Question

In a classic carnival ride, patrons stand against the wall in a cylindrically shaped room. once the room gets spinning fast enough, the floor drops from the bottom of the room! friction between the walls of the room and the people on the ride make them the "stick" to the wall so they do not slide down. in one ride, the radius of the cylindrical room is r = 6.6 m and the room spins with a frequency of 22.1 revolutions per minute. 1) what is the speed of a person "stuck" to the wall?

Answer 1

Given that patrons stand against a cylindrically shaped room and once the room gets spinning enough base of room falls down but patron stick to the wall due to friction force.

Let 'r' be the radius of room and 'f' be the frequency.

Given r= 6.6m and f= 22.1 revolutions per minute.

Speed of person can be found by the formula v = r*w, where r is radius and w = 2πf = 2π*$\frac{22.1}{60 }$ revolutions per second.

Hence speed, $v=r*w = 6.6*2\pi *\frac{22.1}{60}$

                                         = 15.27 m/sec