A stone has a mass of 3.15 g and is wedged into the tread of an automobile tire,

Question

A stone has a mass of 3.15 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 0.941. When the tire surface is rotating at 17.3 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 1.50 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire (in terms of m)

Explanation / Answer

For the stone to fly out, the centripetal force must balance the frictional force.

The formula for centripetal force is Fc = mv2/r

The formula for friction is Ff = Fn

We have two frictional forces to overcome, so we get

mv2/r = 2Fn

(.00315)(17.3)2/r = (2)(.941)(1.5)

r = .334 m

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