Problem 41 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A sports car crosses the bottom of a valley with a radius of curvature equal to \(95 \mathrm{~m}\). At the very bottom, the normal force on the driver is twice his weight. At what speed was the car traveling?

Step-by-Step Solution

Answer

The speed of the car was approximately 30.53 m/s.

Step 1: Identify Given Information

The radius of curvature, \( r \), of the valley is \( 95 \text{ m} \). At the very bottom, the normal force \( F_n \) on the driver is twice his weight \( mg \), where \( m \) is the mass of the driver and \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \)). We need to find the speed \( v \) of the car.

Step 2: Draw Force Diagram and Apply Newton's Second Law

At the bottom of the valley, the forces acting on the driver are the gravitational force \( mg \) directed downwards, and the normal force \( F_n \), directed upwards. According to the problem, \( F_n = 2mg \). According to Newton's Second Law for circular motion, \( F_n - mg = \frac{mv^2}{r} \).

Step 3: Set Up the Equation

Replace \( F_n \) in the equation with \( 2mg \): \[ 2mg - mg = \frac{mv^2}{r} \] Simplifying gives: \[ mg = \frac{mv^2}{r} \]

Step 4: Solve for Velocity

We cancel out the mass \( m \) from both sides since it is non-zero, which gives: \[ g = \frac{v^2}{r} \] Now solve for \( v \) using the given radius \( r = 95 \text{ m} \) and \( g = 9.8 \text{ m/s}^2 \): \[ v^2 = gr \] \[ v = \sqrt{gr} \] Substitute \( g \) and \( r \): \[ v = \sqrt{9.8 \times 95} \text{ m/s} \]

Step 5: Calculate the Result

Perform the calculation: \[ v = \sqrt{931} \approx 30.53 \text{ m/s} \] Thus, the speed of the car is approximately \( 30.53 \text{ m/s} \).

Key Concepts

Normal Force

In physics, the normal force is a type of contact force. It is perpendicular to the surface and acts to support the weight of an object in contact with it. Picture yourself standing on the ground. The ground pushes back up against you with a force. This reaction is called the normal force.

### Important Characteristics of Normal Force

• It is perpendicular to the contact surface.
• It balances the weight of an object when placed on a flat surface.
• Its magnitude can change depending on the motion of the object or the surface.

In the given exercise about the sports car, the normal force at the valley's bottom is twice the weight of the driver. This shows that the car's speed impacts the normal force significantly at the bottom of its motion cycle.

Newton's Second Law

Newton's Second Law describes how the velocity of an object changes when it is subjected to an external force. It is often formulated as the equation:

• Force = mass × acceleration or, in symbols, \( F = ma \).

This law helps us understand how, for example, the speed of the car is connected to the forces acting on it when it moves in circular motion.

### Applying Newton's Second Law to Circular MotionIn the exercise, Newton's Second Law allows us to analyze the forces acting on the driver at the bottom of the valley. Here both gravitational force \( mg \) and normal force \( F_n \) influence the car's movement. The car's circular motion requires that the net force equals the centripetal force needed to keep the car moving in a curve. Hence, we use:

• Net force = \( F_n - mg \) = Centripetal Force \( \frac{mv^2}{r} \).

By setting these forces equal, Newton's Second Law becomes indispensable for solving the problem of finding the car's speed.

Centripetal Force

Centripetal force is the force that keeps an object moving in a circular path. It is directed towards the center of the circle. Without this force, an object would move in a straight line due to inertia.

### Key Attributes of Centripetal Force

• Essential for any kind of circular motion.
• Its magnitude depends on the object's speed, mass, and the radius of the circle.
• In the formula: \( F_c = \frac{mv^2}{r} \), where \( F_c \) is the centripetal force, \( m \) is mass, \( v \) is velocity, and \( r \) is the radius.

For the sports car problem, this force is crucial at the bottom of the valley. Here, the net force acting on the car must match the necessary centripetal force. Hence, knowing this value helps determine the speed of the car, calculated from the equation \( v = \sqrt{gr} \). Understanding centripetal force is key to grasp why the car doesn't slip out of its path while speeding through the valley.