Centripetal Force: Tire Radius Calculation for 5.0 10-3 kg Stone

[bblake2010]

A stone has a mass of 5.0 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 0.71. When the tire surface is rotating at 20 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.

 

[asleight]

A stone has a mass of 5.0 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 0.71. When the tire surface is rotating at 20 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.

$$F_f = 2N\mu_{t|s} = F_c = v^2/r \rightarrow r = v^2/(2N\mu_{t|s})$$.

 

[asleight]

This cannot be true... That tire's huge. I must be missing a key fact.

 

[LowlyPion]

A stone has a mass of 5.0 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 0.71. When the tire surface is rotating at 20 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.

Welcome to PF.

Since the tire is gripping the stone between the treads, the static friction is not dependent on its mass, so much as it is on the pinching force of the treads. Since they give you that force in Newtons and it is applied on each side then the appropriate way to deal with the gripping force is as 2*u*F.

On the other side of the equation - at speed - I think you can consider that at the bottom of the cycle will be the greatest outward force since both gravity and rotational acceleration are acting on the stone.