Technology & Engineering
Linear Momentum Equation
The linear momentum equation, also known as Newton's second law of motion, states that the rate of change of momentum of an object is directly proportional to the force acting on it. Mathematically, it is expressed as F = dp/dt, where F is the force, p is the momentum, and t is time. This equation is fundamental in understanding and analyzing the motion of objects.
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11 Key excerpts on "Linear Momentum Equation"
eBook - PDF- Donald F. Elger, Barbara A. LeBret, Clayton T. Crowe, John A. Roberson(Authors)
- 2019(Publication Date)
- Wiley(Publisher)
156 The Momentum Equation CHAPTER ROAD MAP This chapter presents (a) the Linear Momentum Equation and (b) the angular momentum equation. Both equations are derived from Newton’s second law of motion. CHAPTERSIX FIGURE 6.1 Engineers design systems by using a small set of fundamental equations, such as the momentum equation. (Photo courtesy of NASA.) LEARNING OUTCOMES NEWTON’S SECOND LAW (§6.1) ● Know the main ideas about Newton’s second law of motion. ● Solve problems that involve Newton’s second law by applying the visual solution method. THE Linear Momentum Equation (§6.2 to §6.4) ● List the steps to derive the momentum equation and explain the physics. ● Draw a force diagram and a momentum diagram. ● Explain or calculate the momentum flow. ● Apply the Linear Momentum Equation to solve problems. MOVING CONTROL VOLUMES (§6.5) ● Distinguish between an inertial and noninertial reference frame. ● Solve problems that involve moving control volumes. 6.1 Understanding Newton’s Second Law of Motion The momentum equation, which is the subject of this chapter, is useful for solving many types of engineering problems. For example, the momentum equation is applied to the design of rockets; see Fig. 6.1. The momentum equation is derived from Newton’s second law of motion, which is described next. Body and Surface Forces A force is an interaction between two bodies that can be idealized as a push or pull of one body on another body. A push/pull interaction is one that can cause acceleration. Newton’s third law tells us that forces must involve the interaction of two bodies and that forces occur in pairs. The two forces are equal in magnitude, opposite in direction, and colinear. EXAMPLE. To give examples of force, consider an airplane that is flying in a straight path at constant speed (Fig. 6.2). Select the airplane as the system for analysis. Idealize the air- plane as a particle. Newton’s first law (i.e., force equilibrium) tells us that the sum of forces must balance.
- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2023(Publication Date)
- Wiley(Publisher)
The first definition concerns a familiar word—momentum—that has several meanings in everyday language but only a single precise meaning in physics and engineering. The linear momentum of a particle is a vector quantity p → that is defined as p → = m v → (linear momentum of a particle), (9.3.1) in which m is the mass of the particle and v → is its velocity. (The adjective linear is often dropped, but it serves to distinguish p → from angular momentum, which is introduced in Chapter 11 and which is associated with rotation.) Since m is always a positive scalar quantity, Eq. 9.3.1 tells us that p → and v → have the same direction. From Eq. 9.3.1, the SI unit for momentum is the kilogram-meter per second (kg · m/s). Force and Momentum. Newton expressed his second law of motion in terms of momentum: In equation form this becomes F → net = d p → ____ dt . (9.3.2) In words, Eq. 9.3.2 says that the net external force F → net on a particle changes the particle’s linear momentum p → . Conversely, the linear momentum can be changed only by a net external force. If there is no net external force, p → cannot change. As we shall see in Module 9.5, this last fact can be an extremely powerful tool in solving problems. Manipulating Eq. 9.3.2 by substituting for p → from Eq. 9.3.1 gives, for constant mass m, F → net = d p → ___ dt = d _ dt (m v → ) = m d v → ___ dt = m a → . Thus, the relations F → net = d p → / dt and F → net = m a → are equivalent expressions of Newton’s second law of motion for a particle. The time rate of change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of that force. as the product of the particle’s mass and velocity. 9.3.3 Calculate the change in momentum (magnitude and direction) when a particle changes its speed and direction of travel. 9.3.4 Apply the relationship between a particle’s momentum and the (net) force acting on the particle.
