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. (9-1)

1 2

1 2

1 1 2 2com

1 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 follows:

m x m xx

m mx x

m m

The Center of Mass :

1 1 2 2 3 3 1 1 2 2 3 3com

11 2 3

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

Here is the total mass of

.

all th

.. 1...

e partic

nn n n n

i iin

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

Mm M

n

m 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 th particle ( mass ) has position

n

i

M m m m m

im

com1

vector 1

.n

i ii

i

r m rM

r

(9-2)

com1

com com com com

com

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

ˆ ˆ ˆThe position vector can be written as i j k.The components of are given by the e

n

i ii

r m rM

r x y zr

com com com1 1 1

1 1 1

quat

ions

n n n

i i i i i ii i i

x m x y m y z m zM M M

The center of mass has been defined using the quations given above so that it has the following propeThe center of mass of a system of particles moves as thoughall the system's mass were conc

rty:

entr

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. (9-3)

Solid bodies can be considered as systems with continuous distribution of matter.The sums that are used for the calculation of the center of mass of systems withdis

The Center of Mass for Solid Bodies :

com com com1 1 1

crete distribution of mass become integrals:

The integrals above are rather complicated. A simpler special case is that of

uniform objects in w

h

x xdm y ydm z zdmM M M

com com com1 1 1

ich the mass density is constant and equal to :

In objects with symmetry elements (symmetry point, symmetry line, symmetry plane)it is

x xdV y ydV z zdVV V

dm MdV V

V

not necessary to evaluate the integrals. The center of mass lies on the symmetry element. For example, the com of a uniform sphere coincides with the sphere center.In a uniform rectangular object the com lies at the intersection of the diagonals.

(9-4).CC

O

m1

m3

m2

F1

F2

F2

x y

z

1 2 3,

1 2 3

Consider a system of particles of masses , , ...,

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

n

n

Newton's Second Law for a System n

of Particles :m m m m

r r r r

s is given by

com 1 1 2 2 3 3

com 1 1 2 2 3 3

com 1 1 2 2 3 3 com

... . We take the time derivative of both sides

...

... . Here is the velocity of the

n n

n n

n n

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

com 1 1 2 2 3 3

com 1 1 2 2 3 3 com

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

...

... . Here is the accelera

i

n n

n n

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

tion of the comand is the acceleration of the th particle.ia i

(9-5)

O

m1

m3

m2

F1

F2

F2

x y

z com 1 1 2 2 3 3

com 1 2 3

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

. Here is the net force on the th particle,

... .

n n

i i i i

n

Ma m a m a m a m ai

m a F F i

Ma F F F F

app int

app int app int app int app intcom 1 1 2 2 3 3

co

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

. The above equation takes the form:

...

i

i i i

n n

F

F F F

Ma F F F F F F F F

Ma

app app app app int int int intm 1 2 3 1 2 3

net

... ...

The sum in the first set of parentheses on the RHS of the equation above is

just . The sum in the second set of parentheses

n nF F F F F F F F

F

com net

net , com, net , com,

on the RHS vanishes by virtue of Newton's third law.

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

.

:

x x y y

Ma F

F Ma F Ma

net, com,

z zF Ma (9-6)

com net Ma F net, com,

net, com,

net, com,

x x

y y

z z

F Ma

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 concentrated 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 occurred, 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 follows the sameparabolic trajectory that the rocket would have followed had it not exploded.

(9-7)

mv

pLinear momentum of a particle of mass and velocity

The SI unit for linear momentum is the kg.mis defined a

/s s .

.p mv

p m v

Linear Momentum :

p mv

The time rate of change of the linear momentum of a particle is equal to the magnitude of net force acting on t

Below we will prove the fol

he particle and has the dir

lowing statem

ection of the

ent:

for

net

net

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 c

c

a

.

h

e

n

dpFdt

dp d dvp mv mv m ma Fdt dt dt

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

net dpFdt

(9-8)

O

m1

m3

m2

p1

p2

p3

x y

zIn this section we will extend the definition oflinear momentum to a system of particles. The th particle has mass , velocity , and linearmomentum

i i

i

i m vp

The Linear Momentum of a System of Particles

.

1 2 3 1 1 2 2 3 3 com

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 t

n n n

n

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

com

com com net

hetotal 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 change

M v

dP dP Mv Ma Fdt dt

P

net net

d only by a

net external force . If the net external force is zero, cannot change.F F P

1 2 3 com... nP p p p p Mv

netdP Fdt

(9-9)

We have seen in the previous discussion that the momentum of an object canchange if there is a nonzero external force acting on the object. Such forces exist during the collisio

Collision and Impulse :

n 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 bat.The 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

( ) during the collision.

The magnitude ( ) of the force is plotted versus in fig. a.The force is nonzero only for the time interval .

( ) . Here is the linear momentum of the bal

i f

F tF t t

t t t

dpF t pdt

l,

( ) ( )f f

i i

t t

t t

dp F t dt dp F t dt

(9-10)

( ) = change in momentum

( ) is known as the impulse of the collision.

( ) The magnitude of is equal to the area

under the ver

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

ave

sus plot of fig. a .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

t p J

F

J F

ave

ave

, where .

