At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is 1.2 m. If the coefficient of static friction between a rider's back and the wall is 0.214, what is the minimum rotation speed of the cylindrical room in of revolutions per minute (rpm) if the rider is to remain in place when the floor drops away? Use g=10 m/s².

University Physics Volume 1
18th Edition
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Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter10: Fixed-axis Rotation
Section: Chapter Questions
Problem 47P: A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static...
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At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls
away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is
1.2 m. If the coefficient of static friction between a rider's back and the wall is 0.214, what is the
minimum rotation speed of the cylindrical room in of revolutions per minute (rpm) if the rider is to
remain in place when the floor drops away? Use g=10 m/s².
Transcribed Image Text:At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is 1.2 m. If the coefficient of static friction between a rider's back and the wall is 0.214, what is the minimum rotation speed of the cylindrical room in of revolutions per minute (rpm) if the rider is to remain in place when the floor drops away? Use g=10 m/s².
Expert Solution
Step 1

From Newton's law for the riders to be remain plastered to the wall without falling down is can be obtained by equating vertical equations.

W=Ffmg=μs N

The normal force, the body experience is due to the centripetal force.

N=mω2R

 

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