Chapter 9: Center of Mass and Linear Momentum

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Chapter 9: Center of Mass and Linear Momentum

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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

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Nội dung Text: Chapter 9: Center of Mass and Linear Momentum

  1. 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 (9-1) motion for rockets
  2. The Center of Mass: Consider a system of two particles of masses m1 and m2 at positions x1 and x2 , respectively. We define the position of the center of mass (com) as follows: m1 x1 + m2 x2 xcom = m1 + m2 We can generalize the above definition for a system of n particles as follows: m x + m2 x2 + m3 x3 + ... + mn xn m1 x1 + m2 x2 + m3 x3 + ... + mn xn 1 n xcom = 1 1 m1 + m2 + m3 + ... + mn = M = M ∑m x i =1 i i Here M is the total mass of all the particles M = m1 + m2 + m3 + ... + mn 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 r ( mass mi ) has position vector ri n r 1 r rcom = M ∑ mi ri i =1 (9-2)
  3. n r 1 r The position vector for the center of mass is given by the equation: rcom = M ∑ mi ri i =1 r ˆ The position vector can be written as: rcom = xcomi + ycom ˆ + zcom k ˆ j r The components of rcom are given by the equations: n n n 1 1 1 xcom = M ∑m x i =1 i i ycom = M ∑m y i =1 i i zcom = M ∑m z i =1 i i The center of mass has been defined using the quations given above so that 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 abolve statement will be proved later. An example is given in the figure. A baseball bat is flipped into the air and moves under the influence of the gravitation force. The center of mass is indicated by the black dot. It follows a parabolic path as discussed in Chapter 4 (projectile motion) (9-3) All the other points of the bat follow more complicated paths
  4. The Center of Mass for Solid Bodies (9-4) 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: 1 1 1 xcom = M ∫ xdm ycom = M ∫ ydm zcom = M ∫ zdm The integrals above are rather complicated. A simpler special case is that of dm M uniform objects in which the mass density ρ = is constant and equal to dV V 1 1 1 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. The center of mass lies 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. z Newton's Second Law for a System of Particles Consider a system of n particles of masses m1 , m2 , m3, ..., mn r r r r O and position vectors r1 , r2 , r3 ,..., rn , respectively. x y The position vector of the center of mass is given by: r r r r r Mrcom = m1r1 + m2 r2 + m3r3 + ... + mn rn We take the time derivative of both sides → d r d r d r d r d r M rcom = m1 r1 + m2 r2 + m3 r3 + ... + mn rn → dt dt dt dt dt r m r r r r r Mvcom = m1v1 + m2 v2 + m3v3 + ... + mn vn Here vcom is the velocity of the com F r F m and v is the velocity of the i -th particle. We take the time derivative once more → i m Fd r d r d r d r d r M vcom = m1 v1 + m2 v2 + m3 v3 + ... + mn vn → dt dt dt dt dt r r r r r r Macom = m1a1 + m2 a2 + m3a3 + ... + mn an Here acom is the acceleration of the com r and ai is the acceleration of the i -th particle (9-5)
  6. z r r r r r Macom = m1a1 + m2 a2 + m3 a3 + ... + mn an We apply Newton's second law for the i -th particle: r r r mi ai = Fi Here Fi is the net force on the i -th particle O r r r r r x y Macom = F1 + F2 + F3 + ... + Fn r The force Fi can be decomposed into two components: applied and internal r rapp rint Fi = Fi + Fi The above equation takes the form: rapp rint rapp rint rapp rint rapp rint r ( ) ( ) ( Macom = F1 + F1 + F2 + F2 + F3 + F3 + ... + Fn + Fn → ) ( ) r m rapp rapp rapp rapp rint rint rint rint ( ) ( Macom = F1 + F2 + F3 + ... + Fn + F1 + F2 + F3 + ... + Fn F F ) m r The sum in the first parenthesis on the RHS of the equation above is just Fnet m F The sum in the second parethesis on the RHS vanishes by virtue of Newton's third law. r r The equation of motion for the center of mass becomes: Macom = Fnet In terms of components we have: Fnet , x = Macom , x Fnet , y = Macom, y Fnet , z = Macom, z (9-6)
