PH 221-1D Spring 2013

Center of Mass and Linear Momentum

Lectures 22-23

Chapter 9 (Halliday/Resnick/Walker, Fundamentals of Physics 9th edition)

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 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 motion for rockets

2

1 2

1

1 1 2 2

1 2

2

Consider a system of two particles of masses and at positions and , respectively. We define the position of the center of mass (com) as follow

The Center

s:

of Mass:

comm x m xx

m m

m mx x

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:

Here is the total mass of all the pa

... 1..

rt.

ic

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

MM M

1 2 3

les ...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 -th particle ( mass ) has posi

n

i

M m m m m

im

1

tion vector 1 n

com i ii

i

r

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: The components of are given by the

n

com i ii

com com com com

com

r m rM

r x i y j z kr

1 1 1

1 1 1

equations:

n n n

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 given above so that it has the following propThe center of mass of a system of particles moves as thoughall the system's mass were co

erty:

ncetr

The above statement will be proved la

ated there, and that the vector sum o

ter. An example isgiven in the figur

f all the

e. A bas

external forces were appli

eball bat is flipped into t

ed ther

he a

e

irand 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)All the other points of the bat follow more complicated paths

4

Solid bodies can be considered as systems with continuous distribution of matterThe sums that are used for the calculation of the c

The Center of Mass for Solid Bod

enter of mass of systems withd

ies

iscrete distribution of mass become integrals:

The integrals above are rather complicated.

1 1 1

A simpler special case is that of

uniform objects in wh

ic

com com comx xdm y ydm z zdmM M M

h the mass density is constant and equal to

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

1 1 1

plane)it

is not

com com comx xdV y ydV z zdVV V

dm MdV V

V

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

O

m1

m3m2

F1

F2

F2

x y

z

1 2 3,

1 2 3

Consider a system of particles of masses , , ...,and position vectors , , ,...

Newton's Second Law for

, , respectively.

a System of Particle

The position vector of the center of mas

s

n

n

n m m m mr r r r

s is given by:

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

... We take the time derivative of both sides

...

... Here is the velocity of the c

com n n

com n n

com n n com

Mr m r m r m r m rd 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

1 1 2 2 3 3

1 1 2 2 3 3

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

...

... Here is the acceler

i

com n n

com n n com

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

Ma m a m a m a m a a

ation of the comand is the acceleration of the -th particleia i

7

O

m1

m3m2

F1

F2

F2

x y

z1 1 2 2 3 3

1 2 3

...We apply Newton's second law for the -th particle:

Here is the net force on the -th particle

...

com n n

i i i i

com n

Ma m a m a m a m ai

m a F F i

Ma F F F F

int

int int int int1 1 2 2 3 3

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

The above equation takes the form:

...

i

appi i i

app app app appcom n n

F

F F F

Ma F F F F F F F F

Ma

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

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

app app app appcom n n

net

F F F F F F F F

F

, , , , ,

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 net

net x com x net y com y net z

Ma F

F Ma F Ma F

,

com zMa

8

com netMa F

, ,

, ,

, ,

net x com x

net y com y

net z com z

F MaF 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 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 theexplosion. 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

The

Linear Momentum

SI unit for liis defined as

neal momentum:

is the kg

.m/sp mv

p m v

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

eparticle and has the direc

lowing statem

tion of the f

ent:

orce

In equation form: We will prove this equation using

Newton's second law

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

net

net

dpFdt

dp d dvp mv mv m ma Fdt dt dt

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

netdpFdt

10

O

m1

m3m2

p1

p2

p3

x y

z

In this section we will extedend the definition oflinear momentum to a system of particles. The -th particle ha

The Linear Momentum of a S

s mass , velocity , and linearmomentum

ystem of Particles

i ii m v

ip

1 2 3 1 1 2 2 3 3

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

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

n n n com

n

P p p p p m v m v m v m v Mv

etotal mass of the system and the velocity of the center of mass

The time rate of change of is:

The linear momentum of a system of particles can be changed

com

com com net

M v

dP dP Mv Ma Fdt dt

P

only

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

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

netdP Fdt

11

12

Example. Motion of the Center of Mass

13

We have seen in the previous discussion that the momentum of an object canchange if there is a non-zero external force acting on the object. Such forcesexis

