Map The Gravitron is an amusement park ride in which riders stand against the in

Question

Map The Gravitron is an amusement park ride in which riders stand against the inner wall of a large spinning steel cylinder At some point, the floor of the Graviton drops out, instilling the fear in riders that they will fall a great height. However, the spinning motion of the Gravitron allows them to remain safely inside the ride. Most Gravitrons feature vertical walls but the example shown in the figure has tapered walls of 25.1° According to knowledgeable sources, the coefficient of static friction between typical human clothing and steel ranges between 0.230 to 0.390. In the fiqure, the center of mass of a 54.6 kg rider resides 3.00 m from the axis of rotation As a safety expert inspecting the safety of rides at a county fair, you want to reduce the chances of injury. What minimum rotational speed (expressed in rev/s) is needed to keep the occupants from sliding down the wall during the ride? Number rev/ s What is the maximum rotational speed at which the riders will not slide up the walls of the ride? Number rev/ s

Explanation / Answer

to doing these with the angle given from horizontal that I got most of the angles backwards on my initial attempt. So that I can use the convention I'm familiar with, I'll use
= 90.0º - 23.1º = 66.9º from horizontal

At the minimum speed, the friction force points upslope.
normal Fn = mgcos + m²rsin
and to be conservative calculating the friction force, use µ = 0.220.
friction Ff = µ*Fn = µm*(gcos + ²rsin)

Along incline, weight component = friction force + centripetal component
mgsin = µm*(gcos + ²rsin) + m²rcos mass m cancels
Dropping units for ease ( is in rad/s)
9.8*sin66.9 = 0.22*(9.8cos66.9 + ²*3.00*sin66.9) + ²*3.00*cos66.9
This solves to = 1.67565 rad/s 0.267 rev/s minimum

For the maximum, friction points downslope!
9.8*sin66.9 + 0.22*(9.8cos66.9 + ²*3.00*sin66.9) = ²*3.00*cos66.9
which solves to = 2.67702 rad/s 0.426 rev/s maximum

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