In a popular amusement park ride, a rotating cylinder of radius R = 3.10 m is se

Question

In a popular amusement park ride, a rotating cylinder of radius R = 3.10 m is set in rotation at an angular speed of 4.20 rad/s, as in the figure shown below. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping? Hint: Recall that the magnitude of the maximum force of static friction is equal to ?sn, where n is the normal force

Explanation / Answer

The centripetal acceleration needed to keep an object in circular motion is a = v^2 / r. The velocity relates to the angular velocity as v = omega * r, so a = omega^2 * r. The cylinder acts with the normal force F = m * a = m * omega^2 * r on the rider. The force of static friction is Ff = mu * m * omega^2 * r. To keep the rider from slipping, this should equal the gravitational force pulling the rider down. This is Fg = m * g. Equal the forces to get mu * m * omega^2 * r = m * g -> mu = g / (omega^2 * r)

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