eBook - PDF- Donald F. Elger, Barbara A. LeBret, Clayton T. Crowe, John A. Robertson(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
188 The Momentum Equation CHAPTER ROAD MAP This chapter presents (a) the Linear Momentum Equation and (b) the angular momentum equation. Both equations are derived from Newton’s second law of motion. CHAPTERSIX FIGURE 6.1 Engineers design systems by using a small set of fundamental equations, such as the momentum equation. (Photo courtesy of NASA.) LEARNING OUTCOMES NEWTON’S SECOND LAW (§6.1). ● Know the main ideas about Newton’s second law of motion. ● Solve problems that involve Newton’s second law by applying the visual solution method. THE Linear Momentum Equation (§6.2 to §6.4). ● List the steps to derive the momentum equation and explain the physics. ● Draw a force diagram and a momentum diagram. ● Explain or calculate the momentum flow. ● Apply the Linear Momentum Equation to solve problems. MOVING CONTROL VOLUMES (§6.5). ● Distinguish between an inertial and noninertial reference frame. ● Solve problems that involve moving control volumes. 6.1 Understanding Newton’s Second Law of Motion Because Newton’s second law is the theoretical foundation of the momentum equation, this section reviews relevant concepts. Body and Surface Forces A force is an interaction between two bodies that can be idealized as a push or pull of one body on another body. A push/pull interaction is one that can cause acceleration. Newton’s third law tells us that forces must involve the interaction of two bodies and that forces occur in pairs. The two forces are equal in magnitude, opposite in direction, and colinear. EXAMPLE. To give examples of force, consider an airplane that is flying in a straight path at constant speed (Fig. 6.2). Select the airplane as the system for analysis. Idealize the air- plane as a particle. Newton’s first law (i.e., force equilibrium) tells us that the sum of forces must balance. There are four forces on the airplane: • The lift force is the net upward push of the air (body 1) on the airplane (body 2).
eBook - PDFEngineering Mechanics
Dynamics
- L. G. Kraige, J. N. Bolton(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
We found that the velocity changes could be expressed directly in terms of the work done or in terms of the overall changes in energy. In the next two articles, we will integrate the equation of motion with respect to time rather than displacement. This approach leads to the equations of impulse and momentum. These equations greatly facilitate the solution of many problems in which the applied forces act during extremely short periods of time (as in impact problems) or over specified intervals of time. 3/9 Linear Impulse and Linear Momentum Consider again the general curvilinear motion in space of a particle of mass m, Fig. 3∕11, where the particle is located by its position vector r measured from a fixed origin O. The velocity of the particle is v = r ˙ and is tangent to its path (shown as a dashed line). The resultant ΣF of all forces on m is in the direction of its acceleration v ˙ . We may now write the basic equation of motion for the particle, Eq. 3∕3, as ΣF = mv ˙ = d dt ( mv) or ΣF = G ˙ (3∕ 25) where the product of the mass and velocity is defined as the linear momentum G = mv of the particle. Equation 3 ∕25 states that the resultant of all forces acting on a particle equals its time rate of change of linear momentum. In SI the units of linear momentum mv are seen to be kg ∙ m∕ s, which also equals N ∙ s. In U.S. customary units, the units of linear momentum mv are [lb∕(ft∕ sec 2 )][ft∕ sec] = lb-sec. Because Eq. 3∕ 25 is a vector equation, we recognize that, in addition to the equality of the magnitudes of ΣF and G ˙ , the direction of the resultant force coin- cides with the direction of the rate of change in linear momentum, which is the direction of the rate of change in velocity. Equation 3∕ 25 is one of the most useful and important relationships in dynamics, and it is valid as long as the mass m of the particle is not changing with time. The case where m changes with time is dis- cussed in Art.
