Cars Navigating Horizontal Circular Bends

Expert reviewed • 22 November 2024 • 9 minute read

What is the role of Centripetal Force in Circular Motion?

When a car takes a turn on a flat, horizontal road, it demonstrates a practical application of circular motion. For a car to move in a circular path around a bend, it requires a force directed towards the centre of the curve. This force is called centripetal force, and it’s crucial for maintaining circular motion.

The magnitude of centripetal force is given by:

F_c=\frac{mv^2}{r}

Where:

$F_c$ is the centripetal force
$m$ is the mass of the car
$v$ is the velocity of the car
$r$ is the radius of the circular path

Friction: The Source of Centripetal Force

In the case of a car on a horizontal bend, the centripetal force is provided by friction between the car’s tires and the road surface. Specifically, it’s static friction that keeps the car from skidding outwards.

The maximum static friction force is given by:

F_s=\mu_s N

Where:

$F_s$ is the static friction force
$\mu_s$ is the coefficient of static friction
$N$ is the normal force

On a flat road, the normal force is equal to the weight of the car:

N=mg

Where

g is the acceleration due to gravity (usually measured as $9.8m/s^2$

How are Forces Balanced to Ensure Safe Turns?

For a car to safely navigate a turn without skidding, the centripetal force required must not exceed the maximum static friction available:

F_c\leq F_s

Substituting the equations for centripetal force and static friction:

\frac{mv^2}{r} \leq \mu_s mg

How can you Determine the Maximum Speed for Safe Turning?

By rearranging the above equation, we can determine the maximum speed at which a car can safely navigate a turn:

v_{max} = \sqrt{\mu_s g r}

This equation shows that the maximum safe speed depends on:

The coefficient of static friction $(\mu_s)$
The acceleration due to gravity $(g)$
The radius of the turn $(r)$

What are the Factors Affecting Safe Turning?

Road Conditions

The coefficient of friction varies with road conditions. A dry road typically has a higher coefficient of friction than a wet or icy road, allowing for higher safe speeds.

Tire Condition

Worn tires reduce the effective coefficient of friction, decreasing the maximum safe speed for turning.

Turn Radius

Tighter turns (smaller radius) require lower speeds to navigate safely.

Vehicle Mass

While the mass doesn’t appear in the final equation for maximum speed, it does affect the required centripetal force. Heavier vehicles need more friction force to turn safely.

Practice Question 1

A car approaches a circular bend with a radius of 25 meters. The coefficient of static friction between the tires and the dry road surface is 0.8. What is the maximum speed at which the car can safely navigate this turn?

Identify the given information:

Radius $(r) = 25 m$
Coefficient of static friction $(μ_s) = 0.8$
Acceleration due to gravity $(g) = 9.8 m/s²$

Use the equation for maximum speed:

v_{max} = \sqrt{\mu_s g r}

Substitute the values:

v_{max} = \sqrt{(0.8)(9.8)(25)}\\=\sqrt{196}\\14\;\text{m/s}

Therefore, the maximum safe speed for the car to navigate this turn is approximately 14.0 m/s or 50.4 km/h.

Practice Question 2

A car is traveling at 20 m/s around a horizontal circular bend with a radius of 50 meters. What is the minimum coefficient of static friction required between the tires and the road to prevent the car from skidding?

Identify the given information:

Speed $(v) = 20 m/s$
Radius $(r) = 50 m$
Acceleration due to gravity $(g) = 9.8 m/s²$

Rearrange the maximum speed equation to solve for μs:

v = \sqrt{\mu_s g r}\\ v^2 = μ_sgr\\\mu_s = \frac{v^2}{gr}

Substitute the values:

\mu_s = \frac{(20)^2}{(9.8)(50)}

Calculate:

\mu_s = \frac{400}{490} \approx 0.82

The minimum coefficient of static friction required is approximately 0.82. This is quite high and would require dry, high-quality road conditions and tires.

Understanding Circular Motion

Up Next

Circular Motion on Banked Surfaces

Return to Module 5: Advanced Mechanics