It has been suggested that rotating cylinders about 10 mi long and 5.6 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?
take 5.6/2=2.8
convert 2.8miles into meters=4506
9.8/4506= .002174878
now squaroot that and you wil getthe answer which is
(.0466)
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r=5.6/2 = 2.8mi
convert from mi to m use
imi= 1609.34m
2.8mi? =1609.34*2.8=4506.152m
g=v^2/r , g=9.8, making v subject we get
v^2= 9.8*4506.152
v=210.14m/s
but w=v/r
w= 210.14/4506.152=0.047rad/s
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To find the angular speed required for the centripetal acceleration at the surface of the cylinder to equal the free-fall acceleration, you can follow these steps:
1. Identify the relevant information:
- Length of the cylinder (L) = 10 miles or 16,093 meters
- Diameter of the cylinder (D) = 5.6 miles or 9,012 meters
- Centripetal acceleration (a) = gravitational acceleration (g) = 9.8 m/s^2
2. Convert the length and diameter to meters, as the unit for acceleration is in meters per second squared.
3. Calculate the radius (R) of the cylinder:
- Since the diameter (D) is given, divide it by 2 to find the radius (R):
R = D/2 = 9,012 m / 2 = 4,506 m
4. Determine the tangential velocity (v) of a point on the surface of the cylinder:
- The tangential velocity of a point on the surface of the cylinder is equal to the circumference of the cylinder divided by the time taken to complete one rotation.
- The circumference of a cylinder is given by 2πR, and the time taken to complete one rotation is 24 hours or 86,400 seconds (considering Earth's rotation).
- So, v = Circumference / Time period = 2πR / time period
v = 2π(4,506 m) / 86,400 s
v ≈ 0.329 m/s
5. Calculate the angular speed (ω) of the cylinder:
- Angular speed is defined as the change in angle (θ) per unit of time (t).
- The angle (θ) covered by a point on the surface of the cylinder is equal to its tangential velocity (v) divided by the radius (R).
- So, ω = v/R
ω = 0.329 m/s / 4,506 m
ω ≈ 7.29 x 10^-5 rad/s
Hence, the angular speed required for the centripetal acceleration at the surface of the cylinder to equal the free-fall acceleration is approximately 7.29 x 10^-5 rad/s.