Question

It has been suggested that rotating cylinders about 9 mi long and 5.9 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?

Answer 1

Answer:

$\omega = 4.5\times 10^{-2} rad/s$

Explanation:

Given data:

Rotating cylinder length = 9 mi

diameter of cylinder is 5.9 mi

we know that linear acceleration is given as

a =  r ω^2

where ω is angular velocity

so$\omega = \sqrt{\frac{a}{r}}$

$r = \frac{5.9}[2} \frac{1609 m}{1 mi} = 4.746\times 10^{3} m$

$\omega = \sqrt{\frac{9.80}{4.746\times 10^{3}}}$

$\omega = 4.5\times 10^{-2} rad/s$