Kinetics of Systems of Particles

Kinetics of Systems of Particles

Introduction

• Extends dynamics principles from single particles to general systems of particles.
• Unifies dynamics topics, enabling treatment of rigid and nonrigid systems.
• Rigid body: Solid system of particles with essentially unchanged distances between particles.
• Examples: Machines, land/air vehicles, rockets, spacecraft, moving structures.
• Nonrigid body: Experiences time-dependent changes in shape due to elastic or inelastic deformations.
• Examples: Air/fuel flow through an aircraft engine turbine, burned gases from a rocket motor nozzle, water through a rotary pump.
• Chapter outline includes:
  • Introduction
  • Generalized Newton’s Second Law
  • Work-Energy
  • Impulse-Momentum
  • Conservation of Energy and Momentum
  • Steady Mass Flow
  • Variable Mass
  • Chapter Review

Generalized Newton’s Second Law

• Extends Newton’s second law to a general mass system modeled by n mass particles bounded by a closed surface in space.
• The system is the mass within the envelope, and it must be clearly defined and isolated.
• m is the total system mass is m = \Sigma m_i.
• Representative particle of mass mi has external forces F1, F2, F3, … and internal forces f1, f2, f_3, …
• External forces: contact with external bodies, gravitational/electric/magnetic effects.
• Internal forces: reactions with other mass particles within the boundary.
• Particle mi is located by position vector ri from the nonaccelerating origin O of a Newtonian reference frame.
• The center of mass G of the isolated system of particles is located by the position vector \mathbf{r}, defined as: m\mathbf{r} = \Sigma mi\mathbf{r}i
• Newton’s second law applied to mi: F1 + F2 + F3 + \cdots + f1 + f2 + f3 + \cdots = mi \ddot{\mathbf{r}}_i
• Summing equations for all particles: \Sigma \mathbf{F} + \Sigma \mathbf{f} = \Sigma mi \ddot{\mathbf{r}}i
• \Sigma \mathbf{F} = vector sum of all external forces on the isolated system.
• \Sigma \mathbf{f} = vector sum of all internal forces, which is identically zero due to equal and opposite actions and reactions.
• Differentiating the equation defining \mathbf{r} twice with respect to time: m \ddot{\mathbf{r}} = \Sigma mi \ddot{\mathbf{r}}i, where m has a zero time derivative as long as mass is not entering or leaving the system.
• Substitution into the summation of the equations of motion gives \Sigma \mathbf{F} = m \ddot{\mathbf{r}} which simplifies to \mathbf{F} = m\mathbf{a}
• \mathbf{a} is the acceleration \ddot{\mathbf{r}} of the center of mass of the system.
• This is the generalized Newton’s second law of motion for a mass system, also called the equation of motion of m. It states that the resultant of the external forces on any system of masses equals the total mass of the system times the acceleration of the center of mass.
• \mathbf{a} is the acceleration of the mathematical point representing the mass center for the given n particles.
• For a nonrigid body, this acceleration need not represent the acceleration of any particular particle.
• Equation \mathbf{F} = m\mathbf{a} holds for each instant of time and is therefore an instantaneous relationship.
• Equation \mathbf{F} = m\mathbf{a} had to be proved, as it cannot be inferred directly from the single-particle equation.
• Equation \mathbf{F} = m\mathbf{a} may be expressed in component form using x-y-z coordinates or any convenient coordinate system:
  • \Sigma Fx = max
  • \Sigma Fy = may
  • \Sigma Fz = maz
• Although \mathbf{F} = m\mathbf{a} requires that the acceleration vector \mathbf{a} have the same direction as the resultant external force \Sigma \mathbf{F}, it does not follow that \Sigma \mathbf{F} necessarily passes through G. In general, \Sigma \mathbf{F} does not pass through G.

