Question

It has been suggested that rotating cylinders about 9 mi long and 5.9 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?

Answer 1

`\omega = 4.5* 10^(-2) rad/s`

Step-by-step explanation:

Given data:

Rotating cylinder length = 9 mi

diameter of cylinder is 5.9 mi

we know that linear acceleration is given as

a = r ω^2

where ω is angular velocity

so
`\omega = \sqrt{(a)/(r)}`

`r = \frac{5.9}[2} (1609 m)/(1 mi) = 4.746* 10^(3) m`

`\omega = \sqrt{(9.80)/(4.746* 10^(3))}`

`\omega = 4.5* 10^(-2) rad/s`