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Ch 05: Applying Newton's Laws
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 5, Problem 45a

A small remote-controlled car with mass 1.601.60 kg moves at a constant speed of v=12.0v = 12.0 m/s in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of 5.005.00 m (Fig. E5.455.45). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point AA (bottom of the track)?
Diagram of a remote-controlled car in a vertical circular track with radius 5.00 m, showing points A and B.

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Step 1: Identify the forces acting on the car at point A (bottom of the track). These include the gravitational force (mg, acting downward) and the normal force exerted by the walls of the cylinder (N, acting upward). The car is also undergoing circular motion, so there is a centripetal force directed toward the center of the circle.
Step 2: Write the equation for the net force in the vertical direction at point A. Since the car is moving in a circular path, the net force provides the centripetal force required for circular motion. The equation is: N - mg = m * (υ² / r), where m is the mass of the car, υ is its speed, and r is the radius of the circle.
Step 3: Rearrange the equation to solve for the normal force (N). Add mg to both sides: N = m * (υ² / r) + mg.
Step 4: Substitute the given values into the equation. The mass of the car (m) is 1.60 kg, the speed (υ) is 12.0 m/s, the radius (r) is 5.00 m, and the acceleration due to gravity (g) is approximately 9.80 m/s².
Step 5: Perform the calculations to find the magnitude of the normal force (N). Ensure that you calculate the centripetal force term (m * (υ² / r)) and the gravitational force term (mg) separately before adding them together.

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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 like the remote-controlled car, this force is necessary to keep it moving in a circular trajectory. It can be calculated using the formula F_c = (mv^2)/r, where m is the mass, v is the velocity, and r is the radius of the circular path.
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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 the context of the car at the bottom of the vertical circle, the normal force counteracts both the gravitational force acting on the car and provides the necessary centripetal force to keep it in circular motion.
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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 can be expressed as F_g = mg, where m is the mass of the car and g is the acceleration due to gravity (approximately 9.81 m/s²). At point A, this force acts downward, influencing the net forces acting on the car.
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Textbook Question

A small remote-controlled car with mass 1.601.60 kg moves at a constant speed of v=12.0v = 12.0 m/s in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of 5.005.00 m (Fig. E5.455.45). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point BB (top of the track)?

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Textbook Question

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A small car with mass 0.8000.800 kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.005.00 m (Fig. E5.455.45). If the normal force exerted by the track on the car when it is at the top of the track (point BB) is 6.006.00 N, what is the normal force on the car when it is at the bottom of the track (point AA)?

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