Solving the Mystery of Drop Out Ride: Find Rev/s Needed for Floor Drop Out

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The discussion centers on calculating the required revolutions per second for the Drop Out ride, where riders are held against the wall by friction when the floor drops out. The radius of the ride is 5.5 meters, and the coefficient of static friction is 0.28. Participants express confusion about how to relate centripetal acceleration to the forces acting on the riders. The key equations involve centripetal acceleration and the relationship between linear velocity and angular velocity. The solution requires expressing velocity in terms of angular velocity to find the necessary revolutions per second before the floor can drop.
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Homework Statement


In a ride called Drop Out, riders are spun in horizontal circles of radius 5.5m, which forces them to the outer wall. When they are spinning fast enough, the floor drops out, and they are suspended by friction. The coefficient of static friction is 0.28, how many rev/s must the ride achieve before the floor is allowed to drop out?


Homework Equations


a_{}c = rv^2


The Attempt at a Solution


I'm sort of lost for what to do, I have an FBD with G (down) and Friction (up), I also know the centripetal acceleration will be along the radius, towards the center of the circle, but I do not know how to solve this one.
Thanks in advance.
 
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No, a=v^2/r. If you know the centripetal acceleration, then you know the normal force. You then can relate the frictional force to the normal force, et voila. Please continue.
 
ok so...ac = v^2/5.5
v is unknown though, so we don't really know the centripetal acceleration?
 
You know the radius, and v can be expressed in terms of the angular velocity. Leave it unknown. That's what you want to solve for.
 
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