/*! 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 An engine delivers 175 hp to an ... [FREE SOLUTION] | 91Ó°ÊÓ

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An engine delivers 175 hp to an aircraft propeller at 2400 rev/min. (a) How much torque does the aircraft engine provide? (b) How much work does the engine do in one revolution of the propeller?

Short Answer

Expert verified
(a) The torque is approximately 519.28 Nm. (b) The work done in one revolution is approximately 3261 J.

Step by step solution

01

Convert horsepower to watts

The power output of the engine is given as 175 horsepower (hp). We need to convert this to watts (W) because the standard unit of power in physics is watts. Using the conversion factor, 1 hp = 746 W, we can calculate:\[P = 175 \text{ hp} \times 746 \frac{W}{hp} = 130550 \text{ W}\]
02

Convert revolutions per minute to radians per second

Since angular velocity \( \omega \) is often needed in radians per second (rad/s), we must convert 2400 revolutions per minute (rev/min) appropriately. Use the conversion factor:\[\omega = 2400 \frac{rev}{min} \times \frac{2 \pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}}\]This gives:\[\omega \approx 251.33 \text{ rad/s}\]
03

Find the torque provided by the engine

The torque \( \tau \) provided by the engine can be found using the relationship between power, torque, and angular velocity:\[P = \tau \cdot \omega\]Solving for torque gives:\[\tau = \frac{P}{\omega} = \frac{130550 \text{ W}}{251.33 \text{ rad/s}} \approx 519.28 \text{ Nm}\]
04

Calculate work done in one revolution

Work done in one complete revolution can be calculated using the torque and the angular displacement in radians for one revolution (which is \(2\pi\) radians). Work \(W\) can be described by:\[W = \tau \times \theta\]Thus the work done in one revolution is:\[W = 519.28 \text{ Nm} \times 2\pi \text{ rad} \approx 3261 \text{ J}\]

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Key Concepts

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

Power Conversion
Power conversion is an essential concept in physics and engineering that involves changing the units of power from one system to another. In this exercise, we began with power measured in horsepower (hp), a common unit used in the automotive and aviation industries. However, to perform scientific calculations, it's usually best to convert to watts (W), the standard unit of power in the International System of Units (SI).
  • 1 horsepower is equivalent to 746 watts.
  • Using this conversion, we can determine the power output of engines or motors in terms that are universally understood in physics and engineering fields.
Converting power to watts helps streamline calculations involving torque and angular velocity, both critical to analyzing mechanical systems. Understanding power conversion ensures accuracy and consistency when working with different systems and measures globally. It emphasizes the importance of universal standards in scientific communication.
Angular Velocity
Angular velocity is a measure of the rate of rotation. It specifies how fast an object rotates or revolves relative to another point or axis. For engines and motors, angular velocity is often given in revolutions per minute (rev/min). However, like many rotational quantities, it's usually converted to radians per second (rad/s) during calculations.
  • A full circle equals 2Ï€ radians.
  • To convert rev/min to rad/s, multiply by \(2Ï€ \frac{ rad}{rev}\) and divide by 60 to switch from minutes to seconds.
The angular velocity helps us understand how quickly a system rotates and is crucial for computing torque. In essence, angular velocity bridges the gap between physical rotation and mathematical representations. When computing aspects like torque in our exercise, having the angular velocity in rad/s simplifies the process and aligns with scientific standards.
Work Done
Work done is a vital concept when discussing energy transfer in mechanical systems. In rotational dynamics, work is performed when torque causes an object to rotate through some angle.
The formula for work done by a torque is:
\[W = \tau \times \theta\]
where \(\tau\) is the torque and \(\theta\) is the angular displacement.
In this exercise:
  • The work done in one revolution is found by multiplying the torque by the angle of one revolution, which is \(2\pi\) radians.
  • This tells us the amount of energy transferred through one complete cycle of the propeller.
Understanding work in this context helps illustrate how mechanical engines convert energy into movement. It's a straightforward demonstration of principles that underpin much of mechanical engineering and physics, showing practical applications in real-world systems.
Radian Conversion
Radians are a unit of angular measure used extensively in mathematics and physics. They offer a direct relationship between the linear distance traveled and the radius of the circle in which the motion occurs.
In rotation:
  • One complete revolution is equivalent to \(2\pi\) radians, not 360 degrees.
  • Using radians in calculations often simplifies the mathematics involved in rotational dynamics.
In the exercise, converting revolutions to radians was key for accurate torque calculations. This conversion forms the backbone for linking real-world rotations to mathematical models. By working within the radians framework, one can seamlessly compute other variables like angular velocity and work done. This demonstrates the efficiency and simplicity of using radians over degrees in scientific computations.

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Most popular questions from this chapter

A thin uniform rod has a length of 0.500 \(\mathrm{m}\) and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.400 \(\mathrm{rad} / \mathrm{s}\) and a moment of inertia about the axis of \(3.00 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2} .\) A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is 0.160 \(\mathrm{m} / \mathrm{s}\) . The bug can be treated as a point mass. (a) What is the mass of the rod? (b) What is the mass of the bug?

Tarzan and Jane in the 21 st Century. Tarzan has foolishly gotten himself into another scrape with the animals and must be rescued once again by Jane. The 60.0 -kg Jane starts from rest at a height of 5.00 \(\mathrm{m}\) in the trees and swings down to the ground using a thin, but very rigid, 30.0 -kg vine 8.00 \(\mathrm{m}\) long. She arrives just in time to snatch the 72.0 -kg Tarzan from the jaws of an angry hippopotamus. What is Jane's (and the vine's) angular speed (a) just before she grabs Tarzan and (b) just after she grabs him? (c) How high will Tarzan and Jane go on their first swing after this daring rescue?

While exploring a castle, Exena the Exterminator is spotted by a dragon that chases her down a hallway. Exena runs into a room and attempts to swing the heavy door shut before the dragon gets her. The door is initially perpendicular to the wall, so it must be turned through \(90^{\circ}\) to close. The door is 3.00 \(\mathrm{m}\) tall and 1.25 \(\mathrm{m}\) wide, and it weighs 750 \(\mathrm{N} .\) You can ignore the friction at the hinges. If Exena applies a force of 220 \(\mathrm{N}\) at the edge of the door and perpendicular to it, how much time does it take her to close the door?

When an object is rolling without slipping, the rolling friction force is much less the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3\() .\) When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{z}\) . If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\) , rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal speed \(\omega_{0}\) about the kinetic friction coefficient is \(\mu_{\mathrm{k}}\) (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force. on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0 .\) Rolling without slipping sets in when \(v_{x}=R \omega_{z} .\) Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

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