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Ch 05: Applying Newton's Laws
Chapter 5, Problem 5

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. E5.45). 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 N, what is the normal force on the car when it is at the bottom of the track (point A)?

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1
Identify the forces acting on the car at the top and bottom of the track. At the top, the forces are gravity (downward) and the normal force (also downward). At the bottom, the forces are gravity (downward) and the normal force (upward).
Use the formula for centripetal force, which is required for circular motion: $F_c = \frac{mv^2}{r}$, where $m$ is the mass, $v$ is the velocity, and $r$ is the radius of the circle.
At the top of the track, set up the equation considering the net force acting towards the center of the circle (upward direction as positive): $F_{net,top} = N_{top} + mg = \frac{mv^2}{r}$. Solve this equation to find $v^2$.
At the bottom of the track, set up the equation considering the net force acting towards the center of the circle (downward direction as positive): $F_{net,bottom} = N_{bottom} - mg = \frac{mv^2}{r}$. Use the value of $v^2$ found from the top of the track.
Solve the equation from step 4 to find the normal force at the bottom of the track, $N_{bottom}$. This will give you the required normal force when the car is at the bottom of the track.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Force

Centripetal force is the net force acting on an object moving in a circular path, directed towards the center of the circle. For an object in uniform circular motion, this force is necessary to keep it moving along the curved path. In the context of the car on the track, the centripetal force is provided by the gravitational force and the normal force acting on the car at different points of the track.
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Normal Force

The normal force is the force exerted by a surface to support the weight of an object resting on it, acting perpendicular to the surface. In this scenario, the normal force varies depending on the position of the car on the vertical track. At the top of the track, the normal force and gravitational force together provide the necessary centripetal force, while at the bottom, the normal force must counteract gravity and provide additional centripetal force.
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Gravitational Force

Gravitational force is the attractive force between two masses, calculated using Newton's law of universal gravitation. For the car, this force acts downward and is equal to the product of its mass and the acceleration due to gravity (approximately 9.81 m/s²). Understanding the gravitational force is crucial for determining the net forces acting on the car at different points on the track, particularly how it influences the normal force at the top and bottom of the vertical circle.
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Related Practice
Textbook Question
A 52-kg ice skater spins about a vertical axis through her body with her arms horizontally outstretched; she makes 2.0 turns each second. The distance from one hand to the other is 1.50 m. Biometric measurements indicate that each hand typically makes up about 1.25% of body weight. (b) What horizontal force must her wrist exert on her hand?
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Textbook Question
A small remote-controlled car with mass 1.60 kg moves at a constant speed of υ = 12.0 m/s in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of 5.00 m (Fig. E5.45). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at (a) point A (bottom of the track)

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Textbook Question
A small remote-controlled car with mass 1.60 kg moves at a constant speed of υ = 12.0 m/s in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of 5.00 m (Fig. E5.45). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at (b) point B (top of the track)?

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Textbook Question
The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s). (b) A passenger weighs 882 N at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel?
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Textbook Question
The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s).(c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?
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Textbook Question
The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s). (d) What then would be the passenger's apparent weight at the lowest point?
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