PH 221-3A Fall 2010

Center of Mass and Linear MomentumMomentum

Lecture 15

Chapter 8 (Halliday/Resnick/Walker, Fundamentals of Physics 8th edition)( y , y )

1

Chapter 9Center of Mass and Linear Momentum

In this chapter we will introduce the following new concepts:

-Center of mass (com) for a system of particlesCenter 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 ofcollisions in one and two dimensions and derive the equation of motion for rockets

2

1 2Consider a system of two particles of masses and The Center of Mass:

m m

1 2at positions and , respectively. We define the position of the center of mass (com) as follows:

m x m x

x x

1 1 2 2

1 2com

m x m xxm m

W li h b d fi i i f f i l f ll

1 1 2 2 3 3 1 1 2 2 3 3

11 2 3

...We can generalize the above definition for a system of particles as follows:

... 1...

nn n n n

com i iin

m x m x m x m x m x m x m x m xx m xm m m m

n

M M

11 2 3

Here is the total mass of all the pa..

rt.

icinm m m m

MM M

1 2 3les ...We can further generalize the definition for the center of mass of a system of

nM m m m m

particles in three dimensional space. We assume that the the -th particle ( mass ) has posii

im tion vector 1 n

ir

1

1 n

com i ii

r m rM

3

1

1The position vector for the center of mass is given by the equation:

ˆˆ ˆThe position vector can be written as:

n

com i ii

r m rM

r x i y j z k

The position vector can be written as: The components of are given by the

com com com com

com

r x i y j z kr

1 1 1equations:

n n n

x m x y m y z m z 1 1 1

com i i com i i com i ii i i

x m x y m y z m zM M M

The center of mass has been defined using the equations i b th t it h th f ll i tgiven above so that it has the following prop

The center of mass of a system of particles moves as thoughall the system's mass were co

erty:

ncetrated there, and that the y

The above statement will be proved la

,vector sum o

ter. An example isi i h fi

f all the

b

external forces were appli

b ll b i fli d i

ed ther

h

e

igiven in the figure. A baseball bat is flipped into the airand moves under the influence of the gravitation force. The center of mass is indicated by the black dot. It follows a ce te o ass s d cated by t e b ac dot. t o ows aparabolic path as discussed in Chapter 4 (projectile motion)All the other points of the bat follow more complicated paths

4

Solid bodies can be considered as systems with continuous distribution of matterThe Center of Mass for Solid Bodies

The sums that are used for the calculation of the center of mass of systems withdiscrete distribution of mass become integrals:

1 1 1

The integrals above are rather complicated.

1 1 1

A simpler special case is that of

com com comx xdm y ydm z zdmM M M

uniform objects in which the mass density is constant and equal to

1 1 1dV dV dV

dm MdV V

In objects with symetry elements (symmetry point, symmetry line, symmetry

plane)it is not

com com comx xdV y ydV z zdVV V V

necessary to eveluate the integrals The center of mass lies on the symmetryit 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 centerIn a uniform rectanglular object the com lies at the intersection of the diagonals

.CC 5

6

m1

m

F1 F2

z

1 2 3Consider a system of particles of masses , , ...,Newton's Second Law for a System of Particles

nn m m m m

Om3

m2F2

x y

1 2 3,

1 2 3

y p , , ,and position vectors , , ,..., , respectively. The position vector of the center of mas

n

nr r r r

s is given by:

1 1 2 2 3 3 ... We take the time derivative of both sidescom n nMr m r m r m r m rd d d d d

1 1 2 2 3 3

1 1 2 2 3 3

...

... Here is the velocity of the c

com n n

com n n com

d d d d dM r m r m r m r m rdt dt dt dt dt

Mv m v m v m v m v v

om

1 1 2 2 3 3

and is the velocity of the -th particle. We take the time derivative once more

...

i

com n n

v id d d d dM v m v m v m v m vdt dt dt dt dt

1 1 2 2 3 3 ... Here is the accelercom n n com

dt dt dt dt dtMa m a m a m a m a a ation of the com

and is the acceleration of the -th particleia i

7

m1

m

F1 F2

z1 1 2 2 3 3 ...

