/*! 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 32 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
The aircraft engine provides a torque of approximately 413.72 Nm and does work of about 2600 Joules in one revolution of the propeller.

Step by step solution

01

Calculating Torque

Torque can be derived from the relationship between Power (P), Torque (T) and rotational speed (N) given by the formula \(P = 2 \pi N T / 60\). But before we can use this formula, we need to convert the Power from Horsepower (hp) to Watt (W): knowing that 1 hp is equivalent to 746 W, the Power in Watt becomes \(P = 175 * 746 = 130550 W\). Next, we calculate Torque using the formula: \(T = 60P / (2 \pi N) = 60 * 130550 / (2* \pi * 2400) = 413.72 Nm.\)
02

Calculating Work

We know that one revolution corresponds to a distance of \(2 \pi\) radians. We can use the relation between Work (W), Torque (T) and rotation (in radians) to determine the work done, which is given by \(W = T * 2 * \pi = 413.72 * 2 * \pi = 2600 J.\)
03

Interpretation of Results

The results represent the torque which the engine applies to the aircraft propeller, and the work done by the engine for a complete revolution of the propeller.

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Key Concepts

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

Understanding Torque in Physics
Torque is a fundamental concept in physics which concerns rotational motion. It can be understood as the rotational equivalent of force. Just as force causes an object to move in a straight line, torque causes an object to rotate. The calculation of torque is given by the equation:
\[ T = F \times r \]
where \( T \) is the torque, \( F \) is the force applied, and \( r \) is the distance from the pivot point to the point where the force is applied, or the 'lever arm'.

In the context of an aircraft engine, calculating torque involves determining how much turning force the engine provides to the propeller. As seen in the solution provided, another key relationship involving power \( (P) \), torque \( (T) \), and rotational speed \( (N) \) is employed. Understanding these relationships is crucial, especially when we need to convert power units and work within specific rotational speeds typical for aircraft propellers. It's important to ensure all units are consistent when applying these equations. For example, if power is given in horsepower, it must be converted to watts before using it in the torque formula.
Exploring the Work-Energy Principle
The work-energy principle is a core concept in the study of physics, particularly dynamics, where it relates the work done on an object to its change in energy. Work is defined as the process of energy transfer that occurs when an object is moved over a distance by an external force. The basic formula for work when force is applied to produce displacement is:
\[ W = F \times d \times \text{cos}(\theta) \]
Here, \( W \) is the work done, \( F \) is the force applied, \( d \) is the displacement, and \( \theta \) is the angle between the force and the displacement vector.

However, when dealing with rotational systems, like an aircraft engine propeller, work is calculated by the product of torque and the rotational displacement, which, for one full revolution, is \( 2 \pi \) radians. This exercise showcases the application of the work-energy principle in a practical context where the work done translates into the energy imparted to the plane's propeller for each revolution.
Power to Torque Conversion
Power to torque conversion is a key topic when studying dynamics, especially in machines with rotating components like engines. Power is the rate at which work is done, or energy is transmitted. In rotational motion, power can be expressed as a function of torque and angular velocity. Converting power, measured in watts or horsepower, to torque, requires knowing the speed of the rotating element in revolutions per minute (rpm). The formula that connects these quantities is:
\[ P = \frac{2 \pi N T}{60} \]
where \( P \) represents power, \( N \) is the speed in revolutions per minute, and \( T \) denotes torque. In an aircraft engine scenario, this conversion is essential for translating the engine's power output into the usable torque that actually turns the propeller. This understanding is necessary for optimizing performance and ensuring the engine delivers the necessary force to keep the aircraft moving efficiently. To make the calculation, if power is provided in horsepower, it must first be converted to watts, acknowledging the equivalency of 1 horsepower to 746 watts. This conversion is a critical step in determining the actual torque output of an aircraft engine, as seen in the exercise provided.

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Most popular questions from this chapter

The mechanism shown in Fig. \(\mathbf{P} \mathbf{1 0 . 6 4}\) is used to raise a crate of supplies from a ship's hold. The crate has total mass \(50 \mathrm{~kg} .\) A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius \(0.25 \mathrm{~m}\) and moment of inertia \(I=2.9 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axle. The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius \(0.12 \mathrm{~m},\) the cylinder turns, and the crate is raised. What magnitude of the force \(\vec{F}\) applied tangentially to the rotating crank is required to raise the crate with an acceleration of \(1.40 \mathrm{~m} / \mathrm{s}^{2} ?\) (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel \(0.650 \mathrm{~m}\) in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects \(0.200 \mathrm{~m}\) at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is \(8.00 \mathrm{~kg} ;\) its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at \(5.00 \mathrm{rev} / \mathrm{s}\). Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at \(0.050 \mathrm{rev} / \mathrm{s} ;\) (c) when the shaft is rotating in a horizontal plane about its center at \(0.300 \mathrm{rev} / \mathrm{s}\). (d) At what rate must the shaft rotate in order that it may be supported at one end only?

A uniform solid cylinder with mass \(M\) and radius \(2 R\) rests on a horizontal tabletop. A string is attached by a yoke to a frictionless axle through the center of the cylinder so that the cylinder can rotate about the axle. The string runs over a disk-shaped pulley with mass \(M\) and radius \(R\) that is mounted on a frictionless axle through its center. A block of mass \(M\) is suspended from the free end of the string (Fig. \(\mathbf{P 1 0 . 7 9}\) ). The string doesn't slip over the pulley surface, and the cylinder rolls without slipping on the tabletop. Find the magnitude of the acceleration of the block after the system is released from rest.

A hollow, spherical shell with mass \(2.00 \mathrm{~kg}\) rolls without slipping down a \(38.0^{\circ}\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of static friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to \(4.00 \mathrm{~kg}\) ?

A solid ball is released from rest and slides down a hillside that slopes downward at \(65.0^{\circ}\) from the horizontal. (a) What minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping to occur? (b) Would the coefficient of friction calculated in part (a) be sufficient to prevent a hollow ball (such as a soccer ball) from slipping? Justify your answer. (c) In part (a), why did we use the coefficient of static friction and not the coefficient of kinetic friction?

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