Work-Energy

• Work-energy relation for a single particle applies to a system of two joined particles.
• For a general system, such as that in Fig. 4/1, the work-energy relation for the representative particle of mass mi is (U{1-2})i = \Delta Ti.
  • (U{1-2})i is the work done on mi during an interval of motion by all external forces F1 + F2 + F3 + \cdots and all internal forces f1 + f2 + f_3 + \cdots.
  • The kinetic energy of mi is Ti = \frac{1}{2} mi vi^2, where vi is the magnitude of the particle velocity \mathbf{v}i = \dot{\mathbf{r}}_i.
• For the entire system, the sum of the work-energy equations written for all particles is \Sigma (U{1-2})i = \Sigma \Delta T_i, which may be represented by:
  • U_{1-2} = \Delta T
  • T1 + U{1-2} = T_2
  • Where U{1-2} = \Sigma (U{1-2})i, the work done by all forces (external and internal) on all particles, and \Delta T is the change in the total kinetic energy T = \Sigma Ti of the system.
• For a rigid body or a system of rigid bodies joined by ideal frictionless connections, no net work is done by the internal interacting forces or moments in the connections.
• The work done by all pairs of internal forces, labeled as fi and -fi, at a typical connection is zero since their points of application have identical displacement components while the forces are equal but opposite.
• For a nonrigid mechanical system including elastic members capable of storing energy, part of the work done by external forces goes into changing the internal elastic potential energy V_e.
• If the work done by gravity forces is excluded from the work term and accounted for instead by changes in gravitational potential energy Vg, then equate the work {U'}{1-2} done on the system during an interval of motion to the change \Delta E in the total mechanical energy of the system.
  • {U'}_{1-2} = \Delta E
  • {U'}_{1-2} = \Delta T + \Delta V
  • T1 + V1 + {U'}{1-2} = T2 + V_2
  • Where V = Vg + Ve represents the total potential energy.
• Kinetic Energy Expression
  • The expression is examined for the kinetic energy of the mass system in more detail: T = \Sigma \frac{1}{2} mi vi^2
• By the principle of relative motion, the velocity of the representative particle may be written as: \mathbf{v}i = \mathbf{v} + \dot{\boldsymbol{\rho}}i
  • \mathbf{v} is the velocity of the mass center G
  • \dot{\boldsymbol{\rho}}i is the velocity of mi with respect to a translating reference frame moving with the mass center G.
• Using the identity vi^2 = \mathbf{v}i \cdot \mathbf{v}_i, the kinetic energy of the system can be written as:
  • T = \Sigma \frac{1}{2} mi \mathbf{v}i \cdot \mathbf{v}i = \Sigma \frac{1}{2} mi (\mathbf{v} + \dot{\boldsymbol{\rho}}i) \cdot (\mathbf{v} + \dot{\boldsymbol{\rho}}i)
  • T = \Sigma \frac{1}{2} mi v^2 + \Sigma \frac{1}{2} mi |\dot{\boldsymbol{\rho}}i|^2 + \Sigma mi \mathbf{v} \cdot \dot{\boldsymbol{\rho}}_i
• Because \boldsymbol{\rho}i is measured from the mass center, \Sigma mi \boldsymbol{\rho}i = 0 and the third term is \mathbf{v} \cdot \Sigma mi \dot{\boldsymbol{\rho}}i = \mathbf{v} \cdot \frac{d}{dt} \Sigma (mi \boldsymbol{\rho}_i) = 0.
• Also, \Sigma \frac{1}{2} mi v^2 = \frac{1}{2} v^2 \Sigma mi = \frac{1}{2} m v^2.
• Therefore, the total kinetic energy is: T = \frac{1}{2} m v^2 + \Sigma \frac{1}{2} mi |\dot{\boldsymbol{\rho}}i|^2
  • The total kinetic energy of a mass system equals the kinetic energy of mass-center translation of the system as a whole plus the kinetic energy due to motion of all particles relative to the mass center.

Impulse-Momentum

• Development of the concepts of momentum and impulse is applied to a system of particles.
• From definition, the linear momentum of the representative particle of the system is \mathbf{G}i = mi \mathbf{v}i, where the velocity of mi is \mathbf{v}i = \dot{\mathbf{r}}i.
• The linear momentum of the system is defined as the vector sum of the linear momenta of all of its particles, or \mathbf{G} = \Sigma mi \mathbf{v}i.
• By substituting the relative-velocity relation \mathbf{v}i = \mathbf{v} + \dot{\boldsymbol{\rho}}i and using that \Sigma mi \boldsymbol{\rho}i = m \boldsymbol{\rho} = 0 where \boldsymbol{\rho} is the position vector from the center of mass to itself

\mathbf{G} = \Sigma mi (\mathbf{v} + \dot{\boldsymbol{\rho}}i) = \Sigma mi \mathbf{v} + \frac{d}{dt} \Sigma mi \boldsymbol{\rho}i = \mathbf{v} \Sigma mi + \frac{d}{dt} (0)

