Question

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. The 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

Answer 1

ω = 0.05 rad/s

Step-by-step explanation:

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

`Centripetal Force = Weight\\(mv^(2))/(r) = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\((r\omega)^(2))/(r) = g\\\\\omega^(2) = (g)/(r)\\\\\omega = \sqrt{(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{(9.8\ m/s^(2))/(4023.36\ m)}\\\\`

ω = 0.05 rad/s