# Amusement Park ride - maximum period of rotation

1. Aug 9, 2007

### avsj

1. The problem statement, all variables and given/known data

A 2.8m radius drum (cylinder) rotates such that a person does not fall when the "floor" falls away (imagine the person being pushed against the inside wall of the cylinder, with no bottom to the cylinder)

If the coefficient of friction between the person and the wall is 0.35, what is the maximum period of roation so that a personw ill not fall?

2. Relevant equations

Fg= mg
Ff=uFn

Fc = m4r(pi^2)/T^2
where T is period

3. The attempt at a solution

I am confused with the directions of the forces. Fg will go down the inside wall, Fc is going inwards to the center, and Fn ...? But, Fn should = Fg, so Ff = umg

Then I thought to equate this umg to Fc, and solve for T but I dont get the correct answer, which is 2.0s

THanks,

2. Aug 9, 2007

### avsj

One way which I arrived at the right answer, was to start by setting Ff=Fg, and then setting Fn =Fc, but I am not sure if my reasoning for these directions is correct. In doing this, I got Fn= mg/u which ended up giving me the correct period. So Fn isnt always opposite in direction of Fg? IS it the case that Ff is always parallel to the surface and Fn perpendicular?

THanks

3. Aug 9, 2007

### Staff: Mentor

Realize that the normal force is the centripetal force.

This is correct.
That's right. Fn is the force between two surfaces, in this case the person and the wall. (You might be thinking of the more typical case of a person on the ground.)

This is true.

What might be helpful is to identify the actual forces acting on the person. I count three forces. (Note: "Centripetal Force" doesn't count--specify the actual forces that provide the centripetal force. Centripetal just means "toward the center".)