eBook - ePub- Tapan K. Sengupta, Peter Belobaba, Jonathan Cooper, Allan Seabridge, Peter Belobaba, Jonathan Cooper, Allan Seabridge(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
The time derivative on the right-hand side is the local acceleration component, while the rest of the terms on the right-hand side constitute the convective acceleration. 2.3 Conservation of Linear Momentum: Integral Form First, we derive it in integral form due to its simplicity and its appearance which allows relating to physical causes giving rise to different terms. For the derivation of integral form of momentum equation, one starts from Newton's second law of motion, i.e. If one represents the forces for unit volume and then where the net efflux of momentum through the control surfaces is given by the last term on the right-hand side. 2.4 Conservation of Linear Momentum: Differential Form This is obtained by applying Newton's second law on an infinitesimal control volume representing a fluid particle. The net force acting on the fluid particle is equal to the time rate of change of linear momentum of the fluid particle. As the fluid element moves in space, its shape and volume may change, but its mass is conserved. In an inertial coordinate system (2.9) Writing Equation (2.9) for a fluid element of unit volume (2.10) where is the mass dependent body force. Typical examples encountered in fluid dynamics are Coriolis force, gravitational force and other electrodynamic/magnetic forces appear due to electromagnetic effects. represents all form of surface forces, which act on the surface of an immersed body due to various forms of stress distribution. Typical examples are the pressure and shear forces due to normal and shear stresses appearing as point forces. The surface forces depend on the rate at which the fluid is strained by the present velocity field in the flow. The system of applied forces defines a state of stress and one of the tasks is to relate stress with strain rate. We specifically note that this relationship is an empirical one
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
Key Ideas ● If a system is closed and isolated so that no net external force acts on it, then the linear momentum P → must be constant even if there are internal changes: P → = constant (closed, isolated system). ● This conservation of linear momentum can also be written in terms of the system’s initial momentum and its momentum at some later instant: P → i = P → f (closed, isolated system). Conservation of Linear Momentum Suppose that the net external force F → net (and thus the net impulse J → ) acting on a system of particles is zero (the system is isolated) and that no particles leave or enter the system (the system is closed). Putting F → net in Eq. 9.3.6 then yields dP → /dt = 0, which means that P → = constant (closed, isolated system ). (9.5.1) In words, If the component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change. This result is called the law of conservation of linear momentum and is an extremely powerful tool in solving problems. In the homework we usually write the law as P → i = P → f (closed, isolated system). (9.5.2) In words, this equation says that, for a closed, isolated system, ( total linear momentum at some initial time t i ) = ( total linear momentum at some later time t f ) . Caution: Momentum should not be confused with energy. In the sample prob- lems of this module, momentum is conserved but energy is definitely not. Equations 9.5.1 and 9.5.2 are vector equations and, as such, each is equivalent to three equations corresponding to the conservation of linear momentum in three mutually perpendicular directions as in, say, an xyz coordinate system. Depend- ing on the forces acting on a system, linear momentum might be conserved in one or two directions but not in all directions. However, If no net external force acts on a system of particles, the total linear momentum P → of the system cannot change.
