Question

A curve in a stretch of highway has radius \(R\). The road is banked at angle \(\theta\) to the horizontal. The coefficient of static friction between the tires and road is \(\mu_{\mathrm{s}} .\) What is the fastest speed that a car can travel through the curve?

Short answer

Answer: The maximum speed is given by the formula \(v = \sqrt{\mu_s g \sin{\theta} \cdot R}\).

Step-by-step solution

Identify forces acting on the car

There are three main forces acting on the car: 1. The gravitational force \(F_g = mg\), where \(m\) is the mass of the car and \(g\) is the acceleration due to gravity. 2. The normal force \(F_N\), exerted by the road on the car. It acts perpendicular to the road surface. 3. The static friction force \(F_{friction}\), which prevents the car from slipping as it moves along the curve. It acts parallel to the road surface and has a maximum value of \(F_{max} = \mu_s F_N\).

Set up a coordinate system and find expressions for the net force in each direction

Let's choose a coordinate system such that the positive \(x\)-axis is aligned along the banked curve and the positive \(y\)-axis is perpendicular to the banked curve (i.e., in the vertical direction). Now, we can resolve the forces acting on the car into components along \(x\) and \(y\) directions. In the \(x\)-direction, \(F_{x} = F_{friction}\), which has a maximum value of \(F_{max} = \mu_s F_N\). In the \(y\)-direction, \(F_{y} = F_N - F_g \sin{\theta}\). We can write this as \(F_N = F_g \sin{\theta} + F_{y}\).

Apply Newton's second law for the forces

Now we apply Newton's second law of motion in the \(x\)-direction and the constraint of circular motion in the \(y\)-direction. In the \(x\)-direction, we have \(F_{x} = m a_x\), where \(a_x\) is the acceleration in the \(x\)-direction. Since the car is travelling in a circle of radius \(R\) with speed \(v\), we have \(a_x = \frac{v^2}{R}\). Thus, we get \(F_{max} = m \frac{v^2}{R}\). In the \(y\)-direction, we have \(F_{y} = m a_y\). Since the car is moving in a horizontal circle, it is not accelerating in the vertical direction, so \(a_y = 0\). From step 2, we get \(F_N = F_g \sin{\theta}\).

Solve for the maximum speed

From step 3, we have two equations: 1. \(F_{max} = m \frac{v^2}{R}\) 2. \(F_N = F_g \sin{\theta}\) We also know that \(F_{max} = \mu_s F_N\), so we can write: \(\mu_s F_N = m \frac{v^2}{R}\) Substituting the second equation into the first, we get: \(\mu_s (F_g \sin{\theta}) = m \frac{v^2}{R}\) Since \(F_g = mg\), we can substitute it into the equation: \(\mu_s (mg \sin{\theta}) = m \frac{v^2}{R}\) Now, we can solve for the maximum speed \(v\): \(v^2 = \mu_s g \sin{\theta} \cdot R\) Taking the square root of both sides, we get: \(v = \sqrt{\mu_s g \sin{\theta} \cdot R}\) The fastest speed that a car can travel through the curve is \(v = \sqrt{\mu_s g \sin{\theta} \cdot R}\).