Fundamentals of PhysicsFundamentals of PhysicsMechanicsMechanics(Bilingual Teaching)(Bilingual Teaching)

赵 俊 张昆实赵 俊 张昆实

School of Physical Science and TechnologySchool of Physical Science and TechnologyYangtze UniversityYangtze University

Chapter 9Chapter 9 Systems of Particles Systems of Particles

9-19-1 A Special PointA Special Point

9-29-2 The Central of MassThe Central of Mass

9-39-3 Newton’s Second Law for Newton’s Second Law for

a System of Particle a System of Particle

9-49-4 Linear MomentumLinear Momentum

Chapter 9Chapter 9 Systems of Particles Systems of Particles

9-59-5 The Linear Momentum of a The Linear Momentum of a System of ParticlesSystem of Particles9-69-6 Conservation of Linear Conservation of Linear MomentumMomentum9-7 9-7 Systems with Varying Systems with Varying Mass:A RocketMass:A Rocket9-89-8 External Forces and Internal External Forces and Internal Energy ChangesEnergy Changes

9-1 A Special Point If we toss a ball into the air, it will fol

low a parabolic path. What about a grenade?

9-1 A Special Point Every part of the grenade moves in a different way from every other part except onone speciale special pointpoint 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

we call this special point “the center of mthe center of massass”

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

d

comx

(Fig.9-2a)

2

1 2com

mx d

m m

(9-1)

2 0m 1m

0comx

(2) If , the center of mass lies at the position of

1 0m comx d2m

(3) If , the center of mass shuld be halfway between them.

1 2m m1

2comx d

y

x1m 2m

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

1 1 2 2

1 2com

x m x mx

m m

(9-2)

y

xcomx

1x d2x

1m 2m

(Fig. 9-2b)

o

1 1 2 2

1

1 nn n

com i ii

x m x m x mx m x

M M

(9-3) 1 2( )M m m

For a system of n particles

1 1 2 2com

x m x mx

M

(9-4)

9-2 The Central of MsssIf 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:

i i i ir x i y j z k

(9-6)

The position of the center of mass of a system of particles is given by a position vector:

com com com comr x i y j z k

(9-7)

1

1 n

com i ii

x m xM

(9-5)1

1 n

com i ii

y m yM

1

1 n

com i ii

z m zM

1

1 n

com i ii

x m xM

(9-5)1

1 n

com i ii

y m yM

1

1 n

com i ii

z m zM

Solid Bodies

9-2 The Central of Msss The three scalar equation ofEq.9-5 can be replaced by a single vector equation 1

1 n

com i ii

r m rM

(9-8)

An ordinary solid object contains many particles, can be treated as a continuous distribution of matter. m dm

sum→integrals

( differential mass elements )

the coordinates of the center of mass will be 1

comx xdmM

1

comy ydmM

1

comz zdmM

(9-9)

Substitute

Note: the center of mass of an object need not lie winthin t

he 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

dm M

dV V

( )dm M V dV

(9-10)

into Eq.9-9

1comx xdV

V

1comy ydV

V

1comz zdV

V (9-11)

(2) is the total mass of the system.

(1) is the net force of all external forcesall 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

M

question: If we roll a cue ball at a second billiard ball, How do they move after inpact?

cue ball

the second billiard ball

In fact, what continues to move forwardwhat continues to move forward is the center of mass of the two-ball systemcenter of mass of the two-ball system.

net comF Ma

(9-14)(Newton’s Second Law )

Note:netF

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

coma

, ,net x com xF Ma , ,net y com yF Ma , ,net z com zF Ma (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

0netF

0coma

Two balls collide

internal forces don’t contribute to the net force

0netF

0coma

Thus, the center of mass must still move forward after the collision with the same speed and in the same direction.

net comF Ma

(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

1 1 2 2 3 3com n nMr m r m r m r m r

1 1 2 2 3 3com n nMa m a m a m a m a

Newton’s Second Law applies not only to a system of system of particlesparticles but also to a solid bodya solid body.

