/*! 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 9 The diameters of the main rotor ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The diameters of the main rotor and tail rotor of a single-engine helicopter are \(7.60 \mathrm{~m}\) and \(1.02 \mathrm{~m}\), respectively. The respective rotational speeds are \(450 \mathrm{rev} / \mathrm{min}\) and \(4138 \mathrm{rev} / \mathrm{min}\). Calculate the speeds of the tips of both rotors. Compare these speeds with the speed of sound, \(343 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The speed of the main rotor's tip is \(179.26 \, \mathrm{m/s}\) and the tail rotor's tip is \(220.59 \, \mathrm{m/s}\). Both these speeds are lower than the speed of sound, which is \(343 \, \mathrm{m/s}\)

Step by step solution

01

Conversion of Rotational Speed

Convert the rotational speeds from revolutions per minute to radians per second. This can be done using the fact that 1 revolution is \(2\pi\) radians and 1 minute is 60 seconds. The conversions are: For main rotor, \(\omega_1 = 450 \mathrm{rev/min} \cdot \frac {2\pi \mathrm{rad}} {1 \mathrm{rev}} \cdot \frac {1\mathrm{min}} {60 \mathrm{s}} = 47.12 \, \mathrm{rad/s}\) For tail rotor, \(\omega_2 = 4138 \mathrm{rev/min} \cdot \frac {2\pi \mathrm{rad}} {1 \mathrm{rev}} \cdot \frac {1 \mathrm{min}} {60 \mathrm{s}} = 433.11 \, \mathrm{rad/s}\)
02

Calculation of Linear Speeds

Apply the formula of linear velocity, \(v = r \cdot \omega\), to calculate the speeds of the tips of both rotors. Remember to convert the diameter to radius by halving it. For main rotor, \(v_1 = r_1 \cdot \omega_1\), substituting the values we have \(v_1 = \frac{7.60 m}{2} \times 47.12 \mathrm{rad/s} = 179.26 \, \mathrm{m/s}\) For tail rotor, \(v_2 = r_2 \cdot \omega_2\), substituting the values we have \(v_2 = \frac{1.02 m}{2} \times 433.11 \mathrm{rad/s} = 220.59 \, \mathrm{m/s}\)
03

Comparision with Speed of Sound

Compare the calculated speeds with the speed of sound (343 m/s). The speeds of the main rotor and tail rotor tips are lower than the speed of sound.

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.

Helicopter Physics
Understanding the physics behind a helicopter is crucial in grasping how these incredible machines operate. A helicopter utilizes rotary wings or rotors to provide lift, which allows it to hover, take off, and land vertically. The main rotor, typically the larger one, is responsible for lifting the helicopter vertically by generating a low-pressure region above it. The smaller tail rotor counteracts the torque effect created by the main rotor's spin, allowing for directional control.

Both these rotors spin at significant speeds; their rotational speeds are measured in revolutions per minute (rev/min). While the main rotor generates lift, the tail rotor provides stability, making it an indispensable part of helicopter design. Without it, the helicopter would uncontrollably spin in the opposite direction of the main rotor. Understanding the dynamics and mechanics of these rotors helps explain how helicopters can perform their unique aerial maneuvers.
Speed of Sound Comparison
When we talk about the speed of sound, we refer to how quickly sound waves travel through a medium, typically air, at standard atmospheric conditions. This speed is approximately 343 meters per second (m/s) at sea level and room temperature (20°C or 68°F). The speed of sound changes with the medium and temperature; it increases in warmer air and decreases in colder conditions due to variations in air density.

Comparing rotor speeds with the speed of sound provides insight into a helicopter's performance and safety. For instance, if the tip speed of a helicopter's rotor exceeds the speed of sound, it can create supersonic shock waves. This phenomenon can cause excessive noise, increased aerodynamic drag, and potential structural damage to the rotor blades. In our example exercise, both the main and tail rotor speeds are below the speed of sound, which keeps the helicopter in a safe and efficient operational range.
Linear Velocity Calculation
Linear velocity refers to the speed at which a point on a rotating object travels along a straight path. In the context of helicopter rotors, this is the speed of the rotor tips as they cut through the air. Calculating this involves determining the product of the rotor's angular velocity (in radians per second) and the radius of the rotor.

To find the linear velocity, you need to first convert the rotational speed from revolutions per minute (rev/min) to radians per second (rad/s) using the conversion factor \(1 \, \text{rev} = 2\pi \, \text{radians}\) and \(1 \, \text{minute} = 60 \, \text{seconds}\). Once you have the rotor's angular velocity, you calculate the linear velocity using \(v = r \cdot \omega\), where \(v\) is the linear velocity, \(r\) is the radius of the rotor, and \(\omega\) is the angular velocity.

In the solved example, the main rotor's tips travel at approximately 179.26 m/s, and the tail rotor's tips reach 220.59 m/s, illustrating the high-speed mechanics involved in helicopter flight. Converting and understanding these values not only aids in mechanical calculations but also enhances safety and performance assessments.

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

Two satellites are in circular orbits around the Earth. Satellite A is at an altitude equal to the Earth's radius, while satellite \(B\) is at an altitude equal to twice the Earth's radius. What is the ratio of their periods, \(T_{\mathrm{B}} / T_{\mathrm{A}}\) ?

radial and tangential components) just before it is released.One method of pitching a softball is called the "windmill" delivery method, in which the pitcher's arm rotates through approximately \(360^{\circ}\) in a vertical plane before the 198-gram ball is released at the lowest point of the circular motion. An experienced pitcher can throw a ball with a speed of \(98.0 \mathrm{mi} / \mathrm{h}\). Assume the angular acceleration is uniform throughout the pitching motion and take the distance between the softball and the shoulder joint to be \(74.2 \mathrm{~cm}\). (a) Determine the angular speed of the arm in rev/s at the instant of release. (b) Find the value of the angular acceleration in \(\mathrm{rev} / \mathrm{s}^{2}\) and the radial and tangential acceleration of the ball just before it is released. (c) Determine the force exerted on the ball by the pitcher's hand (both

A \(50.0\)-kg child stands at the rim of a merry-go-round of radius \(2.00 \mathrm{~m}\), rotating with an angular speed of \(3.00 \mathrm{rad} / \mathrm{s}\). (a) What is the child's centripetal acceleration? (b) What is the minimum force between her feet and the floor of the carousel that is required to keep her in the circular path? (c) What minimum coefficient of static friction is required? Is the answer you found reasonable? In other words, is she likely to stay on the merry-go-round?

A car rounds a banked curve where the radius of curvature of the road is \(R\), the banking angle is \(\theta\), and the coefficient of static friction is \(\mu\). (a) Determine the range of speeds the car can have without slipping up or down the road. (b) What is the range of speeds possible if \(R=100 \mathrm{~m}, \theta=10^{\circ}\), and \(\mu=0.10\) (slippery conditions)?

Because of Earth's rotation about its axis, a point on the equator has a centripetal acceleration of \(0.0340 \mathrm{~m} / \mathrm{s}^{2}\), whereas a point at the poles has no centripetal acceleration. (a) Show that, at the equator,the gravitational force on an object (the object's true weight) must exceed the object's apparent weight. (b) What are the apparent weights of a \(75.0-\mathrm{kg}\) person at the equator and at the poles? (Assume Earth is a uniform sphere and take \(g=9.800 \mathrm{~m} / \mathrm{s}^{2}\).)

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.