Angular Work and Power
Angular work and power are concepts that relate to rotational motion. Angular work is the product of the torque applied to an object and the resulting angular displacement, while angular power is the rate at which work is done. These concepts are important in understanding the transfer of energy and the performance of rotating systems.
5 Key excerpts on "Angular Work and Power"
eBook - ePub- A. D. Johnson(Author)
- 2017(Publication Date)
- Routledge(Publisher)
...Work, energy and power 5 5.1 AIMS To introduce the concepts of work, energy and power. To define the equations used to analyse work, energy and power. To explain how work, energy and power are applied to both linear and angular motion. To define the relationship between potential energy and kinetic energy. To define efficiency in terms of power. 5.2 Work A body which possesses energy, perhaps due to its motion or its position, has the capacity to do work. Before energy or power can be discussed it is, therefore, important to understand what is meant by the terms ‘to do work’ or ‘work done’. Work is done when the point of application of a force is exerted over a distance and can be defined as follows: When a force of 1 N is exerted over a distance of 1 m, then a value of 1 N m of work has been expended. Since the unit of N m is used to define a moment or a torque, the unit of work is termed the joule and designated J. work done = force × distance w = F × s (J) (5.1) Example 5.1 At constant velocity a car is opposed by a resistive force of 150N. Find the work done in moving the car through a distance of 2 km. Solution The ‘work done’ can be found by using equation (5.1) : w = F × s Now F = 150 N and s = 2000m so that w = 150 × 2000 = 300 000 J = 300 kJ 5.2.1 Work represented by a diagram Figure 5.1 shows a graph whose vertical axis represents force and whose horizontal axis represents distance. The area enclosed represents the total work done. Here, the point of application of a force of 50 N moves through a distance of 12 m, giving a total value of w of 600 J. Work done may also be represented by the area enclosed under a curve on a graph. Consider a spring which is gradually extended by applying a force of increasing magnitude. If the extension of the spring and the load are plotted on a force/extension graph, as in Figure 5.2, a straight line can be plotted. The area under the graph line represents the work done. Fig. 5.1 Work represented by area. Fig...
eBook - ePub- W. Bolton(Author)
- 2015(Publication Date)
- Routledge(Publisher)
...Consider an object which, starting from rest, is set rotating with an angular acceleration α as a result of a torque T produced by a tangential force F (Figure 27.13) The work done by the force is Tθ. But T = Iα and thus the work done = Iαθ. As a result of this rotation from rest through angle θ, the object has an angular velocity ω. Using ω 2 = + 2 αθ, we have ω 2 = 0 + 2 αθ and so the work done = ½ Iω 2. This is the energy transferred to the body as a result of the work done and is called the angular kinetic energy. Thus, for angular motion: angular kinetic energy = ½ Iω 2 Example What is the angular kinetic energy of a clutch plate, mass of 40 kg and a radius of gyration of 160 mm, rotating at 10 rev/s? The moment of inertia I = mk 2 = 40 × 0.160 2 kg m 2 and so the angular kinetic energy is ½× 40 × 0.160 2 × (2 π × 10) 2 = 2021 J. Example A solid cylinder, density of 7000 kg/m 3, of diameter 25 mm and length 25 mm is allowed to roll down an inclined plane, the plane being at an elevation of 20° to the horizontal. Determine the linear velocity of the cylinder when it has rolled a distance of 1.2 m. The loss in potential energy of the cylinder in rolling down the slope must equal the gain in linear kinetic energy plus the gain in rotational kinetic energy. The loss in potential energy is mgh with m being ¼ π × 0.025 2 × 0.025 × 7000 = 0.0859 kg and h = 1.2 sin 20° = 0.410 m. The loss in potential energy is 0.0859 × 9.81 × 0.410 = 0.345 J. The gain in linear kinetic energy is ½× 0.0859 v 2. The gain in rotational kinetic energy is ½ Iω 2, where I = ½ mr 2 = ½ × 0.0859 × 0.0125 2 = 6.71 × 10 −6 kg m 2 and ω = v / r = v /0.0125. Thus the gain in rotational energy is ½× 6.71 × 10 −6 × (v /0.0125) 2 = 0.0215 v 2...
