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

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Homework Help Overview

The problem involves a stone with a mass of 5.0 x 10-3 kg that is wedged into the tread of an automobile tire. The scenario describes the conditions under which the stone flies out of the tread while the tire rotates at a speed of 20 m/s. The coefficient of static friction and the normal force acting on the stone are provided, with the goal of determining the radius of the tire.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between static friction and the centripetal force required to keep the stone in place. Some question the validity of the calculated radius based on the provided parameters, while others explore the implications of the normal force and friction on the stone's motion.

Discussion Status

The discussion is ongoing, with participants offering insights into the mechanics of the problem. Some have suggested that the gripping force is influenced by the normal force applied by the tire tread, while others express skepticism about the feasibility of the calculated tire size based on the given data.

Contextual Notes

Participants note that the problem may lack certain key facts or assumptions that could affect the outcome, such as the specific geometry of the tire tread or additional forces acting on the stone.

bblake2010
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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.
 
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bblake2010 said:
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}).
 
This cannot be true... That tire's huge. I must be missing a key fact.
 
bblake2010 said:
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.
 

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