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A jet pilot puts an aircraft with a constant speed into a vertical circular loop. (a) Which is greater, the normal force exerted on the seat by the pilot at the bottom of the loop or that at the top of the loop? Why? (b) If the speed of the aircraft is \(700 \mathrm{~km} / \mathrm{h}\) and the radius of the circle is \(2.0 \mathrm{~km}\), calculate the normal forces exerted on the seat by the pilot at the bottom and top of the loop. Express your answer in terms of the pilot's weight.

Short Answer

Expert verified
The normal force is greater at the bottom. At the bottom: \(28.74m\) and at the top: \(9.12m\).

Step by step solution

01

Understanding Forces in Circular Motion

In vertical circular motion, when the jet is at the bottom of the loop, both gravity and the centripetal force are acting down towards the center of the loop, and the normal force must exert an upward force to provide the difference. At the top, gravity acts in the same direction as the centripetal force, reducing the normal force needed.
02

Determining Greater Normal Force

The normal force at the bottom of the loop is greater than at the top because at the bottom, it needs to counteract both the gravitational force and provide enough centripetal force to keep the pilot moving in a circle.
03

Converting Speed to Meters per Second

First, convert the speed from kilometers per hour to meters per second:\[ 700 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} \approx 194.44 \text{ m/s} \]
04

Calculating Centripetal Force

The centripetal force required is \( F_c = \frac{mv^2}{r} \). Substituting \(v = 194.44 \text{ m/s}\) and \(r = 2000 \text{ m}\), we get:\[ F_c = \frac{m (194.44)^2}{2000} \approx \, \frac{m \times 37860.33}{2000} = 18.93m \text{ N} \]
05

Calculating Normal Force at Bottom

The normal force at the bottom of the loop is given by:\[ N_{bottom} = F_c + mg \]Thus, substituting the forces, we have:\[ N_{bottom} = 18.93m + mg \]Since \(g = 9.81 \text{ m/s}^2\):\[ N_{bottom} = 18.93m + 9.81m = 28.74m \text{ N} \]
06

Calculating Normal Force at Top

The normal force at the top of the loop is given by:\[ N_{top} = F_c - mg \]Thus, we have:\[ N_{top} = 18.93m - mg \]With \(g = 9.81 \text{ m/s}^2\):\[ N_{top} = 18.93m - 9.81m = 9.12m \text{ N} \]

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Key Concepts

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

Centripetal Force
In circular motion, the centripetal force is crucial as it keeps an object moving in a circle. This force always points towards the center of the circle. It is not an individual force, but the net force directed inwards. Without this force, objects would move off in straight lines rather than curves.
Centripetal force can come from gravity, tension, friction, or in this exercise, the normal force exerted by the seat on the pilot. When an aircraft moves in a vertical loop, the combination of forces provides the necessary centripetal force to maintain the curved path. It's important to remember that the magnitude of centripetal force changes depending on the object's speed and the loop’s radius.
  • Formula for centripetal force: \[ F_c = \frac{mv^2}{r} \]
  • \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius of the loop.
Normal Force
The normal force is a support force that acts perpendicular to the surface of contact.
In circular motion, especially in vertical loops, the normal force changes depending on the position of the object in the loop. At the bottom of the loop, the normal force is greater because it has to not only counteract gravity but also supply additional force for the centripetal motion.
Conversely, at the top of the loop, the normal force is less because gravity itself helps provide some of the centripetal force needed. Thus, the normal force doesn't have to work as hard against gravity.
  • Bottom of the loop: \( N_{bottom} = F_c + mg \)
  • Top of the loop: \( N_{top} = F_c - mg \)
This difference explains why the normal force is greater at the bottom than at the top.
Vertical Loops
Vertical loops are fascinating maneuvers where an object moves in a circle oriented perpendicular to the earth's surface. Though thrilling, they involve complex physics. As an object moves through the loop, different forces act differently depending on its position around the loop.
In a vertical loop, an object experiences highest forces at the bottom, where both gravity and centripetal force work against each other. This requires a larger normal force. At the top of the loop, gravity assists the centripetal force, reducing the need for a high normal force.
The speed of an object in a vertical loop is vital. If too slow, the object won’t complete the loop. If too fast, it could exceed structural limits.
Gravity in Circular Motion
In circular motion, gravity plays a unique role, especially in vertical loops. Unlike linear motion, gravity assists the motion at the top and acts against it at the bottom.
Understanding gravity's influence ensures safety and control in loops. At the loop's crest, gravity acts downward, reducing the normal force. At the bottom, it acts opposite the required centripetal motion, increasing the burden on the normal force.
Key points to remember:
  • Gravity assists at the loop's top, reducing normal force.
  • Against at the bottom, magnifying the required normal force.
Balanced forces ensure smooth, secure motion through vertical loops.

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Most popular questions from this chapter

Two objects are attracting each other with a certain gravitational force. (a) If the distance between the objects is halved, the new gravitational force will (1) increase by a factor of 2,(2) increase by a factor of 4,(3) decrease by a factor of 2,(4) decrease by a factor of \(4 .\) Why? (b) If the original force between the two objects is \(0.90 \mathrm{~N},\) and the distance is tripled, what is the new gravitational force between the objects?

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