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

Question

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 22.8o. According to knowledgeable sources, the coefficient of static friction between typical human clothing and steel ranges between 0.230 to 0.390. In the figure, 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? What is the maximum rotational speed at which the riders will not slide up the walls of the ride?

Explanation / Answer

A: The tapered walls make this looked like a banked turn problem.

= 90.0º - 22.8º = 67.2º 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*sin67.2 = 0.22*(9.8cos67.2 + ²*3.00*sin67.2) + ²*3.00*cos67.2
This solves to = 1.444 rad/s 0.2300 rev/s minimum

For the maximum, friction points downslope!
9.8*sin67.2 + 0.22*(9.8cos67.2 + ²*3.00*sin67.2) = ²*3.00*cos67.2
which solves to = 2.67702 rad/s 0.426 rev/s maximum

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