Fundamentals of Physics Mechanics (Bilingual Teaching) 赵 俊 张昆实 School of Physical Science and Technology Yangtze University

Chapter 9 Systems of Particles 9-1 A Special Point 9-2 The Central of Mass 9-3 Newton’s Second Law for a System of Particle a System of Particle 9-4 Linear Momentum

Chapter 9 Systems of Particles 9-5 The Linear Momentum of a System of Particles System of Particles 9-6 Conservation of Linear Momentum Momentum 9-7 Systems with Varying Mass:A Rocket Mass:A Rocket 9-8 External Forces and Internal Energy Changes Energy Changes

9-1 A Special Point If we toss a ball into the air, it will follow a parabolic path. What about a grenade?

9-1 A Special Point one specialpoint Every part of the grenade moves in a different way from every other part except one special point of the grenade,which still moves in a simple parabolic path : 1) The grenade’s total mass were concentrated there 2) The gravitational force on the grenade acted only there the center of mass we call this special point “the center of mass”

Discuss: (1) if, the center of mass lies at the position of System of Two Particles The position of the center of mass of this two can be difined as: 9-2 The Central of Mass (Fig.9-2a) (9-1) (2) If, the center of mass lies at the position of (3) If, the center of mass shuld be halfway between them.

For a more generalized situation, in which the coordinate system has been shifted leftward. Note: in spite of the shift of the coordinate system, the center of mass is still the same distance from each particle. 9-2 The Central of Msss (9-2) (Fig. 9-2b) o (9-3) For a system of n particles (9-4)

9-2 The Central of Msss 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: (9-6) The position of the center of mass of a system of particles is given by a position vector: (9-7) (9-5)

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

Substitute Note: the center of mass of an object need not lie winthin the object. 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. Since evaluating integrals for most common objects would be diffcult, we consider only uniform objects (uniform density ) 9-2 The Central of Msss (9-10) into Eq.9-9 (9-11)

(2) is the total mass of the system. all external forces (1) is the net force of all external forces acted on the system. The center of mass moves like a particle whose mass is equal to the total mass of the system. Then,the motion of it will be governed by Eq.(9-14) 9-3 Newton’s Second Law for a System of Particles question: If we roll a cue ball at a second billiard ball, How do they move after inpact? cue ball the second billiard ball what continues to move forward center of mass of the two-ball system In fact, what continues to move forward is the center of mass of the two-ball system. (9-14) (Newton’s Second Law ) Note:

Eq.9-14 is equivalent to three equations along the three coordinate axes. (3) is the acceleration of the center of mass of the system. 9-3 Newton’s Second Law for a System of Particles (9-15) 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 Two balls collide internal forces don’t contribute to the net force Thus, the center of mass must still move forward after the collision with the same speed and in the same direction. (9-14)

For a system of n particles, Differentiating Eq.9-17 with respect to time leads to Differentiating Eq.9-16 with respect to time gives 9-3 Newton’s Second Law for a System of Particles system of particlesa solid body Newton’s Second Law applies not only to a system of particles but also to a solid body. Proof of Equation 9-14 (9-16) (9-17) (9-18) We can rewrite Eq.9-18 as (9-19) (9-8) From

 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. The linear momentum of a particle is a vector, defined as and have the same direction, the SI unit for momentum is kilogram-meter per second. 9-4 Linear Momentum (9-22) Express Newton’s Second Law of Motion in terms of momentum: (9-23) Substituting for from Eq.9-22 gives Equivalent expressions

The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass.  The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass. The equation above gives us another way to define 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 the vector sum of the individual particles’ linear momenta. Compare it with Eq.9-17, we see the linear momentum of the system 9-5 The Linear Momentum of a System of Particles (9-24) (linear momentum, system of particles) (9-25) (9-17)

9-5 The Linear Momentum of a System of Particles the time derivative Take the time derivative of Eq.9-25, we find (9-26) Comparing Eq.9-14 and 9-26 allows us to write Newton’s second law for a system of particles in the equivalent form (system of particles) (9-27) (9-14) (Newton’s Second Law )

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  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 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 9-6 Conservation of Linear Momentum =constant (closed,isolated system) (9-29) Eq.9-29 can also be written as (closed,isolated system) (9-30) For a closed, isolated system total linear momentum at some later time total linear momentum at some initial time =

 If the component of the external force on a closed system is zero along an axis then the component of the linear momentum of the system along that axis cannot change. 9-6 Conservation of Linear Momentum Depending on the forces acting on a system, linear momentum might be conserved in one or two directions but not in all directions.

9-7 Systems with Varying Mass: A Rocket a rocket Now let us study a special system: a rocket

the mass changes in a rocket consider the rocket and its ejected combustion products the total mass still constant. 9-7 Systems with Varying Mass: A Rocket Finding the Acceleration Watching in a inertial reference frame ； In deep space, no gravitational or atmospheric drag forces At an arbitrary time : An time interval later For a closed 、 isolated system, the linear momentum must be conserved. (9-38) (a) (b) ( 漆安慎 P103~105)

9-7 Systems with Varying Mass: A Rocket Eq.9-38 can be simplified by using the relative speed (9-39) inertial reference frame Earth absolute velocity: velocity of rocket 绝对速度 relative to inertial frame 绝对速度 relative to inertial frame relative velocity: velocity of rocket 相对速度 relative to exhaust products 相对速度 relative to exhaust products convected velocity: velocity of products 牵连速度 relative to inertial frame 牵连速度 relative to inertial frame ( 漆安慎 P52)

9-7 Systems with Varying Mass: A Rocket Eq.9-38 can be simplified by using the relative speed (9-39) Substituting it into (9-40) (9-41) Replace by, where is the mass rate of fuel comsuption (first rocket equation)(9-42) the thrust of the rocket engine (9-38) yields:

Finding the Velocity 9-7 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 (9-40)From (second rocket equation) (9-43) The advantage of multistage rockets !

9-8 External Forces and Internal Energy Changes A ice skater pushes herself away from a railing, the force from the railing accelerates her, increasing her speed until she leaves the railing. Her kinetic energy is increased via the force, but there are still two major differences. 1 Previously, each part of an object moved rigidly in the same direction. Here, the skater’s arm does not move like the rest of her body. 2 Previously, energy was transferred between the object and its environment via an external force. Here, the energy is transferred internally via the external force. Internal biochemical energy(in muscles Internal biochemical energy(in muscles) kinetic energy

If a change in its gravitational potential energy exists, Now, relate the external force to the internal energy transfer. For a displacement, a change in its kinetic energy 9-8 External Forces and Internal Energy Changes (9-44) (9-45) The left side here is the mechanical energy ( external force, change in ) (9-46) Since does’t transfer energy to or from the system,so (9-47) (9-48)

four-wheel drive four frictional forces external force Fig A vehicle accelerates to the left using four-wheel drive. The road exerts four frictional forces (two of them shown) on the bottom surfaces of the tires. Taken together, these four forces make up the net external force acting on the car. 9-8 External Forces and Internal Energy Changes Substituting Eq.9-46 into the equation above (external force, internal energy change) (9-49) If is not constant in magnitude, replace with for the average magnitude of. Internal energy stored in the fuel kinetic energy

9-8 External Forces and Internal Energy Changes Proof of Equation 9-44 During displacement, velocity changes from to, then (9-50) Multiplying both sides of Eq.9-50 by the mass M and rearranging yield (9-51) Substituting the product for the product according to the Newton’s second law