In a \"Rotor-ride\" at a carnival, people are rotated in a cylindrically walled

Question

In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled "room." (See the figure below.) The room radius is 4.5 m, and the rotation frequency is 0.6 revolutions per second when the floor drops out.

In a Rotor-ride at a carnival, people are rotated in a cylindrically walled room. (See the figure below.) The room radius is 4.5 m, and the rotation frequency is 0.6 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that the people will not slip down?

Explanation / Answer

Friction must balance the weight

Friction = uFn and Fn in this case is centripetal, so

umv2/r = mg (mass cancels)

We can get v from the angular speed

w = .6rev/s = 3.77 rad/s

v = wr = 3.77(4.5) = 16.96 m/s

u(16.96)2/(4.5) = (9.8)

u = .153

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