Physics, 02.12.2019 18:31 malika61

In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-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 0.60 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70.what is the minimum rotational frequency, in rpm, for which the ride is safe? a previous answer was given with the formulas and answer, but i don't understand how i would come up with the formula %u03bcsn=mgto begin with. here's the reply from previous post: for minimum speed when the floor may be removed the friction is thelimiting force which is equal to %u03bcsn%u03bcn=mgbut n=mvv/rso %u03bcsmvv/r=mghence v=%u221arg/%u03bcs=%u221a2.59.8/0. 6=6.39 m/secso angular velocity=%u03c9=v/r=2.55rad/secrota tional frequency=%u03c9/2%u03c0so rotational frequency=0.4060 rot/secin rpm the value is 0.4060*60=24.36 rpm

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In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow...

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