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1 year ago

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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?

1 answer:

taurus [48]1 year ago

7 0

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

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

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b

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c

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d

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The free body for this question is shown on the second uploaded image

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