Center of Mass and Linear

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1 PH 221-3A Fall 2010 Center of Mass and Linear Momentum Lecture 15 Chapter 8 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition) 1

2 Chapter 9 Center of Mass and Linear Momentum In 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 particles We will derive the equation of motion for the center of mass, and discuss the principle of conservation of linear momentum Finally we will use the conservation of linear momentum to study collisions in one and two dimensions and derive the equation of motion for rockets 2

3 The Center of Mass: Consider a system of two particles of masses m and m at positions x and x, respectively. We define the position of the center of mass (com) as follows: x com mx m mx m We can generalize the above definition i i for a system of n particles as follows: x mxmx mx... mx mxmx mx... mx 1 n n n n n com mx i i m 1 m 2 m m n M M i 1 Here M is the total mass of all the particles M m1m2 m3... m n We can further generalize the definition for the center of mass of a system of particles in three dimensional space. We assume that the the i-th particle ( mass mi ) has position vector r i 1 n r mr com M i 1 i i 3

The position vector for the center of mass is given by the equation: The position vector can be written as: r ˆ ˆ ˆ com xcomi ycom jzcomk The components of rcom are given by the equations: n n n 1 1

mass of a system of particles moves as though all the system's mass were concetrated there, and that the vector sum of all the external forces were applied there The above statement will be proved

4 The position vector for the center of mass is given by the equation: The position vector can be written as: r ˆ ˆ ˆ com xcomi ycom jzcomk The components of rcom are given by the equations: n n n xcom mx i i ycom mi yi zcom mz i i M M M i1 i1 i1 r com 1 M n i1 mr i i The center of mass has been defined using the equations given above so that t it has the following property: The center of mass of a system of particles moves as though all the system's mass were concetrated there, and that the vector sum of all the external forces were applied there The above statement will be proved later. An example is given in the figure. A baseball bllbat is flipped dinto the air and moves under the influence of the gravitation force. The center of mass is indicated by the black dot. It follows ows a parabolic path as discussed in Chapter 4 (projectile motion) All the other points of the bat follow more complicated paths 4

5 The Center of Mass for Solid Bodies Solid bodies can be considered as systems with continuous distribution of matter The sums that are used for the calculation of the center of mass of systems with discrete distribution of mass become integrals: xcom xdm ycom ydm zcom zdm M M M The integrals above are rather complicated. A simpler special case is that of dm M uniform objects in which the mass density is constant and equal to dv V xcom xdv ycom ydv zcom zdv V V V In objects with symetry elements (symmetry point, symmetry line, symmetry plane) it is not necessary to eveluate the integrals. Thecenterofmassliesonthesymmetry on the symmetry element. For example the com of a uniform sphere coincides with the sphere center In a uniform rectanglular object the com lies at the intersection of the diagonals C. C 5

6 6

7 m 1 F 1 F 2 m 2 F 2 O x z m 3 y Newton's Second Law for a System of Particles Consider a system of particles of masses,,..., and position vectors r1, r2, r3,..., r n, respectively. The position vector of the center of mass is given by: n m 1 m 2 m 3, m n Mrcom m11 r m2r2 m33 r... mnrn We take the time derivative of both sides d d d d d M rcom m1 r1m2 r2 m3 r3... mn rn dt dt dt dt dt Mvcom m1v 1m2v2 m3v3... mnvn Here vcom is the velocity of the com and v is the velocity of the i-th particle. We take the time derivative once more i d d d d d M vcom m1 v1m2 v2 m3 v3... mn vn dt dt dt dt dt Macom m1a 1m2a2 m3a3... mnan Here acom is the acceleration of the com and a is the acceleration of the i-th particle i 7

8 m 1 F 1 F 2 m 2 F 2 O x z Ma m a m a m a m a com n n We apply Newton's second law for the i-th particle: m 3 ma i i Fi Here Fi is the net force on the i-th particle y Macom F1F 2 F3... Fn The force Fi can be decomposed into two components: applied and internal app int Fi Fi Fi The above equation takes the form: app int app int app int app int Macom F1 F1 F2 F2 F3 F3... Fn Fn Ma F app F app F app... F app F int F int F int... F int com n n The sum in the first parenthesis on the RHS of the equation above is just F net The sum in the second parethesis on the RHS vanishes by virtue of Newton's third law. The equation of motion for the center of mass becomes: In terms of components we have: Fnet, x Macom, x Fnet, y Macom, y Fnet, z Ma com, z Ma com F net 8

Ma com F net F F F Ma net, x com, x Ma net, y com, y Ma net, z com, z The equations above show that the center of mass of a system of particles moves as though all the system's mass were concetrated

