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Posted by on Monday, April 30, 2012 at 2:26pm.

A stone has a mass of 3.40 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 0.916. When the tire surface is rotating at 18.8 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 2.50 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire (in terms of m).

  • physics - , Monday, April 30, 2012 at 2:42pm

    Friction force F(fr) = μ•N = 0.916•2.5 = 2.29 N.
    F(total) = 2•F(fr) = 2•2.29 = 4.58 N.
    F(total) = F(centripet) = mv^2/R.
    R = mv^2/F(total) = 3.4•10^-3•(18.8)^2/4.58 = 0.262 m

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