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It has been suggested that rotating cylinders about 10 mi long and 5.0 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

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rokettojanpu

Centripetal acceleration is given by:

a = v²/r

a = centripetal acceleration, v = speed of cylinder at surface, r = radius of cylinder

This equation relates the speed of the cylinder at its surface to its angular velocity:

v = rω

v = speed at surface, r = radius, ω = angular velocity

Make a substitution:

a = (rω)²/r

a = rω²

Given values:

a = 9.81m/s², r = (5.0mi)/2 = 2.5mi = 4023m

Plug in and solve for ω:

9.81 = 4023ω²

ω = 0.049rad/s

nuhulawal20

For the centripetal acceleration at the cylinder's surface to equal the free-fall acceleration on Earth, the angular speed must be 0.049rad/s

Given the data in the question;

  • Length of the cylinder; [tex]l = 10mi = 16093.4m[/tex]
  • Diameter of the cylinder; [tex]d = 5.0 mi = 8046.72m[/tex]
  • Radius; [tex]r = \frac{d}{2} = \frac{8046.72m}{2} = 4023.36m[/tex]

To determine the angular speed, we use the expression for Centripetal Acceleration:

[tex]a_c = rw^2[/tex]

Where r is the radius, ω is the angular velocity and [tex]a_c[/tex] is the centripetal acceleration.

From the question, the centripetal acceleration at its surface equals the free-fall acceleration on Earth

Hence, centripetal acceleration; [tex]a_c[/tex] = 9.8m/s²

So we make ''ω" the subject of the formula

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

We substitute in our given values into the equation

[tex]w = \sqrt{\frac{9.8m/s^2}{4023.36m} } \\\\w = 0.049 rad/s[/tex]

Therefore, for the centripetal acceleration at the cylinder's surface to equal the free-fall acceleration on Earth, the angular speed must be 0.049rad/s

Learn more; https://brainly.com/question/14745232

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