Question

It has been suggested that rotating cylinders about 14.5 mi long and 4.78 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

Angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth is 0.0466 rad/s.

Length of the cylinder, L = 9 mi

= (14.5mi)(1609.344 m / 1 mi)

= 23,335.48 m

Diameter of the cylinder, D = 4.78 mi

= (4.78 mi)(1609.344 m / 1 mi)

= 7692.645 m

Radius of the cylinder, r = D / 2

= ( 7692.645 m) / 2

=  3846.32 m

Centripetal acceleration is given by, ac = v^2 / r

We have the relation between linear velocity (v) and the angular velocity(ω) as

 v = r ω

Then, ac = v2 / r

= (rω)2 / r

ac = ω^2 r

If the centripetal acceleration is equals the free fall acceleration of earth, then

ω^2 r = g

ω=0.0466 rad/s

• Angular acceleration is the term used to describe the rate of change in angular velocity. If the angular velocity is constant, the angular acceleration is constant.

To know more about  angular velocity visit : brainly.com/question/12446100

#SPJ1