  7. r r (9-7) Macom = Fnet Fnet , x = Macom, x Fnet , y = Macom, y Fnet , z = Macom, 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
  8. v m Linear Momentum p r r Linear momentum p of a particle of mass m and velocity v r r r r is defined as: p = mv 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 r r dp In equation form: Fnet = We will prove that this equation using dt Newton's second law r r r r dp d r dv r r p = mv → = ( mv ) = m = 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 r r dp Fnet = (9-8) dt
  9. z 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 r O y i -th particle has mass mi , velocity vi , and linear x r momentum pi We define the linear momentum of a system of n particles as follows: r r r r r r r r r r P = p1 + p2 + p3 + ... + pn = m1v1 + m2 v2 + m3v3 + ... + mn vn = Mvcom The linear momentum of a system of particles is equal to the product of the r total mass M of the syetsm and the velocity vcom of the center of mass m r p p r dP d r r r = ( Mvcom ) = Macom = Fnet m The time ratemof change of P is: dt dt p r The linear momentum P of a system of particles can be changed only r r r by a net external force Fnet . If the net external force Fnet is zero P cannot change r r r r r r r dP r P = p1 + p2 + p3 + ... + pn = Mvcom = Fnet (9-9) dt
  10. 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 force 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 r The ball is acted upon by a force F (t ) during the collision The magnitude F (t ) of the force is plotted versus t in fig.a The force is non-zero only for the time interval ti < t < t f r r dp r F (t ) = Here p is the linear momentum of the ball dt tf tf r r r r dp = F (t )dt → ∫ dp = ∫ F (t )dt (9-10) ti ti
  11. tf tf tf r r r r r r ∫ dp = ∫ F (t )dt ti ti ∫ dp = p f − pi = ∆p = change in momentum ti tf r r ∫ F (t )dt is known as the impulse J of the collision ti r tf r r J = ∫ F (t )dt The magnitude of J is equal to the area ti r r under the F versus t plot of fig.a → ∆p = J In many situations we do not know how the force changes with time but we know the average magnitude Fave of the collision force. The magnitude of the impulse is given by: J = Fave ∆t where ∆t = t f − ti Geometrically this means that the the area under the ∆p = J F versus t plot (fig.a) is equal to the area under the J = Fave ∆t Fave versus t plot (fig.b) (9-11)
  12. Series of Collisions Consider a target which collides with a steady stream of r identical particels of mass m and velocity v along the x-axis A number n of the particles collides with the terget 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 = −n∆p The average force on the target is: J −n∆p n Fave = = = − m∆v 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 = −v − v = −2v (9-12)
  13. z Conservation of Linear Momentum r Consider a system of particles for which Fnet = 0 r dP r r = Fnet = 0 → P = Constant O y dt x r If no net external force acts on a system of particles the total linear momentum P cannot change  total linear momentum   total linear momentum  at some initial time t  = at some later time t   i   f  The conservation of linear momentum is an importan principle in physics. m m It also provides a powerful rule we can use to solve problems in mechanics such as p p m collisions. p r Note 1: In systems in which Fnet = 0 we can always apply conservation of linear momentum even when the internal forces ared vely 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 (9-13)
