In a popular amusement park ride, a rotating cylinder of radius 4.00 m is set in rotation at an angular speed of 5.00 rad/s. 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 µn, where n is the normal force - in this case, the force causing the centripetal acceleration.)

asked by Elisa

16 years ago

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The friction force M R w^2 * u must equal or exceed the weight M g.

Therefore u > g/(R w^2)

u is the coefficient of friction and w is the angular velocity. R is the radius of the cylinder.

answered by drwls

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Question

In a popular amusement park ride, a rotating cylinder of radius 4.00 m is set in rotation at an angular speed of 5.00 rad/s. 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 µn, where n is the normal force - in this case, the force causing the centripetal acceleration.)

1 answer

The friction force M R w^2 * u must equal or exceed the weight M g.

Therefore u > g/(R w^2)

u is the coefficient of friction and w is the angular velocity. R is the radius of the cylinder.

Ask a New Question or answer this question.

drwls · Answer

The friction force M R w^2 * u must equal or exceed the weight M g. Therefore u > g/(R w^2) u is the coefficient of friction and w is the angular velocity. R is the radius of the cylinder.