Question

It has been suggested that rotating cylinders about 17.0 mi long and 4.99 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 = 49.86*10^(-3)rad/s`

Step-by-step explanation:

We start converting to SI units,

A mile = 1609m

`L=17mi=27000m`

`D=4.99mi=7884m`

We know that the expression, which can relate linear acceleration and angular velocity is given by,

`a_c = r\omega^2`

Where
`\omega` is the angular velocity

r=radius

`a_c =` linear acceleration,

Re-arrange for \omega,

`\omega = \sqrt{(a_c)/(r)}`

Our acceleration is equal to the gravity force, so replacing,

`\omega = \sqrt{(9.8)/((7884/2))}`

`\omega = 49.86*10^(-3)rad/s`