Estimating Work on the Ferris Wheel: George Washington Gale Ferris Jr.

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The discussion revolves around estimating the work required for the machinery of the Ferris wheel designed by George Washington Gale Ferris Jr. The original poster attempts to calculate work using two methods but arrives at zero, questioning the validity of this result given the wheel's operation. Participants suggest that the key to understanding the work involved lies in the loading process, indicating that once the wheel is fully loaded, it performs no additional work during the ride. The consensus is that the work is primarily associated with the initial loading rather than the constant motion of the wheel. Thus, the focus should be on the mechanics of loading the passengers rather than the ride itself.
jonlevi68
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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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Did you post this problem twice? I just replied to the same problem somewhere else.

This is probably where it belongs

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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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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