In a classic carnival ride patrons stand against the wall


Solve the all parts of given question:

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 = 6.2 m and the room spins with a frequency of 23.4 revolutions per minute.

Part A: What is the speed of a person "stuck" to the wall?

Part B: What is the normal force of the wall on a rider of m = 49 kg?

Part C: What is the minimum coefficient of friction needed between the wall and the person?

Part D: What is the minimum coefficient of friction needed between the wall and the person?

Part E: Which of the following changes would decrease the coefficient of friction needed for this ride?

Part F: To be safe, the engineers making the ride want to be sure the normal force does not exceed 2 times each person's weight - and therefore adjust the frequency of revolution accordingly. What is the minimum coefficient of friction now needed?

I need help to find the minimum coefficient of friction needed between the wall and the person.

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Physics: In a classic carnival ride patrons stand against the wall
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