We apply Newton's second law for the -th particle:com n nMa m a m a m a m a

i

Om3

m2F2

x y1 2 3

Here is the net force on the -th particle

...i i i i

com n

m a F F i

Ma F F F F

int

The force can be decomposed into two components: applied and internal

The above equation takes the form:i

appi i i

F

F F F

int int int int1 1 2 2 3 3 ...app app app app

com n nMa F F F F F F F F

Ma

int int int int1 2 3 1 2 3... ...app app app app

com n nF F F F F F F F

The sum in the first parenthesis on the RHS of the equation above is just The sum in the second parethesis on the RHS van

netF

ishes by virtue of Newton's third law.

The equation of motion for the center of mass becomes: In terms of components we have:

com netMa F

, , , , ,

In terms of components we have: net x com x net y com y net zF Ma F Ma F ,

com zMa

8

com netMa F

, ,net x com xF MaF Ma

, ,

, ,

net y com y

net z com z

F Ma

F Ma

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 pointthe influence of gravity on a parabolic path (projectile motion). At a certain pointthe rocket explodes into fragments. If the explosion had not occured, the rocketwould have continued to move on the parabolic trajectory (dashed line). The forcesof 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

l i Thi h h f f h f f l hexplosion. This means that the center of mass of the fragments folows the sameparabolic trajectory that the rocket would have followed had it not exploded 9

mv

p Linear momentum of a particle of mass and velocity Linear Momentum

p m v

The SI unit for liis defined as

neal momentum:

is the kg

.m/sp mv

p mv

The time rate of change of the linearmomentum of a particle is equal to the magnitude of net force acting on thBelow we will prove the fol

eti l d h th di

lowing statem

ti f th f

ent:

particle and has the direction of the force

In equation form: We will prove this equation usingnetdpFdt

Newton's second law

netdp d dvp mv mv m ma Fdt dt dt

This equation is stating that the linear momentum of a particle can be changedt dt dt

donly by an external force. If the net external force is zero, the linear momentumcannot change

netdpFdt

10

m1p1 p3

z

In this section we will extedend the definition ofThe Linear Momentum of a System of Particles

Om3

m2p2

x y

linear momentum to a system of particles. The -th particle has mass , velocity , and linear

moment mi ii m v

momentum ip

We define the linear momentum of a system of particles as follows:n

P M

1 2 3 1 1 2 2 3 3... ...The linear momentum of a system of particles is equal to the product of th

n n n comP p p p p m v m v m v m v Mv e

total mass of the system and the velocity of the center of masscomM v

y y

The time rate of change of is:

h li f f i l b h d

com

com com netdP dP Mv Ma Fdt dt

lThe linear momentum of a system of particles can be changed P only

by a net external force . If the net external force is zero cannot changenet netF F P

dP

1 2 3 ... n comP p p p p Mv

netdP Fdt

11

Example. Motion of the Center of Mass

12

13

We have seen in the previous discussion that the momentum of an object canh if th i t l f ti th bj t S h f

Collision and Impulse

change if there is a non-zero external force acting on the object. Such forcesexist 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, y g , y p gmomentum of the colliding objects.

Consider the collision of a baseball with a baseball batThe collision starts at time when the ball touches the batand ends at when the two objects separate

i

f

tt

The ball is acted upon by a force (F ) during the collisionThe magnitude ( ) of the force is plotted versus in fig.aThe force is non-zero only for the time interval i f

tF t t

t t t The force is non zero only for the time interval

( ) Here is the linear momentum of the ball

i ft t tdpF t pdt

d ( ) ( )f f

i i

t t

t t

p F t dt dp F t dt

14

( ) = change in momentumf f ft t t

f idp F t dt dp p p p ( ) change in momentum

( ) is known as the impulse of the collision

i i i

f

f it t t

t

dp F t dt dp p p p

F t dt J

( ) p

( ) The magnitude of is equal to the area

i

f

t

t

J F t dt J

under the vit

F

ersus plot of fig.a

In many situations we do not know how the force changest p J

In many situations we do not know how the force changeswith time but we know the average magnitude of thecollision force. The magnitude of the impulse is given by:

