A stone has a mass of 8.0 10-3 kg 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 0.83. When the tire surface is rotating at 17 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 1.8 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire.
http://i21.photobucket.com/albums/b272/Cajunboiler/PICTURE.gif
2006-10-16 14:48:33 · 2 answers · asked by blah 1 in Science & Mathematics ➔ Physics
The static friction force acting on the stone is:
F = uN, and this force is also the source of the centripetal force on the stone.
F = ma = mv^2/r
Therefore, setting uN = mv^2/r, we get r = mv^2/uN
But this is not the radius of the tire. It is the diameter of the tire because the tire is not rotating about the center of the tire, since the tire is rolling on the ground and thus the tire is rotating about the bottom point of the tire.
So radius of the tire, R=(1/2)r=mv^2/2uN=0.77m
2006-10-16 15:09:56 · answer #1 · answered by PhysicsDude 7 · 0⤊ 0⤋
http://en.wikipedia.org/wiki/Coefficient_of_friction
2006-10-17 00:42:02 · answer #2 · answered by Surya M. 3 · 0⤊ 0⤋