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.

A. What minimum rotational speed (expressed in rev/s) is needed to keep the occupants from sliding down the wall during the ride?

B. What is the maximum rotational speed at which the riders will not slide up the walls of the ride?

In this problem, the rider undergoes centripetal (that is, radial) acceleration. Newton's second law remains unchanged, but the more familiar linear equation of motion becomes, for the radial direction,a(subscript r)=rw^2

For the case where you are asked to find the minimum angular speed needed to keep the rider from sliding down, note first that the frictional force must point up along the wall, since the intended motion of the system is down along the wall. (See the upper figure on the right.) Since Newton's second law is a vector equation, we can write it for any direction. The figure on the right defines two directions: The vertical y-direction and the radial r-direction. Sum the forces acting along the vertical direction, then sum the forces acting along the radial direction, noting that ar = r?2 in Newtons second law. The situation where you are asked to find the maximum angular speed in which the riders will not slide up the wall is solved similarly, although in this situation the frictional force points down along the ramp wall, as shown in the lower figure.

Explanation / Answer

A)   minimum rotational speed , w

=> w2R = usgcos(theta)

=>   w2 * 3 = 0.250 *9.8 * cos(23.2)

=>    minimum rotational speed , w   = 0.8663 rad/sec

                                                           = 0.1379 rev/sec

B)    w2R = usgcos(theta)

=>   w2 * 3 = 0.390 *9.8 * cos(23.2)

=>    maximum rotational speed , w   = 1.0821 rad/sec

                                                           = 0.1723 rev/sec

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