9 Ma com F net F F F Ma net, x com, x Ma net, y com, y Ma net, z com, z The equations above show that the center of mass of a system of particles moves as though all the system's mass were concetrated there, and that the vector sum of all the external forces were applied there. A dramatic example is given in the figure. In a fireworks display a rocket is launched and moves under the influence of gravity on a parabolic path (projectile motion). At a certain point the rocket explodes into fragments. If the explosion had not occured, the rocket would have continued to move on the parabolic trajectory (dashed line). The forces of the explosion, even though large, are all internal and as such cancel out. The only external force is that of gravity and this remains the same before and after the explosion. This means that the center of mass of the fragments folows the same parabolic trajectory that the rocket would have followed had it not exploded 9

10 m p v mv Linear Momentum p Linear momentum p of a particle of mass m and velocity v is defined as: p mv The SI unit for lineal momentum is the kg.m/s Below we will prove the following statement: The time rate of change of the linear momentum of a particle is equal to the magnitude of net force acting on the particle and has the direction of the force dp In equation form: Fnet We will prove this equation using dt Newton's second law dp d dv p mv mvm ma Fnet dt dt dt This equation is stating that the linear momentum of a particle can be changed only by an external force. If the net external force is zero, the linear momentum cannot change F net dp dt 10

11 z m 1 p 1 p 3 m 2 p 2 x O m 3 y The Linear Momentum of a System of Particles In this section we will extedend the definition of linear momentum to a system of particles. The i-th particle has mass mi, velocity v i, and linear momentum m p i We define the linear momentum of a system of n particles as follows: P p1 p2 p3... pn mv 1 1 m2v2 m3v3... mnvn Mv com The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity v com of the center of mass dp d The time rate of change of P is: Mvcom Macom Fnet dt dt The linear momentum P of a system of particles il can be changed only by a net external force F. If the net external force F is zero P cannot change net P p p p p Mv n com net dp Fnet dt 11

12 Example. Motion of the Center of Mass 12

13 13

Collision and Impulse We have seen in the previous discussion that the momentum of an object can change if there is a non-zero external lforce acting on the object.

These forces act for a brief time interval, they are large, and they are responsible for the changes in the linear momentum of the colliding objects.

14 Collision and Impulse We have seen in the previous discussion that the momentum of an object can change if there is a non-zero external lforce acting on the object. Such forces exist during the collision of two objects. These forces act for a brief time interval, they are large, and they are responsible for the changes in the linear momentum of the colliding objects. Consider the collision of a baseball with a baseball bat The collision starts at time ti when the ball touches the bat and ends at t f when the two objects separate The ball is acted upon by a force F ( t) during the collision The magnitude Ft ( ) of the force is plotted versus tin fig.a The force is non-zero only for the time interval t i t t f dp Ft ( ) Here pis the linear momentum of the ball dt d tf tf p Ftdt () dp Ftdt () t i t i 14

dp F () tdt dp p p p = change in momentum t t t f f f t t t i i i t f F ()

magnitude of J is equal to the area t i under the F versus t plot of fig.

but we know the average magnitude F ave of the collision force.

Geometrically this means that the the area under the F t l t (fi ) i l t

15 dp F () tdt dp p p p = change in momentum t t t f f f t t t i i i t f F () tdt is known as the impulse J of the collision t i t f J F( t) dt The magnitude of J is equal to the area t i under the F versus t plot of fig.a p J f In many situations we do not know how the force changes with time but we know the average magnitude F ave of the collision force. The magnitude of the impulse is given by: J F t where t t t ave f i Geometrically this means that the the area under the F t l t (fi ) i l t th d th F versus ave t plot (fig.b) p J versus plot (fig.a) is equal to the area under the J Favet 15 i

16 Collisions. Impulse and Momentum 16

17 The Impulse-Momentum Theorem 17

F ave Series of Collisions Consider a target which collides with a steady stream of identical particles of mass m and velocity v along the x-axis A number n of the particles collides with the target

18 F ave Series of Collisions Consider a target which collides with a steady stream of identical particles of mass m and velocity v along the x-axis A 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 with the target. During each collision a momentum change p is imparted on the target. The Impulse on the target during the time interval t is: J np The average force on the target is: J n p n Fave mv Here v is the change in the velocity t t t of each particle along the x-axis due to the collision with the target m m Fave v Here is the rate at which mass collides with the target t t If the particles stop after the collision then v 0 v v If the particles bounce backwards then v vv 2v 18

