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 μsn, where n is the normal force—in this case, the force causing the centripetal acceleration.

μs =

asked by jojo

8 years ago

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mu * normal force = m g

normal force = m omega^2 R

so
m g = mu m omega^2 R

mu = g/(omega^2 R)
or as we could have guessed
mu = gravity acceleration/centripetal acceleration

mu = 9.81/[ 4.2^2 * 3.1 ]

answered by Damon

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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 μsn, where n is the normal force—in this case, the force causing the centripetal acceleration.
μs =

Damon · Answer

mu * normal force = m g normal force = m omega^2 R so m g = mu m omega^2 R mu = g/(omega^2 R) or as we could have guessed mu = gravity acceleration/centripetal acceleration mu = 9.81/[ 4.2^2 * 3.1 ]