Problem 34 - Physics for Scientists and Engineers with Modern Physics

Question

(I) What is the maximum speed with which a \(1200-\mathrm{kg}\) car can round a turn of radius \(80.0 \mathrm{~m}\) on a flat road if the coefficient of friction between tires and road is \(0.65 ?\) Is this result independent of the mass of the car?

Step-by-Step Solution

Answer

The maximum speed is approximately 22.6 m/s and is independent of the car's mass.

Step 1: Identify the known quantities

We are given the following values:- Mass of the car, \( m = 1200 \; \text{kg} \)- Radius of the turn, \( r = 80.0 \; \text{m} \)- Coefficient of friction, \( \mu = 0.65 \).We need to find the maximum speed of the car and determine if this speed depends on the mass of the car.

Step 2: Determine the forces involved

The frictional force provides the necessary centripetal force for the car to round the turn without skidding. The maximum frictional force can be expressed as:\[ F_f = \mu \cdot F_N \]where \( F_N \) is the normal force, equal to the gravitational force \( F_g = m \cdot g \), with \( g = 9.8 \; \text{m/s}^2 \). Therefore:\[ F_f = \mu \cdot m \cdot g \]

Step 3: Express the centripetal force required

The centripetal force required to keep the car moving in a circle of radius \( r \) is given by:\[ F_c = \frac{m \cdot v^2}{r} \]where \( v \) is the speed of the car.

Step 4: Set the frictional force equal to the centripetal force

For the car to round the turn without skidding, the maximum frictional force must equal the centripetal force:\[ \mu \cdot m \cdot g = \frac{m \cdot v^2}{r} \]

Step 5: Solve for the maximum speed

Since we set the frictional and centripetal forces equal, the mass \( m \) cancels out:\[ \mu \cdot g = \frac{v^2}{r} \]Rearranging for \( v \), we get:\[ v = \sqrt{\mu \cdot g \cdot r} \]Substituting the known values:\[ v = \sqrt{0.65 \cdot 9.8 \cdot 80.0} \approx \sqrt{509.6} \approx 22.6 \; \text{m/s} \]

Step 6: Conclusion on the mass dependency

The derivation shows that the mass \( m \) cancels out in the equation for maximum speed \( v \). Therefore, the maximum speed is independent of the mass of the car.

Key Concepts

Centripetal Force

In circular motion, centripetal force acts as the inward force required to keep an object moving in a circular path.
The term "centripetal" comes from Latin words meaning "center seeking." This force doesn't exist by itself but is instead provided by other forces, like gravitational or frictional forces, depending on the scenario.

For a car moving around a turn, the centripetal force is provided by the friction between the tires and the road. Without this force, the car would continue in a straight line due to inertia. The magnitude of the necessary centripetal force can be calculated using the formula:

• \( F_c = \frac{m \cdot v^2}{r} \)

This equation tells us that centripetal force is directly proportional to the mass \( m \), the square of the velocity \( v \), and inversely proportional to the radius \( r \) of the path.
It highlights that a higher speed or a sharper turn (smaller radius) increases the demand for centripetal force. On a straight road, there wouldn’t be any centripetal force required since the path isn’t curved.

Coefficient of Friction

The coefficient of friction \( \mu \) is a dimensionless number representing the frictional force between two surfaces.
When a car takes a turn on a flat road, the static friction between the tires and the road provides the necessary centripetal force. This means the force that prevents the tires from sliding off the road.

• The maximum frictional force can be expressed by \( F_f = \mu \cdot F_N \)

where \( F_N \) is the normal force, which for a flat surface, is equal to the gravitational force on the car. The friction here is static because it's preventing sliding and isn't dependent on relative motion at that moment.

The coefficient of friction depends on the surface materials and their texture.
A higher value of \( \mu \) indicates a stronger grip, allowing for higher speeds in turns without skidding. In this problem scenario, \( \mu = 0.65 \) suggests a moderate grip level, typical for a dry asphalt road.

Mass Dependency in Circular Motion

You might think that a heavier car would need more force to turn, but in terms of maximum speed achievable safely around a turn, mass doesn’t play a direct role.
The mass dependency cancels out when calculating the speed because it appears in both the frictional force and the centripetal force equations.

From our previous steps, this is shown as follows:

• \( \mu \cdot m \cdot g = \frac{m \cdot v^2}{r} \)

Here the mass \( m \) is present on both sides of the equation, allowing it to cancel out:

• \( \mu \cdot g = \frac{v^2}{r} \)

This result implies that, for a fixed coefficient of friction and road curvature, the mass of the car doesn't affect the maximum speed for a given turn. However, in situations where additional factors like aerodynamic drag or tire deformation play a role, mass might still have indirect effects.