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The cylindrical space station in Figure Q6.26, 200 m in diameter, rotates in order to provide artificial gravity of g for the occupants. How much time does the station take to complete one rotation?


a) 3 s
b)20 s
c)28 s
d)32 s

The cylindrical space station in Figure Q626 200 m in diameter rotates in order to provide artificial gravity of g for the occupants How much time does the stat class=

Respuesta :

Answer:  T = 2π/ω

Where T is the period of rotation and ω is the angular velocity.

In this case, we are given that the station provides artificial gravity of g for the occupants. The acceleration due to gravity, g, is related to the angular velocity, ω, through the equation:

g = ω²r

Where r is the radius of the cylindrical space station.

We are given the diameter of the station, which is 200 m. The radius, r, is half the diameter:

r = 200 m/2 = 100 m

Substituting the value of the radius into the equation for g, we have:

g = ω²(100 m)

Since g is the acceleration due to gravity on Earth (9.8 m/s²), we can rearrange the equation to solve for ω:

ω = √(g/100 m)

Plugging in the value of g, we get:

ω = √(9.8 m/s²/100 m) = √0.098 rad/s

Now, we can substitute the value of ω into the equation for the period of rotation:

T = 2π/ω = 2π/√0.098 rad/s

Using a calculator, we can evaluate this expression to find the period of rotation:

T ≈ 20.01 s

Therefore, the correct answer is b) 20 s.

Explanation:

msm555

Answer:

b)20 s

Explanation:

The time it takes for the cylindrical space station to complete one rotation can be determined using the formula for the period [tex]\sf (T)[/tex] of a rotating object:

[tex] \boxed{\boxed{\sf T = \dfrac{2\pi}{\omega}}} [/tex]

where:

  • [tex]\sf T [/tex] is the period,
  • [tex]\sf \omega [/tex] is the angular velocity.

The angular velocity [tex]\sf ( \omega )[/tex] is related to the gravitational acceleration [tex]\sf ( g )[/tex] and the radius [tex]\sf ( r )[/tex] of the rotating object by the formula:

[tex]\sf \omega = \sqrt{\dfrac{g}{r}} [/tex]

Given that the diameter of the cylindrical space station is 200 m, the radius [tex]\sf ( r )[/tex] is half of the diameter:

[tex]\sf r = \dfrac{200}{2} = 100 \, \text{m} [/tex]

Now, we can calculate [tex]\sf \omega [/tex] and then use it to find the period [tex]\sf T [/tex].

[tex]\sf \omega = \sqrt{\dfrac{g}{r}} [/tex]

[tex]\sf \omega = \sqrt{\dfrac{g}{100}} [/tex]

The value of [tex]\sf g [/tex] is approximately [tex]\sf 9.8 \, \text{m/s}^2 [/tex].

[tex]\sf \omega = \sqrt{\dfrac{9.8}{100}} [/tex]

[tex]\sf \omega = \sqrt{0.098} [/tex]

Now, substitute [tex]\sf \omega [/tex] back into the formula for [tex]\sf T [/tex]:

[tex]\sf T = \dfrac{2\pi}{\omega} [/tex]

[tex]\sf T = \dfrac{2\pi}{\sqrt{0.098}} [/tex]

[tex]\sf T = \dfrac{ 6.283185307 }{0.3130495168} [/tex]

[tex]\sf T \approx 20.07089923 [/tex]

[tex]\sf T \approx 20 \textsf{ (rounded to nearest whole number)}[/tex]

So, the station takes approximately 20s to complete one rotation.

Therefore, the answer is b) 20 s

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