Question

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?

Answer 1

Answer:

Explanation:

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