aveF

J where

Geometrically this means that the the area under thel t (fi ) i l t th d th

ave f iF t t t t

F t

p J versus plot (fig.a) is equal to the area under theversus plot (fig.b)

ave

F tF t

p J

aveJ F t 15

Collisions. Impulse and Momentum

16

The Impulse-Momentum Theorem

17

Consider a target which collides with a steady stream ofSeries of Collisions

Fave Consider a target which collides with a steady stream ofidentical particles of mass and velocity along the -axism v x

A number of the particles collides with the target during a time interval .n tA number of the particles collides with the target during a time interval . Each particle undergoes a change in momentum due to the collision withthe target. During each collision a momentum

n tp

change is imparted on thep target. The Impulse on the target during the time interval is:

The average force on the target is:t

J n pJ n p n

Here is the chanaveJ n p nF m v vt t t

ge in the velocity

of each particle along the -axis due to the collision with the target x

Here is the rate at which mass collides with the target

If the particles stop after the

avem mF vt t

collision then 0v v v

If the particles bounce backwards then 2v v v v 18

m1p1 p3

z

Consider a system of particles for which 0

Conservation of Linear Momentum

netF

Om3

m2p2

x y0 Constantnet

dP F Pdt

If l f f i l h l li P

total linear momentumtotal linear momentum

If no net external force acts on a system of particles the total linear momentumcannot change

P

total linear momentumtotal linear momentumat some later time at some initial

The

time

cfi tt

onservation 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.

I t i hi hN t 1 F

0 l l ti f li In systems in which Note 1: netF 0 we can always apply conservation of linearmomentum even when the internal forces are very large as in the case of colliding objects g j

We will encounter problems (e.g. inelastic collisNote 2: ions) in which the energyis not conserved but the linear momentum is 19

1 2Consider two colliding objects with masses and ,Momentum and Kinetic Energy in Collisions

m m

1 2 1 2initial velocities and and final velocities and ,

respectivelyi i f fv v v v

If the system is isolated i.e. the net force 0 linear momentum is conservedThe conervation of linear momentum is true regardless of the the collision typeThis is a powerful rule that allows us t

netF

o determine the results of a collision withoutThis 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: and .

elasticinelastic

A collision is if there is no loss of elast kinetic eic nergy i.e.

A collision is if kinetic energy is lost during the collision due to conversioni t th f f I thi h

inelastici fK K

K K

into other forms of energy. In this case we have:

A special case of inelastic collisions

af iK K

re known as. In these collisions the two colliding objects stick togethercompletely inlelastic . In these collisions the two colliding objects stick together

and they move as a single body. In these collisions the loss of kinetcomp

ic eletely inlela

nergy is max

stic

imum20

In these collisions the linear momentium of the colliding One Dimensional Inelastic Collisions

1 2 1 2

1 1 1 2 1 1 1 2

objects is conserved i i f f

i i f f

p p p p

m v m v m v m v

In these collisions the two colliding objects stick togetherOne Dimensional Completely Inelastic Collisio sn

2

and move as a single body. In the figure to the leftwe show a special case in which 0.iv

1 1 1 2im v mV m V

m

11

1 2

The velocity of the center of mass in this collision

imV v

m m

1 2 1 1

1 2 1 2 1 2

is

In the picture to the left we show some freeze frames

i i icom

P p p m vvm m m m m m

In the picture to the left we show some freeze-framesof a totally inelastic collition 21

