Calculating Tire Radius Given Mass, Friction & Velocity

[lalalah]

1. A stone has a mass of 3.92 g and is wedged into the tread of an automobile tire, as the drawing shows. The coefficient of static friction between the stone and each side of the tread channel is 0.864. When the tire surface is rotating at 18.4 m/s, the stone flies out of the tread. The magnitude FN of the normal force that each side of the tread channel exerts on the stone is 2.37 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire (in terms of m).



2. v = sq.root(\mu_s*g*r), \SigmaF_c = m(v²/r), a_c = v²/r, v = [2\pir]/T



3. link to picture of the stone wedged in the tire.
http://www.flickr.com/photos/20949091@N00/2911847669/
as it is wedged, there are two normal forces acting directly opposite each to the other on either side of the stone. Even though the two normal forces should cancel out, I still want to use the 2.37 N in the v = sq.root(m_u*r*g) by substituting "g" for (F_n/m)

I am assuming that the velocity is 18.4 m/s. Is this correct?

So...

velocity = 18.4 m/s = sq.root((m_u)*(F_n/m)*radius_tire)
18.4 m/s = sq.root(0.864*(2.37N/.00392kg)*radius _tire)
squaring both sides gets:
338.56 = 522.37 * radius
radius = .648 m

is there something I'm missing? my intuition tells me this answer is too big!

 

[Kurdt]

A simpler way to do it is to set the centrifugal force equal to the friction force on the stone. Remember there are two friction forces. That is F=2\mu N.

 

[lalalah]

ah i see.
so in this case, i can say there are two friction forces because they do not cancel each other out?

so it should be...
(mv^2)/r = 2 mu *mg
(.00392*18.4^2)/r = 2*0.864*.00392*9.8
1.327/r = 0.0664
r= 19.98 m
?

 

[lalalah]

wait let me try that again

Fn = 2.37
(mv^2)/r = 2 mu *Fn
(.00392*18.4^2)/r = 2*0.864*2.37
1.327/r = 4.095
r = .324

so it's the first answer divided by two! I'm starting to get the idea of how to use forces now... thank you so much for your help, i appreciate it!

 

[Kurdt]

Your first answer was ok except you hadn't included the other friction force, hence why it was twice as much. The other method is a bit more intuitive.