eBook - PDFElementary Heat Transfer Analysis
Pergamon Unified Engineering Series
- Stephen Whitaker, Thomas F. Irvine, James P. Hartnett(Authors)
- 2014(Publication Date)
- Pergamon(Publisher)
(4.2-22) 4.2 The Laws of Mechanics In t h e analysis of a c o n t i n u o u s m e d i u m t h e r e are t w o l a w s of m e c h a n i c s [4], t h e linear m o m e n t u m principle ( E u l e r ' s first l a w of m e c h a n i c s ) a n d t h e angular m o m e n t u m principle ( E u l e r ' s s e c o n d law of m e c h a n i c s ) . T h e s e m a y b e stated as {the time rate of change of linear momentum of a body} = {the force acting on the body}, (4.2-1) a n d {the time rate of change of angular momentum of a body} = {the torque acting on the body}. (4.2-2) It m u s t b e r e m e m b e r e d t h a t for a c o n t i n u u m t h e s e t w o l a w s s t a n d as s e p a r a t e f u n d a m e n t a l p o s t u l a t e s . T h i s is p r o b a b l y c o n t r a r y to w h a t t h e s t u d e n t h a s e n c o u n t e r e d in earlier p h y s i c s o r d y n a m i c s c o u r s e s ; h o w e v e r , t h e p r o b l e m s e n c o u n t e r e d in t h o s e c o u r s e s generally did n o t deal explicitly w i t h s h e a r f o r c e s a n d c o n s e q u e n t l y a s o m e w h a t simpler p o i n t of v i e w c o u l d b e f o l l o w e d . t tSee Essays in the History of Mechanics, Chap. V, Whence the Law of Moment of Momentum, by C. Truesdell, Springer-Verlag N e w York Inc., 1968. 96 The Basic Equations of Momentum and Energy Transfer T h e s y m b o l i c f o r m of E q . 4.2-1 is -g f P v d V = p g d V + f t ( n ) d A . L>1 Jr m (t) Jr m (t) JSD m (t) (4.2-3) H e r e w e h a v e split t h e f o r c e acting o n t h e b o d y into b o d y f o r c e s ( p g r e p r e s e n t s t h e b o d y force p e r unit v o l u m e ) a n d s u r f a c e f o r c e s . T h e t e r m t (n ) r e p r e s e n t s t h e s u r f a c e f o r c e p e r unit a r e a , a n d is called t h e stress vector.
- Allan D. Kraus, James R. Welty, Abdul Aziz(Authors)
- 2011(Publication Date)
- CRC Press(Publisher)
14 Newton’s Second Law of Motion Chapter Objectives • To introduce the control volume relationship for linear momentum. • To provide applications of the momentum theorem. • To consider the control volume relationship for the moment of momentum. • To use applications taken from a number of physical situations to show applications of the moment of momentum relationship. 14.1 Introduction The third fundamental law to be considered is Newton’s second law . As in Chapter 13, we will develop control volume expressions, which, in this case, will be related to both linear and angular motion. The basic expressions will then be applied to a number of physical situations. 14.2 Linear Momentum Newton’s second law of motion may be stated as The time rate of change of momentum of a system is equal to the net force on the system and takes place in the direction of the net force. This statement is notable in two ways. First, it is cast in a form that includes both magni-tude and direction and is, therefore, a vector expression. Second, it refers to a system rather than a control volume. As we know by now, a system is a fixed collection of mass, whereas a control volume is a fixed region in space that encloses a different mass of fluid (or system) at different times. The transformation of the second law statement from a system to a control volume point of view is dealt with in numerous texts. We will presume the correctness of the following word equation: ⎧ ⎨ ⎩ Net force acting on the control volume ⎫ ⎬ ⎭ = ⎧ ⎨ ⎩ Rate of momentum out of the control volume by mass flow ⎫ ⎬ ⎭ − ⎧ ⎨ ⎩ Rate of momentum into the control volume by mass flow ⎫ ⎬ ⎭ + ⎧ ⎨ ⎩ Rate of accumulation of momentum within the control volume ⎫ ⎬ ⎭ (14.1) 441 442 Introduction to Thermal and Fluid Engineering m e Control Volume m i . . FIGURE 14.1 A general control volume and flow field. The control volume shown in Figure 14.1 has the same general features that are shown in Figure 13.1.
