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Chapter 6: Problem 48

\(\infty \Rightarrow \sqrt{\text { A roller-coaster car at an amusement park has a mass }}\) of \(1200 \mathrm{~kg}\) when fully loaded with passengers. As the car passes over the top of a circular hill of radius \(18 \mathrm{~m}\), assume that its speed is not changing. At the top of the hill, what are the (a) magnitude \(F_{N}\) and (b) direction (up or down) of the normal force on the car from the track if the car's speed is \(v=11 \mathrm{~m} / \mathrm{s} ?\) What are (c) \(F_{N}\) and (d) the direction if \(v=14 \mathrm{~m} / \mathrm{s}\) ?

Short Answer

Expert verified
(a) \( F_{N} = 3706.67 \, \text{N} \), (b) Down; (c) \( F_{N} = 0 \, \text{N} \), (d) None (car leaves the track).

Step by step solution

01

Understanding the Forces

At the top of the hill, the only forces acting on the roller-coaster car are the gravitational force (weight) acting downward and the normal force exerted by the track. Due to circular motion, the net force on the car must equal the centripetal force required to keep it moving in a circle.
02

Calculating Gravitational Force

The gravitational force, or weight of the car, is calculated using the equation \[ F_{g} = m imes g, \]where \( m = 1200 \, \text{kg} \) is the mass of the car and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Thus,\[ F_{g} = 1200 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 11760 \, \text{N}. \]
03

Centripetal Force Equation

At the top of the hill, the net force, which in this case is the centripetal force, is given by\[ F_{c} = \frac{m v^2}{r}, \]where \( v \) is the speed of the car and \( r = 18 \, \text{m} \) is the radius of the hill.
04

Calculating Normal Force at \( v = 11 \, \text{m/s} \)

We use the net force (centripetal) equation:\[ F_{c} = F_{g} - F_{N}, \]where \( F_{N} \) is the normal force. At \( v = 11 \, \text{m/s} \), solve the equation:\[ \frac{1200 \, (11)^2}{18} = 11760 - F_{N}. \]Calculate \( F_{c} = 8053.33 \, \text{N} \). Therefore, \[ 8053.33 = 11760 - F_{N} \Rightarrow F_{N} = 3706.67 \, \text{N}. \]
05

Determine Direction for \( v = 11 \, \text{m/s} \)

The normal force acts upward against gravity, but since it's calculated from the excess of gravitational force causing centripetal force, it's considered downward relative to gravity at this point.
06

Calculating Normal Force at \( v = 14 \, \text{m/s} \)

Similarly, use the net force equation:\[ \frac{1200 \, (14)^2}{18} = 11760 - F_{N}. \]Calculate \( F_{c} = 13066.67 \, \text{N} \). Thus,\[ 13066.67 = 11760 - F_{N} \Rightarrow F_{N} = -1306.67 \, \text{N}. \]
07

Determine Direction for \( v = 14 \, \text{m/s} \)

Since \( F_{N} \) is negative, it implies the normal force disappears, and the roller-coaster does not maintain contact with the track, indicating it would begin to leave the track. Thus, the force is not exerted.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
When an object moves in a circular path, such as a roller-coaster car on a hill, it experiences a force that keeps it on that path. This force is called centripetal force. Without centripetal force, the object would fly off in a straight line due to inertia. In our roller-coaster example, the centripetal force ensures the car follows the circular path over the hill.

Centripetal force can be calculated using the formula \[ F_{c} = \frac{m v^2}{r}, \] where:
  • \( m \) is the mass of the object (1200 kg for the roller-coaster),
  • \( v \) is the speed of the object, and
  • \( r \) is the radius of the circular path.
At different speeds, different amounts of centripetal force are needed to keep the roller-coaster on track. It directly affects how much of the other forces, like the normal force, act on the roller-coaster at any given moment.
Normal Force
The normal force is a force exerted by a surface to support the weight of an object resting on it. It's always perpendicular to the surface. In our roller-coaster scenario, this is the force exerted by the track on the car.

At the top of the hill, the normal force adjusts based on the centripetal force and gravitational force. When the car moves at 11 m/s, the normal force is calculated by rearranging the formula \[ F_{c} = F_{g} - F_{N}, \] where:
  • \( F_{c} \) is the centripetal force,
  • \( F_{g} \) is the gravitational force, calculated as \( m \times g \), and
  • \( g \) is the acceleration due to gravity (9.8 m/s虏).
The direction of the normal force at 11 m/s is downwards as it counterbalances the gravitational force for maintaining the circular path. However, at 14 m/s, the normal force becomes negative, meaning the car begins to lose contact with the track due to the greater demand for centripetal force.
Gravitational Force
Gravitational force, also known as weight, is the force with which Earth pulls an object towards its center. It acts downward on the roller-coaster car.

To compute the gravitational force acting on the car, we use the equation \[ F_{g} = m \times g, \] where:
  • \( m \) is the mass of the car, and
  • \( g \) is the acceleration due to gravity (9.8 m/s虏).
For our roller-coaster car with a mass of 1200 kg, the gravitational force is \[ F_{g} = 1200 \times 9.8 = 11760 \text{ N}. \]
This force remains constant, while the normal force fluctuates, showing how the distribution between forces varies with speed in circular motion.

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Most popular questions from this chapter

A \(100 \mathrm{~N}\) force, directed at an angle \(\theta\) above a horizontal floor, is applied to a \(25.0 \mathrm{~kg}\) chair sitting on the floor. If \(\theta=0^{\circ}\), what are (a) the horizontal component \(F_{h}\) of the applied force and (b) the magnitude \(F_{N}\) of the normal force of the floor on the chair? If \(\theta=30.0^{\circ}\), what are (c) \(F_{h}\) and \((\) d \() F_{N}\) ? If \(\theta=60.0^{\circ}\), what are (e) \(F_{h}\) and (f) \(F_{N}\) ? Now assume that the coefficient of static friction between chair and floor is \(0.420\). Does the chair slide or remain at rest if \(\theta\) is \((\mathrm{g}) 0^{\circ},(\mathrm{h}) 30.0^{\circ}\), and (i) \(60.0^{\circ} ?\)

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