Centripetal force, same as normal force on the "wall"
= mrω²
Static friction
=μs*N
=μsmrω²
For the people to be "glued" on the wall, static friction must exceed the weight of the riders due to gravity.
Therefore,
μsmrω²
≥ mg
Using
r=5.7m,
ω=0.35*2π radians / s
and
cancelling common factor m,
μsrω²
≥ g
we get
μ≥g/(rω²)
≥ 9.8 / (5.7*(3.5*2π)²)
≥ 0.356 approximately
Please Help I have no idea how to approach this problem
In a "Rotor-ride" at a carnival, people rotate in a vertical cylindrically walled "room."
If the room radius was 5.7m , and the rotation frequency 0.35 revolutions per second when the floor drops out, what minimum coefficient of static friction keeps the people from slipping down?
1 answer
1 answer