An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it
rotates around its axis. Once the Rotor reaches its operating speed, the floor drops but the riders remain pinned against the wall of the cylinder. Suppose the cylinder makes 26.0 rev/min and has a radius of 3.70 m. 1) What is the coefficient of static friction between the wall of the cylinder and the backs of the riders
While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.
This acceleration is produced by the centripetal force.
Now, this force is not a different type of force, is the net force acting on the riders in this direction.
Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.
In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.
When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:
where μs= coefficient of static friction (our unknown)
As we have already said Fn = Fc.
The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:
Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:
Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:
The force applied by the competitor is littler than the heaviness of the barbell. At the point when the barbell quickens upward, the power applied by the competitor is more prominent than the heaviness of the barbell. When it decelerates upward, the power applied by the competitor is littler than the heaviness of the barbell.
