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When looping the loop, the Blue Angels stunt pilot of the U.S. Navy fy their jet aircraft along a vertical circle of diame ter \(1000 \mathrm{~m}\) (Fig, 4.35). At the tep of the circic, the speed is \(350 \mathrm{lom} / \mathrm{h}\) s at the boetom of the circle, the speed is \(620 \mathrm{~km} / \mathrm{h}\). What is the centripetal acceleration at the top? At the bottom? In the reference frame of one of these aircraft, what is the acecleration thut the pilot fcels it the top and at the botworn; i.e, what is the acceleration relative to the aincrift of a serall body, woch as a coin, released by the pilot?

Short Answer

Expert verified
Centripetal: top is 18.91 m/s², bottom is 59.35 m/s². Pilot feels: top is 9.11 m/s², bottom is 69.15 m/s².

Step by step solution

01

Convert Speeds from km/h to m/s

First, we need to convert the velocities given in km/h to m/s. The formula to convert km/h to m/s is to multiply by \( \frac{1}{3.6} \). \( 350 \text{ km/h} = 350 \times \frac{1}{3.6} = 97.22 \text{ m/s} \). Similarly, \( 620 \text{ km/h} = 620 \times \frac{1}{3.6} = 172.22 \text{ m/s} \).
02

Calculate the Radius of the Circle

The diameter of the circle is given as \(1000\) m, so the radius \(r\) is half of that. \( r = \frac{1000}{2} = 500 \text{ m} \).
03

Compute Centripetal Acceleration at the Top

The formula for centripetal acceleration is \( a_c = \frac{v^2}{r} \). At the top, the speed \( v = 97.22 \text{ m/s} \). Thus, \( a_{c_{top}} = \frac{(97.22)^2}{500} = 18.91 \text{ m/s}^2 \).
04

Compute Centripetal Acceleration at the Bottom

Using the same formula \( a_c = \frac{v^2}{r} \), at the bottom, the speed \( v = 172.22 \text{ m/s} \). Thus, \( a_{c_{bottom}} = \frac{(172.22)^2}{500} = 59.35 \text{ m/s}^2 \).
05

Determine Acceleration Felt by Pilot at the Top

At the top of the circle, the centripetal acceleration is directed towards the center of the circle. Therefore, the acceleration felt by the pilot \( a_{felt_{top}} = a_{c_{top}} - g \), where \( g = 9.8 \text{ m/s}^2 \). Thus, \( a_{felt_{top}} = 18.91 - 9.8 = 9.11 \text{ m/s}^2 \).
06

Determine Acceleration Felt by Pilot at the Bottom

At the bottom of the circle, the centripetal acceleration and gravitational acceleration both point upward. Therefore, the acceleration felt by the pilot \( a_{felt_{bottom}} = a_{c_{bottom}} + g \). Thus, \( a_{felt_{bottom}} = 59.35 + 9.8 = 69.15 \text{ m/s}^2 \).

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Key Concepts

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

Circular Motion
In circular motion, an object moves along a circular path. The path can be various shapes of circles, but the key concept is that the motion is around some axis at a constant radius. This axis is where the centripetal force acts, continuously pulling the object toward the center, preventing it from flying out in a straight line. Without this force, perhaps due to friction or another force like gravity, the object would simply move off in a tangent to the circle.

When a pilot flies in a vertical loop, they experience changes in speed and force as they move from the top to the bottom and back again. The required centripetal acceleration must be sufficient to keep the plane inside the loop without falling out or stalling. Each point in the loop has its own velocity, affecting the centripetal force required.
Pilot Experience
A pilot in a stunt flying situation experiences unique forces due to circular motion, especially in a vertical loop. This is not just due to speed but also because of the changing directions of gravitational and centripetal forces. At the top of the loop, gravity works with the centripetal force to pull the pilot downward, making them feel lighter. However, at the bottom, these forces oppose each other, leading to maximum pressure felt by the pilot.

The net effect is defined by the pilot's perceived acceleration, a combination of the actual centripetal acceleration and the acceleration due to gravity. These experiences are often described as "g-forces." At the top, the pilot may feel normal or slightly lighter (fewer g-forces), but at the bottom, they may experience increased g-forces. These forces are key for a stunting pilot to manage, ensuring safe and effective loop execution.
Velocity Conversion
Velocity conversion is essential for understanding different units of measurement, especially when dealing with international aviation data often presented in mixed units. The speed of the aircraft was initially given in kilometers per hour (km/h). For our calculations, we converted this to meters per second (m/s), as the formula for centripetal acceleration is simpler in these SI units.

The conversion from km/h to m/s involves dividing the speed value by 3.6. This conversion is vital before inserting values into physics formulas because it provides consistency, reducing errors and ensuring the calculations align with scientific standards. Accurate unit conversion is an essential skill in physics and engineering applications, especially when dealing with rapid speed changes like those experienced in aviation loops.
Gravitational Force
Gravitational force plays a critical role in the experience of pilots during circular motion. It's the force exerted by the Earth, pulling aviation objects towards it at approximately 9.8 m/s². In the context of circular motion, this force impacts the net acceleration felt by the pilot.

When at the top of a loop, gravitational force is working alongside centripetal force, effectively reducing the overall force the pilot feels. Whereas at the bottom of the loop, gravity opposes the centripetal force, increasing the felt acceleration significantly.
  • Gravitational force can either add to or subtract from the centripetal force, depending on the pilot's position in the loop.
  • Understanding the interplay of these forces is critical for pilot safety and aircraft control.
This balancing act between gravity and centripetal acceleration is what gives pilots their unique aerial experience, necessitating rigorous training for safe stunt performances.

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