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

Answer:

$\omega = 49.86*10^{-3}rad/s$

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,

$\omega = \sqrt{\frac{a_c}{r}}$

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

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

$\omega = 49.86*10^{-3}rad/s$