In an old-fashioned amusement park, passengers sit inside a hollow steel cylinder 5.5 {rm m} in diameter, with their backs against the wall. The cylinder begins to rotate around a vertical axis. Then the floor on which the passengers are standing suddenly collapses! If all goes well, the passengers will “stick” to the wall and not slip. The garment has a coefficient of static friction against steel in the range of 0.65 to 1.0 and a kinetic coefficient in the range of 0.40 to 0.70. A sign next to the entrance reads “No children under 30 {rm kg} allowed”.
Consider the forces acting on a passenger during this ride
The weight acts downward
A normal force acts inward
The friction acts against the weight and is therefore upwards
In the vertical direction, the passenger is expected to stay in place against the wall.
Therefore
F-W=0
In the horizontal direction, the passenger moves in a uniform circular motion once the ride reaches speed. Thus, the centripetal acceleration is inward, to which the horizontal forces must be added:
N = m*a_centripetal
To quantify centripetal acceleration in terms of angular velocity, use
a_centripetal = r*omega^2
By definition of radius,
r = d/2
Friction is based on the normal force:
F = mu*N
Let’s summarize our system of equations:
N = mdomega^2/2
F = mu*N
F = m*g
Solve for omega:
N = F/m
N = m*g/mu
mg/mu = md*omega^2/2
2g/(mud) = omega^2
omega = sqrt (2g/mud)
note: omega will be in radians/s
Connect data:
g = 9.8 m/s^2; m=0.65; d=5.5m;
We accept the lowest coefficient of static friction, because if a passenger is glued to the wall, it is the friction that he undergoes.
Results:
omega = 12.878 rad/s
In RPM:
omega = 123 RPM
Note: to do the worst case, if the passenger slips by accident, consider the smallest coefficient of kinetic friction, which is mu = 0.4
The result for this version is omega = 16.42 rad/s, or omega = 157 RPM.