Center Of Mass And Linear Momentum

PH 221-1D Spring 2013Center of Mass and LinearMomentumLectures 22-23Chapter 9(Halliday/Resnick/Walker, Fundamentals of Physics 9th edition)1

Chapter 9Center of Mass and Linear MomentumIn this chapter we will introduce the following new concepts:-Center of mass (com) for a system of particles-The velocity and acceleration of the center of mass-Linear momentum for a single particle and a system of particlesWe will derive the equation of motion for the center of mass, anddiscuss the principle of conservation of linear momentumFinally we will use the conservation of linear momentum to studycollisions in one and two dimensions and derive the equation ofmotion for rockets2

The Center of Mass:Consider a system of two particles of masses m1 and m2at positions x1 and x2 , respectively. We define theposition of the center of mass (com) as follows:xcom m1 x1 m2 x2m1 m2We can generalize the above definition for a system of n particles as follows:xcomm x m2 x2 m3 x3 . mn xn m1 x1 m2 x2 m3 x3 . mn xn1 1 1 m1 m2 m3 . mnMMn m xi 1Here M is the total mass of all the particles M m1 m2 m3 . mnWe can further generalize the definition for the center of mass of a system ofparticles in three dimensional space. We assume that the the i -th particle ( mass mi ) has position vector ri1 rcom M mi rini 13i i

1 The position vector for the center of mass is given by the equation: rcom M The position vector can be written as: rcom xcomiˆ ycom ˆj zcom kˆ The components of rcom are given by the equations:xcom1 Mn m xi 1i iycom1 Mn m yi 1iizcom1 M mr iini 1n m zi 1i iThe center of mass has been defined using the equationsgiven above so that it has the following property:The center of mass of a system of particles moves as thoughall the system's mass were concetrated there, and that thevector sum of all the external forces were applied thereThe above statement will be proved later. An example isgiven in the figure. A baseball bat is flipped into the airand moves under the influence of the gravitation force. Thecenter of mass is indicated by the black dot. It follows aparabolic path as discussed in Chapter 4 (projectile motion)All the other points of the bat follow more complicated paths4

The Center of Mass for Solid BodiesSolid bodies can be considered as systems with continuous distribution of matterThe sums that are used for the calculation of the center of mass of systems withdiscrete distribution of mass become integrals:111xdmy ydmz zdmcomcom MMMThe integrals above are rather complicated. A simpler special case is that ofdmMis constant and equal touniform objects in which the mass density dVV111xcom xdVycom ydVzcom zdVVVVIn objects with symetry elements (symmetry point, symmetry line, symmetry plane)xcom it is not necessary to eveluate the integrals. The center of mass lies on the symmetryelement. For example the com of a uniform sphere coincides with the sphere centerIn a uniform rectanglular object the com lies at the intersection of the diagonalsC.C5

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zF1m2m1m3F2xNewton's Second Law for a System of ParticlesConsider a system of n particles of masses m1 , m2 , m3, ., mn and position vectors r1 , r2 , r3 ,., rn , respectively.F2OyThe position vector of the center of mass is given by: Mrcom m1r1 m2 r2 m3r3 . mn rn We take the time derivative of both sides d d d d d M rcom m1 r1 m2 r2 m3 r3 . mn rn dtdtdtdtdt Mvcom m1v1 m2 v2 m3v3 . mn vn Here vcom is the velocity of the com and vi is the velocity of the i -th particle. We take the time derivative once more d d d d d M vcom m1 v1 m2 v2 m3 v3 . mn vn dtdtdtdtdt Macom m1a1 m2 a2 m3 a3 . mn an Here acom is the acceleration of the com and ai is the acceleration of the i -th particle7

Macom m1a1 m2 a2 m3 a3 . mn anzF1m2m1F2m3F2xOyWe apply Newton's second law for the i -th particle: mi ai Fi Here Fi is the net force on the i -th particle Macom F1 F2 F3 . Fn The force Fi can be decomposed into two components: applied and internal app intFi Fi FiThe above equation takes the form: app int app int app int app int Macom F1 F1 F2 F2 F3 F3 . Fn Fn app app app app int int int int Macom F1 F2 F3 . Fn F1 F2 F3 . Fn The sum in the first parenthesis on the RHS of the equation above is just Fnet The sum in the second parethesis on the RHS vanishesby virtue of Newton's third law. The equation of motion for the center of mass becomes: Macom FnetIn terms of components we have:Fnet , x Macom, xFnet , y Macom, yFnet , z Macom , z8

Macom FnetFnet , x Macom, xFnet , y Macom, yFnet , z Macom, zThe equations above show that the center of mass of a system of particlesmoves as though all the system's mass were concetrated there, and that thevector sum of all the external forces were applied there. A dramatic example isgiven in the figure. In a fireworks display a rocket is launched and moves underthe influence of gravity on a parabolic path (projectile motion). At a certain pointthe rocket explodes into fragments. If the explosion had not occured, the rocketwould have continued to move on the parabolic trajectory (dashed line). The forcesof the explosion, even though large, are all internal and as such cancel out. Theonly external force is that of gravity and this remains the same before and after theexplosion. This means that the center of mass of the fragments folows the sameparabolic trajectory that the rocket would have followed had it not exploded 9

vLinear Momentummp p mv Linear momentum p of a particle of mass m and velocity v is defined as: p mvThe SI unit for lineal momentum is the kg.m/sBelow we will prove the following statement: The time rate of change of the linearmomentum of a particle is equal to the magnitude of net force acting on theparticle and has the direction of the force dpIn equation form: Fnet We will prove this equation usingdtNewton's second law dp d dv p mv mv m ma Fnetdt dtdtThis equation is stating that the linear momentum of a particle can be changedonly by an external force. If the net external force is zero, the linear momentumcannot change dpFnet dt10

