Problem 42 - University Physics with Modern Physics

Question

A small car with mass 0.800 kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.00 m (Fig. E5.42). If the normal force exerted by the track on the car when it is at the top of the track (point \(B\) is 6.00 \(\mathrm{N},\) what is the normal force on the car when it is at the bottom of the track (point \(A ) ?\)

Step-by-Step Solution

Answer

The normal force at the bottom is approximately 21.70 N.

Step 1: Identify Forces at the Top

At the top of the track, the forces acting on the car are its weight and the normal force. At point B, the normal force is given as 6.00 N. The weight of the car is calculated using the formula \( F_g = mg \), where \( m = 0.800 \) kg and \( g = 9.81 \) m/s².

Step 2: Apply Newton's Second Law to the Top

The net force acting towards the center of the circular path at point B (top) is the difference between the gravitational force and the normal force. Using Newton's second law, we set up the equation: \( mg + N_{B} = \frac{mv^2}{r} \), where \( v \) is the speed of the car.

Step 3: Solve for Speed at Top of the Track

Plug the known values into the equation: \( (0.800)(9.81) + 6.00 = \frac{0.800v^2}{5.00} \). Simplifying this gives: \( 7.848 + 6.000 = \frac{0.800v^2}{5.00} \), hence \( 13.848 = \frac{0.800v^2}{5.00} \). Solve for \( v \) to find \( v^2 = \frac{13.848 imes 5.00}{0.800} \).

Step 4: Calculate Speed at Top

Solve the equation \( v^2 = \frac{13.848 imes 5.00}{0.800} \) to find \( v \). You get \( v^2 = 86.55 \), thus \( v = \sqrt{86.55} \approx 9.30 \) m/s.

Step 5: Identify Forces at the Bottom

At the bottom of the track (point A), both the weight and the normal force work towards the center of the circle. The equation for the net centripetal force at the bottom is \( N_{A} - mg = \frac{mv^2}{r} \).

Step 6: Solve for Normal Force at the Bottom

Substitute the values for \( m \), \( g \), \( v^2 \), and \( r \) into the equation \( N_{A} - mg = \frac{mv^2}{r} \). Therefore, \( N_{A} = \frac{0.800 imes 86.55}{5.00} + 0.800 imes 9.81 \).

Step 7: Calculate Normal Force at Bottom

The equation simplifies to \( N_{A} = 13.848 + 7.848 \). Thus, \( N_{A} = 21.696 \). The normal force at the bottom is approximately \( 21.70 \) N.

Key Concepts

Normal Force

Normal force is an important concept, especially in situations involving circular motion. It is the force exerted by a surface to support the weight of an object in contact with it. This force acts perpendicular to the surface.

In our exercise, as the car moves along the vertical circular track, the normal force changes at different points. At the top of the track, the normal force is less because part of the centripetal force needed is provided by gravity. Here, the normal force equals 6.00 N. At the bottom of the track, the normal force increases because it must counteract gravity and provide the centripetal force needed for the car's circular path. Calculations show it reaches approximately 21.70 N.

The variation in normal force demonstrates its role in balancing gravitational force and providing centripetal force, ensuring the car maintains its motion on the track.

Centripetal Force

Centripetal force is essential for any object moving in a circular path. It is the force that acts towards the center of the circle, keeping the object in motion along a curved path. Without this force, the object would move in a straight line due to inertia.

In the case of the car on the track, centripetal force is provided by the combination of gravitational force and the normal force. At the top of the track, the gravitational force helps in providing the necessary centripetal force, while at the bottom, more of the centripetal force must come from the normal force. This is a clear demonstration of how these forces work together to maintain circular motion.

The equation for centripetal force is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is mass, \( v \) is velocity, and \( r \) is the radius of the circular path. This formula helps in calculating how much of the normal and gravitational forces are devoted to the circular motion.

Gravitational Force

Gravitational force is the attraction between two masses. It is most commonly experienced as the weight of an object, which is the force exerted by gravity on that object. This force always acts downward, towards the center of the Earth.

In circular motion exercises like our car on a track, gravitational force plays dual roles. At the top of the track, it acts in the same direction as the centripetal force, reducing the need for normal force. At the bottom, however, gravitational force opposes the direction of centripetal force, meaning the normal force must increase to maintain circular motion.

Calculations for gravitational force simplify to \( F_g = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximated as \( 9.81 \text{ m/s}^2 \)). In our exercise, this becomes a crucial part in balancing forces at different positions of the track.