In an old-fashioned amusement park ride, passengers stand inside a 4.7-m-diamete

Question

In an old-fashioned amusement park ride, passengers stand inside a 4.7-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.65 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed."

What is the minimum angular speed, in rpm, for which the ride is safe

Explanation / Answer

Friction Force as a result of Centripetal Force (Ff) = Gravity Force (Fg)

Friction Force = Mu * Normal Force and Normal Force in this case = Centripetal Force

Ff = mu * M * V^2/r

where, mu: Cofficient of Friction

M: Mass

V: Tangential Velocity

r: Radius

We want a Tangential Velocity that is sufficient for the minimum mu.

We want no sliding. So, we can ignore Kinetic Friction,

Ff = Fg

mu*M*V^2/r = M*g

V^2 = g*r/mu

(wr)^2 = gr/mu

w = sqrt(g/(mu*r))

= sqrt(9.8/(0.4*2.35))

= 3.22 rad/s = 30.75 rpm answer

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