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制作 张昆实 Yangtze University

Bilingual Mechanics. Chapter 5 Center of Mass and Linear Momentum. 制作 张昆实 Yangtze University. 制作 张昆实 赵 俊 Yangtze University. Chapter 5 Center of Mass and Linear Momentum. 5-1 What Is Physics? 5-2 The Center of Mass

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制作 张昆实 Yangtze University

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  1. BilingualMechanics Chapter 5 Center of Mass and Linear Momentum 制作 张昆实 Yangtze University 制作 张昆实 赵 俊 Yangtze University

  2. Chapter 5Center of Mass and Linear Momentum 5-1 What Is Physics? 5-2 The Center of Mass 5-3 Newton’s Second Law for a System of Particles 5-4 Linear Momentum 5-5 The Linear Momentum of a System of Particles 5-6 Collision and Impulse

  3. Chapter 5Center of Mass and Linear Momentum 5-7 Conservation of Linear Momentum 5-8 Momentum and Kinetic Energy in Collisions 5-9 Inelastic Collisions in One Dimension 5-10 Elastic Collisions in One Dimension 5-11 Collisions in Two Dimensions 5-12 Systems with Varying Mass: A Rocket

  4. ★ We discuss how the complicated motion of a system of objects can be simplified if we determine a special point of the system thecenter of mass of that system. 5-1 What Is Physics ★Analyzing complicated motion requires simplification via an understanding of physics. ★ In this chapter we learn what is physics through studying linear momentum and collisions.

  5. 5-2 The Center of Mass If we toss a small ball into the air, it will follow a parabolic path. What about a grenade?

  6. 5-2 The Center of Mass Every part of the grenade moves in a different way exceptone specialpoint, the center of mass,of the grenade,which still movesin a parabolic path . The center of mass of a system of particles is the point that moves as though • all the system’s mass were concentrated there and (2) all external forces were applied there.

  7. (Fig.5-2a) 5-2 The Center of Mass Systems of Particles The position of the center of mass (com) of this two-particle system can be difined as: (5-1) Discuss (1) if , the center of mass lies at the position of (2) If, the center of mass lies at the position of (3) If, the center of mass shuld be halfway between them.

  8. o (Fig. 5-2b) 5-2 The Center of Mass For a more generalized situation, in which the coordinate system has been shifted leftward. (5-2) Note:in spite of the shift of the coordinate system, the center of mass is still the same distance from each particle. (5-3) For a system of n particles (5-4)

  9. (5-5) (5-6) (5-7) 5-2 The Center of Mass If particles are distributed in three dimensions, the center of mass must be identified by three coordinates: Define the center of mass with the language of vectors. The position of a particle is given by a position vector: The position of the center of mass of a system of particles is given by a position vector:

  10. (5-8) (5-5) (5-9) 5-2 The Center of Mass The three scalar equation of Eq.5-5 can be replaced by a single vector equation Solid Bodies An ordinary solid object contains many particles, can be treated as a continuous distribution of matter. ( differential mass elements ) sum→integral the coordinates of the center of mass will be

  11. Substitute into Eq.5-9 gives (5-11) 5-2 The Center of Mass We consider only uniform objects (uniform density ) (5-10) We can bypass one or more of these integrals if an object has a point, a line, or a plane of symmetry. The center of mass of such an object then lies at that point, or that line, or in that plane. Note: the center of mass of an object need not lie winthin the object.

  12. the second billiard ball (5-14) (Newton’s Second Law) (1)is the net force of all external forces acted on the system. cue ball (2) is the total mass of the system. Note: 5-3Newton’s Second Law for a System of Particles If we roll a cue ball at a secondbilliard ball that is at rest, How do they move after inpact? In fact, what continues to move forward is thecenter of mass of the two-ball system. The center of mass moves like a particlewhose mass is equal to the total mass of the system. Then,the motion of it will be governed by Eq.(5-14)

  13. (5-14) (5-15) 5-3 Newton’s Second Law for a System of Particles (3)is the acceleration of the center of mass of the system. Eq.5-14 is equivalent to three equations along the three coordinate axes. Go back and examine the behavior of the billiard balls. no net external force acts on the system The cue ball has begun to roll internal forces don’t contribute to the net force Two balls collide Thus, the center of mass must still move forward after the collision with thesame speed and in the same direction.

