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Ch 08: Dynamics II: Motion in a Plane
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 08: Dynamics II: Motion in a PlaneProblem 37
Chapter 8, Problem 37

A car can just barely turn a corner on an unbanked road at 45 km/h on a dry sunny day. What is the car's maximum cornering speed on a rainy day when the coefficient of static friction has been reduced by 50%?

Verified step by step guidance
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Step 1: Begin by understanding the relationship between the maximum speed a car can take a corner and the coefficient of static friction. The centripetal force required for the car to turn is provided by the frictional force, which is given by \( F_{friction} = \mu_s \cdot m \cdot g \), where \( \mu_s \) is the coefficient of static friction, \( m \) is the mass of the car, and \( g \) is the acceleration due to gravity.
Step 2: The centripetal force is also expressed as \( F_{centripetal} = \frac{m \cdot v^2}{r} \), where \( v \) is the speed of the car and \( r \) is the radius of the turn. Equating the frictional force to the centripetal force gives \( \mu_s \cdot m \cdot g = \frac{m \cdot v^2}{r} \). Notice that the mass \( m \) cancels out, simplifying the equation to \( \mu_s \cdot g = \frac{v^2}{r} \).
Step 3: Solve for \( v \), the maximum speed, using the equation \( v = \sqrt{\mu_s \cdot g \cdot r} \). This equation shows that the maximum speed depends on the coefficient of static friction, the radius of the turn, and the acceleration due to gravity.
Step 4: On a dry sunny day, the car's maximum speed is given as 45 km/h. Convert this speed to meters per second for consistency in units: \( 45 \, \text{km/h} = \frac{45 \cdot 1000}{3600} \, \text{m/s} \). Use this speed to calculate the original coefficient of static friction \( \mu_s \) by rearranging the formula \( \mu_s = \frac{v^2}{g \cdot r} \).
Step 5: On a rainy day, the coefficient of static friction is reduced by 50%, so \( \mu_s \) becomes \( 0.5 \cdot \mu_s \). Substitute this reduced coefficient into the formula \( v = \sqrt{\mu_s \cdot g \cdot r} \) to find the new maximum speed. Ensure all units are consistent during calculations.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Force

Centripetal force is the net force required to keep an object moving in a circular path, directed towards the center of the circle. For a car turning a corner, this force is provided by the friction between the tires and the road. The maximum speed at which the car can turn without skidding depends on the available frictional force, which is influenced by the road conditions.
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Coefficient of Friction

The coefficient of friction is a dimensionless scalar value that represents the frictional force between two surfaces. It varies with the materials in contact and the conditions (e.g., dry vs. wet surfaces). A lower coefficient indicates reduced friction, which affects the maximum speed a vehicle can achieve while turning, as seen when the road is wet and the coefficient is halved.
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Kinematic Equations for Circular Motion

Kinematic equations for circular motion relate the speed of an object moving in a circle to the radius of the circle and the forces acting on it. The maximum speed for a car turning on a flat surface can be calculated using the formula v = √(μgR), where μ is the coefficient of friction, g is the acceleration due to gravity, and R is the radius of the turn. Changes in the coefficient of friction directly affect the maximum speed.
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Related Practice
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