The Roundup Ride: Experience the Thrill of 14.56m/s^2!

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The discussion revolves around calculating the force exerted on a rider at the top of The Roundup amusement park ride. The ride features a rotating ring with a diameter of 17.0m, and the rider has a mass of 52.0kg. The rider's velocity is calculated to be 11.126 m/s, leading to a centripetal acceleration of 14.56 m/s². The initial attempt incorrectly calculated the resultant force without considering the gravitational force acting on the rider. Ultimately, the correct approach involves recognizing that the normal force and gravitational force both act downwards at the top of the ride, leading to the need for a revised calculation of the normal force.
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Homework Statement



In an amusement park ride called The Roundup, passengers stand inside a 17.0m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

Suppose the ring rotates once every 4.80s . If a rider's mass is 52.0kg , with how much force does the ring push on her at the top of the ride?

Homework Equations


1.v=2*pi*r/T
2.Fr=mg+n=Fnet=m*Ac=m*v^2/r

The Attempt at a Solution



What I do is T=4.80 m=52.0kg

v=formula#1=11.126m/s
Ac= formula#2=14.56m/s^2
Fr=m*Ac= 757,12N

The answer is wrong but I don't know why...
 
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The centripetal force, you are finding, is the resultant force on the rider. At the top of the ride, what are the forces acting and their directions?
 
n and Fg pointing downwards?
 
ZzZerozZ said:
n and Fg pointing downwards?

Right, I think the question wants you to find n. Your second equation n+Fg=Fr, find n.
 
OO.. yea thanks a lot :D
 

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