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

AI Thread Summary
To find the coefficient of friction (mu) for a carousel where individuals are pinned against the wall, it must be defined in relation to mass (m), gravitational acceleration (g), radius (r), and velocity (v). The centripetal force required to keep the individuals in circular motion is provided by the frictional force, which equals the normal force. The equations indicate that mu can be expressed as mu = (gr)/(v^2). However, the user notes that this expression lacks a square root, which they believe should be present based on the answer choices. The discussion revolves around correctly deriving mu with the necessary variables and ensuring it aligns with the expected format.
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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