Question

An amusement park ride consists of a large vertical cylinder of radius 4 m, that rotates about its axis with a constant speed fast enough that a 600 N person inside is held up against the wall when the floor drops away as show in the figure. If the normal force acting on the person is 1500 N, what is the speed of the cylinder necessary to keep the person from falling?

Answer 1

the friction force is just able to counter gravitational force
that normal force which contributes to centripetal force is related to the friction force.
do you need further help? I believe u should be able to write eqs from this

Answer 2

From the weight of the person (600N) you can get her mass in Kg :\[W=mg \rightarrow m=\frac{ W }{ g }\]
The centrifugal force is given by:\[F_c=m·a_c=m \frac{ v^2 }{ R }\rightarrow v=\sqrt{\frac{ R·Fc }{ m }}=\sqrt{\frac{ R·F_C·g }{ W }}\]