The Linear Momentum of a System of ParticlesIn this section we will extedend the definition ofzm1p1m2p3m3p2xOylinear momentum to a system of particles. The i -th particle has mass mi , velocity vi , and linear momentum piWe define the linear momentum of a system of n particles as follows: P p1 p2 p3 . pn m1v1 m2v2 m3v3 . mn vn MvcomThe linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity vcom of the center of mass dP d The time rate of change of P is: Mvcom Macom Fnetdt dt The linear momentum P of a system of particles can be changed only by a net external force Fnet . If the net external force Fnet is zero P cannot change P p1 p2 p3 . pn Mvcom dP Fnetdt11

Example. Motion of the Center of Mass12

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Collision and ImpulseWe have seen in the previous discussion that the momentum of an object canchange if there is a non-zero external force acting on the object. Such forcesexist during the collision of two objects. These forces act for a brief timeinterval, they are large, and they are responsible for the changes in the linearmomentum of the colliding objects.Consider the collision of a baseball with a baseball batThe collision starts at time ti when the ball touches the batand ends at t f when the two objects separate The ball is acted upon by a force F (t ) during the collisionThe magnitude F (t ) of the force is plotted versus t in fig.aThe force is non-zero only for the time interval ti t t f dp F (t ) Here p is the linear momentum of the balldttftf dp F (t )dt dp F (t )dttiti14

tf dp F (t )dttftititf dp p f pi p change in momentumti F (t )dt is known as the impulse J of the collisiontfti tf J F (t )dt The magnitude of J is equal to the areati under the F versus t plot of fig.a p JIn many situations we do not know how the force changeswith time but we know the average magnitude Fave of thecollision force. The magnitude of the impulse is given by:J Fave t where t t f ti p JJ Fave tGeometrically this means that the the area under theF versus t plot (fig.a) is equal to the area under theFave versus t plot (fig.b)15

Collisions. Impulse and Momentum16

The Impulse-Momentum Theorem17

FaveSeries of CollisionsConsider a target which collides with a steady stream of identical particles of mass m and velocity v along the x -axisA number n of the particles collides with the target during a time interval t.Each particle undergoes a change p in momentum due to the collision withthe target. During each collision a momentum change p is imparted on thetarget. The Impulse on the target during the time interval t is:J n pThe average force on the target is:J n pnFave Here v is the change in the velocity m v t t tof each particle along the x-axis due to the collision with the target m mFave v Hereis the rate at which mass collides with the target t tIf the particles stop after the collision then v 0 v vIf the particles bounce backwards then v v v 2v18

Conservation of Linear Momentumzm1p1m2p3m3p2Oy Consider a system of particles for which Fnet 0 dP Fnet 0 P Constantdt If no net external force acts on a system of particles the total linear momentum Pxcannot change total linear momentum total linear momentum at some initial time t at some later time tfi The conservation of linear momentum is an importan principle in physics.It also provides a powerful rule we can use to solve problems in mechanics such ascollisions. Note 1: In systems in which Fnet 0 we can always apply conservation of linearmomentum even when the internal forces are very large as in the case ofcolliding objectsNote 2: We will encounter problems (e.g. inelastic collisions) in which the energyis not conserved but the linear momentum is19

Momentum and Kinetic Energy in CollisionsConsider two colliding objects with masses m1 and m2 , initial velocities v1i and v2i and final velocities v1 f and v2 f ,respectively If the system is isolated i.e. the net force Fnet 0 linear momentum is conservedThe conervation of linear momentum is true regardless of the the collision typeThis is a powerful rule that allows us to determine the results of a collision withoutknowing the details. Collisions are divided into two broad classes: elastic andinelastic.A collision is elastic if there is no loss of kinetic energy i.e. Ki K fA collision is inelastic if kinetic energy is lost during the collision due to conversioninto other forms of energy. In this case we have: K f KiA special case of inelastic collisions are known ascompletely inlelastic. In these collisions the two colliding objects stick togetherand they move as a single body. In these collisions the loss of kinetic energy20is maximum

One Dimensional Inelastic CollisionsIn these collisions the linear momentium of the colliding objects is conserved p1i p2i p1 f p2 f m1v1i m1v2i m1v1 f m1v2 fOne Dimensional Completely Inelastic CollisionsIn these collisions the two colliding objects stick togetherand move as a single body. In the figure to the left we show a special case in which v2i 0. m1v1i m1V m2V m1V v1im1 m2The velocity of the center of mass in this collision p1i p2im1v1iP is vcom m1 m2 m1 m2 m1 m2In the picture to the left we show some freeze-framesof a totally inelastic collision21

Inelastic Collision22

The Principle of Conservation of Linear Momentum23

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One-Dimensional Elastic CollisonsConsider two colliding objects with masses m1 and m2 , initial velocities v1i and v2i and final velocities v1 f and v2 f ,respectivelyBoth linear momentum and kinetic energy are conserved.Linear momentum conservation: m1v1i m1v2i m1v1 f m1v2 f22m1v12i m1v22i m1v1 f m2 v2 f Kinetic energy conservation:2222We have two equations and two unknowns, v1 f and v2 f(eqs.1)(eqs.2)If we solve equations 1 and 2 for v1 f and v2 f we get the following solutions:v1 f v2 fm1 m22m2v1i v2im1 m2m1 m2m2 m12m1v1i v2i m1 m2m1 m225