A stone has a mass of 3.34 g and is wedged into the tread of an automobile tire,

Question

A stone has a mass of 3.34 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 0.993. When the tire surface is rotating at 18.0 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 1.83 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire (in terms of m)

Explanation / Answer

We will need the stone's mass in kg, so let's do that conversion first. m = 5.71 g * (1 kg/1000 g) = ??? Next calculate the total friction holding the stone in the tread. The tread on one side provides Ff = mu*FN = 0.789*2.06 N = ??? The friction on the 2 sides add, so the total friction, Tf, that was overcome when the tire surface was rotating at 13.0 m/s is Tf = 2*Ff = ??? So when the tire surface was rotating at 13.0 m/s, the stone needed more centripetal force than the friction was able to provide. Centripetal force, Fc, is given by Fc = m*v^2/r So the limiting condition is represented by the equation Fc = m*v^2/r = Tf Plug in the values of m, v, and Tf and solve for r.

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