The Principle of Conservation of Linear Momentum

22

23

1 2Consider two colliding objects with masses and ,One-Dimensional Elastic Collisons

m m

1 2 1 2initial velocities and and final velocities and ,

respectivelyi i f fv v v v

1 1 1 2 1 1 1 2

Both linear momentum and kinetic energy are conserved. Linear momentum conservation: (eqs.1)i i f fm v m v m v m v

2 22 21 1 2 21 1 1 2Kinetic energy conservation: (eqs.2)

2 2 2 2We have tw

f fi i m v m vm v m v

o equations and two unknowns andv vWe have tw

1 2 2

1 2

1 2

o equations and two unknowns, and

If we solve equations 1 and 2 for and we get the following solution

2

s:f f

f f

m m m

v v

v v

1 2 21 1 2

1 2 1 2

1 2 12 1 2

2

f i i

f i i

v v vm m m m

m m mv v v

2 1 2

1 2 21f i im m m m

24

2

The sSpecial C

ubstitutease of elastic Col

0 in thelisions-Stationary

two solutions for Tar

andget 0i

v vvv

2 1 2The substitute 0 in the two solutions for and i f fv v v

1 2 2 1 22 m mv vm m mv v v

1 1 2

1 2 1 2

1 2 12 1 2

1 11 2

12 1

2 2

f i i

f i

f i

f ii

v vm m

m

v v vm m m m

m m mv v vm m m m

v vm m

1 2 1 1 22

Below we examine several special cases for which we know the outcomeof the collision from experience

m m m m m m

1. Eq au

1 21 1 1

1 2

0

l masses

f i im m m mv v vm m m m

m m m

m m

v1i

v1f = 0 v2f

v2i = 0x

1 2

12 1 1 1

1 2

2 2

f

f i i i

m m m mm mv v v v

m m m m

m mx

The two colliding objects have exchanged velocit esi25

12 1

2

2. A mass <1 ive target mm mm

v2i = 0v1i

1

1 2 21 1 1 1

1

f i i i

mm m mv v v vmm m

m1 m2

1i

v2f

x

x 11 2

2

1

1

2

mm mm

m

m1m2

v1fx

21 12 1 1 1

11 2 2

2

22 2

1f i i i

mm mv v v vmm m mm

2m

Body 1 (small mass) bounces back along the incoming path with its speed practically unchanged. B d 2 (l ) f d ith ll

1

2

Body 2 (large mass) moves forward with a very small

speed because <1 mm2

26

v1i2

1 2 <1 2. A massive projectile mm m

m1

m2 v1fv2f

v2i = 0x

1

2

1 2 1

1

mm

m m mv v v v

m1

m2

v2fx 1 1 1 1

21 2

1

1

2 2

f i i iv v v vmm mm

m

12 1 1 1

21 2

1

2 2 21

f i i imv v v vmm m

m

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

27

Elastic Collision

28

Inelastic Collision

29

In this section we will remove the restriction that thellidi bj t l i I t d

Collisions in Two Dimensions

colliding objects move along one axis. Instead we assumethat the two bodies that participate in the collisionmove in 1 2the -plane. Their masses are and xy m m1 2py

1 2 1 2

1 2 1 2

The linear momentum of the sytem is conserved: If the system is elastic the kinetic energy is also conserved:

i i f f

i i f f

p p p pK K K K

1 2 1 2

2

y gyWe assume that is stationary and that after the co

i i f f

m

1 2 1

llision particle 1 and particle 2 move at angles and with the initial direction of motion of m

1 1 1 1

In this case the conservation of momentum and kinetic energy take the form:axis: i fx m v m v 1 2 2 2cos cos (eqs.1) axis: 0 sin sin (eqs 2)

fm vy m v m v

1 1 1 2 2 2

2 2 21 1 1 2 2 2

axis: 0 sin sin (eqs.2)1 1 1 (eqs.3) We have three equations and seven variables:2 2 2

f f

i f f

y m v m v

m v m v m v

1 2 1 1 2Two masses: , three speeds: , , ai f fm m v v v 1 2nd two angles: , . If we know the values of four of these parameters we can calculate the remaining three

30

'

P rob lem 7 2 . T w o 2 .0 kg b od ie s , A an d B c o llid e . T h e ve loc itie s b e fo re th e co llis ion a re ˆ ˆ ˆ ˆ(1 5 3 0 ) m /s an d ( 1 0 5 .0 ) m /s . A fte r th e co llis ion ,

ˆ ˆ( 5 .0 2 0 ) m /s . W h a t a re (a ) th e f in aA B

A

v i j v i j

v i j

l ve loc ity o f B an d (b ) th e ch an gein th e to ta l k in e tic en e rgy ( in c lu d in g s ign )?