• Which simplifies to: \mathbf{G} = m \mathbf{v}
  • The linear momentum of any system of constant mass is the product of the mass and the velocity of its center of mass.
• The time derivative of \mathbf{G} is m \dot{\mathbf{v}} = m \mathbf{a}, which by \mathbf{F} = m \mathbf{a} is the resultant external force acting on the system. Thus: \Sigma \mathbf{F} = \dot{\mathbf{G}}
  • Which has the same form as for a single particle.
  • States that the resultant of the external forces on any mass system equals the time rate of change of the linear momentum of the system.
  • It is an alternative form of the generalized second law of motion.
  • In general, \mathbf{F} does not pass through the mass center G.
  • In deriving this equation, it was differentiated with respect to time and assumed that the total mass is constant. Thus, the equation does not apply to systems whose mass changes with time.
• Angular Momentum
  • The angular momentum of the general mass system is determined about the fixed point O, about the mass center G, and about an arbitrary point P, which may have an acceleration \mathbf{a}P = \ddot{\mathbf{r}}P.
  • About a Fixed Point O: The angular momentum of the mass system about the point O, fixed in the Newtonian reference system, is defined as the vector sum of the moments of the linear momenta about O of all particles of the system and is given by \mathbf{H}O = \Sigma (\mathbf{r}i \times mi \mathbf{v}i)
  • The time derivative of the vector product is: \dot{\mathbf{H}}O = \Sigma (\dot{\mathbf{r}}i \times mi \mathbf{v}i) + \Sigma (\mathbf{r}i \times mi \dot{\mathbf{v}}_i)
  • The first summation vanishes since the cross product of two parallel vectors \dot{\mathbf{r}}i and mi \mathbf{v}_i is zero.
  • The second summation is \Sigma (\mathbf{r}i \times mi \mathbf{a}i) = \Sigma (\mathbf{r}i \times \mathbf{F}_i), which is the vector sum of the moments about O of all forces acting on all particles of the system.
  • This moment sum \Sigma \mathbf{M}_O represents only the moments of forces external to the system, since the internal forces cancel one another and their moments add up to zero.
  • \Sigma \mathbf{M}O = \dot{\mathbf{H}}O
    • The resultant vector moment about any fixed point of all external forces on any system of mass equals the time rate of change of angular momentum of the system about the fixed point.
    • This equation does not apply if the total mass of the system is changing with time.
  • About the Mass Center G: The angular momentum of the mass system about the mass center G is the sum of the moments of the linear momenta about G of all particles and is: \mathbf{H}G = \Sigma \boldsymbol{\rho}i \times mi \dot{\mathbf{r}}i
  • Rewriting the absolute velocity \dot{\mathbf{r}}i as (\dot{\mathbf{r}} + \dot{\boldsymbol{\rho}}i) so that \mathbf{H}G becomes: \mathbf{H}G = \Sigma \boldsymbol{\rho}i \times mi (\dot{\mathbf{r}} + \dot{\boldsymbol{\rho}}i) = \Sigma \boldsymbol{\rho}i \times mi \dot{\mathbf{r}} + \Sigma \boldsymbol{\rho}i \times mi \dot{\boldsymbol{\rho}}i
  • The first term can be rewritten as \dot{\mathbf{r}} \times \Sigma mi \boldsymbol{\rho}i, which is zero because \Sigma mi \boldsymbol{\rho}i = 0 by definition of the mass center.
  • Thus, \mathbf{H}G = \Sigma \boldsymbol{\rho}i \times mi \dot{\boldsymbol{\rho}}i
    • The expression of equation \mathbf{H}G = \Sigma \boldsymbol{\rho}i \times mi \dot{\mathbf{r}}i is called the absolute angular momentum because the absolute velocity \dot{\mathbf{r}}_i is used.
    • The expression of equation \mathbf{H}G = \Sigma \boldsymbol{\rho}i \times mi \dot{\boldsymbol{\rho}}i is called the relative angular momentum because the relative velocity \dot{\boldsymbol{\rho}}_i is used.
    • With the mass center G as a reference, the absolute and relative angular momenta are identical.
    • Differentiating \mathbf{H}G = \Sigma \boldsymbol{\rho}i \times mi \dot{\mathbf{r}}i with respect to time gives:
      \dot{\mathbf{H}}G = \Sigma \dot{\boldsymbol{\rho}}i \times mi (\dot{\mathbf{r}} + \dot{\boldsymbol{\rho}}i) + \Sigma \boldsymbol{\rho}i \times mi \ddot{\mathbf{r}}_i
  • The first summation is expanded as \Sigma \dot{\boldsymbol{\rho}}i \times mi \dot{\mathbf{r}} + \Sigma \dot{\boldsymbol{\rho}}i \times mi \dot{\boldsymbol{\rho}}. The first term may be rewritten as \dot{\mathbf{r}} \times \Sigma mi \dot{\boldsymbol{\rho}}i = -\dot{\mathbf{r}} \times \frac{d}{dt} \Sigma mi \boldsymbol{\rho}i , which is zero from the definition of the mass center. The second term is zero because the cross product of parallel vectors is zero.
  • MG = the sum of all external moments about point G.
  • \Sigma \mathbf{M}G = \dot{\mathbf{H}}G
• About an Arbitrary Point P: The angular momentum about an arbitrary point P (which may have an acceleration \ddot{\mathbf{r}}P) will now be expressed in terms of: \mathbf{H}P = \Sigma {\boldsymbol{\rho}'}i \times mi \dot{\mathbf{r}}i = \Sigma (\boldsymbol{\rho} + {\boldsymbol{\rho}}i) \times mi \dot{\mathbf{r}}i
  \mathbf{H}P = {\boldsymbol{\rho}} \times mv + \mathbf{H}G
• \mathbf{H}P = \mathbf{H}G + {\boldsymbol{\rho}} \times m\mathbf{v}
  • The absolute angular momentum about any point P equals the angular momentum about G plus the moment about P of the linear momentum m\mathbf{v} of the system considered concentrated at G.
• Using the principle of moments to represent the resultants of the external forces with a resultant force \Sigma \mathbf{F} through G and a couple \Sigma \mathbf{M}_G
• The sum of the moments about P of all forces external to the system must equal the moment of their resultants.
  • \Sigma \mathbf{M}P = \Sigma \mathbf{M}G + {\boldsymbol{\rho}} \times \Sigma \mathbf{F}
  • \Sigma \mathbf{M}P = \dot{\mathbf{H}}G + {\boldsymbol{\rho}} \times m\mathbf{a}
    • This enables you to write the moment equation about any convenient moment center P.
• Using the momentum relative to P:
  \mathbf{H}{Prel} = \Sigma {\boldsymbol{\rho}'}i \times mi {\dot{\boldsymbol{\rho}}'}i
• Moment equation about P:
  \Sigma \mathbf{M}P = (\dot{\mathbf{H}}P){rel} + {\boldsymbol{\rho}} \times m\mathbf{a}P
• The form of equation \Sigma \mathbf{M}P = (\dot{\mathbf{H}}P){rel} + {\boldsymbol{\rho}} \times m\mathbf{a}P is convenient when a point P whose acceleration is known is used as a moment center.
• The equation reduces to \Sigma \mathbf{M}P = (\dot{\mathbf{H}}P)_{rel} if:
  • \mathbf{a}_P = 0
  • \boldsymbol{\rho} = 0
  • \boldsymbol{\rho} and \mathbf{a}P are parallel with \mathbf{a}P directed toward or away from G

