How Does Physics Explain Forces on The Roundup Amusement Park Ride?

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The discussion revolves around calculating the forces acting on riders in The Roundup amusement park ride, which features a 16.0 m-diameter rotating ring. Participants are tasked with determining the force exerted by the ring on a 54.0 kg rider at both the top and bottom of the ride, as well as identifying the maximum rotation period to prevent riders from falling off. Key equations include centripetal force and angular velocity calculations, with suggestions for using the relationship between rotation period and angular velocity. There is some confusion regarding the calculations, prompting requests for clarification and assistance. The thread emphasizes the importance of correctly applying physics principles to solve the problems presented.
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Homework Statement



In an amusement park ride called The Roundup, passengers stand inside a 16.0 m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

Part A: Suppose the ring rotates once every 4.10 s. If a rider's mass is 54.0 kg, with how much force does the ring push on her at the top of the ride?

Part B: Suppose the ring rotates once every 4.10 s. If a rider's mass is 54.0 kg, with how much force does the ring push on her at the bottom of the ride?

Part C: What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?


Homework Equations



Fr=nr+(Fg)r=n+mg=mv^2/r
w=v/r
1 rev=2pi radians
Might be more needed..

The Attempt at a Solution


4.10s/1 rev=1 rev/2pi radians=6.44 rad/s=w
w=v/r=6.44rad/s/8m=51.52m/s
Then do I plug it into the Fr equation? I don't have the acceleration..
Any help or how to think about it would be advantageous.
Thanks.
 
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Those calcs don't look right to me.
You could do w = 1 rev/4.1s = 2*pi/4.1 = 1.53 radians/s
Or just work with v = 2*pi*r/T.
The n+mg=mv^2/r looks good.
 
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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