Col

t d

lision and I

uring the co

mpulse

llision of two objects. These forces act for a brief time interval, they are large, and they are responsible for the changes in the linearmomentum 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

The ball is acted upon by a force (

i

f

tt

F

) during the collisionThe magnitude ( ) of the force is plotted versus in fig.aThe force is non-zero only for the time interval

( ) Here is the linear momentum of the ball

i f

tF t t

t t tdpF t pdt

d

( ) ( )f f

i i

t t

t t

p F t dt dp F t dt

14

( ) = change in momentum

( ) is known as the impulse of the collision

( ) The magnitude of is equal to the area

under the v

f f f

i i i

f

i

f

i

t t t

f it t t

t

t

t

t

dp F t dt dp p p p

F t dt J

J F t dt J

F

ersus plot of fig.aIn 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:

ave

t p J

F

J

where

Geometrically this means that the the area under the versus plot (fig.a) is equal to the area under the

versus plot (fig.b)

ave f i

ave

F t t t t

F tF t

p J

aveJ F t 15

16

Collisions. Impulse and Momentum

17

The Impulse-Momentum Theorem

Consider a target which collides with a steady stream ofidentical particles of mass and velocity

Series of

alon

Co

g

llision

the -a

s

xism v x

A 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 thetarget. The Impulse on the target during the time interval is:

The average force on the target is:

Here is the chanave

pt

J n pJ 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

Here is the rate at which mass collides with the target

If the particles stop after the

ave

xm mF vt t

collision then 0If the particles bounce backwards then 2

v v vv v v v

Fave

18

O

m1

m3m2

p1

p2

p3

x y

z

Consider a system of particles for which 0

0 Const

Conservation of Linear Momentum

ant

net

net

F

dP F Pdt

total linear momentumtotal linear momentumat some later time at some initial

If no net external force acts on a system of particles the total linear momentum cannot change

The

time

cfi tt

P

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.

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

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

1 2

1 2 1 2

Consider two colliding objects with masses and ,initial velocities and and final velocities and ,

respective

Momentum and Kinetic Energy in Collisions

ly

i i f f

m mv 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 without knowing the details. Collisions are divided into two broad classes: and . A collision is if there is no

elasticin

loss of elastic

elast kinetic eic nergy i.e.

A collision is if kinetic energy is lost during the collision due to conversioninto other forms of energy. In this case we have:

A special case of ine

i

l

ne

astic collis

lastic

ions

a

i f

f i

K K

K K

re known as . 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

1 2 1 2

1 1 1 2 1 1 1 2

In these collisions the linear momentium of the colliding objects is conser

One Dimensional Inelastic Collisio

ved

ns

i i f f

i i f f

p p p p

m v m v m v m v

2

In these collisions the two colliding objects stick togetherand move as a single body. In the figure to the

One Dimensional Completely I

left we show a special case

nelastic Collis

in which

ns

0.

io

iv

1 1 1 2

11

1 2

1 2 1 1

1 2 1 2 1 2

The velocity of the center of mass in this collision

is

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

i

i

i i icom

m v mV m VmV v

m m

p p m vPvm m m m m m

collision 21

Inelastic Collision

22

23

The Principle of Conservation of Linear Momentum

24

1 2

1 2 1 2

Consider two colliding objects with masses and ,initial velocities and and final v

One-Dimensional

elocities and ,

respectivel

Elas

y

tic Collisons

i i f f

m mv v v v

1 1 1 2 1 1 1 2

2 22 21 1 2 21 1 1 2

Both linear momentum and kinetic energy are conserved. Linear momentum conservation: (eqs.1)

