In a classic carnival ride, patrons stand against the wall in a cylindrically sh
ID: 1329152 • Letter: I
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 = 7.4 m and the room spins with a frequency of 21.1 revolutions per minute.
1)What is the speed of a person “stuck” to the wall?
2)
What is the normal force of the wall on a rider of m = 52 kg?
3)
What is the minimum coefficient of friction needed between the wall and the person?
4)
If a new person with mass 104 kg rides the ride, what minimum coefficient of friction between the wall and the person would be needed?
5)
Which of the following changes would decrease the coefficient of friction needed for this ride?
increasing the rider's mass
increasing the radius of the ride
increasing the speed of the ride
increasing the acceleration due to gravity (taking the ride to another planet)
6)
To be safe, the engineers making the ride want to be sure the normal force does not exceed 2.2 times each persons weight - and therefore adjust the frequency of revolution accordingly. What is the minimum coefficient of friction now needed?
Explanation / Answer
1)
the formula for linear speed is
v = r w = 7.4 m ( 21.1 rev/min) ( 2 pi / 60 ) = 16.34 m/s
2)
the normal force of the wall on a rider is
F = mv^2 / R
= 52( 16.34)^2 / 7.4
= 1876.18 N
3)
the minimum coefficient of friction needed between the wall and the person
F = fs
mg = mu . N
mu = 52 ( 9.8 )/ 1876.18
= 0.27
4)
mu = 104 ( 9.8 )/ 1876.18
= 0.54
Just the same = the increase in reaction force ( N ) would balance the need for a greater friction force needed to support the weight.