/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 115 At the bottom \(A\) of the verti... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 115

At the bottom \(A\) of the vertical inside loop, the magnitude of the total acceleration of the airplane is \(3 g\). If the airspeed is \(800 \mathrm{km} / \mathrm{h}\) and is increasing at the rate of \(20 \mathrm{km} / \mathrm{h}\) per second, calculate the radius of curvature \(\rho\) of the path at \(A\)

Short Answer

Expert verified
The radius of curvature \(\rho\) is approximately 1677 meters.

Step by step solution

01

Understanding the total acceleration

The total acceleration at point A is given as the sum of the centripetal acceleration and the tangential acceleration. Given that the magnitude of the total acceleration is \(3g\), where \(g\) is the acceleration due to gravity (\(9.81 \text{ m/s}^2\)), the total acceleration at A is \( 3 \times 9.81 \text{ m/s}^2 = 29.43 \text{ m/s}^2 \).
02

Converting units for airspeed

The airspeed is given in \(\text{km/h}\). Convert it to \(\text{m/s}\) by dividing by \(3.6\): \( 800 \text{ km/h} = \frac{800}{3.6} \text{ m/s} \approx 222.22 \text{ m/s} \).
03

Calculate tangential acceleration

The rate at which the airspeed is increasing is given as \(20 \text{ km/h per second}\). Convert this to \(\text{m/s}^2\) by dividing by \(3.6\): \(20 \text{ km/h per second} = \frac{20}{3.6} \text{ m/s}^2 \approx 5.56 \text{ m/s}^2 \).
04

Define and calculate centripetal acceleration

The total acceleration is the vector sum of the tangential and centripetal accelerations. Since we know the magnitude of total acceleration, we use the relation \( a_t = \sqrt{a_{tan}^2 + a_{cen}^2} \), where \( a_{tan} \) is the tangential acceleration. From Step 1, \( a_t = 29.43 \text{ m/s}^2 \). Solve for centripetal acceleration \(a_{cen}\) using: \[ a_{cen} = \sqrt{a_t^2 - a_{tan}^2} \].Calculate: \[ a_{cen} = \sqrt{(29.43)^2 - (5.56)^2} \approx \sqrt{866.72} \approx 29.44 \text{ m/s}^2 \].
05

Calculate the radius of curvature

Use the formula for centripetal acceleration, \( a_{cen} = \frac{v^2}{\rho} \), where \(v\) is the speed. We rearrange this to solve for \(\rho\): \( \rho = \frac{v^2}{a_{cen}} \).Plug in \(v = 222.22 \text{ m/s}\) and \(a_{cen} = 29.44 \text{ m/s}^2\): \[ \rho = \frac{(222.22)^2}{29.44} \approx \frac{49382.17}{29.44} \approx 1677 \text{ meters} \].

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Key Concepts

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

Total Acceleration
Total acceleration is a crucial concept in dynamics as it describes the overall change in velocity of an object that is moving along a curved path. It combines both linear and rotational dynamics effects.
The formula for total acceleration is:
  • \( a_t = \sqrt{a_{tan}^2 + a_{cen}^2} \)
where:
  • \( a_{tan} \) is the tangential acceleration.
  • \( a_{cen} \) is the centripetal acceleration.
In this exercise, the total acceleration is given as a multiple of gravitational acceleration \( g \). In many physics problems like this, especially in loops or circuits, it's common to compare accelerations with \( g \) to express them in physically intuitive terms.
When you compute such a problem, converting all given speeds and changes in speeds from \( \text{km/h} \) to \( \text{m/s} \) is vital. This ensures that units are consistent, simplifying calculations and reducing errors later on.
Centripetal Acceleration
Centripetal acceleration is the acceleration that keeps an object moving in a circular path. It is directed towards the center of the curvature of the path. In dynamics, understanding centripetal acceleration is essential when dealing with circular motion.
Centripetal acceleration can be calculated through the formula:
  • \( a_{cen} = \frac{v^2}{\rho} \)
where:
  • \( v \) is the velocity of the object along the path.
  • \( \rho \) is the radius of curvature.
This acceleration results from the change in direction of the velocity vector, even if its magnitude (speed) remains constant. Therefore, it only acts perpendicular to the velocity. In this problem, you use the known total acceleration and tangential acceleration to solve for the centripetal component using the Pythagorean theorem in the context of vector components.
Radius of Curvature
The radius of curvature \( \rho \) is a measure of how tightly a curve is bent at a particular point, and it plays a significant role in curvilinear motion. Understanding the radius of curvature helps in analyzing the path that the object takes, and its effects on the centripetal acceleration.
To find the radius of curvature when given the centripetal acceleration, use the relationship:
  • \( \rho = \frac{v^2}{a_{cen}} \)
This relationship shows the dependency between the speed of the object, the centripetal acceleration, and how sharply it is turning. If an object is moving faster while maintaining the same curvature, the centripetal acceleration must increase to keep the object on its path. Conversely, maintaining the same velocity on a path with a decreasing radius of curvature implies a higher centripetal acceleration.
Knowing \( \rho \) is essential for navigation and control in aviation, as it affects the maneuverability of the aircraft along the defined path.

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Most popular questions from this chapter

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