It has been suggested that rotating cylinders about 12 mi long and 4.1 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?
To find the angular speed required for the rotating cylinder, we need to equate the centripetal acceleration (ac) at its surface with the free-fall acceleration on Earth (g).
1. First, let's determine the radius of the rotating cylinder. The diameter given is 4.1 miles, which means the radius (r) will be half of that:
r = 4.1 mi / 2 = 2.05 mi
2. Next, let's convert the radius from miles to meters since SI units are commonly used in physics.
1 mile is approximately equal to 1609 meters, so the radius in meters (R) is:
R = 2.05 mi x 1609 m/mi = 3298.45 m
3. The centripetal acceleration at the surface of the rotating cylinder is given by:
ac = ω^2 * R
where ω represents the angular speed in radians per second.
4. The free-fall acceleration on Earth is approximately 9.8 m/s^2 (rounded to one decimal place).
5. Now, we can equate the centripetal acceleration to the free-fall acceleration:
ω^2 * R = g
6. Rearranging the equation to solve for ω:
ω = √(g/R)
7. Substituting the known values:
ω = √(9.8 m/s^2 / 3298.45 m)
8. Calculating ω:
ω = √(0.002971 rad/s^2)
Thus, the angular speed needed for the rotating cylinder is approximately 0.055 rad/s (rounded to three decimal places).