Conservation of Energy and Momentum

• Under certain conditions, there's no net change in the total mechanical energy or momentum of a system during an interval of motion.
• Conditions are treated separately as follows.
• Conservation of Energy
  • A mass system is conservative if it does not lose energy due to internal friction forces or inelastic members.
  • If no work is done on a conservative system by external forces (other than gravity or potential forces), none of the system's energy is lost.
  • For this case, {U'}{1-2} = 0, where {U'}{1-2} is the work done by external forces, and we can write:
    • \Delta T + \Delta V = 0
    • T1 + V1 = T2 + V2
      • Expresses the law of conservation of dynamical energy.
      • The total energy E = T + V is a constant, so that E1 = E2.
      • This law holds only in the ideal case where internal kinetic friction is negligibly small.
• Conservation of Momentum
  • If the resultant external force \Sigma \mathbf{F} acting on a conservative or nonconservative mass system is zero for a certain interval of time, then \dot{\mathbf{G}} = 0, so that during this interval: \mathbf{G}1 = \mathbf{G}2
    • Expresses the principle of conservation of linear momentum.
    • In the absence of an external impulse, the linear momentum of a system remains unchanged.
  • If the resultant moment about a fixed point O or about the mass center G of all external forces on any mass system is zero, then:
    • (\mathbf{H}O)1 = (\mathbf{H}O)2
    • (\mathbf{H}G)1 = (\mathbf{H}G)2
      • Expresses the principle of conservation of angular momentum for a general mass system in the absence of an angular impulse.
      • If there is no angular impulse about a fixed point (or about the mass center), the angular momentum of the system about the fixed point (or about the mass center) remains unchanged.
      • Either equation may hold without the other.
• Basic laws of Newtonian mechanics hold for measurements made relative to a set of axes that translate with a constant velocity.
• These equations are valid provided all quantities are expressed relative to the translating axes.
• Common applications of these laws are specific mass systems such as rigid and nonrigid solids and certain fluid systems, as further discussed in the following articles.

