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 \(\mu\), ss 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 \(\mu_{s}=0.700\)

Short answer

The optimal speed for a banked turn with a radius of curvature of 400m, an angle of 45 degrees, and an iced-over surface is approximately 88.4 m/s. The maximum speed when there is friction with a coefficient of 0.7 is approximately 131.4 m/s, and the minimum speed is approximately 44.7 m/s.

Step-by-step solution

(a) Determine the condition for an optimal speed

When the track is iced over, there is no friction between the tires and the track. In this case, the optimal speed is when the car is in equilibrium and only the gravitational force and the normal force are acting on it. We need to find the centripetal force that keeps the car moving in a circle with radius R and analyze the vertical and horizontal components of the gravitational force and normal force.

(a) Use free-body diagram and Newton's second law to find a relationship between forces

Analyze the forces acting on the car by drawing a free-body diagram. There is the normal force \(N\) acting perpendicular to the plane, and the weight of the car \(mg\) acting downward. Decompose the normal force and the gravitational force into horizontal and vertical components. The vertical component of the normal force is \(N \sin(\theta)\) and the horizontal component is \(N \cos(\theta)\). The centripetal force is given by \(mv^2/R\), where v is the speed of the car. Using Newton's second law, we can set up the force balance equations: - For vertical force components: \(N \sin(\theta) = mg\) - For horizontal force components: \(N \cos(\theta) = \dfrac{mv^2}{R}\)

(a) Solve for the optimal speed in the icy condition

Divide both equations to eliminate the normal force, resulting in \(\tan(\theta) = \dfrac{v^2}{gR}\). Then, solve for the optimal speed \(v\) using the provided angle θ and radius R: \(v = \sqrt{gR\tan(\theta)}\).

(b) Determine the maximum and minimum speeds when there is friction

When the surface is ice-free, the friction between the tires and the track play a role in determining the maximum and minimum speeds at which the turn can be taken. If there is friction, the horizontal component of the normal force, \(N \cos(\theta)\), adds to (or subtracts from) the frictional force \(\mu_s N\). The centripetal force equation with friction becomes: \(\dfrac{mv^2}{R} = \pm \mu_s N - N \cos(\theta)\) (with '+' sign for the maximum speed case and '-' for the minimum speed case).

(b) Calculate the maximum and minimum speeds with friction

Start with the already obtained vertical force components equation: \(N \sin(\theta) = mg\). Multiply both sides of this equation by the coefficient of friction, obtaining \(\mu_s N \sin(\theta) = \mu_s mg\). Add and subtract this equation from the modified centripetal force equation described above, and solve for the maximum and minimum speeds \(v\): - For maximum speed: \(v_{max} = \sqrt{\dfrac{Rg(\mu_s + \cos^2(\theta))}{\cos(\theta)}}\) - For minimum speed: \(v_{min} = \sqrt{\dfrac{Rg(\mu_s - \cos^2(\theta))}{\cos(\theta)}}\)

(c) Evaluate the results for given values

Given the values of \(R = 400m\), \(\theta = 45^{\circ}\), and \(\mu_s = 0.7\), plug these values into the formulas for the optimal speed (in the icy condition), maximum speed, and minimum speed, and calculate the results: 1. Optimal speed (icy condition): \(v = \sqrt{(9.8 m/s^2)(400m)(\tan(45^{\circ}))}\) 2. Maximum speed (friction): \(v_{max} = \sqrt{\dfrac{(400m)(9.8 m/s^2)(0.7+0.5)}{\cos(45^{\circ})}}\) 3. Minimum speed (friction): \(v_{min} = \sqrt{\dfrac{(400m)(9.8 m/s^2)(0.7-0.5)}{\cos(45^{\circ})}}\) Calculate the speeds, and the exercise is complete.

Key concepts

Centripetal Force

Centripetal force is crucial to understand when examining the physics of a banked curve. It is an inward force that keeps an object moving in a circular path. Its magnitude is given by the formula \( F_c = \frac{mv^2}{R} \), where \( m \) is the mass of the object, \( v \) is the tangential speed, and \( R \) is the radius of the curve. Without centripetal force, an object would continue in a straight line due to inertia. In the case of a car on a banked curve with no friction, as on an icy track, the only forces providing this centripetal force are the horizontal component of the normal force exerted by the track's surface.

In the exercise, we see that the optimal speed is achieved when only the gravitational force and the normal force act on the car. This simplified condition allows us to find the speed at which the car can navigate the curve without relying on friction - a precarious situation that would be rare and risky in real driving conditions, but is useful to understand the purely physical interaction between the car and the banked surface.

Friction in Circular Motion

Friction plays a significant role when dealing with circular motion on practical surfaces. It is the force that resists the relative motion or tendency of such motion of two surfaces in contact. In a banked curve scenario, friction can either help or hinder the centripetal force needed to keep the car on the track.

When a car takes a turn on a track, the frictional force, given by \( F_f = \$ \mu \$ N \), comes into play, where \( \$ \mu \$ \) is the coefficient of friction and \( N \) is the normal force. This force can add to (in the case of the maximum speed) or subtract from (in the case of the minimum speed) the horizontal force component of the normal force, affecting the speeds at which the car can safely navigate the turn. The balance between friction and the car's velocity becomes critical to ensure that the car doesn't slip or skid off the track.

Optimal Speed on Banked Curve

The optimal speed on a banked curve, especially one that is iced over with negligible friction, is determined by the point where the forces are in equilibrium. This balance ensures that the car stays on its circular path without sliding up, down, or off the banked surface.

The remarkable aspect of this optimal speed is that it can theoretically be reached without the use of friction—reliant solely on the banking angle of the curve and gravitational force. The formula derived in the exercise, \( v = \sqrt{gR\tan(\theta)} \), is a direct result of this balance, showing that the optimal speed increases with a steeper angle of banking (\theta) and a larger radius (\theta). It is a valuable illustration of how banking can be designed to counteract the need for friction and maintain a certain speed in curves.

Newton's Second Law in Circular Motion

Newton's second law is the fundamental principle behind the analysis of forces in a banked curve. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass, expressed as \( F = ma \). In the context of circular motion, this acceleration is centripetal and directed towards the center of the circular path.

In the exercise, we decompose forces into their vertical and horizontal components to apply Newton's second law distinctly in each direction. The car's weight acts downward, providing the vertical force, and the horizontal force comes from the track pushing perpendicularly to the car. When there's no friction, these forces must be aligned perfectly to provide the necessary centripetal force—resulting in the optimal speed for which a formula is derived. This application of Newton's second law to circular motion showcases the interplay between linear and centripetal acceleration, pivotal for understanding how objects move on curved paths.