Question

t has been suggested that rotating cylinders aboutIt has been suggested that rotating cylinders about 15.5 mi long and 5.75 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth? rad/s 18.5 mi long and 4.52 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth? rad/s

Answer 1

The angular speed required for the second rotating cylinder to have a centripetal acceleration at its surface equal to the free-fall acceleration on Earth is approximately 0.05195 radians per second.

Step-by-step explanation:

To find the angular speed required for the rotating cylinder to have a centripetal acceleration at its surface equal to the free-fall acceleration on Earth, we can use the following formula:

Centripetal acceleration (ac) at the surface of the cylinder = Free-fall acceleration on Earth (g)

The centripetal acceleration at the surface of a rotating cylinder can be calculated using the formula:

ac = ω^2 * r

where:

ac = Centripetal acceleration

ω = Angular speed (in radians per second)

r = Radius of the cylinder (distance from the center to the surface)

The free-fall acceleration on Earth is approximately 9.81 m/s².

Given the dimensions of the cylinder:

Length (L) = 15.5 mi = 15.5 mi * 1609.34 m/mi ≈ 24901.187 m

Diameter (D) = 5.75 mi = 5.75 mi * 1609.34 m/mi ≈ 9252.981 m

Radius (r) = D/2 ≈ 9252.981 m / 2 ≈ 4626.491 m

Now, we need to find the angular speed (ω) that satisfies the condition ac = g:

ω^2 * r = g

Let's plug in the values and solve for ω:

ω^2 = g / r

ω^2 = 9.81 m/s² / 4626.491 m

ω^2 ≈ 0.0021207 s⁻²

Now, solve for ω:

ω ≈ √(0.0021207 s⁻²)

ω ≈ 0.04607 rad/s

So, the angular speed required for the rotating cylinder to have a centripetal acceleration at its surface equal to the free-fall acceleration on Earth is approximately 0.04607 radians per second.

For the second cylinder with dimensions:

Length (L) = 18.5 mi = 18.5 mi * 1609.34 m/mi ≈ 29766.79 m

Diameter (D) = 4.52 mi = 4.52 mi * 1609.34 m/mi ≈ 7270.6808 m

Radius (r) = D/2 ≈ 7270.6808 m / 2 ≈ 3635.3404 m

We use the same formula as before:

ω^2 * r = g

ω^2 = g / r

ω^2 = 9.81 m/s² / 3635.3404 m

ω^2 ≈ 0.0026976 s⁻²

Now, solve for ω:

ω ≈ √(0.0026976 s⁻²)

ω ≈ 0.05195 rad/s

So, the angular speed required for the second rotating cylinder to have a centripetal acceleration at its surface equal to the free-fall acceleration on Earth is approximately 0.05195 radians per second.