A cylindrical space colony 8.00 km in diameter and 30.0 km long has been proposed as living quarters for future space explorers. Such a habitat would have cities, land, and lakes on the inside surface and air and clouds in the center. All this would be held in place by the rotation of the cylinder about the long axis. How fast would such a cylinder have to rotate to produce a 1-g gravitational field at the walls of the cylinder

Answers

Answer 1

Answer:

ω = 0.05 rad/s

Explanation:

In order to produce the acceleration equal to the acceleration due to gravity at the surface of Earth, the centripetal acceleration must be equal to the value of g:

[tex]a_c = g\\g = \frac{v^2}{r}\\\\but,\ v=r\omega\\therefore,\\\\g = \omega^2r\\\\\omega = \sqrt{\frac{g}{r}}[/tex]

where,

ω = angular speed = ?

g = acceleration due to gravity on the surface of the Earth = 9.81 m/s²

r = radius of cylinder = 8 km/2 = 4 km = 4000 m

Therefore,

[tex]\omega = \sqrt{\frac{9.81\ m/s^2}{4000\ m}}[/tex]

ω = 0.05 rad/s

Related Questions

A spinning disc with a mass of 2.5kg and a radius of 0.80m is rotating with an angular velocity of 1.5 rad/s. A ball of clay with unknown mass is dropped onto the disk and sticks to the very edge causing the angular velocity of the disk to slow to 1.13 rad/s. What is the mass of the ball of clay

Answers

Answer:

M = 1.90 Kg

Explanation:

Given data: mass = 2.5 Kg

radius R = 0.8 m

angular velocity ω = 1.5 rad/s

Angular momentum L =0.5×Iω^2

Where, I is the moment of inertia of the spinning disc.

I = 0.5MR^2

I = 0.5×2.5×0.8^2

I = 0.8 Kg/m^2

Then L = 0.5×0.8×1.5^2 = 0.8×2.25 = 0.9 Kg-m^2/sec

Let unknown mass be M

New mass of disc = (2.5+M) Kg, R = 0.8 m

New I = 0.5(2.5+M)(0.8)^2

Since, angular momentum is conserved

Angular momentum before = angular momentum after

0.5×0.5(2.5+M)(0.8)^2×(1.13)^2 = 0.9

Solving for M we get

0.204304(2.5+M)=0.9

M = 1.90 Kg

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