Question

A speedway turn, with radius of curvature \(R\), is banked at an angle \(\theta\) above the horizontal. a) What is the optimal speed at which to take the turn if the track's surface is iced over (that is, if there is very little friction between the tires and the track)? b) If the track surface is ice-free and there is a coefficient of friction \(\mu_{s}\) between the tires and the track, what are the maximum and minimum speeds at which this turn can be taken? c) Evaluate the results of parts (a) and (b) for \(R=400 . \mathrm{m}\), \(\theta=45.0^{\circ},\) and \(\mu_{\mathrm{s}}=0.700 .\)

Short answer

Based on the given values for the radius (R), angle (θ), and coefficient of friction (μ_s), the optimal speed when the track is iced and there is no friction between the tires and the track is approximately 88.2 m/s. When friction is present, the maximum speed the car can achieve is around 154.6 m/s, while the minimum speed is approximately 69.3 m/s.

Step-by-step solution

Analyze the forces acting on the car without friction

First, we need to consider the forces acting on the car when it moves in a circle with a banked track. In this case, we have only two forces acting on the car: the gravitational force (mg) acting downward, and the normal force (N) acting perpendicular to the track surface.

Write equations for the radial and vertical directions of the motion

In the radial direction, we have the centripetal force, which is given by \(\frac{mv^2}{R}\). Also, we can write the normal force as \(N\cos(\theta)\) and the gravitational force as \(mg\sin(\theta)\). So, for the radial direction, we have: \(\frac{mv^2}{R} = N\cos(\theta)\) In the vertical direction, the car does not move, so the net force must be balanced. Therefore, we have: \(N\sin(\theta) = mg\)

Solve the equations to find the optimal speed (v)

Solve the equations from step 2 to get the optimal speed (v). First, divide the radial equation by the vertical equation: \(\frac{mv^2/R}{N\sin(\theta)} = \frac{N\cos(\theta)}{mg}\) thus, we have \(v^2 = Rg\tan(\theta)\) Now, solve for v: \(v = \sqrt{Rg\tan(\theta)}\) This is the optimal speed when there is no friction between the tires and the track. #b) Calculate the maximum and minimum speeds when there is friction#

Analyze the forces acting on the car with friction

When friction is present, we have to consider the frictional force (f) as well. The normal force (N) and gravitational force (mg) are the same as before. This time, the frictional force acts horizontally toward the center of the circle.

Write equations for the radial and vertical directions of the motion

In the radial direction, we have \(N\cos(\theta) + f = \frac{mv^2}{R}\), and in the vertical direction, we have \(N\sin(\theta) = mg - f\tan(\theta)\). Notice that this time, the frictional force also contributes to the vertical force balance.

Solve the equations to find the maximum and minimum speeds

We will solve for the maximum and minimum speeds by maximizing and minimizing the frictional force, respectively. For the maximum speed, we have: \(f_{max} = \mu_sN\) \(v_{max}^2 = Rg\tan(\theta) + Rg\frac{\mu_s}{\cos(\theta)}\) \(v_{max} = \sqrt{Rg\tan(\theta) + Rg\frac{\mu_s}{\cos(\theta)}}\) Similarly, for the minimum speed, we will have: \(v_{min}^2 = Rg\tan(\theta) - Rg\frac{\mu_s}{\cos(\theta)}\) \(v_{min} = \sqrt{Rg\tan(\theta) - Rg\frac{\mu_s}{\cos(\theta)}}\) #c) Evaluate the results for given values#

Substitute the given values into the formulas

Now, let's plug the given values for R, θ, and μ_s into the formulas derived in parts (a) and (b) for the optimal speed (v), maximum speed (v_max), and minimum speed (v_min). We have: \(R = 400 \thinspace m, \thinspace \theta= 45.0^\circ, \thinspace \mu_s = 0.700\) \(v = \sqrt{400 \cdot 9.81 \cdot \tan(45^\circ)}\) \(v_{max} = \sqrt{400 \cdot 9.81 \cdot \tan(45^\circ) + 400 \cdot 9.81 \cdot \frac{0.700}{\cos(45^\circ)}}\) \(v_{min} = sqrt{400 \cdot 9.81 \cdot \tan(45^\circ) - 400 \cdot 9.81 \cdot \frac{0.700}{\cos(45^\circ)}}\)

Calculate the results

Calculate the optimal speed (v), maximum speed (v_max), and minimum speed (v_min) using the formulas in Step 1: \(v \approx 88.2 \thinspace m/s\) \(v_{max} \approx 154.6 \thinspace m/s\) \(v_{min} \approx 69.3 \thinspace m/s\) The optimal speed for the iced over track is around 88.2 m/s, while the maximum and minimum speeds for the ice-free track are approximately 154.6 m/s and 69.3 m/s, respectively.

Key concepts

Centripetal Force

Understanding the role of centripetal force is critical when analyzing the physics of a banked curve. Centripetal force is not an independent force but refers to the 'center-seeking' component of force that keeps an object moving in a circular path. It's always directed towards the center of the circle. The formula for centripetal force is given by \( F_c = \frac{mv^2}{R} \),where \(F_c\) is the centripetal force, \(m\) is the mass of the object, \(v\) is the speed of the object, and \(R\) is the radius of curvature of the path.

In the context of a car on a banked track, the centripetal force is provided by the horizontal component of the normal force when there is no friction. If friction is present, it also contributes to centripetal force. It's essential for vehicles taking a turn on a banked curve because it allows the vehicle to take the turn at higher speeds without skidding outward. Without sufficient centripetal force, the car would not follow the curved path and would tend to move in a straight line due to inertia, potentially leading to an accident.

Frictional Force

Frictional force plays a pivotal role in real-world driving conditions. It arises due to the interaction between the surface of the track and the car's tires. The formula for the maximum frictional force that can occur without slippage is given by \(f_{max} = \mu_sN\), where \(f_{max}\) is the maximum static frictional force, \(\mu_s\) is the coefficient of static friction, and \(N\) is the normal force exerted by the track on the car.

In a banked curve, friction can help the car stay on the track by providing additional centripetal force. When the surface is icy, friction is nearly negligible, and the optimal speed depends only on gravity and the banking angle. However, on a dry track, friction becomes a crucial factor that allows a vehicle to take the curve at speeds higher or lower than the optimal speed without skidding. The maximum speed is limited by the static friction before it starts sliding, while the minimum speed is where the frictional force is needed to keep the car moving inwards and prevent it from slipping down the slope.

Normal Force

The normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface which, in the case of a banked curve, would not be vertically upwards but rather at an angle from the horizontal. The magnitude of the normal force can be altered by an object's weight (mass times the gravitational acceleration) and the angle of the bank.

When a car is on a banked curve, the normal force gets divided into two components — the vertical component balances the weight of the car, while the horizontal component provides the necessary centripetal force (in the absence of friction). A correct understanding of the normal force is essential for calculating both the optimal speed on a frictionless banked curve and the maximum and minimum speeds on a curve where friction is present. It ensures the necessary forces are met for a car to remain on the path without veering off the track or slipping.