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 23.2o. According to knowledgeable sources, the coefficient of static friction between typical human clothing and steel ranges between 0.250 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? Both answers need to be in REV/S (revolutions per second)

Explanation / Answer

Force parallel = 54.6 * 9.8 * cos 232.2 = 535.08 * cos 23.2
Force perpendicular = 535.08 * sin 23.2
Ff = 0.39 * (N + 535.08 * sin 23.2) = 0.39 * N + 208.962 * sin 23.2

0.39 * N + 208.962 * sin 23.2 = 535.08 * cos 23.2
0.39 * N = 535.08 * cos 23.2 - 208.962 * sin 23.2
N = 1052.0633N
Since the riders are moving in a horizontal circle, the centripetal force must be equal to the horizontal component of the net force.

Horizontal component = (1372 * cos 23.2 - 535.08 * sin 23.2) * cos 23.2 = 1050.26 N

Fc = 54.6 * v^2/3 = 18.2 * v^2
18.2 * v^2 = 1050.26
v = 7.6 m/s

As the cylinder rotates one time, the rider moves a distance that is equal to the circumference of the cylinder.

C = 2 * pi * 3 = 6 * Pi
This is approximately 18.85 meters. To determine the time for the cylinder to rotate one time, divide the circumference by the velocity.

Time =distance / velocity = 18.85 /7.6 = 2.480 s
This is approximately 2.48 seconds
Rotational speed 1 rev/2.48
This is approximately 0.403 rev/s.

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

At the minimum speed, the friction force points upslope.
normal Fn = mgcos(theta) + mw^2rsin(theta)
and to be conservative calculating the friction force, use u = 0.250.
friction Ff = u*Fn = um*(gcos(theta) + w^2rsin(theta))

weight component = friction force + centripetal component
mgsin(theta) = um*(gcos(theta) + w2rsin(theta)) + mw^2rcos(theta)
Dropping units for ease (w is in rad/s)
9.8*sin23.2 = 0.25*(9.8cos23.2 + w^2*3.00*sin23.2) + w^2*3.00*cos23.2
This solves to w = 2.01297 rad/s = 0.3199 rev/sec

For the maximum, friction points downslope!
9.8*sin23.2 + 0.39*(9.8cos23.2 + w^2*3.00*sin23.2) = w^2*3.00*cos23.2
which solves to w= 3.61159 rad/s

w = 3.61 rad/s

w= 0.57 rev/s

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