The angular speed such a cylinder must have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth is 0.046 rad/s.
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Angular speed of the cylinder</h3>
The angular speed such a cylinder must have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth is calculated as follows;
a = v²/r
v² = ar
(ωr)² = ar
ω²r² = ar
ω²r = a
ω² = a/r
ω = √(a/r)
where;
- a is centripetal acceleration = acceleration due to gravity, g
- r is radius of the cylinder = 5.66 m / 2 = 2.83 mi = 4554.4 m
- ω is angular speed
ω = √(g/r)
ω = √(9.8/4554.4)
ω = 0.046 rad/s
Thus, the angular speed such a cylinder must have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth is 0.046 rad/s.
Learn more about angular speed here: brainly.com/question/6860269
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