eBook - ePub- John Bird(Author)
- 2019(Publication Date)
- Routledge(Publisher)
...It next shows how the energy and work done can be calculated from these terms. It then derives the expression which relates torque to the product of mass moment of inertia and the angular acceleration. The expression for kinetic energy due to rotation is also derived. These expressions are then used for calculating the power transmitted from one shaft to another, via a belt. This work is very important for calculating the power transmitted in rotating shafts and other similar artefacts in many branches of engineering. Science and Mathematics for Engineering. 978-0-367-2O475-4, © John Bird. Published by Taylor & Francis. All rights reserved. 28.2 Couple and torque When two equal forces act on a body as shown in Figure 28.1, they cause the body to rotate, and the system of forces is called a couple. The turning moment of a couple is called a torque T. In Figure 28.1, torque = magnitude of either force × perpendicular distance between the forces, i.e. T = Fd Figure 28.1 The unit of torque is the newton metre, N m. When a force F newtons is applied at a radius r metres from the axis of, say, a nut to be turned by a spanner, as shown in Figure 28.2, the torque T applied to the nut is given by: T = Fr N m Figure 28.2 Problem 1. Determine the torque when a pulley wheel of diameter 300 mm has a force of 80 N applied at the rim Torque T = Fr, where force F = 80 N and. radius r = 3 0 0 2 = 1 5 0 m m = 0. 1 5 m Hence, torque T = (80)(0.15) = 12 N m Problem 2. Determine the force applied tangentially to a bar of a screw jack at a radius of 800 mm, if the torque required is 600 N m Torque T = force × radius, from which force = t o r q u e r a d i u s = 6 0 0 N m 8 0 0 × 1 0 - 3 m = 750 N Problem 3. The circular hand-wheel of a valve of diameter 500 mm has a couple applied to it composed of two forces, each of 250 N...
eBook - ePub- William Bolton(Author)
- 2012(Publication Date)
- Routledge(Publisher)
...Determine the work done for that half rotation. Figure 5.9 Example The work done is the area under the graph with the angle having to be converted into radians. Thus: work done = ½ 200 × π/4 + 200 × π/4 + ½ 200 × π/2 = 393 J Revision 7 Determine the work done per revolution for a shaft for which the torque rises at a constant rate from 0 at 0° to 30 N m at 90°, is then constant at 30 N m from 90° to 270°, then decreases at a constant rate to 0 at 360°. 5.2.4 Power Power is the rate at which energy is transferred. With the energy in units of joule and time in seconds then power is in J/s or watt (W). For an object on which work is done for a time t and results in a displacement s in that time in the direction of the force, the power P developed for a constant force F is: But the average velocity v = s/t and so: P = Fv [8] For an object which is rotated by a constant torque T through an angle θ in time t : But the angular velocity ω = θ / t and so: P = Tω [9] If the rotation is at n revolutions per second then ω = 2π n and so equation [ 9 ] can be written as: P = 2π nT [10] Example The locomotive of a train exerts a constant force of 120 kN on a train while pulling it at 40 km/h along a level track. What is the power? Power = Fv = 120 × 10 3 × 40 × 10 3 /3600 = 1.3 MW Example What power will be required for a pump to extract water from a mine at 5 m 3 /s and pump it through a vertical height of 20 m. Water has a density of 1000 kg/m 3. Take g as 9.8 m/s 2. A volume of 5 m 3 of water has a mass of 5000 kg and a weight of 5000g N. This weight of water has to be moved through a distance of 20 m in 1 s, i.e. an average velocity of 20 m/s. Hence: Power = Fv = 5000 × 9.8 × 20 = 980 × 10 3 W = 980 kW. Example The shaft of a car rotates at 1000 rev/min and transmits power of 40 kW...
eBook - ePub- J Jones, J Burdess, J Fawcett(Authors)
- 2012(Publication Date)
- Routledge(Publisher)
...For example, the power of an engine is the rate at which the engine can do work on another system. The units of power are Js −1, Nms −1 or in fundamental units, kgm 2 s −3. Power is often quoted in Watts (W) so that 1W = 1 Js −1 = 1 Nms −1 = 1 kgm 2 s −3. Using eqn (5.21) power may also be expressed as The term δr / δt is the magnitude of the velocity vector ν of the point of application of the force. in which θ can be interpreted as the angle between the force vector F and the velocity vector ν of the point of application of the force, as shown in Fig. 5.16. F IG. 5.16 When F and ν are given in terms of their components F x, F y and ν x, ν y eqn (5.3) shows that the power P can also be written in the form Similarly if the point of application of the force moves with ν = ds / dt along a curve C as shown in Fig. 5.17 then eqn (5.8) gives the power P as F IG. 5.17 The power is therefore determined by the tangential component of the force F t, and the velocity ν of the point of application of the force along its curve. From example 5.2 the work δW done by a moment, or torque, M can be written as The rate at which work is done by the moment is therefore For example, in a case where the moment M is applied to the end of a shaft, will be the angular velocity of the shaft. Example 5.4 Figure 5.18(a) shows the torque characteristic of a motor in which the torque M varies with its rotational speed ω such that Let us find how the power output of the motor varies with speed. From eqn (5.27) F IG. 5.18 The speed at which maximum power is developed can be obtained by differentiating eqn (ii) with respect to ω and equating the result to zero, i.e. Thus, at maximum power, The graph of P against ω is as shown in Fig. 5.18(b). When a motor with such a torque characteristic is used to drive a steady load, the size of the motor can be minimised by ensuring that the motor runs at a speed of ω 0 /2 i.e. at its maximum power condition...
