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9 February, 15:44

On a popular amusement park ride, the riders stand along the inside wall of a large cylinder with a diameter of 8.27 m. The cylinder rotates and when it reaches a rotational frequency of 0.66 revolutions per second the floor drops out. What is the minimum coefficient of static friction needed between a rider and the cylinder wall to prevent the rider from moving down?

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  1. 9 February, 16:50
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    D = 8.27 m ⇒ R = D / 2 = 8.27 m / 2 = 4.135 m

    ω = 0.66 rev/sec = (0.66 rev/sec) * (2π rad/1 rev) = 4.1469 rad/s

    We can apply the equation

    Ff = W ⇒ μ*N = m*g (I)

    then we have

    N = Fc = m*ac = m * (ω²*R)

    Returning to the equation I

    μ*N = m*g ⇒ μ*m*ω²*R = m*g ⇒ μ = g / (ω²*R)

    Finally

    μ = (9.81 m/s²) / ((4.1469 rad/s) ²*4.135 m) = 0.1379
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