Question

A curve in a highway has radius of curvature \(120 \mathrm{m}\) and is banked at \(3.0^{\circ} .\) On a day when the road is icy, what is the safest speed to go around the curve?

Short answer

The safest speed is approximately 7.84 m/s.

Step-by-step solution

Understanding the Problem

We are tasked with finding the safest speed to drive around a banked curve with a specific angle and without any friction (as the road is icy). The parameters given are: radius of curvature \( R = 120 \mathrm{m} \) and banking angle \( \theta = 3.0^{\circ} \). The absence of friction means we will only consider the centripetal force and the gravitational component influenced by the banking.

Setting Up the Equation

For a banked curve without friction, the normal force's component provides the necessary centripetal force. Balancing forces allows us to derive: \[ v = \sqrt{Rg \tan \theta} \] where \( g = 9.8 \mathrm{m/s^2} \) is the acceleration due to gravity.

Calculate the Tangent of the Banking Angle

We first need \( \tan(\theta) \) for the banking angle \( \theta = 3.0^{\circ} \). Convert \( \theta \) to radians: \( \theta = \frac{3.0 \times \pi}{180} \approx 0.05236 \) radians. Then calculate \( \tan(0.05236) \approx 0.05241 \).

Applying the Formula

Substitute \( R = 120 \mathrm{m} \), \( g = 9.8 \mathrm{m/s^2} \), and \( \tan(\theta) = 0.05241 \) into the equation: \[ v = \sqrt{120 \times 9.8 \times 0.05241} \]

Calculate the Safe Speed

Perform the calculation: \[ v = \sqrt{120 \times 9.8 \times 0.05241} \approx \sqrt{61.5888} \approx 7.84 \mathrm{m/s} \].

Key concepts

Centripetal Force

Centripetal force is a fundamental concept when it comes to objects moving in a circular path. It is the force required to keep a body moving in that path rather than flying off in a straight line. For a vehicle making a turn on a curved road, this force acts towards the center of the curve, continuously changing the vehicle's direction. The centripetal force can be calculated using the formula:

• \( F_c = \frac{mv^2}{R} \)

Here, \( m \) is the mass of the vehicle, \( v \) is its velocity, and \( R \) is the radius of curvature of the path. In the context of a banked curve, the components of gravity and normal force combine to provide this necessary centripetal force. Thus, understanding how they interact is crucial for calculating the safe speed on a banked curve.

Radius of Curvature

The radius of curvature is the radius of the circle that best fits the curved path a vehicle follows. When driving on a banked curve, such as a highway ramp, its radius of curvature denotes how sharp the turn is. A smaller radius implies a sharper turn, which requires more centripetal force to navigate safely.
In this case, the radius of curvature is \(120 \text{ m}\). This value is crucial because it directly affects the formula used to determine the vehicle's safe speed:

• \( v = \sqrt{Rg \tan \theta} \)

As this radius increases, the resulting speed at which the vehicle can safely navigate the curve also increases, showing the importance in determining safe driving parameters.

Banking Angle

The banking angle \( \theta \) is the incline of the road compared to the horizontal plane. It plays a key role in how a vehicle can safely round a curved path, especially when the road conditions are icy or slick. In the absence of friction, the banking angle helps provide the required centripetal force to keep the vehicle safely on its path.

• The formula \( \tan(\theta) \) allows us to calculate the needed component of the gravitational force along with the banking to maintain the necessary centripetal force.

For this scenario, the banking angle is \(3.0^{\circ}\). It's a relatively slight incline but enough to significantly influence the dynamics of the vehicle's motion around the curve.

Gravitational Component

The gravitational component affects how a vehicle navigates a banked curve. With the gravitational force acting downwards, its effect on a slope can be broken into components parallel and perpendicular to the road surface. For banked roads, this component provides part of the centripetal force needed for the vehicle to navigate the curve safely, especially when friction is negligible, as on an icy road.

• The gravitational component can be expressed as \( mg \sin(\theta) \), where \( m \) is mass, \( g \) is gravity, and \( \theta \) is the banking angle.

This component, combined with the normal force, creates the net force opposing the vehicle's inertial tendency to maintain straight-line motion.

Acceleration Due to Gravity

Acceleration due to gravity, represented as \( g \), is a constant value \(9.8 \text{ m/s}^2\) on Earth's surface. It influences all objects in free-fall and plays an integral role in the dynamics of banked curves.
This gravitational force is utilized when calculating the centripetal force needed for a vehicle to safely traverse a curve. In the equation

• \( v = \sqrt{Rg \tan \theta} \)

where \( v \) is the velocity, \( R \) is the radius of curvature, and \( \theta \) is the banking angle, gravity is key to determining the maximum safe speed that allows a vehicle to maintain its path without skidding off. This highlights gravity's crucial role in the design of safe, banked roadways.