Proof of Equation 9-14

(9-16)

1 1 2 2 3 3com n nMv m v m v m v m v (9-17)

(9-18)

We can rewrite Eq.9-18 as

1 2 3com n netMa F F F F F (9-19)

1

1 n

com i ii

r m rM

(9-8)From

The The time rate of changetime rate of change of the momentum of particle i of the momentum of particle is equal to the net force acting on the particle and is in ts equal to the net force acting on the particle and is in the direction of that force.he 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

p p mv

(9-22)

p

v

Express Newton’s Second Law of Motion in terms of momentum:

net

dpF

dt

(9-23)

Substituting for from Eq.9-22 givesp

( )net

dp d dvF mv m ma

dt dt dt

Equivalent

expressions

The linear momentum of a system of particles is equal to The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocthe product of the total mass M of the system and the velocity of the center of mass.ity 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

p

1 2 3

1 1 2 2 3 3

n

n n

p p p p p

m v m v m v m v

(9-24)

comp Mv(linear momentum, system of particles) (9-25)

1 1 2 2 3 3com n nMv m v m v m v m v (9-17)

9-5 The Linear Momentum of a System of Particles

Take the time derivativethe time derivative of Eq.9-25, we find

comcom

dvdpM ma

dt dt

(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

net

dpF

dt

(system of particles) (9-27)

net comF Ma

(9-14)(Newton’s Second Law )

If no net external force acts on a system of particles,If no net external force acts on a system of particles, the total linear momentum of the system cannot the total linear momentum of the system cannot change. change. Law of conservation of linear momentumLaw 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

p

=constant (closed,isolated system)p

0dp

dt

(9-29)

Eq.9-29 can also be written as

i fp p(closed,isolated system) (9-30)

For a closed, isolated system

total linear momentum

at some later time

total linear momentum

at some initial time it=

jt

0netF

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

Now let us study a special system: a rocketa 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 : t p MvAn time interval laterdt

( )( )p dMU M dM v dv For a closed 、 isolated system, the linear momentum must be conserved.

( )( )Mv dMU M dM v dv (9-38)

Mv

x(a)

(b)

xv dv

dMU

M dMt dt

t

(( 漆安慎 漆安慎 P103~105)P103~105)

9-7 Systems with Varying Mass: A Rocket

relvEq.9-38 can be simplified by using the relative speed

relv dv v U

relU v dv v (9-39)

xv dv

dMU

M dMt dt

inertial reference frame

Earth

absolute velocityabsolute velocity: velocity of rocket : velocity of rocket

绝对速度绝对速度 relative to inertial framerelative to inertial frame

relative velocityrelative velocity: velocity of rocket : velocity of rocket

相对速度 相对速度 relative to exhaust productrelative to exhaust productss

convected velocityconvected velocity: velocity of products : velocity of products

牵连速度牵连速度 relative to inertial framerelative to inertial frame

abs rel conv v v

( 漆安慎 P52)

9-7 Systems with Varying Mass: A Rocket

relvEq.9-38 can be simplified by using the relative speed

relU v dv v relv dv v U (9-39)

Substituting it into

reldMv Mdv

rel

dM dvv M

dt dt

(9-40)

(9-41)

Replace by , where is the mass rate of fuel

comsuption

dM

dt

relRv Ma(first rocket equation)

(9-42)

relRv the thrust of the rocket engine

R R

( )( )Mv dMU M dM v dv (9-38)

yields:

T Ma

Finding the Velocity

9-7 Systems with Varying Mass: A Rocket

rel

dMdv v

M

f f

i i

v M

relv M

dMdv v

M

Now, let us study how the velocity of a rocket changes as it consums its fuel.

Integrating leads to

reldMv Mdv (9-40)From

ln if i rel

f

Mv v v

M (second rocket equation) (9-43)

The advantage of multistage rockets !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.

F

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 muscleInternal biochemical energy(in muscless)

kinetic energykinetic energy

F

0v

v

d

x

com

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

d

U

cosK Fd (9-44)

K

cosK U Fd (9-45)

The left side here is the mechanical energy mecE

cosmecE Fd ( external force, change in ) mecE (9-46)

Since does’t transfer energy to or from the system,so 0E F

int 0mecE E (9-47)

int mecE E (9-48)

Fig. 9-14 A vehicle accelerates to the left using four-wheel drivefour-wheel drive. The road exerts four ffour frictional forcesrictional forces (two of them shown) on the bottom surfaces of the tires. Taken together, these four forces make up the net external forexternal forcece 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)

int cosE Fd (9-49)

If is not constant in magnitude, replace with

for the average magnitude of .

F

F F avgF

1F

2F

Internal energy stored in the fuel kinetic energyInternal 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 , thend

0v v

2 20 2 xv v a d (9-50)

Multiplying both sides of Eq.9-50 by the mass M and rearranging yield

2 20

1 1

2 2 xMv Mv Ma d (9-51)

Substituting the product for the product according to the Newton’s second law

cosF xMa

cosK Fd