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At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders \"plastered\" against the wall. Suppose the radius of the room is 3.20 m and the speed of the wall is 10.4 m/s when the floor falls away. What is the minimum coefficient of static friction that must exist between 72 kg rider's back and the wall, if the rider is to remain in place when the floor drops away?

Vishal Lalchand Rohra , 9 Years ago
Grade
anser 1 Answers
Prithvi Raj

Last Activity: 9 Years ago

This thing is simply a rotor and it work for same priciple as death well.
Imagin the rider who is now in the rotating round wall, now lets draw his free bosy diagram.
!
  1. Weight “mg” = 72 * (9.8 or just 10)   acting downward 
  2. He has a tendency to slide down so friction acting upward = f
  3. Normal reaction by the wall = N   which is provided by centripetel force = mv2/r
Now just balence forces,
N = mv2/r  from that you will get “N”
and by eqating mg = f
                      mg = u * N  (u is coefficent of static friction 
                      m=72kg, g=9.8, N=from solving above eq
same case for a death well where a person drives bicycle in a vertical surface insted of wall the man is moving thats all 

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