The formula for the frequency of rotation of a spaceship, to create the desired artificial gravity, is as follows:

f = (1/2π)√(a/r)

where,

f = frequency of rotation = ?

a = artificial gravity required = 0.5 g

g = acceleration due to gravity on surface of Earth = 9.8 m/s²

r = radius of ship = 35 mm/2 = 17.5 mm = 17.5 x 10⁻³ m

Therefore,

f = (1/2π)√[(0.5)(9.8 m/s₂)/(17.5 x 10⁻³ m)]

f = 2.66 Hz

Now, for the time required for one revolution, is given as:

Time Period = T = 1/f

T = 1/2.66 Hz

T = 0.38 s

Answer 2

Answer:

The time needed is $T = 16.8 s$

Explanation:

From the question we are told that

      The magnitude of the stimulated acceleration due gravity is  $a = 0.5 g$

        The diameter of the spaceship is  $d = 35m$

       

Generally the force acting on the spaceship is  

       $F = ma$

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

      $F = \frac{mv^2}{r}$

Thus  

       $ma = \frac{mv^2}{r}$

=>    $\frac{v^2}{r} = a$

      Generally the speed of this spaceship is mathematically represented as

      $v = \frac{2 \pi}{T}$

=>    $v^2 = [\frac{2\pi}{T}] ^2$

=>     $\frac{\frac{4\pi^2 r^2}{T^2} }{r} = 0.5g$

=>       $\frac{4 \pi^2 r }{T^2} = 0.5 g$

=>         $T = \sqrt{ \frac{4\pi^2 r}{0.5g}}$

substituting values

          $T = \sqrt{ \frac{4* (3.142)^2 *(35)}{0.5 * 9.8}}$

         $T = 16.8 s$