A center pivot irrigation system has a pivot to outer wheel track length of 1000 feet and it takes 28 minutes to travel a 250-foot staked course. What is the center pivot's rotation time for one complete revolution?

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Multiple Choice

A center pivot irrigation system has a pivot to outer wheel track length of 1000 feet and it takes 28 minutes to travel a 250-foot staked course. What is the center pivot's rotation time for one complete revolution?

Explanation:
The rotation time is found by treating the outer wheel path as a circle with radius equal to the distance from the center to the outer wheel track. With a radius of 1000 feet, the 250-foot staked arc corresponds to an angle θ = s/r = 250/1000 = 0.25 radians. This arc is traversed in 28 minutes, so the angular speed is ω = θ/t = 0.25/28 radians per minute. A full revolution covers 2π radians, so the time for one complete turn is T = 2π / ω = 2π / (0.25/28) = 224π minutes ≈ 703.7 minutes, which is about 11 hours and 43 minutes. The closest option is 11 hours 43 minutes.

The rotation time is found by treating the outer wheel path as a circle with radius equal to the distance from the center to the outer wheel track. With a radius of 1000 feet, the 250-foot staked arc corresponds to an angle θ = s/r = 250/1000 = 0.25 radians. This arc is traversed in 28 minutes, so the angular speed is ω = θ/t = 0.25/28 radians per minute. A full revolution covers 2π radians, so the time for one complete turn is T = 2π / ω = 2π / (0.25/28) = 224π minutes ≈ 703.7 minutes, which is about 11 hours and 43 minutes. The closest option is 11 hours 43 minutes.

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