1) A car moving at 12.0 m/s tries to negotiate a corner in a circular roadway of radius 15.0 m. The roadway is flat. a) Draw a free body diagram for the car as it makes this circular turn. b) Derive the algebraic expression for the coefficient of friction between the wheels and the road if the car is to not skid? c) How does the mass of the car affect the coefficient of friction? d) Calculate the coefficient of friction.

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### Physics Problem on Circular Motion and Friction

**Problem Statement:**

1. A car moving at \(12.0 \, \text{m/s}\) tries to negotiate a corner in a circular roadway of radius \(15.0 \, \text{m}\). The roadway is flat.
   
   a) Draw a free body diagram for the car as it makes this circular turn.

   ![Free Body Diagram](image-url)
   *(The diagram should show the car turning on a flat road with forces acting on it.)*
   
   In the figure accompanying this prompt, a car is illustrated negotiating a circular turn. The car follows a curved path, indicting its direction of motion.

   b) Derive the algebraic expression for the coefficient of friction between the wheels and the road if the car is to not skid.

   c) How does the mass of the car affect the coefficient of friction?

   d) Calculate the coefficient of friction.

**Detailed Explanation:**

- **Part a:** The free body diagram of the car should show:
  - The gravitational force acting downwards (\(mg\)).
  - The normal force acting upwards (opposite to \(mg\)).
  - The frictional force acting towards the center of the circular path (providing centripetal force).
  - While the actual image isn't displayed here, ensure your diagram correctly represents this.

- **Part b:** To derive the expression for the coefficient of friction, we use the concept that the frictional force must provide the necessary centripetal force for circular motion. 

  The centripetal force \(F_c\) needed to keep the car moving in a circle of radius \(r\) at speed \(v\) is given by:
  \[
  F_c = \frac{mv^2}{r}
  \]
  Where:
  - \(m\) = mass of the car
  - \(v\) = speed of the car
  - \(r\) = radius of the circular path

  The frictional force \(F_f\) providing this centripetal force is:
  \[
  F_f = \mu N
  \]
  Where:
  - \(\mu\) = coefficient of friction
  - \(N\) = normal force

  On a flat road, the normal force \(N\) equals the gravitational force (\(mg\)). Therefore:
  \[
Transcribed Image Text:### Physics Problem on Circular Motion and Friction **Problem Statement:** 1. A car moving at \(12.0 \, \text{m/s}\) tries to negotiate a corner in a circular roadway of radius \(15.0 \, \text{m}\). The roadway is flat. a) Draw a free body diagram for the car as it makes this circular turn. ![Free Body Diagram](image-url) *(The diagram should show the car turning on a flat road with forces acting on it.)* In the figure accompanying this prompt, a car is illustrated negotiating a circular turn. The car follows a curved path, indicting its direction of motion. b) Derive the algebraic expression for the coefficient of friction between the wheels and the road if the car is to not skid. c) How does the mass of the car affect the coefficient of friction? d) Calculate the coefficient of friction. **Detailed Explanation:** - **Part a:** The free body diagram of the car should show: - The gravitational force acting downwards (\(mg\)). - The normal force acting upwards (opposite to \(mg\)). - The frictional force acting towards the center of the circular path (providing centripetal force). - While the actual image isn't displayed here, ensure your diagram correctly represents this. - **Part b:** To derive the expression for the coefficient of friction, we use the concept that the frictional force must provide the necessary centripetal force for circular motion. The centripetal force \(F_c\) needed to keep the car moving in a circle of radius \(r\) at speed \(v\) is given by: \[ F_c = \frac{mv^2}{r} \] Where: - \(m\) = mass of the car - \(v\) = speed of the car - \(r\) = radius of the circular path The frictional force \(F_f\) providing this centripetal force is: \[ F_f = \mu N \] Where: - \(\mu\) = coefficient of friction - \(N\) = normal force On a flat road, the normal force \(N\) equals the gravitational force (\(mg\)). Therefore: \[
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