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 larpe 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 25.6degree. According to knowledgeable sources, the coefficient of static friction between typical human dothing and steel ranges between 0.220 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

I am using angle relative to horizontal,

= 90.0º - 25.6 º = 64.4º…………… 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)

For the maximum, µ=0.22
9.8*sin64.4 = 0.22*(9.8cos64.4 + ²*3.00*sin64.4) + ²*3.00*cos64.4
This solves to = 2.04 rad/s 0.267 rev/s ……………. maximum

For the minimum, µ=0.39
9.8*sin66.9 + 0.39*(9.8cos66.9 + ²*3.00*sin66.9) = ²*3.00*cos66.9
which solves to = 1.75 rad/s 0.426 rev/s …………. minimum

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