Steady Mass Flow

• The momentum relation developed is a direct means of analyzing the action of mass flow where a change of momentum occurs.
• The dynamics of mass flow is important in fluid machinery: turbines, pumps, nozzles, air-breathing jet engines, and rockets.
• This treatment is not meant to replace the study of fluid mechanics but presents basic momentum principles and equations used in fluid mechanics.
• Mass flow occurs during steady-flow conditions where the rate at which mass enters a given volume equals the rate at which mass leaves the same volume.
• The volume may be enclosed by a rigid container (fixed or moving), blades in a gas turbine, or a bend in a pipe through which a fluid flows at a steady rate.
• The design of such fluid machines depends on analyzing the forces and moments associated with the corresponding momentum changes of the flowing mass.
• Analysis of Flow Through a Rigid Container
  • Consider a rigid container where mass flows steadily at the rate m' through the entrance section of area A_1.
  • Mass leaves the container through the exit section of area A_2 at the same rate, so there is no accumulation or depletion of mass within.
  • The velocity of the entering stream is v1 normal to A1, and that of the leaving stream is v2 normal to A2.
  • If \rho1 and \rho2 are the respective densities of the two streams, conservation of mass requires: \rho1 A1 v1 = \rho2 A2 v2 = m'
  • To describe the forces, isolate either the mass of fluid within the container or the entire container and the fluid within it.
  • The system isolated consists of the fixed structure of the container and the fluid within it at a particular instant of time.
  • Account for all external forces applied to this system, including: forces exerted on the container at points of attachment to other structures, forces acting on the fluid due to static pressure at positions A1 and A2, and the weight of the fluid and structure.
  • The resultant \Sigma \mathbf{F} of all these external forces must equal \dot{\mathbf{G}}, the time rate of change of the linear momentum of the isolated system.
• Incremental Analysis
  • The expression for \dot{\mathbf{G}} may be obtained with incremental analysis.
  • The system at time t includes the container, the mass within it, and an increment \Delta m about to enter during time \Delta t.
  • At time t + \Delta t, the same total mass includes the container, the mass within it, and an equal increment \Delta m that exits in time \Delta t.
  • The linear momentum of the container and mass between sections A1 and A2 remains unchanged during \Delta t.
  • The change in momentum of the system in time \Delta t is: \Delta \mathbf{G} = (\Delta m)\mathbf{v}2 - (\Delta m)\mathbf{v}1 = \Delta m (\mathbf{v}2 - \mathbf{v}1)
  • Division by \Delta t and passage to the limit yields: \dot{\mathbf{G}} = m' \Delta v
    • \dot{\mathbf{G}} = m' \Delta \mathbf{v}
      • Establishes the relation between the resultant force on a steady-flow system and the corresponding mass flow rate and vector velocity increment.
        • Alternatively, the time rate of change of linear momentum is the vector difference between the rate at which momentum leaves the system and the rate at which momentum enters.
        • Thus, \dot{\mathbf{G}} = m'\mathbf{v}2 - m'\mathbf{v}1 = m'\Delta \mathbf{v}, which agrees with the previous result.
  • This equation includes a rigid body (the structural container) and particles in motion (the flow of mass).
  • By defining the system's boundary with constant mass within for steady-flow conditions, we can use the equation's generality.
• Angular Momentum in Steady-Flow Systems
  • The resultant moment of all external forces about a fixed point O equals the time rate of change of angular momentum of the system about O.
  • For steady flow in a single plane: MO = m'(v2 d2 - v1 d_1)
  • When velocities of incoming and outgoing flows are not in the same plane:
    • \Sigma \mathbf{M}O = m'(\mathbf{d}2 \times \mathbf{v}2 - \mathbf{d}1 \times \mathbf{v}_1)
      • \mathbf{d}1 and \mathbf{d}2 are position vectors to the centers of A1 and A2 from a fixed coordinate system.
    • The mass center G may be used as a moment center.
  • These equations relate external forces to resultant changes in momentum and are independent of the flow path and momentum changes internal to the system.
  • This analysis also applies to systems moving with constant velocity, noting that the basic relations \Sigma \mathbf{F} = \dot{\mathbf{G}} and \Sigma \mathbf{M}O = \dot{\mathbf{H}}O apply.
  • The only restriction is that the mass within the system remains constant with respect to time. Three examples of the analysis of steady mass flow are given in the following sample problems, which illustrate the application of the principles.