/*! 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 30 A model of a helicopter rotor ha... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A model of a helicopter rotor has four blades, each \(3.40 \mathrm{~m}\) in length from the central shaft to the tip of the blade. The model is rotated in a wind tunnel at 550 rev \(/\) min. (a) What is the linear speed, in \(\mathrm{m} / \mathrm{s},\) of the blade tip? (b) What is the radial acceleration of the blade tip, expressed as a multiple of the acceleration \(g\) due to gravity?

Short Answer

Expert verified
(a) The linear speed is approximately 195.8 m/s. (b) The radial acceleration is about 1150 times gravity.

Step by step solution

01

Convert revolutions per minute to radians per second

The first step is to convert the rotational speed from revolutions per minute (rev/min) to radians per second (rad/s). The formula to convert is:\[\omega = \text{angular speed in rad/s} = \left( \frac{550 \text{ rev/min} \times 2\pi \text{ rad/rev}}{60 \text{ sec/min}} \right)\]Calculating this gives:\[\omega = \frac{550 \times 2\pi}{60} \approx 57.6 \text{ rad/s}\]
02

Find the linear speed of the blade tip

Now that we have the angular speed, we can find the linear speed of the blade tip. The linear speed \(v\) is given by the formula:\[v = r \cdot \omega\]where \(r = 3.40 \, \text{m}\) is the radius (length of one blade). Using the values we have:\[v = 3.40 \times 57.6 \approx 195.8 \, \text{m/s}\]
03

Compute radial acceleration

The radial (or centripetal) acceleration \(a_r\) is given by the formula:\[a_r = \frac{v^2}{r}\]Using our previously calculated linear speed \(v = 195.8 \, \text{m/s}\) and \(r = 3.40 \, \text{m}\), we find:\[a_r = \frac{(195.8)^2}{3.40} \approx 11280.36 \, \text{m/s}^2\]
04

Express radial acceleration as a multiple of g

Given that the acceleration due to gravity \(g\) is approximately \(9.81 \, \text{m/s}^2\), we need to express the radial acceleration as a multiple of \(g\):\[\text{Multiple of } g = \frac{a_r}{g} = \frac{11280.36}{9.81} \approx 1150\]Therefore, the radial acceleration is approximately 1150 times the acceleration due to gravity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Speed
Angular speed is a measure of how fast an object rotates or revolves compared to a reference point, often a circle's center. It is generally expressed in radians per second (rad/s).
To convert angular speed from revolutions per minute (rev/min) to radians per second, we use a simple conversion where one complete revolution equals \(2\pi\) radians. Therefore, the formula is:
  • \(\omega = \left( \frac{\text{revolutions/minute} \times 2\pi \, \text{rad/revolution}}{60 \, \text{seconds/minute}} \right)\)
Plugging in the numbers for a helicopter rotor rotating at 550 rev/min, the angular speed becomes about 57.6 rad/s. Think of it as the speed at which any point on the rotor moves around the central shaft, measured in radians over each second.
Linear Speed
Linear speed refers to how fast a point moves along a straight line. For an object moving in a circular path, like blades of a helicopter rotor, linear speed can be determined from its angular speed.
The formula for finding linear speed \(v\) is:
  • \(v = r \cdot \omega\)
Where \(r\) is the radius (length from the center to the tip of the blade) and \(\omega\) is angular speed.
For a blade tip of 3.40 m, rotating with an angular speed of 57.6 rad/s, the linear speed calculates as approximately 195.8 m/s. This means that the blade tip is moving nearly 196 meters every second, forming a tangent line to the circle at that instant.
Centripetal Acceleration
Centripetal acceleration is experienced by an object moving in a circular path, pointing toward the circle's center. This acceleration is crucial for an object to maintain its circular motion and is defined by the formula:
  • \(a_r = \frac{v^2}{r}\)
Where \(v\) is the linear speed, and \(r\) is the radius. Using the linear speed of 195.8 m/s and a radius of 3.40 m, we compute the centripetal acceleration as approximately 11280.36 m/s2.
To relate this to something more familiar, it is important to express it as a multiple of gravitational acceleration \(g\), which is about 9.81 m/s2. Therefore, \(a_r\) is approximately 1150 times \(g\). This enormous value highlights the immense force a rotor tip experiences during rotation.
Conversion of Units
Converting units aids in expressing measurements in the most practical units for comprehension and analysis. In physics, differing units often need conversion to perform calculations correctly.
For rotational motion:
  • To go from revolutions per minute to radians per second, multiply by \(\left(\frac{2\pi}{60}\right)\).
  • For expressing acceleration as a multiple of \(g\), divide the calculated acceleration by 9.81 m/s2.
These unit conversions ensure that calculations are seamless and understandable. Recognizing how to convert appropriately allows one to interpret and solve real-world problems effectively using consistent measurement systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A curving freeway exit has a radius of \(50.0 \mathrm{~m}\) and a posted speed limit of \(35 \mathrm{mi} / \mathrm{h}\). What is your radial acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) if you take this exit at the posted speed? What if you take the exit at a speed of \(50 \mathrm{mi} / \mathrm{h} ?\)

A velocity vector has a magnitude of \(25.0 \mathrm{~m} / \mathrm{s}\). If its \(y\) component is \(-13.0 \mathrm{~m} / \mathrm{s},\) what are the possible values of its \(x\) component?

A bottle rocket can shoot its projectile vertically to a height of \(25.0 \mathrm{~m}\). At what angle should the bottle rocket be fired to reach its maximum horizontal range, and what is that range? (You can ignore air resistance.)

A test rocket is launched by accelerating it along a \(200.0-\mathrm{m}\) incline at \(1.25 \mathrm{~m} / \mathrm{s}^{2}\) starting from rest at point \(A\) (Figure 3.38 ). The incline rises at \(35.0^{\circ}\) above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the maximum height above the ground that the rocket reaches, and (b) the greatest horizontal range of the rocket beyond point \(A\).

A particle starts from rest at the origin with an acceleration vector that has magnitude \(4 \mathrm{~m} / \mathrm{s}^{2}\) and direction \(30^{\circ}\) above the positive \(x\) axis. (a) What are the components of its velocity vector \(20 \mathrm{~s}\) later? (b) What is the particle's position at that time?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.