  14. (5-8) Proof of Equation 5-14 (5-17) 5-3Newton’s Second Law for a System of Particles Newton’s Second Law applies not only to a system of particles but also to a solid body. From For a system of n particles, (5-16) Differentiating Eq.5-16 with respect to time gives Differentiating Eq.5-17 with respect to time leads to (5-18) We can rewrite Eq.5-18 as (5-19)

  15. (5-22) Express Newton’s Second Law of Motion in terms of momentum: • The time rate of change of the momentum of particle is equal to the net force acting on the particle and is in the direction of that force. (5-23) Equivalent expressions Substituting for from Eq.5-22 gives 5-4Linear Momentum The linear momentum of a particle is a vector quantity , defined as and have the same direction, the SI unit for momentum is kilogram-meter per second .

  16. Compare it with Eq.5-17, we see the linear momentum of the system (5-17) (linear momentum, system of particles) (5-25) The equation above gives us another way to define thelinear momentum of a system of particles: • The linear momentum of a system of particles is equal to the product of the total mass M of the systemand the velocity ofthe center of mass. 5-5 The Linear Momentum of a System of Particles Now consider a system of n particles. The system as a whole has a total linear momentum , which is defined to be thevector sum of the individual particles’ linear momenta. (5-24)

  17. (5-26) (5-14) (Newton’s Second Law) (5-27) (system of particles) 5-5 The Linear Momentum of a System of Particles Take the time derivative of Eq.5-25, we find Comparing Eq.5-14 and 5-26 allows us to write Newton’s second law for a system of particles in the equivalent form

  18. 5-6 Collision and Impulse ★In everyday language, a collision occurs when an objects crashinto each other. Example:tennis ball contacts with racket the collision ofa ball with a bat ★ A collision is an isolated event in which two or more bodies (the colliding bodies) exert relatively strong forces on each other for a relatively short time.definition

  19. (5-28) Integrating Eq. 5-28 from to : (5-29) 5-6 Collision and Impulse During a head-on collision: A third law force pair , ★Single collision The change of the linear momentum depends on: The forces and the action time Apply Newton’s second law to ball R

  20. (5-29) (5-30) (5-35) J rectangle 5-6 Collision and Impulse ★is the impulse, which is a measure of both themagnitude and the duration of the collision force.

  21. (5-29) (5-30) (5-31) (5-32) ( linear momentum-Impulse theorem ) The change in an object’s momentum is equal to the impulse on the object. (5-33) (5-34) 5-6 Collision and Impulse From Eqs. 5-29 and 5-30 Component form

  22. Target A steady stream of identical projectilescollides with a fixed target. Find the average Projectiles forceon the target during the bombardment. The total change in momentum for projectiles in is : The impulseon the target in : (5-36) Combining Eq. 5-35 and 5-36: (5-37) 5-6 Collision and Impulse Series of collisions

  23. The rate at which the projectiles collide with the target Target Projectiles In , an amount of mass collide with the target, so The rate at which the mass collides with the target (5-40) If If 5-6 Collision and Impulse Series of collisions (5-37)

  24. (5-42) =constant (closed,isolated system) • If no net external force acts on a system of particles, the total linear momentum of the system cannot change. Law of conservation of linear momentum Eq.5-42 can also be written as (5-43) (closed,isolated system) For a closed, isolated system total linear momentum at some later time total linear momentum at some initial time = 5-7Conservation of Linear Momentum Suppose that the net external force acting on a system of particles is zero (isolated), and that no particles leave or enter the system(closed)，Then

  25. 5-7Conservation of Linear Momentum • If the component of the net external force on a closed system is zeroalong an axis then the component of the linear momentum of the system along that axiscannot change.

  26. 1.Elastic collision: is a special type of collision in which the kinetic energe of the system of colliding bodies is conserved. 2.Inelastic collision: after the collision the kinetic energe of the system is not conserved. 3.Completely inelastic collision: after the collision the bodies stick together and have the same final velocity ( the greatest loss of kinetic energe). 5-8Momentum and Kinetic Energy in Collisions Discussion collisions in closed(no mass enters or leaves them) and isolated(no net external forces act on the bodies within them ) systems. Kinatic Energy（in collisions）

  27. The law of conservation of linear momentum In a closed , isolatedsystemcontaining a collision,the linear momentum of each colliding body may change butthe total linear momentum of the systemcannotchange, whether the collision is elastic or inelastic. 5-8Momentum and Kinetic Energy in Collisions Regardless of the details of the impulses in a collision; Regardless of what happens to the kinetic energy of the system, The total linear momentum of a closed, isolatedsystemscannot change. (no external force !) Linear Momentum（in collisions）

