Calculating Revolutions for Weightlessness on a 15m Ferris Wheel

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To achieve weightlessness at the top of a 15m diameter Ferris wheel, the required centripetal acceleration must equal gravitational acceleration. The radius of the wheel is 7.5m, and the formula for centripetal acceleration is v²/r, where v is the linear velocity. After calculating the necessary velocity to create this condition, it was determined that the Ferris wheel would need to complete approximately 11 revolutions per minute for passengers to feel weightless. The discussion highlights the importance of understanding the relationship between radius, velocity, and the forces acting on passengers. This calculation is essential for designing amusement rides that provide specific sensations.
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How many revolutions per minute would a 15m diameter Ferris wheel need to make for the passengers to feel "weightless" at the topmost point of the trip?

Please help me with that question, as I don't even know where to start.
 
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Consider the forces on a person at the top of the wheel.
 
Can someone please give me a bit more information because I;m confused at what to do when I'm only given the radius.
 
Actually after doing a few more problems like this, I think I've gotten the hang of it. For this question I got an answer of 11 revolutions per minute. Is this correct?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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