  14. Momentum and Kinetic Energy in Collisions Consider two colliding objects with masses m1 and m2 , r r r r initial velocities v1i and v2i and final velocities v1 f and v2 f , respectively r If the system is isolated i.e. the net force Fnet = 0 linear momentum is conserved The conervation of linear momentum is true regardless of the the collision type This is a powerful rule thet 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. K i = K f A collision is inelastic if kinetic energy is lost during the collision due to conversion into other forms of energy. In this case we have: K f < K i 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 (9-14)
  15. One Dimensional Inelastic Collisions In these collisions the linear momentium of the colliding r r r r objects is comserved → p1i + p2i = p1 f + p2 f r r r r m1v1i + m1v2i = m1v1 f + m1v2 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 r a special case in which v2i = 0. → m1v1i = m1V + m1V → m1 V= v1i m1 + m2 The velocity of the center of mass in this collision r r r r r P p1i + p2i m1v1i is vcom = = = m1 + m2 m1 + m2 m1 + m2 In the picture to the left we show some freeze-frames of a totally inelastic collition (9-15)
  16. One-Dimensional Elastic Collisons Consider two colliding objects with masses m1 and m2 , r r r r initial velocities v1i and v2i and final velocities v1 f and v2 f , respectively Both linear momentum and kinetic energy are conserved. Linear momentum conservation: m1v1i + m1v2i = m1v1 f + m1v2 f (eqs.1) 2 2 m1v12i m1v2i m1v1 f m2 v2 f 2 Kinetic energy conservation: + = + (eqs.2) 2 2 2 2 We have two equations and two unknowns, v1 f and v1 f If we solve equations 1 and 2 for v1 f and v1 f we get the following solutions: m1 − m2 2m2 v1 f = v1i + v2i m1 + m2 m1 + m2 2m1 m − m1 v2 f = v1i + 2 v2i m1 + m2 m1 + m2 (9-16)
  17. Special Case of elastic Collisions-Stationaly Target ( v2i = 0 ) The substitute v2i = 0 in the two solutions for v1 f and v1 f m1 − m2 2m2 m − m2 v1 f = v1i + v2i → v1 f = 1 v1i m1 + m2 m1 + m2 m1 + m2 2m1 m − m1 2m1 v2 f = v1i + 2 v2i → v2 f = v1i m1 + m2 m1 + m2 m1 + m2 Below we examine several special cases for which we know the outcome of the collision form experience 1. Equal masses m1 = m2 = m x m1 − m2 m−m m m v1 f = v1i = v1i = 0 m1 + m2 m+m x 2m1 2m m m v2 f = v1i = v1i = v1i m1 + m2 m+m The two colliding objects have exchanged velocities (9-17)
  18. m1 2. A massive target m2 ? m1 → =1 m2 x m1 m1 −1 m2 m − m2 m v1 f = 1 v1i = 2 v1i ≈ −v1i x m1 + m2 m1 m1 +1 m2 m2 m  2 1  v2 f = 2m1 v1i =  m2  v ≈ 2  m1  v m1 1i   1i m1 + m2 +1  m2  m2 Body 1 (small mass) bounces back along the incoming path with its speed v v =0 practically unchanged. Body 2 (large mass) moves forward with avery small v v m1 speed because =1 m2 (9-18)
  19. m2 2. A massive projectile m1 ? m2 → =1 m1 x m2 m2 1− m1 m1 − m2 m1 v = v1i = v ≈v x 1f m1 + m2 m2 1i 1i m2 1+ m1 m1 2m1 2 v2 f = v1i = v ≈ 2v1i m1 + m2 m2 1i 1+ m1 v Body 1 (large mass) keeps on going scarcely slowed by the collision . Body 2 (small mass) charges ahead at twice the speed of body 1 v =0 v v (9-19)
  20. Collisions in Two Dimensions (9-20) 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 m1 and m2 r r r r The linear momentum of the sytem is conserved: p1i + p2i = p1 f + p2 f If the system is elastic the kinetic energy is also conserved: K1i + K 2i = K1 f + K 2 f We assume that m2 is stationary and that after the collision particle 1 and particle 2 move at angles θ1 and θ 2 with the initial direction of motion of m1 In this case the conservation of momentum and kinetic energy take the form: x − axis: m1v1i = m1v1 f cos θ1 + m2 v2 f cos θ 2 (eqs.1) y − axis: 0 = − m1v1 f sin θ1 + m2 v2 f sin θ 2 (eqs.2) 1 1 1 m1v12i = m1v2 f + m2 v2 f (eqs.3) We have three equations and seven variables: 2 2 2 2 2 Two masses: m1 , m2 three speeds: v1i , v1 f , v2 f and two angles: θ1 , θ 2 . If we know the values of four of these parameters we can calculate the remaining three
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