  28. befor after (5-50) The motion is one-dimensional, component form: (5-51) 5-9Inelastic Collisions in One Dimension One Dimensional Inelastic collision Two colliding bodies form a closed , isolatedsystem along the x axis. The law of conservation of linear momentum before、 after collision

  29. befor Projectile Target after collision (5-52) (5-53) 5-9 Inelastic Collisions in One Dimension Before the collisionbody 2 is at rest, body 1 moves directly toward it. After the collision, the stuck-together bodiesmove withthe same velocity . One-Dimensional Completely Inelastic Collision The law of conservation of linear momentum before、 after collision

  30. Find out (5-54) (5-55) (left side=constant) (5-56) (constant) 5-9Inelastic Collisions in One Dimension In a closed, isolatedsystem, the velosity of the center of mass of the systemcannot be changed by a collision for there is no net external forceto change it. Velocity of Center of Mass Collision ! Completely inelastic collision

  31. befor Projectile Target after Total kinetic energy after the collision Total kinetic energy beforethe collision = 5-10 Elastic Collisions In One Dimension We can approximate some of the everyday collisions as being elastic; we can approximatethat the total kinetic energy of the cilli-ding bodies is conserved. ☞ In an elastic collision, the kinetic energy of each colliding bodymay change, but the total kinetic energy of the systemdoes not change.

  32. befor Projectile Target after (5-63) ( linear momentum ) If the collisionis also elastic, the total kinetic energy of the systemis also conserved. ☞ (5-64) ( kinetic energy ) 5-10 Elastic Collisions In One Dimension Before the collisionbody 2 is at rest, body 1 moves toward it(head on collision). Stationary Target Assume this two body system is closed and Isolated. Then the net linear momentum of the system is conserved.

  33. If know , Then can find out Rewrite Eq.5-63 (5-65) Rewrite Eq.5-64 (5-63) ( linear momentum ) (5-66) (5-67) (5-68) (5-64) ( kinetic energy ) 5-10 Elastic Collisions In One Dimension Stationary Target

  34. (5-67) (5-68) 1.Equal masses: If Eqs. (5-67) and (5-68) reduce to and 2.A massive target: If Eqs.(5-67)and(5-68) reduce to and (5-69) 3.A massive projectile: If Eqs.(5-67)and(5-68) reduce to and (5-70) 5-10 Elastic Collisions In One Dimension Stationary Target Discussion: A few special situations Changing vilocities ! bounds back

  35. To work outand rewritethese conservation equations: (5-73) (5-74) (5-75) (5-76) 5-10 Elastic Collisions In One Dimension 1.Conservationof the linear momentum Moving Target 2.Conservation of the kinetic energy head on elastic collision

  36. Rewrite Eq.5-77 for componentsalong thex、y axis (x axis) (5-79) (y axis) (5-80) (5-81) 5-11 Collisions in Two Dimensions glancing collision Two dimension (not head-on) elastic collision System: closed + isolated Conservation equations: (5-77) (5-78)

  37. 5-12 Systems with Varying Mass: A Rocket Now let us study a special system: a rocket

  38. 5-12 Systems with Varying Mass: A Rocket Most of the mass of a rocket on its launching pad is fuel, all of which will eventually be burned and ejected from the nozzle of the rocket engine Consider the rocket and its ejected combustion products as a system, the total mass is still constant. China´s manned spacecraft Shenzhou-7 blasts off

  39. Finding the Acceleration Watching in a inertial reference frame; In deep space, nogra- vitional or atmospheric drag forces At an arbitrary time : An timeinterval later For a closed、isolatedsystem, the linear momentum must be conserved. (5-82) (5-83) 5-12 Systems with Varying Mass: A Rocket (a) (b)

  40. velocity of products relative to frame velocity of rocket relative to frame absolute velocity convected velocity For the one- dimensional motion (5-84) 5-12 Systems with Varying Mass: A Rocket inertial reference frame Earth Eq.5-83 can be simplified by using the relative speed velocity of rocket relative to products = + relative velocity

  41. Substituting it into (5-83) yields: (5-85) (5-86) Replace by , where is the mass rate of fuel comsuption (first rocket equation) (5-87) the thrust of the rocket engine (5-84) 5-12 Systems with Varying Mass: A Rocket

  42. Finding the Velocity From (5-85) (second rocket equation) (5-88) The advantage of multistage rockets ! 5-12 Systems with Varying Mass: A Rocket Now, let us study how the velocity of a rocket changes as it consums its fuel. Integrating leads to

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