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A 834 kg automobile is moving at a maximum speed of 33 m/s on a level circular track of radius 362 m. What is the coefficient of friction?

A 834 kg automobile is moving at a maximum speed of 33 m/s on a level circular track of radius 362 m. What is the coefficient of friction?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Problem Statement: Calculating the Coefficient of Friction** An 834 kg automobile is moving at a maximum speed of 33 m/s on a level circular track with a radius of 362 m. What is the coefficient of friction? **Explanation and Solution Approach:** To find the coefficient of friction (\( \mu \)), we need to use the formula for centripetal force and the force of friction: 1. **Centripetal Force (\( F_c \)):** - Formula: \( F_c = \frac{mv^2}{r} \) - Where \( m \) is mass (834 kg), \( v \) is velocity (33 m/s), and \( r \) is radius (362 m). 2. **Force of Friction (\( F_f \)):** - This force is also the centripetal force required to keep the automobile moving in a circle. - Formula: \( F_f = \mu mg \) - Where \( \mu \) is the coefficient of friction, \( m \) is mass, and \( g \) is acceleration due to gravity (9.8 m/s²). 3. **Equating the Forces:** - Since the frictional force provides the necessary centripetal force for circular motion, \( \mu mg = \frac{mv^2}{r} \). 4. **Solve for \( \mu \):** - Cancel out the mass \( m \) from the equation, and solve for \( \mu \). - \( \mu = \frac{v^2}{rg} \). By substituting the known values into the equation, we can determine the coefficient of friction required for the automobile to safely navigate the circular track.
Transcribed Image Text:**Problem Statement: Calculating the Coefficient of Friction** An 834 kg automobile is moving at a maximum speed of 33 m/s on a level circular track with a radius of 362 m. What is the coefficient of friction? **Explanation and Solution Approach:** To find the coefficient of friction (\( \mu \)), we need to use the formula for centripetal force and the force of friction: 1. **Centripetal Force (\( F_c \)):** - Formula: \( F_c = \frac{mv^2}{r} \) - Where \( m \) is mass (834 kg), \( v \) is velocity (33 m/s), and \( r \) is radius (362 m). 2. **Force of Friction (\( F_f \)):** - This force is also the centripetal force required to keep the automobile moving in a circle. - Formula: \( F_f = \mu mg \) - Where \( \mu \) is the coefficient of friction, \( m \) is mass, and \( g \) is acceleration due to gravity (9.8 m/s²). 3. **Equating the Forces:** - Since the frictional force provides the necessary centripetal force for circular motion, \( \mu mg = \frac{mv^2}{r} \). 4. **Solve for \( \mu \):** - Cancel out the mass \( m \) from the equation, and solve for \( \mu \). - \( \mu = \frac{v^2}{rg} \). By substituting the known values into the equation, we can determine the coefficient of friction required for the automobile to safely navigate the circular track.
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