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

Question

De 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, 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 24.9 degree. According to knowledgeable sources, the coefficient of static between typical human clothing and steel between 0.230 to 0.390. In the fi 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 3.09 rev/s What is the maximum rotational speed at which the riders will not slide up the walls of the ride? Number 1.74 rev/s

Explanation / Answer

Downslope gravitational acceleration:
a_g = gcos = 9.8m/s² * cos24.9º = 8.889 m/s²

normal acceleration:
a_n = gsin + ²rsin = 9.8m/s² * sin24.9º + ²*3m*cos24.9º, so
a_n = 4.126m/s² + ² * 2.721m
Then the friction acceleration (use the lower coefficient for safety) is
a_f = 0.23 * a_n = 0.949m/s² + 0.626m * ²
and this, too, points downslope.

Upslope, we have a component of the centripetal acceleration:
a_c = ²rsin = ² * 3m * sin24.9º = ² * 1.263m

Balancing these accelerations, we have
² * 1.263m = 0.949m/s² + 0.626m * ² + 8.889 m/s²
Gather like terms:
² * (1.263 - 0.626)m = (0.949 + 8.889)m/s²
² = 9.838m/s² / 0.637m = 15.444 rad/s²
= 3.929 rad/s * 1rev/2rads = 0.625 rev/s

For minimum

friction Ff = µ*Fn = µm*(gcos + ²rsin)

9.8*sin65.1 = 0.23*(9.8cos65.1 + ²*3.00*sin65.1) + ²*3.00*cos65.1

= 2.05 rad/s 0.326 rev/s minimum

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