Amusement park ride rotating about y.axis

[aeromat]

Homework Statement

An amusement park ride consists of a large
cylinder that rotates around a vertical axis.
People stand on a ledge inside. When the
rotational speed is high enough, the ledge
drops away and people “stick” to the wall.
If the period of rotation is 2.5 s and the radius
is 2.5 m, what is the minimum coefficient
of friction required to keep the riders from
sliding down?

Homework Equations

Rotational motion equations
Centripetal force/acceleration

The Attempt at a Solution

All I really could manage to pick out from this eq. is the 'T'-period is 2s, the radius is 2.5m. I am not too sure as to what is providing the centripetal force. I think it is the frictional force that is required to keep them in circular motion, while the gravity applies in areas where they are the top of the ride. For this question, I am lost as to what to do.

 

[PhanthomJay]

The centripetal force is always the net force in the centripetal direction, toward the center of the curve. In this case, the centripetal force is a horizontal force. Identify what the horizontal contact force is, and solve for it using the cenrtripetal force equation in the x direction. Then look in the y direction to see what forces (contact, gravity) act in that direction.

 

[kjohnson]

The centripetal force is what's keeping the person going in a circle, but it is friction that will act vertically to oppose gravity. Since friction is $$\mu$$N where 'N' is your normal force, what is your normal force equal to...

 

[aeromat]

My normal force is equal to gravity; (m)(g)

 

[PhanthomJay]

Normal forces act perpendicular to the objects on which they act...since there is nothing under the person's feet to provide a normal force, there is no normal force in the vertical direction, just a gravity and friction force. Look in the horizontal direction to find the normal force.

 

[aeromat]

But in terms of the minimum coefficient of friction, it will be when they are at the bottom of the y-axis? I think it makes sense that the maximum will be when they reach the peak of the y-axis, because friction and gravity together will make them slip down.

 

[PhanthomJay]

But in terms of the minimum coefficient of friction, it will be when they are at the bottom of the y-axis? I think it makes sense that the maximum will be when they reach the peak of the y-axis, because friction and gravity together will make them slip down.

No, in this problem, this ride is moving in a horizontal circle, not a vertical circle. The rider is always at elevation y = 0 no matter where that person is during the motion. The centripetal force, which is the normal force here of the walls of the ride acting on the person's back, is in the horizontal direction. The friction and weight force are in the vertical y direction. If there is no slipping in the y direction, what can you say about these 2 forces acting in the y direction?