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 Gravitron drops out, instilling the fear in riders that they will fall a great height. However, the spinning motion of the Gravitrons feature vertical walls, but the example shown in the figure has tapered walls of 25.5 degrees.

According to knowledgeable sources, the coefficient of static friction between typical human clothing and steel ranges between 0.220 to 0.390. In the figure, the center of the mass of a 54.6 kg rider resides 3.00 m from the axis of rotation.

As a sfety 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 slidding 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 (expressed in rev/s)?

Explanation / Answer

a)Balancing friction force with the weight in the critical condition (using u=0.22),

mgCos(theta)-mv^2/r*Sin(theta)=u*mv^2/rCos(theta)

v^2*(umCos(25.5)/r+mSin(25.5)/r)=mgCos(theta)

v=sqrt(54.6*9.81*Cos(25.5)/(0.22*54.6Cos(25.5)/3+54.6Sin(25.5)/3)=6.50

This is the minimum velocity scenario.

f=6.50/(3*2pi)=0.3445rev/s

b)mv^2/r*Sin(theta)-mgCos(theta)=u*mv^2/rCos(theta)

v^2/r*(mSin(theta)-umCos(theta))=mgCos(theta)

v=sqrt(rgCos(theta)/(Sin(theta)-uCos(theta)))=sqrt((3*9.81*Cos(25.5))/(Sin(25.5)-0.22*Cos(25.5)))=10.70m/s

f=10.7/3=3.5672rad/s=0.5677rev/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.