1. What is the speed of the rider? 2. What is the normal force on a rider of 52
ID: 1655200 • Letter: 1
Question
1. What is the speed of the rider?
2. What is the normal force on a rider of 52 kg?
3. What is the minimum coefficient of friction needed between the wall and person?
4. If a new person with mass 104 kg rides the ride, what min coefficient of friction between the wall and person is needed?
5. To be safe, the engineers want to be sure the normal force does not exceed 2.3 times each persons weight--and therefore adjust the frequency of revolution accordingly. What is the new minimum coefficient of friction required?
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-7.6 m and the room spins with a frequency of 21.1 revolutions per minute.Explanation / Answer
1. w = 21.1 rev/min
w = 21.1 x 2pi rad / 60 s = 2.21 rad/s
v = w r = 2.21 x 7.6
v = 16.8 m/s
2. Normal force = m * centripetal acc.
N = m v^2 / R = 52 x 16.8^2 / 7.6
N = 1929.5 N ........Ans
3. us = f / N
in vertical f- m g = 0
f = m g
us = (52 x 9.81) / (52 x 37.1)
us = 0.26
4. us = g / centripetal acc.
hence independent of mass.
So coefficient of friction will be same.
5. now N = 2.3 m g
us = f / N = 1/2.3 = 0.43 .......Ans
for frequency,
N = m w^2 r
2.3 x 9.81 = w^2 (7.6)
w = 1.723 rad/s
w (in rpm) = 16.5 rpm
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