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 20

Chapter 8, Problem 20

The weight of passengers on a roller coaster increases by 50% as the car goes through a dip with a 30 m radius of curvature. What is the car's speed at the bottom of the dip?

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Understand the problem: The weight of the passengers increases by 50% at the bottom of the dip due to the centripetal force acting on them. This means the normal force (apparent weight) is 1.5 times the gravitational force. We need to find the speed of the roller coaster car at this point.

Write the forces acting on the passengers at the bottom of the dip: The normal force \( F_n \) and the gravitational force \( F_g = m g \) act upward and downward, respectively. The net force provides the centripetal force \( F_c \), which is \( \frac{m v^2}{r} \), where \( m \) is the mass, \( v \) is the speed, and \( r \) is the radius of curvature.

Set up the equation for the forces: At the bottom of the dip, the normal force is greater than the gravitational force. The net force is \( F_n - F_g = F_c \). Substituting \( F_c \), we get \( F_n - m g = \frac{m v^2}{r} \).

Relate the normal force to the apparent weight: Since the apparent weight is 1.5 times the gravitational force, \( F_n = 1.5 m g \). Substitute this into the equation: \( 1.5 m g - m g = \frac{m v^2}{r} \).

Simplify and solve for \( v \): Combine terms on the left-hand side to get \( 0.5 m g = \frac{m v^2}{r} \). Cancel \( m \) from both sides (assuming \( m \neq 0 \)) and solve for \( v \): \( v = \sqrt{0.5 g r} \). Substitute \( g = 9.8 \, \text{m/s}^2 \) and \( r = 30 \, \text{m} \) to find the speed.

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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 required to keep an object moving in a circular path, directed towards the center of the circle. In the context of the roller coaster, as the car goes through the dip, the gravitational force and the normal force from the track provide the necessary centripetal force to maintain circular motion. The relationship between these forces is crucial for determining the speed of the car at the bottom of the dip.

Weight and Normal Force

Weight is the force exerted by gravity on an object, calculated as the product of mass and gravitational acceleration. At the bottom of the dip, the normal force acting on the passengers increases due to the additional centripetal force required for circular motion. The problem states that the weight increases by 50%, indicating that the normal force is greater than the gravitational force, which is essential for calculating the car's speed.

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. In this scenario, the relationship between speed, radius of curvature, and forces acting on the roller coaster can be analyzed using these equations. By applying the principles of circular motion and the forces involved, one can derive the speed of the car at the bottom of the dip, which is essential for solving the problem.

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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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