19 z m 1 p 1 p 3 m 2 p 2 x O m 3 y Conservation of Linear Momentum Consider a system of particles for which Fnet 0 dp F net 0 P Constant dt If no net external force acts on a system of particles the total llinear momentum P cannot change total linear momentum total linear momentum at some initial time t i at some later time t f 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 as collisions. Nt Note 1: In systems in which h F net 0 we can always apply conservation of linear momentum even when the internal forces are very large as in the case of colliding objects Note 2: We will encounter problems (e.g. inelastic collisions) in which the energy is not conserved but the linear momentum is 19

20 respectively If the system is isolated i.e. the net force F Momentum and Kinetic Energy in Collisions Consider two colliding objects with masses m1 and m2, initial velocities v and v and final velocities v and v, 1i 2i 1f 2 f net 0 linear momentum is conserved The conervation of linear momentum is true regardless of the the collision type This is a powerful rule that allows us to determine the results of a collision without knowing the details. Collisions are divided into two broad classes: elastic and inelastic. A collision is elastic if there is no loss of kinetic energy i.e. A collision is inelastic if kinetic energy is lost during the collision due to conversion it into other forms of energy. In this case we have: K f K i K i K A special case of inelastic collisions are known as completely inlelastic. In these collisions the two colliding objects stick together and they move as a single body. In these collisions the loss of kinetic energy is maximum f 20

21 One Dimensional Inelastic Collisions In these collisions the linear momentium of the colliding objects is conserved p1 i p2i p1f p2 f mv mv mv mv 1 1i 1 2i 1 1f 1 2 f One Dimensional Completely Inelastic Collisions In these collisions the two colliding objects stick together and move as a single body. In the figure to the leftwe show a special case in which v2i 0. mv 1 1i mv 1 mv 2 m1 V v1 i m1 m2 The velocity of the center of mass in this collision P p1 i p2i mv 1 1i is vcom m1m2 m1m2 m1m2 In the picture to the left we show some freeze-framesframes of a totally inelastic collition 21

22 The Principle of Conservation of Linear Momentum 22

23 23

24 One-Dimensional Elastic Collisons Consider two colliding objects with masses m1 and m2, initial velocities v and v and final velocities v and v, respectively 1i 2i 1f 2 f Both linear momentum and kinetic energy are conserved. Linear momentum conservation: mv mv mv mv (eqs.1) 1 1i 1 2i 1 1f 1 2 f mv 1 1i mv mv f mv i 2 2 f Kinetic energy conservation: (eqs.2) We have tw o equations and two unknowns, v and v f 1i 2i m1m2 m1m2 1f 2 f 1f 2 f If we solve equations 1 and 2 for v and v we get the following solutions: m m 2m v v v 2m m m v v v f 1 2 m1m i i 2 m1 m2 24

25 Special Case of elastic Collisions-Stationary Target v 0 The substitute v 0inthe two solutions for v and v 2i 1f 2 f 2i m m 2m m m v v v v f 1i 2i 1f 1i m1m2 m1m2 m1 m2 2m m m 2m v v v v f 1i 2i 2 f 1i m1 m2 m1 m2 m1 m2 Below we examine several special cases for which we know the outcome of the collision from experience 1. Equal masses m m m 1 2 m m mm v v v f 1i 1i m1 m2 m m 2m 2m v v v v 1 2 f 1i 1i 1i m1m2 mm The two colliding objects have exchanged velocities 0 v v v 1i v 2i = 0 m m v 1f = 0 m m v 2f x x 25

26 v 1i m 1 m 2 v 2i = 0 x v 1f m 1 m 2 v 2f 1 2. A mass ive target m2 m1 <1 m2 m1 1 m m m v v v v 1 m f 1i 1i 1i x m m 1 m2 1 m 1 2 2m m 1 2 m 1 v2 f v1 i v1 i 2 v1 i m m 1 m2 1 1 m2 m 2 Body 1 (small mass) bounces back along the incoming path with its speed practically unchanged. Bd Body 2(l (large mass) moves forward with a very small m 1 speed because <1 m 2 2 m 26

27 v 2 1i 2. A massive projectile m1 m2 <1 v 2i = 0 x m 1 m m m m1 m m 2 1 m m 2 v m 1f 1 v f v v v v x m 1 m 2 f i i i 2 2 m1 2 v v v 2v m m 1 m2 2 1 m 2 f 1i 1i 1i Body 1 (large mass) keeps on going scarcely slowed by the collision. Body 2 (small mass) charges ahead at twice the speed of body m m 1 27

