Problem 44 - University Physics with Modern Physics

Question

A flat (unbanked) curve on a highway has a radius of 220.0 \(\mathrm{m} .\) A car rounds the curve at a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) . (a) What is the minimum coefficient of friction that will prevent sliding? (b) Suppose the highway is icy and the coefficient of friction between the tires and pavement is only one-third what you found in part (a). What should be the maximum speed of the car so it can round the curve safely?

Step-by-Step Solution

Answer

(a) Minimum coefficient of friction is 0.291. (b) Maximum speed on ice is approximately 14.69 m/s.

Step 1: Understand the Forces Involved

When a car is rounding a curve, the primary force that keeps it on the path is the frictional force. This force provides the necessary centripetal force to keep the car moving in a circle. We need to find out what this necessary centripetal force is to prevent the car from sliding.

Step 2: Calculate the Centripetal Force Required

The formula for centripetal force is given by:\[ F_c = \frac{mv^2}{r} \]where \( m \) is the mass of the car, \( v \) is the velocity (given as 25.0 m/s), and \( r \) is the radius of the curve (220.0 m). Since we don't have the car's mass \( m \), we will express it in terms of \( m \).

Step 3: Relate Centripetal Force to Friction Force

The frictional force \( F_f \) that prevents sliding is given by:\[ F_f = \mu N \]where \( \mu \) is the coefficient of friction, and \( N \) is the normal force. On a flat road, the normal force \( N = mg \), the weight of the car. Thus:\[ F_f = \mu mg \]To prevent sliding, the frictional force must be equal to the centripetal force:\[ \mu mg = \frac{mv^2}{r} \]

Step 4: Solve for the Coefficient of Friction \( \mu \)

Cancel out the mass \( m \) from both sides of the equation since it is not zero:\[ \mu g = \frac{v^2}{r} \]Now solve for \( \mu \):\[ \mu = \frac{v^2}{rg} \]Substitute \( v = 25.0 \) m/s, \( r = 220.0 \) m, and \( g = 9.8 \) m/s²:\[ \mu = \frac{(25.0)^2}{220.0 \times 9.8} \approx 0.291 \]

Step 5: Determine the Reduced Coefficient Due to Icy Conditions

With icy conditions, the coefficient of friction is one-third the calculated coefficient in part (a):\[ \mu_{icy} = \frac{1}{3} \times 0.291 \approx 0.097 \]

Step 6: Calculate the Maximum Speed with Reduced Friction

Using the relationship between frictional force and centripetal force on ice, solve for velocity \( v \):\[ \mu_{icy} mg = \frac{mv_{max}^2}{r} \]Solving for \( v_{max} \):\[ v_{max} = \sqrt{\mu_{icy}rg} \]Substitute \( \mu_{icy} = 0.097 \), \( r = 220.0 \) m, and \( g = 9.8 \) m/s²:\[ v_{max} = \sqrt{0.097 \times 220.0 \times 9.8} \approx 14.69 \] m/s.

Key Concepts

Coefficient of Friction

The coefficient of friction, denoted as \( \mu \), is a dimensionless number that represents the amount of friction between two surfaces. It plays a crucial role in determining how easily one object can slide over another. In our problem, it's vital for calculating how much frictional force is needed to prevent a car from sliding while taking a curve on a highway.

• It is calculated by the ratio of the frictional force to the normal force: \( \mu = \frac{F_f}{N} \).
• There are two main types of friction coefficients to be aware of: static friction (before sliding begins) and kinetic friction (once sliding has started).

For the car to round the curve without slipping, a minimum coefficient of friction is required. This minimum value ensures that the friction can provide the necessary centripetal force. In the exercise, we calculated that at a speed of 25.0 m/s and a curve radius of 220.0 m, the coefficient of friction must be around 0.291. If the road becomes icy, this coefficient drops significantly, making it harder to maintain grip without sliding.

Frictional Force

Frictional force is the resistance force that acts opposite to the direction of motion of a body. It is a result of the interaction between two surfaces in contact. In our scenario, this is the force that must provide the necessary centripetal force to keep the car moving in a circle.

• The frictional force is given by the formula: \( F_f = \mu N \), where \( N \) is the normal force, and \( \mu \) is the coefficient of friction.
• For a car on a curve, this force must equal the centripetal force needed to keep the car on its circular path.
• The frictional force must be strong enough to counteract the car's tendency to slide outward due to inertia.

Changes in friction, such as those due to icy conditions, decrease the frictional force, presenting a risk of sliding unless adjustments are made to the speed or curve radius. In part (b) of our problem, the iced road conditions reduce the friction force by one-third, impacting the maximum speed the car can safely maintain.

Normal Force

The normal force is a support force exerted upon an object that is in contact with another stable object. In the context of a car on a flat road, the normal force is equal to the gravitational force acting on the car, keeping it from falling through the road surface.

• The normal force \( N \) is often equal to the gravitational force on a flat surface: \( N = mg \).
• It acts perpendicular to the surface contact, hence its name "normal" (meaning perpendicular).
• In the calculation of frictional force, the normal force is vital, as it directly influences the maximum friction that can be produced: \( F_f = \mu N \).

In our case with the highway curve, understanding the normal force helps us determine how much friction can potentially be generated to keep the car on its path. The magnitude of the normal force is unchanged by elaborate conditions like icy roads, though the effect of friction changes dramatically with variations in the coefficient of friction.