eBook - PDFEngineering Dynamics
A Comprehensive Introduction
- N. Jeremy Kasdin, Derek A. Paley(Authors)
- 2011(Publication Date)
- Princeton University Press(Publisher)
In general the tangential components of the velocities are constant and the normal components obey the law of conservation of total linear momentum, m P v P (t 2 ) + m Q v Q (t 2 ) = m P v P (t 1 ) + m Q v Q (t 1 ). The normal components also obey the coefficient-of-restitution equation, e = v P (t 2 ) − v Q (t 2 ) v Q (t 1 ) − v P (t 1 ) . . In a collision between a particle and a fixed surface, the tangential components are constant, but the total linear momentum is no longer conserved for the normal components. The coefficient-of-restitution equation is e = − v P (t 2 ) v P (t 1 ) . LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM 237 . For a variable-mass system with mass flowing in a steady stream at a rate ˙ m , the equation of motion for the center of mass is m G I a G/O = F ( ext ) cv + ˙ m B v in /G − B v out /G , where B v in /G and B v out /G are the inflow and outflow velocities. . For a variable-mass system with mass flowing in at a rate ˙ m in and out at a rate ˙ m out , the equation of motion for the center of mass is m G (t) I a G/O = F G − ˙ m out B v out /G + ˙ m in B v in /G , where B v out /G is the relative velocity of the outflowing mass and B v in /G is the relative velocity of the inflowing mass. 6.6 Notes and Further Reading Although Newton formulated the basic laws of mechanics, it was Euler who showed they could be applied to collections of particles independently and first developed the techniques of multiparticle systems we presented here. (He also defined the center of mass and formulated the N -body problem.) In modern texts it forms the basis of the development of rigid-body motion, which we discuss in Chapter 9. Almost all introductory texts have some discussion of multiparticle systems and impacts.
eBook - PDF- Ronald L. Panton(Author)
- 2013(Publication Date)
- Wiley(Publisher)
There are three major independent dynamical laws in continuum mechanics: the continuity equation, momentum equation, and energy equation. These laws are formulated in the first part of this chapter. There are several additional laws that may be derived from the momentum equation. The first of these governs kinetic energy, the second angular momentum, and the third vorticity. The first two are treated in this chapter; the law governing vorticity is introduced in Chapter 13. The last law we study in this chapter is the second law of thermodynamics. In the final sections we give the integral or global forms of the laws and jump conditions that apply across discontinuities. 5.1 CONTINUITY EQUATION The equation derived in this section has been called the continuity equation to emphasize that the continuum assumptions (the assumption that density and velocity may be defined at every point in space) are prerequisites. The continuum assumption is, of course, a foundation for all the basic laws. The physical principle underlying the equation is the 74 5.1 Continuity Equation 75 Figure 5.1 Continuity equation for a material region. Surface velocity is equal to fluid velocity. conservation of mass. It may be stated in terms of a material region as follows: The time rate of change of the mass of a material region is zero. The mass of the material region (MR) is computed by integrating the density over the region. Thus, in mathematical terms we have dM MR dt = d dt MR ρ dV = 0 (5.1.1) The bounding surface of the material region is moving with the local fluid velocity v i (Fig. 5.1).
eBook - ePub- Richard E. Meyer(Author)
- 2012(Publication Date)
- Dover Publications(Publisher)
CHAPTER 2Momentum Principle and Ideal Fluid
10. Conservation of Linear Momentum
Postulate VI (Momentum Principle). A vector field f and a tensor field with components p if are defined on the closure of any fluid domain Ωt with regular boundary surface ∂Ωt so that(10.1)where n is the unit outward normal on ∂Ωt .The left-hand side of (10.1) represents the rate of change of the linear momentum of the body of fluid which occupies the domain Ωt , and the right-hand side therefore represents the resultant external force exerted on this body of fluid. The postulate states that external force can be exerted on the fluid in two ways: by a stress acting on the surface of the fluid and by a body force ρ f acting directly on each volume element. The most common body forces are gravity, centrifugal and Coriolis forces (to be considered only in Chapter 5), and electromagnetic forces (not considered in this book). Since the postulate applies also to every subdomain of a fluid domain, p ij n j represents the i -component of force per unit area exerted across an area element with “outward” unit normal n by the fluid on the “outer” side of the area element upon the fluid on its “inner” side.It should be emphasized that this postulate is more fundamental than those of Sections 1 and 2, in the sense that Postulates V, VI, and VIII (in contrast to Postulates I to IV) hold for all of continuum mechanics.It may also be noted that the momentum principle does not require the pointwise definition of f i and p ij . Indeed, the values of these quantities (and of the density) at a point are not directly observable, and only the integrals occurring in Postulates V and VI have strict physical meaning. As noted in Section 3, however, ρ and p ij
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