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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 \(A\))?

Short Answer

Expert verified
The normal force at the bottom of the track is approximately 21.70 N.

Step by step solution

01

Understand the forces at point B

At point B, the forces acting on the car are the gravitational force pointing downward and the normal force exerted by the track pointing downward. Since the car is in a vertical circular motion, the net force acting on it is the centripetal force required to keep it moving in a circle. Therefore, we can write the equation: \\[ F_{net,B} = F_n + mg = \frac{mv^2}{r} \] \where \( F_n = 6.00 \text{ N} \), \( m = 0.800 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( r = 5.00 \text{ m}. \)
02

Solve for velocity at point B

To find the velocity \( v \) at point B, we rearrange the equation: \\[ \frac{mv^2}{r} = F_n + mg \] \Substitute the given values: \\[ \frac{(0.800)v^2}{5.00} = 6.00 + 0.800 \times 9.81 \] \Solving for \( v^2 \): \\[ 0.160v^2 = 6.00 + 7.848 \] \\[ 0.160v^2 = 13.848 \] \\[ v^2 = \frac{13.848}{0.160} \approx 86.55 \] \Thus, \\[ v \approx \sqrt{86.55} \approx 9.30 \text{ m/s}. \]
03

Analyze forces at point A

At the bottom of the track, the direction of the normal force is upward, while the gravitational force still points downward. Therefore, the net centripetal force equation is: \\[ F_{net,A} = F_n - mg = \frac{mv^2}{r} \] \We know \( v^2 \) from previous calculation as \( 86.55 \).
04

Solve for normal force at point A

Insert the known values into the centripetal force equation to solve for the normal force \( F_n \) at point A: \\[ F_n - mg = \frac{mv^2}{r} \] \\[ F_n = \frac{mv^2}{r} + mg \] \\[ F_n = \frac{0.800 \times 86.55}{5.00} + 0.800 \times 9.81 \] \\[ F_n = \frac{69.24}{5.00} + 7.848 \] \\[ F_n = 13.848 + 7.848 \] \\[ F_n \approx 21.696 \text{ N} \] \So, the normal force exerted by the track on the car at the bottom is approximately \( 21.70 \text{ N}. \)

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Key Concepts

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

Circular Motion
When we talk about circular motion, we're describing the movement of an object along the circumference of a circle. This kind of motion is fascinating because it involves both an angular component and a linear component.
In most questions involving circular motion, like the one in the exercise, we're interested in how forces keep an object moving in a circle without spiraling out. Circular motion is governed by the requirement of a centripetal force. This force acts towards the center of the circle and is essential to maintain the circular path of an object. Without it, the object would move off in a straight line due to inertia. Key Characteristics:
  • Constant Speed: Though the speed stays the same, the object is constantly changing direction.
  • Centripetal Force: This force could be due to gravity, tension, or in the case of our exercise, a combination of gravitational and normal forces.
  • Acceleration: Even with constant speed, there is acceleration due to the constant change in direction.
In our example, the car's speed at the top and bottom of the circle plays a crucial role. The forces must adjust to maintain this speed and keep the car on its curved trajectory.
Normal Force
Normal force is the force perpendicular to the surface with which an object is interacting. It's a concept that's crucial in determining how objects move or stay still in relation to that surface. In our vertical circle problem, normal force varies depending on the position of the car along the track. For instance, at the top of the circle, both gravitational force and normal force point downward.
In contrast, at the bottom, the normal force points upward, counteracting gravity. Characteristics of Normal Force:
  • Reaction Force: It's always perpendicular to the surface, reacting to other forces acting on the object.
  • Variable Strength: As seen in the exercise, the magnitude can vary based on the object's position within a vertical circular motion.
  • Dependent on Other Forces: It works with gravitational force to achieve the necessary centripetal force.
In this exercise, determining the normal force at different points in the circle helps illustrate how forces change depending on position, speed, and the need to maintain circular motion.
Vertical Circle
Understanding how objects move in a vertical circle is key to grasping the forces at play in our problem. Imagine a loop-the-loop on a roller coaster; that's a common example of a vertical circular motion. A vertical circle introduces the complexity of gravity acting in different ways as the object travels around the circle. At the top of the circle, gravity aids in the centripetal force by pulling towards the center. Meanwhile, at the bottom, it opposes the upward centripetal direction needed to keep the object on course. Characteristics:
  • Changing Force Dynamics: The balance of forces varies with position.
  • Potential Energy Changes: As the object moves up and down, potential energy decreases and increases respectively.
  • Kinetic Energy Roles: Speed is sometimes lowest at the top and fastest at the bottom due to energy exchange.
In our problem, the car's elevation and speed changes mean understanding vertical circles is crucial to accurately calculate forces at any point along its path.

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Most popular questions from this chapter

A 2.00-kg box is moving to the right with speed 9.00 m/s on a horizontal, frictionless surface. At \(t =\) 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude \(F(t) = (6.00 N/s^2)t^2\). (a) What distance does the box move from its position at \(t =\) 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at \(t =\) 3.00 s?

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