How to find the coefficient of friction in a rotor-ride?

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SUMMARY

The coefficient of friction (μ) for a carousel moving in a circle can be defined using the equation μ = (gr)/(v²), where g represents the acceleration due to gravity, r is the radius of the carousel, and v is the velocity of the carousel. This relationship ensures that individuals with mass m remain pinned against the wall without sliding. The derivation involves equating centripetal force and gravitational force, leading to the conclusion that μ must be expressed in terms of these variables.

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  • Understanding of centripetal force and its relation to circular motion
  • Knowledge of basic physics concepts such as force, mass, and gravity
  • Familiarity with algebraic manipulation of equations
  • Concept of friction and its role in motion
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  • Study the derivation of centripetal force in circular motion
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broadwayboo
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Homework Statement


I have no values at all, so it's just going to be the variables in the relevant equations. I need to define mu using those variables.
A carousel is moving in a circle at velocity v and has a radius r. It is moving fast enough that each person with mass m is pinned against the wall and not moving. What does mu (the coefficient of friction) have to be in order for the people to not move? Define it in terms of m, g, r, or v.
I know that it has a square root involved somewhere because all the answer choices have square roots.

Homework Equations


friction=mu(normal force)
centripetal Force= mv^2/r
Force of Gravity=gm

The Attempt at a Solution


Centripetal force=normal force=force of friction/ mu
force of friction/ mu=mv^2/r
force of friction= force of gravity (because friction pushes up with the same amount as gravity pushes down)
mg/ mu= mv^2/r
g/ mu =v^2/r
(cross multiply)
mu*v^2= gr
mu=(gr)/(v^2)
But I know this can't be right because there's no square root.
 
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Hi b! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

What are the answer choices offered?
 
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