Question

A flat (unbanked) curve on a highway has a radius of 170.0 m. A car rounds the curve at a speed of 25.0 m/s. (a) What is the minimum coefficient of static friction that will prevent sliding? (b) Suppose that the highway is icy and the coefficient of static friction between the tires and pavement is only one- third of what you found in part (a). What should be the maximum speed of the car so that it can round the curve safely?

Short answer

(a) Minimum coefficient of static friction: 0.375. (b) Maximum speed on ice: 14.5 m/s.

Step-by-step solution

Understanding the Forces

To solve this problem, we need to consider that for a car to round a curve without sliding, the static frictional force must be at least equal to the required centripetal force to keep the car moving in a circle. This condition prevents sliding.

Calculate Required Centripetal Force

The centripetal force necessary to keep the car moving in a circle is given by the expression \( f_c = \frac{mv^2}{r} \), where \( m \) is the mass of the car, \( v \) is the velocity (25 m/s), and \( r \) is the radius of the curve (170 m). Although the mass isn’t specified, it will cancel out later.

Express Static Friction Force

The static frictional force \( f_s \) can be expressed as \( f_s = \mu_s F_n = \mu_s mg \), where \( \mu_s \) is the coefficient of static friction, \( F_n \) is the normal force, and \( mg \) is the weight of the car.

Set Up the Equation for Static Friction

For the car not to slide, the frictional force must equal the centripetal force: \( \mu_s mg = \frac{mv^2}{r} \). Cancelling \( m \) from both sides, we solve for \( \mu_s \): \( \mu_s = \frac{v^2}{rg} \).

Calculate Minimum Coefficient of Static Friction

Substitute \( v = 25 \text{ m/s} \), \( r = 170 \text{ m} \), and \( g = 9.8 \text{ m/s}^2 \) into the equation \( \mu_s = \frac{v^2}{rg} \): \( \mu_s = \frac{(25)^2}{170 \times 9.8} \approx 0.375 \).

Adjust for Reduced Coefficient

Given that the icy situation provides a coefficient of static friction that is one-third of \( 0.375 \), the effective \( \mu_{ice} = \frac{0.375}{3} = 0.125 \).

Calculate Maximum Safe Speed on Ice

To find the maximum speed \( v_{max} \) that the car can travel on ice, we use \( \mu_{ice} mg = \frac{mv_{max}^2}{r} \). Solving for \( v_{max} \), we get \( v_{max} = \sqrt{\mu_{ice} rg} \). Substitute \( \mu_{ice} = 0.125 \): \( v_{max} = \sqrt{0.125 \times 170 \times 9.8} \approx 14.5 \text{ m/s} \).

Key concepts

Centripetal Force

When a car makes a turn on a circular track, it requires a force to keep it moving along that curved path. This force, known as centripetal force, acts towards the center of the circle. It's responsible for changing the direction of the car's velocity, allowing it to continue in a circular motion.
To compute the centripetal force (\[ f_c \]), we use the formula \[ f_c = \frac{mv^2}{r} \]. In this equation, \( m \) represents the mass of the car, \( v \) is its speed, and \( r \) is the radius of the curve. Although the mass doesn't directly affect the determination of the required coefficient of friction, it's crucial to note that this force essentially compels the car to seek the center of the circle as it moves around the curve.
In the exercise, the centripetal force is counteracted by the static frictional force coming from the contact between the car's tires and the road. If the frictional force weren't present or sufficient, the car would fail to follow the curve and slide away from the circular path.

Coefficient of Friction

The coefficient of friction (\( \mu \)) is a value that quantifies the frictional forces between two surfaces. In this exercise, we are dealing with the static coefficient of friction, which applies when there isn’t relative motion between the contact surfaces yet.
The static frictional force (\( f_s \)) is expressed as \[ f_s = \mu_s F_n \], where \( F_n = mg \) is the normal force, or the perpendicular force that a surface exerts on an object. For the car not to slide off the curve, this static frictional force must be at least equal to the centripetal force. Thus, we have \( \mu_s mg = \frac{mv^2}{r} \), leading to \( \mu_s = \frac{v^2}{rg} \).
In scenarios where surfaces may be less conducive to friction, such as an icy road, the coefficient drops. Consequently, the speed at which the car can handle the turn safely decreases. This concept is crucial, not just for understanding physics problems, but also for real-world driving safety.

Circular Motion

Circular motion refers to the movement of an object along a circular path. In our problem, the car travels around a curved section of a highway. Maintaining this circular motion requires constant acceleration towards the center, known as centripetal acceleration. It simplifies to \( a_c = \frac{v^2}{r} \), resulting in the necessary centripetal force.
For real-world applications, the role of static friction in providing enough grip to allow vehicles to complete such curves cannot be overstated. The static friction must match or exceed the force needed to change the vehicle's straightforward velocity into a circular path.

• Vehicles must adjust speed, as determined by \( v = \sqrt{\mu r g} \), to safely maneuver around centers with differing coefficients of friction.
• The concept demonstrates the delicate balance necessary for achieving safe transit during turns.

By understanding the dynamics of circular motion, predictions can be made regarding how alterations in speed or surface condition will affect the path and safety of the transit. Such insights aid in designing safer roadways and vehicles.