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

versus plot (fig. b).

f it t t t

F tF t

p J

(9-11)aveJ F t

Consider a target that collides with a steady stream ofidentical particles of mass and velocity along the -axis.m v x

Series of Collisions

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

ave

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 change in the ve

pt J n p

J n p nF m vt t t

v

ave

locity 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 collision

xmF vt

mt

, then 0 .If the particles bounce backwards, then 2 .

v v vv v v v

(9-12)

Fave

O

m1

m3

m2

p1

p2

p3

x y

z

net

net

Consider a system of particles for which 0

0 Constant

F

dP F Pdt

Conservation of Linear Momentum :

total linear momentumtotal linear momentumat some later time at some initial time

If no net external force acts on a system of particles, the total linear momentum

cannot change.

Thfi tt

P

e conservation of linear momentum is an important principle in physics. It also provides a powerful rule we can use to solve problems in mechanics suc

Note 1:

h as collisions.

In systems in which F

net 0 we can always apply conservation of linear momentum even when the internal forces are very large as in the case of colliding objects.

We will encounter problems (e.g., inelasticNote 2 co s: lli

ions) in which the energy is not conserved but the linear momentum is. (9-13)

1 2

1 2 1 2

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

respectively.i i f f

m mv v v v

Momentum and Kinetic Energy in Collisions

netIf the system is isolated, i.e., the net force 0, linear momentum is conserved.The conservation of linear momentum is true regardless of the collision type.This is a powerful rule that allows us

F

to determine the results of a collision without knowing the details. Collisions are divided into two broad classes:

and . A collision is if there is elastic in

no loss oelasticela f kinetic stic energy, i.e.,

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

A special case

.

i

of inelastic collisi

nelas

tic.

i f

f i

K K

K K

ons are known as . In these collisions the two colliding objects stick together and they move as a single body. In these collisions the loss of kinetic

co

e

mpletely in

nergy is ma

elastic

ximum.(9-14)

1 2 1 2

1 1 1 2 1 1 1 2

In these collisions the linear momentum of the colliding objects is conserved .i i f f

i i f f

p p p p

m v m v m v m v

One - Dimensional Inelastic Collisions :

2

In these collisions the two colliding objects stick togetherand move as a single body. In the figure to the left we show a special case in which 0iv

One - Dimensional Completely Inelastic Collisions :

1 1 1 1

11

1 2

1 2 1 1com

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 inelasti

i

i

i i i

m v mV mVmV v

m m

p p m vPvm m m m m m

c collision.(9-15)

1 2

1 2 1 2

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

respectively.i i f f

m mv v v v

One - Dimensional Elastic Collisions

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: (eq. 1)

Kinetic energy conservation: (eq. 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 1

1 1

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

(9-16)

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

2Special Case of Elastic Collisions - Stationary Target v = 0 :i

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. E

erience.q

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

11 2

12 1 1 1

1 2

2

0

2 2

The two colliding objects have exchanged velocit

al

ie

masses

s.

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

(9-17)

12 1

2

1

1 2 21 1 1 1

11 2

2

1

21 12 1 1 1

11 2 2

2

1

1

1

22

2. A massive targ

2

et

1

f i i i

f i i i

mm mm

mm m mv v v v

mm mm

mmm mv v v v

mm m mm

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

(9-18)

v2i = 0

m1

m1

m2

m2

v1i

v1f

v2f

x

x

m1

m1

m2

m2

v1i

v1f v2f

v2i = 0x

x

21 2

1

2

1 2 11 1 1 1

21 2

1

12 1 1 1

21 2

1

12. A massive projectil e

1

1

2 2 21

f i i i

f i i i

mm mm

mm m mv v v v

mm mm

mv v v vmm mm

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.

(9-19)

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

Collisions in Two Dimensions :

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

1 2 1 2

1 2 1 2

2

The linear momentum of the system is conserved: .

If the system is elastic the kinetic energy is also conserved: .

We assume that is stationary and that after the

i i f f

i i f f

p p p p

K K K K

m

1 2 1

1 1 1

collision 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

m

x m v m

1 1 2 2 2

1 1 1 2 2 2

2 2 21 1 1 2 2 2

1 2 1 1

cos cos (eq. 1)

axis: 0 sin sin (eq. 2)

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

f f

f f

i f f

i f

v m v

y m v m v

m v m v m v

m m v v v

2 1 2; and two angles: , . If we know

the values of four of these parameters we can calculate the remaining three.f

(9-20)

rel

A rocket of mass and speed ejects mass backwards

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

constant speed relative to the rocket.

M vdMdt

v

Systems with Varying Mass : The Rocket :

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

( ) (eq. 1).

Here

a b t t dt

p t p t dt Mv UdM M dM v dv

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

rel

rel

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

U vU v dv v U

Mdv dM

rel.v (9-21)

rel

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

dMdt

(eq. 3). Here is a constant positive number.R R

re

rel rel

l

We divide both sides of eq.(2) by

Here is the rocket's acceleration.We use equat

(First rocket equation 2 to determine the rocket's speed as a function

ion . o

f

)

dv dMdt M v Rvdt

Mat

v aRd

rel rel

rel rel rel

rel (Second rocket eq

time :

. We integrate both sides

ln

uation)ln

ln

ln

f f

i i

f i

i f

if

v M

v M

M M if i M M

f

if

t

dM dMdv v dv vM M

Mv v v M v M

Mv v v

M

vM

vf

vi Mi/Mf

O(9-22)