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Sergio039 [100]
3 years ago
7

It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T

he acceleration of gravity is 9.8 m/s 2 . What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on
Physics
1 answer:
tekilochka [14]3 years ago
7 0

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\

<u>ω = 0.05 rad/s</u>

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3 years ago
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In fact:

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