An amusement park ride consists of a 15 m diameter rotating drum. The axis of ro

Question

An amusement park ride consists of a 15 m diameter rotating drum. The axis of rotation is vertical with respect to the earth. The riders stand against the inside wall of the drum, which then starts to rotate. From the perspective of the riders, centrifugal force pushes them against the wall. Once a critical speed is obtained, the floor drops out and the riders do not fall downward; they are "stuck" to a wall. Assumer that the heaviest rider has a mass of 250 kg, the contact area of the body with the wall is 1.2 m^2 and the static coefficient of friction is 0.7. Calculate the smallest angular speed of the rotating drum where it is safe to let the floor drop out.

Explanation / Answer

The force of friction must balance the weight, and the normal forec is centripetal, so...

umv^2/r = mg (mass cancels)

v^2 = qr/u

v^2 = (9.8)(7.5)/(.7)

v = 10.25 m/s

Then w = v/r

w = (10.25)/(7.5)

w = 1.37 rad/s

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