# Physics Quick!

A tire 0.650 m in radius rotates at a constant rate of 240 rev/min. Find the speed of a small stone lodged in the tread of the tire (on its outer edge). (Hint: In one revolution, the stone travels a distance equal to the circumference of its path, 2¥ðr.)

What is the acceleration of the stone?

and its direction?
a)away from the center of the path
b)toward the center of the path
c)tangent to the path

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1. Angular speed of wheel = 240*(2 pi)/60 = 25.13 rad/s = w
Speed of stone (relative to axle) = R*w

The speed of the stone relative to the ground depends upon its height above the ground.
Centripetal acceleration of stone = R*w^2 = 411 m/s^2, , toward the center.

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