Ch 08: Dynamics II: Motion in a Plane

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 08: Dynamics II: Motion in a Plane

Problem 21

Chapter 8, Problem 21

The normal force equals the magnitude of the gravitational force as a roller-coaster car crosses the top of a 40-m-diameter loop-the-loop. What is the car's speed at the top?

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Identify the forces acting on the roller-coaster car at the top of the loop. The forces are the gravitational force \( F_g = m g \) (acting downward) and the normal force \( F_N \) (also acting downward). At the top of the loop, the centripetal force required to keep the car in circular motion is provided by the sum of these two forces.

Write the equation for the centripetal force \( F_c \) at the top of the loop: \( F_c = F_g + F_N \). Since the problem states that the normal force equals the gravitational force, substitute \( F_N = F_g \) into the equation.

Express the centripetal force in terms of the car's speed \( v \) and the radius of the loop \( r \): \( F_c = \frac{m v^2}{r} \). Substitute \( F_c = F_g + F_N = 2 F_g = 2 m g \) into this equation.

Solve for the car's speed \( v \): \( \frac{m v^2}{r} = 2 m g \). Cancel \( m \) from both sides (assuming the car's mass is nonzero), and rearrange to get \( v^2 = 2 g r \).

Substitute the radius of the loop \( r \) (which is half the diameter, \( r = \frac{40}{2} = 20 \ \text{m} \)) and the acceleration due to gravity \( g = 9.8 \ \text{m/s}^2 \) into the equation \( v^2 = 2 g r \). Take the square root to find \( v \).

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Key Concepts

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

Normal Force

The normal force is the perpendicular force exerted by a surface against an object in contact with it. In the context of circular motion, such as a roller-coaster car at the top of a loop, the normal force acts upward against the weight of the car, which is directed downward due to gravity. At the top of the loop, the normal force and gravitational force must balance to maintain circular motion.

Gravitational Force

Gravitational force is the attractive force between two masses, calculated using Newton's law of universal gravitation. For an object near the Earth's surface, this force can be simplified to the weight of the object, given by the equation F_gravity = m * g, where m is mass and g is the acceleration due to gravity (approximately 9.81 m/s²). At the top of the loop, this force acts downward and influences the car's motion.

Centripetal Force and Speed

Centripetal force is the net force required to keep an object moving in a circular path, directed towards the center of the circle. At the top of the loop, the centripetal force is provided by the difference between gravitational force and normal force. The speed of the roller-coaster car can be determined using the centripetal force equation, F_c = m * v²/r, where v is the speed and r is the radius of the loop.

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