28 Elastic Collision 28

29 Inelastic Collision 29

30 Collisions in Two Dimensions In this section we will remove the restriction that the colliding objects move along one axis. Instead we assume that the two bodies that participate in the collision move in the xy -plane. Their masses are m 1 and m 2 The linear momentum of the sytem is conserved: p p p p 1i 2i 1f 2 f If the system is elastic the kinetic energy is also conserved: K K K K 2 1 i 2 i 1 f 2 f We assume that m is stationary and that after the collision particle 1 and particle 2 move at angles and with the initial direction of motion of m axis: 1 1i 1 1f 1 mv 2 2 f In this case the conservation of momentum and kinetic energy take the form: x mv mv cos cos (eqs.1) y axis: 0 mv sin m v sin (eqs.2) 1 1f f mv 1 1i mv 1 2 f mv 2 2 f (eqs.3) We have three equations and seven variables: Two masses: m1, m2 three speeds: v1 i, v1f, v2 f and two angles: 1, 2. If we know the values of four of these parameters we can calculate the remaining three 30

31 Problem 72. Two 2.0 kg bodies, A and B collide. The velocities before the collision are ˆ ˆ v (15 30 ) m/s and ( 10 ˆ 5.0 ˆ A i j vb i j) m/s. After the collision, v ( 5.0iˆ 20 ˆj) m/s. What are (a) the final velocity of B and (b) the change ' A in the total kinetic energy (including sign)? (a) Conservation of linear momentum implies mv mv mv' mv'. A A B B A A B B Since m A = m B = m = 2.0 kg, the masses divide out and we obtain v (15i ˆ 30ˆj)m/s ( 10ˆi 5j)m/s ˆ ( 5ˆi 20ˆ B va vb va j)m/s (10 ˆi 15 ˆj) m/s. (b) The final and initial kinetic energies are c h J K 1 mv mv f ' A ' B ( 2. 0) ( 5) Ki mva mvb ( 2. 0) c15 30 ( 10) 5 h J The change kinetic energy is then K = J (that is, 500 J of the initial kinetic energy is lost). 31

Systems with Varying Mass: The Rocket A rocket of mass M and speed v ejects mass backwards dm at a constant rate. The ejected material is expelled at a dt constant speed vrel relative to the rocket.

32 Systems with Varying Mass: The Rocket A rocket of mass M and speed v ejects mass backwards dm at a constant rate. The ejected material is expelled at a dt constant speed vrel relative to the rocket. T hus the rocket loses mass and accelerates forward. We will use the conservation of linear momentum to determine the speed v of the rocket In figures (a) and (b) we show the rocket at times t and t dt. If we assume that there are no external forces acting on the rocket, linear momentum is conserved p( t) p t dt Mv dmu M dm v dv (eqs.1) Here dm is a negative number because the rocket's mass decreases with time t U is the velocity of fthe ejected gases with respect to the inertial reference frame in which we measure the rocket's speed v. We use the transformation equation for velocities (Chapter 4) to express U in terms of v rel which is measured with respect to the rocket. U vdvvrel We substitute U in equation 1 and we get: 32 Mdv dmvrel

Using the conservation of linear momentum we derived the equation of motion for the rocket Mdv dmv rel (eqs.

3) Here R is a constant positivenumber, dt the positive mass rate of fuel consumprion. dv dm We devide both sides of eqs.

33 Using the conservation of linear momentum we derived the equation of motion for the rocket Mdv dmv rel (eqs.2) We assume that material is ejected from the rocret's nozzle at a constant rate dm RR (eqs.3) Here R is a constant positivenumber, dt the positive mass rate of fuel consumprion. dv dm We devide both sides of eqs.(2) by dt M vrel Rvr el Ma Rvre l dt dt (First rocket equation) Here a is the rocket's acceleration, Rv the thrust of the rocket engine. We use equation 2 to determine the rocket's speed as function of time f f dm dm dv vrel We integrate both sides dv vrel M M f v M M M i f i rel M rel rel i M f M f i v v v ln M ln v ln rel v M v M v f M i vf vi vrel ln (Second rocket equation) v i M i /M f M 33 f i M i O

34 Problem 78. A 6090 kg space probe moving nose-first toward Jupiter at 105 m/s relative to the Sun fires its rocket engine, ejecting 80.0 kg of exhaust at a speed of 253 m/s relative to the space probe. What is the final velocity of the probe? M i 6090 kg vf vi vrel ln 105 m/s (253 m/s) ln 108 m/s. M f 6010 kg 34

Center of Mass and Linear Momentum

Center of Mass and Linear Momentum PH 221-2A Fall 2014 Center of Mass and Lnear Momentum Lectures 14-15 Chapter 9 (Hallday/Resnck/Walker, Fundamentals of Physcs 9 th edton) 1 Chapter 9 Center of Mass and Lnear Momentum In ths chapter we