(adsbygoogle = window.adsbygoogle || []).push({}); 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([tex]\mu[/tex]_{s}*g*r), [tex]\Sigma[/tex]F_{c}= m(v^{2}/r), a_{c}= v^{2}/r, v = [2[tex]\pi[/tex]r]/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!

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# Homework Help: Find the radius of a car tire, given the mass of wedged stone & tire rotation in m/s

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