(a) Conservation of linear momentum implies

m v m v m v m vA A B B A A B B ' ' .

Since mA = mB = m = 2.0 kg, the masses divide out and we obtain g,

ˆ ˆ ˆ ˆ ˆ ˆ (15i 30j) m/s ( 10 i 5j) m/s ( 5i 20 j) m/s

ˆ ˆ(10 i 15 j) m/s .B A B Av v v v

(b) The final and initial kinetic energies are

K mv mv 1 1 1 2 0 5 20 10 15 8 0 102 2 2 2 2 2 2' ' ( ) ( )c h JK mv mv

K mv mv

f A B

i A B

2 2 22 0 5 20 10 15 8 0 10

12

12

12

2 0 15 30 10 5 13 102 2 2 2 2 2 3

' ' ( . ) ( ) .

( . ) ( ) .

c h

c h

J

J .

The change kinetic energy is then K = –5.0 102 J (that is, 500 J of the initial kinetic energy is lost). 31

A rocket of mass and speed ejects mass backwards Systems with Varying Mass: The Rocket

M v

at a constant rate . The ejected material is expelled at a

constant speed relative to the rocket Tl

dMdt

v hus the rocket losesconstant speed relative to the rocket. Trelv hus the rocket loses mass and accelerates forward. We will use the conservation of linear momentum to determine the speed of the rocket v

In figures (a) and (b) we show the rocket at times and . If we assume thatthere are no external forces acting on the rocket, linear momentum is conserved

t t dt

( ) (eqs.1)Herp t p t dt Mv dMU M dM v dv

e is a negative number because the rocket's mass decreases with time i th l it f th j t d ith t t th i ti l f f

dM tU is the velocity of the ejected gases with respect to the inertial reference framein which we measure the rocket's speed . WU

v e use the transformation equation for velocities (Chapter 4) to express in terms of which is measured with respect relU v( p ) p pto the rocket. We substitute in equation 1 and we get:

rel

relU v dvM

vdv

U

reldMv32

Using the conservation of linear momentum we derived the equation of motion for the rocket

(eqs.2) We assume that material islMdv dMv (eqs.2) We assume that material is ejected from the rocret's nozzle at a constant rate

relMdv dMv

dM (eqs.3) Here is a constant positive number,R R dt

(eqs.3) Here is a constant positive number,

the positive mass rate of fuel consumprion.

R R

We devide both sides of eqs (2) by dv dMdt M v Rv Ma Rv We devide both sides of eqs.(2) by

Here is the rocket's acceleration, the thrust of the rocket engine

We use e (First rocket equation)

quation 2 to det

e

rerel r lel

rel

dt M v Rvdt dt

a Rv

Ma Rv

rmine the rocket's speed as function of timerocket engine. We use equation 2 to determine the rocket s speed as function of time

We integrate both sides f fv M

rel relv M

dM dMdv v dv vM M

vf

ln ln ln

i i

f i

i f

v M

M M if i rel rel relM M

f

Mv v v M vv MM

f

(Second rocket equation)ln if i rel

f

Mv v vM

vi Mi/Mf

O33

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

h t t d f 253 / l ti t th b Wh t i th fi lexhaust at a speed of 253 m/s relative to the space probe. What is the final velocity of the probe?

rel6090 kgln 105 m/s (253 m/s) ln 108 m/s.6010 kg

if i

f

Mv v vM

34