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Transcript
ENES100-0702 Introduction to Engineering Design R. J. Phaneuf Fall 2002 (Click here to return to list of ENES100 documents) Forces, Work and Energy A force is a push or pull which alters the state of motion of an object. From Newton’s observations we know that the net force applied to an object is proportional to its acceleration, that is, the change in its velocity per unit time: F ma (1). This is referred to as Newton’s second law. You are most likely familiar also with the concept of work. The work done by a force is defined as : W F r (2), that is, by the dot product of the force and the change in position. Notice that only that component along the displacement does work. The component of a force perpendicular to the displacement does no work. If the force is opposite the motion (for example frictional forces always oppose motion) then it does negative work. Work and Kinetic Energy Now consider the work done by a constant horizontally-applied force in accelerating an object, initially at rest to a final velocity vx = vf, over a time period t: W F r Fx max (3) To calculate x we note that the average velocity <v> = ½vf, since for a constant applied force the acceleration is also constant, x v t 1 v f t 2 (4), so that the work can be written: 1 W ma v f t 2 (5), where v f at (6), or a vf (7) t and substituting this into equation (5) finally yields: v 1 1 2 W m f v f t mv f t 2 2 (8) The quantity ½ mv2 has a special significance. We associate it with the energy of a moving object, the kinetic energy, KE = ½ mv2 (9). Thus the work done in accelerating the object from its original condition at rest is equal to the kinetic energy. More generally, the work done in accelerating an object with an initial horizontal speed v0 to a final horizontal speed vf is the difference in kinetic energy: W = KE = ½ mvf2 – ½ mv02 . (10) Rotational Motion Even if the position of the center of mass of an object is fixed, it can often be set into rotational motion if the application of force results in a torque about the axis of rotation: r F . (11) It is possible to describe the rotational motion of the object using an equation analogous to Newton’s second law: I (12) where is the angular acceleration, and I is the moment of inertia of the object about its rotational axis. The angular acceleration is equal to the time rate of change of the angular velocity, which in turn is equal to the time rate of change of the angular position: t (13) , t (14) where is a vector whose magnitude is the change in the angular position of the object and whose direction is given by the right hand rule. Note that the units for angle are radians, with 2π radians in one rotation. Thus the angular velocity ω is measured in radians/s. There is a related quantity, the rotational frequency which is proportional to the angular velocity: (1rotation / 2radian ) , (15) which is measured in rotations/s or more commonly rotations/minute (RPM): [ RPM ] (1rotation / 2radians )60s / 1min . (16) The moment of inertia for an object made up of a collection of masses mi displaced from the rotational axis by ri can be written as: I mi ri , 2 (17) i which would be replaced by an integral for a continuous distribution of mass. For an object with a uniform mass density ρ=M/V, the moment of inertia can be easily determined. For example for a disk of radius R, Idisk=MR2/2. For a ring of inner radius Ri and outer radius Ro, I ring M Ri Ro / 2 , and for a sphere of radius R, I sphere 2 2 2 MR 2 . 5 For an object rotating with constant angular velocity ω, the associated kinetic energy can be written by analogy to equation (9): KErot 1 2 I 2 (18) For the case of an object both rotating and translating KEtotal KEtrans KErot 1 2 1 2 mv I . 2 2 (19) Work and Potential Energy Now consider a second simple case: that of a vertical force applied on an object, initially moving upward with a velocity vy. If the applied force is equal and opposite to that exerted by gravity, the velocity will be constant. The work done by the applied force in moving the object from its initial position y0 to its final position yf is given by W F r mg( y f y0 ) . (20) In this case there is no change in velocity, and thus no change in kinetic energy. Where did the work go? We recognize that an object at a height H = yf –y0 initially at rest would gain kinetic energy if released, due to the action of gravity. There must be energy stored in raising the height of an object above the ground. We refer to this stored energy as potential energy. From the above expression for the work we define the difference in gravitational potential energy as PE = mg(y – y0). (21) There are many other examples of forces which allow potential energy to be stored-the Hooke’s law force associated with deflection of a spring is another familiar example. For forces of this type, in the absence of applied external forces the sum of the kinetic and potential energies is conserved KE + PE = Constant, (22) so that these forces are referred to as conservative forces. The second type of force is referred to as dissipative, since these forces remove energy from a system. Friction, and viscous fluid forces are examples. This does not mean that energy is not conserved, however. Dissipative forces cause heating of the objects within the system and their surroundings. If we keep track of the heat flow into the surroundings, and consider the entire system including the surroundings, energy is always conserved. We can write down the most general form of the work-energy relation WappliedWdisKE + PE, (14) where Wdis is always negative, and equal in magnitude to the heat flow into the objects of the system (increasing their so called internal energy) and surroundings. Wapp = KE + PE – Wdis (15) Power and Efficiency The rate at which a force supplies work is referred to as the power: W F r P F v . t t (16) If the work is instead supplied by a torque, it can be written in an analogous form: P (17) From equation (15) we see that the power which must be generated in moving an object, goes into increasing its kinetic energy, increasing its potential energy, and in doing work against dissipative forces. The overall input power Pin, which must be supplied by applied forces and/or torques is given by: Pin KE PE Wdis Pout Pdis t t t where the output power is defined as Pout dissipative forces is defined as Pdis (18) KE PE , and the power supplied against t t Wdis . t It is possible to define the efficiency of a process in which power is used to increase the kinetic and potential energy of an object, as well as to counter dissipative forces: Pout Pout Pin Pout Pdis (19) Note that since Wdis is always negative, the power supplied against dissipation is always positive, and thus the efficiency of a process is less than or equal to 1. (Click here to return to list of ENES100 documents)