Problem 4 - College Physics

Question

I A flat (unbanked) curve on a highway has a radius of \(220 \mathrm{~m}\). A car successfully rounds the curve at a speed of \(35 \mathrm{~m} / \mathrm{s}\) but is on the verge of skidding out. (a) If the coefficient of static friction between the car's tires and the road surface were reduced by a factor of \(2,\) with what maximum speed could the car round the curve? (b) Suppose the coefficient of friction were increased by a factor of 2; what would be the maximum speed?

Step-by-Step Solution

Answer

(a) The maximum speed is approximately 24.56 m/s. (b) The maximum speed is approximately 49.13 m/s.

Step 1: Calculate the Original Friction Force

First, determine the force of static friction that allows the car to round the curve without skidding. We start by knowing that the centripetal force required to keep the car on the curve is provided by the frictional force. The centripetal force is given by \[ F_c = \frac{mv^2}{r} \]where \( m \) is the mass of the car, \( v = 35 \, \text{m/s} \), and \( r = 220 \, \text{m} \). Since the car is on the verge of skidding, the static frictional force is at its maximum, \[ f_s = \mu_s m g \]where \( \mu_s \) is the static friction coefficient and \( g \approx 9.8 \, \text{m/s}^2 \). Set these equal:\[ \mu_s m g = \frac{mv^2}{r} \]

Step 2: Simplify the Equation

Simplify the equation for \( \mu_s \) by canceling the mass \( m \) on both sides, which gives:\[ \mu_s g = \frac{v^2}{r} \] Solving for \( \mu_s \), we find:\[ \mu_s = \frac{v^2}{rg} \]Plug in the given values:\[ \mu_s = \frac{35^2}{220 \times 9.8} \approx 0.56 \]

Step 3: Part (a): Reduced Coefficient by a Factor of 2

If the coefficient of static friction is reduced by a factor of 2, it becomes:\[ \mu'_s = \frac{0.56}{2} = 0.28 \]Substitute back into the centripetal force equation to find the new maximum speed \( v' \):\[ \mu'_s g = \frac{v'^2}{r} \]\[ 0.28 \times 9.8 = \frac{v'^2}{220} \]Solve for \( v' \):\[ v'^2 = 0.28 \times 9.8 \times 220 \]\[ v' \approx \sqrt{603.36} \approx 24.56 \, \text{m/s} \]

Step 4: Part (b): Increased Coefficient by a Factor of 2

If the coefficient of static friction is increased by a factor of 2, it becomes:\[ \mu''_s = 2 \times 0.56 = 1.12 \]Substitute into the centripetal force formula to find the new maximum speed \( v'' \):\[ \mu''_s g = \frac{v''^2}{r} \]\[ 1.12 \times 9.8 = \frac{v''^2}{220} \]Solve for \( v'' \):\[ v''^2 = 1.12 \times 9.8 \times 220 \]\[ v'' \approx \sqrt{2413.76} \approx 49.13 \, \text{m/s} \]

Key Concepts

Centripetal Force

When a car moves around a curve, it requires a force to keep it moving along that circular path. This essential force is what we call the centripetal force. In the context of a car on a highway curve, this force is provided by static friction between the car's tires and the road surface. The formula for centripetal force is \(F_c = \frac{mv^2}{r}\), where \(m\) is the mass of the object, \(v\) is its velocity, and \(r\) is the radius of the circle the car is negotiating. This force must be directed towards the center of the circle.

• It is always perpendicular to the velocity of the car.
• Without sufficient centripetal force, the car would not be able to stay on the curve.

These principles are vital for understanding how vehicles maintain traction while cornering, especially at higher speeds.

Static Friction

Static friction is what prevents the car from skidding as it rounds the curve. Unlike kinetic friction which acts on moving objects, static friction acts on objects that are stationary relative to each other. This is why even though the car is moving, the point of contact between the road and the tires is not sliding, using static friction. \(f_s = \mu_s mg\) shows the maximum static friction force, where \(\mu_s\) is the coefficient of static friction, \(m\) is mass, and \(g\) is the acceleration due to gravity (approximately \(9.8 \text{m/s}^2\)).

• When static friction is at its maximum, it equals the centripetal force.
• The higher the coefficient of static friction, the faster a car can travel without skidding.

In real-world scenarios, things like tire tread and road conditions will affect this coefficient, influencing how safely a vehicle can navigate curves.

Centripetal Acceleration

Centripetal acceleration refers to the rate of change of the velocity of an object moving in a circular path. Despite the speed of the object being constant, its direction changes continuously, which means it is accelerating. The formula for centripetal acceleration is \(a_c = \frac{v^2}{r}\), where \(v\) is the linear speed of the object and \(r\) is the radius. This acceleration points towards the center of the circle the object is traveling along. Key highlights include:

• Centripetal acceleration is what keeps the car moving in a circle, rather than flying off tangentially.
• It is directly proportional to the square of the velocity; hence if the speed doubles, the acceleration increases fourfold.

Recognizing the role of centripetal acceleration helps in designing safe curves and optimizing speeds for vehicles on roads or tracks.

Kinematics

Kinematics is a core concept in physics that deals with the motion of objects. In the context of cars on a curve, it encompasses the study of their velocity, acceleration, and how different forces act upon them. Basic kinematic equations are used to describe these motions and calculate various parameters without direct regard to the forces causing the motion. Here are some kinematic principles relevant to our problem:

• Speed describes how fast an object is moving, while velocity includes both speed and direction.
• Acceleration is the rate of change of velocity. For objects in uniform circular motion, centripetal acceleration is the primary concern.

With a solid grasp of kinematics, we can predict the behavior of objects in motion, make calculations regarding distances and speeds, and understand how and why objects move the way they do. Kinematics serves as a fundamental building block for more complex topics, including dynamics and mechanics.