in an old fassion amusement park ride, passengers stand inside a 5.0 m diameter
ID: 1741438 • Letter: I
Question
in an old fassion amusement park ride, passengers stand inside a 5.0 m diameter hollow steel cylinder with there backs against the wall. the cylinder begins to rotate along a vertical axis. then the floor on which the passengers are standing suddenly drops away. if all goes well the passengers will stick to the side of the wall and not fall. clothing has a static coefficent of friction against the steel at a range of .6 to one and a kinetic coefficient in the range of .4 to .7. A sign next to the exit reads "no children under 30kg" what is the minimum angular speed, in rpm, for which the ride is safeExplanation / Answer
Diameter is 5m so R is 2.5 m. Since the person is laying against the wall and is in a cylinder we can assume that normal force is going toward the circle.
Normal force is going towards the circle / gravitational force is going dowm / and friction force is going opposite of gravitational force and perpendicular to the normal force.
Since we are looking for the minimal angular speed in which the passenger will not slide, we will use the smallest static friction coefficient (0.6).
Friction force is antiparallel to the gravitational force and the net force of the y-component is zero hence : Fs-FG = 0
Fs=FG
usFn=mg
Fn= mg/us
We also know that normal force is going towards the circle and the x-component has acceleration.
Fn= mw2r
mg/s = mw2r
w= 2.55 rad per sec
multiply omega with 60 then divide by 2 to get 24.4 RPM
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