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

[ZzZerozZ]

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...

 

[rock.freak667]

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?

 

[ZzZerozZ]

n and Fg pointing downwards?

 

[rock.freak667]

n and Fg pointing downwards?

Right, I think the question wants you to find n. Your second equation n+Fg=Fr, find n.

 

[ZzZerozZ]

OO.. yea thanks a lot :D