Question

A car starts from rest and accelerates around a flat curve of radius \(R=36 \mathrm{~m}\). The tangential component of the car's acceleration remains constant at \(a_{\mathrm{t}}=3.3 \mathrm{~m} / \mathrm{s}^{2},\) while the centripetal acceleration increases to keep the car on the curve as long as possible. The coefficient of friction between the tires and the road is \(\mu=0.95 .\) What distance does the car travel around the curve before it begins to skid? (Be sure to include both the tangential and centripetal components of the acceleration.)

Short answer

Answer: The car travels approximately 51.56 meters around the curve before it begins to skid.

Step-by-step solution

Calculate the frictional force

The frictional force can be calculated using the following formula: \(F_f = \mu m_{car}g\) Since we are not given the mass of the car, we cannot calculate the exact value of the frictional force, but we can relate the centripetal and tangential components of the force acting on the car.

Calculate the total force and acceleration on the car

The total force on the car is a vector sum of centripetal and tangential forces: \(F_t = F_c + F_t = \mu m_{car}g\) The centripetal force is given by \(F_c = m_{car}a_c\), where \(a_c\) is the centripetal acceleration, and the tangential force is given by \(F_t = m_{car}a_t\). Hence, we can rewrite the total force as: \(a = \sqrt{a_c^2 + a_t^2}\), where \(a\) is the total acceleration.

Find the critical speed

At the critical speed point, the frictional force becomes equal to the centripetal force needed to keep the car on the curve. Using this condition, we can find the critical speed using the following equation: \(v^2 = \mu Rg\) Substituting the given values, we get: \(v^2 = 0.95 \times 36 \times 9.81 \approx 337.22 \Rightarrow v \approx 18.37 \mathrm{m/s}\)

Calculate travel time

We will now use the critical speed and the tangential acceleration to find the time it takes for the car to reach the critical speed: \(v = a_t t \Rightarrow t = \frac{v}{a_t}\) Substituting the given values, we get: \(t = \frac{18.37}{3.3} \approx 5.57 \mathrm{s}\)

Calculate the distance traveled around the curve

To find the distance traveled around the curve, we use the following kinematic equation: \(s = \frac{1}{2}a_t t^2\) Substituting the values, we get: \(s = \frac{1}{2} \times 3.3 \times (5.57)^2 \approx 51.56 \mathrm{m}\) So, the car travels approximately 51.56 meters around the curve before it begins to skid.

Key concepts

Centripetal Acceleration

When an object moves in a circular path, it experiences a force that keeps it moving in that circle; this force is always directed towards the center of the circle, hence the term centripetal, which means 'center seeking'. This force causes an acceleration called centripetal acceleration.

Centripetal acceleration, denoted by the symbol \(a_c\), is given by the formula \(a_c = \frac{v^2}{R}\) where \(v\) is the velocity of the object, and \(R\) is the radius of the circular path.

In the context of the car problem, as the car speeds up, the velocity increases, and since the curve's radius \(R\) is constant, the centripetal acceleration must also increase in order to maintain the circular path. This explains why centripetal acceleration grows with increasing speed. When the car's centripetal acceleration surpasses the frictional force between the tires and the road, the car will begin to skid.

Coefficient of Friction

The coefficient of friction is a measure that describes how much frictional resistance is present between two surfaces in contact. It is a dimensionless quantity represented by the Greek letter \(\mu\). The value of \(\mu\) depends on the types of material in contact and can range from near zero (for very slippery surfaces) to greater than one (for very sticky surfaces).

In our car problem, the coefficient of friction between the tires and the road is 0.95, a relatively high number, indicating a good amount of grip. This coefficient is crucial as it determines the maximum centripetal force that can be sustained without skidding, which is directly tied to the frictional force \(F_f\) that can be calculated as \(F_f = \mu m_{car}g\), where \(m_{car}\) is the mass of the car and \(g\) is the acceleration due to gravity.

The implications of the coefficient of friction go beyond the maximum centripetal force; they also affect stopping distances, the ability to accelerate, and overall safety during motion.

Tangential Acceleration

While centripetal acceleration is directed towards the center of the circular path, tangential acceleration refers to the acceleration along the direction of motion - tangent to the curve. It represents how quickly the speed of the object is increasing or decreasing.

The formula for tangential acceleration, represented as \(a_t\), is simply the change in velocity over time along the tangential direction. For uniform acceleration, \(a_t = \frac{\Delta v}{\Delta t}\).

In the given car problem, the car starts from rest and accelerates around the curve with a constant tangential acceleration \(a_t = 3.3 \mathrm{m/s^2}\), which means the speed increases steadily. This tangential component determines how quickly the car can reach the critical speed at which the centripetal force equals the maximum frictional force available. The tangential acceleration is crucial for calculating the distance traveled before skidding occurs, using the kinematic equation \(s = \frac{1}{2}a_t t^2\), as seen in the solution.