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?

Respuesta :

Answer:

[tex]\omega = 49.86*10^{-3}rad/s[/tex]

Explanation:

We start converting to SI units,

A mile = 1609m

[tex]L=17mi=27000m[/tex]

[tex]D=4.99mi=7884m[/tex]

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

[tex]a_c = r\omega^2[/tex]

Where [tex]\omega[/tex] is the angular velocity

r=radius

[tex]a_c =[/tex] linear acceleration,

Re-arrange for \omega,

[tex]\omega = \sqrt{\frac{a_c}{r}}[/tex]

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

[tex]\omega = \sqrt{\frac{9.8}{(7884/2)}}[/tex]

[tex]\omega = 49.86*10^{-3}rad/s[/tex]