Solving the Ferris Wheel Problem: Estimating Work Requirement

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The discussion centers on estimating the work required for a Ferris wheel to rotate passengers. The original Ferris wheel, designed by George Washington Gale Ferris Jr., has specific parameters, including a diameter of 76 meters and a loading process involving 36 cars. The author explores two methods for calculating work but finds both yield zero, leading to confusion about the implications of constant angular speed. The conclusion suggests that the work is primarily associated with the loading process, as the ride itself does not require additional work once the wheel is in motion. Overall, the focus is on understanding the work dynamics involved in loading versus the rotational movement of the Ferris wheel.
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I've been thinking about this problem for the last half hour, but can't seem to come up with a way of finding a solution. Maybe some of you can help.

"George Washington Gale Ferris. Jr., a civil engineering graduate from RPI, built the original Ferris wheel. The wheel carried 36 wooden cars, each holding up to 60 passengers, around a circle 76m in diameter. The cars were loaded 6 at a time, and once all 36 cars were full, the wheel made a complete rotation at constant angular speed in about 2 min. Estimate the amount fo work that was required of the machinery to rotate the passengers alone."

I approached this problems through two methods:

1) W = (1/2)(I)(wf^2 - wi^2), where I is the inertia, wf is the final angular velocity and wi is the initial angular velocity, and

2) W = t(thetaf - thetai), where t is the torque, thetaf is the final angle and thetai is the initial angle.

But both attempts give me zero (which, for some reason, doesn't seem liek the right answer). If the ferris wheel is traveling at a constant angular speed, then shouldn't wi = wf, and therefore W = 0? And, if it completes a revolution, shouldn't thetaf = thetai and, also, W = 0?

Thanks for any help.
 
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I think you need to focus on the loading process. Once the wheel is loaded it should be finished working. The 2 min ride is work free.
 
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