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?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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