Kinetic energy conservation: (eqs.2)2 2 2 2

We have tw

i i f f

f fi i

m v m v m v m v

m v m vm v m v

1 2 21 1 2

1 2 1 2

1 2 12 1 2

1 2 2

1 2

1 2

1

o equations and two unknowns, and

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

2

2

s:f f

f i i

f i i

f f

m m mv v vm m m m

m m mv v vm m

v v

m

v v

m

25

2

2 1 2The sSpecial C

ubstitutease of elastic Col

0 in the lisions-Stationary

two solutions for Tar

and get 0i

i f fv vvv

1 2 21 1 2

1 2 1 2

1 2

1 2 12 1 2

1 2 1

1 11 2

12 1

1 22

2

2

Below we examine several special cases for which we know the outcomeof the collision from exp1. Eq

eriencea

2

u

f i i

f i

f i

f ii

m mv vm m

m

m m mv v vm m m m

m m mv v vm m m m

v vm m

1 21 1 1

1 2

12 1 1 1

1

1 2

2

0

2 2

The two colliding objects have exchanged veloci

l masses

t es

i

f i i

f i i i

m m m mv v vm m m m

m mv v

m m

v vm m m m

m

m

m

m

m

v1i

v1f = 0 v2f

v2i = 0x

x

26

12 1

2

1

1 2 21 1 1 1

11 2

2

1

21 12 1 1 1

11 2 2

2

2. A mass <1

1

1

ive target

22 2

1

f i i i

f i i i

mm mm

mm m mv v v vmm m

m

mmm mv v v vmm m m

m

1

2

Body 1 (small mass) bounces back along the incoming path with its speed practically unchanged. Body 2 (large mass) moves forward with a very small

speed because <1 mm

v2i = 0

m1

m1

m2

m2

v1i

v1f

v2f

x

x

27

m1

m1

m2

m2

v1i

v1fv2f

v2i = 0x

x

21 2

1

2

1 2 11 1 1 1

21 2

1

12 1 1 1

21 2

1

<1

1

1

2 2 2

2. A massive projectile

1

f i i i

f i i i

mm mm

mm m mv v v vmm m

mmv 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

28

29

Elastic Collision

In this section we will remove the restriction that thecolliding objects move along one axis. Instead we assumethat the two bodies that participa

Collisions

te in the c

in Two

ollisi

Dimensio

onmov

ns

e in 1 2the -plane. Their masses are and xy m m

1 2 1 2

1 2 1 2

2

The linear momentum of the sytem is conserved: If the system is elastic the kinetic energy is also conserved: We assume that is stationary and that after the co

i i f f

i i f f

p p p pK K K K

m

1 2 1

1 1 1 1

llision particle 1 and particle 2 move at angles and with the initial direction of motion of In this case the conservation of momentum and kinetic energy take the form:

axis: i f

m

x m v m v

1 2 2 2

1 1 1 2 2 2

2 2 21 1 1 2 2 2

1 2 1 1 2

cos cos (eqs.1) axis: 0 sin sin (eqs.2)

1 1 1 (eqs.3) We have three equations and seven variables:2 2 2Two masses: , three speeds: , , a

f

f f

i f f

i f f

m vy m v m v

m v m v m v

m 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

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

K mv mv

f A B

i A B

12

12

12

2 0 5 20 10 15 8 0 10

12

12

12

2 0 15 30 10 5 13 10

2 2 2 2 2 2 2

2 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

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

cons

Systems with Varying M

tant speed re

ass:

lati

Th

ve

e Rocket

to the rocket. Trel

M vdMdt

v 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

( ) (eqs.1)Her

t t dt

p t p t dt Mv dMU M dM v dv

e is a negative number because the rocket's mass decreases with time is the velocity of the ejected gases with respect to the inertial reference frame

in which we measure the rocket's speed . W

dM tU

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

rel

rel

U vU v dv

Mv

dvU

reldMv32

Using the conservation of linear momentum we derived the equation of motion for the rocket (eqs.2) We assume that material is ejected from the rocret's nozzle at a constant rate

relMdv dMv

dMdt

(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

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

dv dMdt M v Rvdt dt

a Rv

Ma Rv

rmine the rocket's speed as function of time

We integrate both s

(Second rocket equation)

ides

ln ln l

ln

n

f f

i i

f i

i f

v M

rel relv M

M M if i rel rel re

if i rel

f

lM Mf

dM dMdv v dv vM M

Mv v v M

Mv v v

v

M

v MM

vf

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 exhaust 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