CHAPTER 4 DYNAMICS OF A SYSTEM OF PARTICLES · PDF file1 CHAPTER 4 DYNAMICS OF A SYSTEM OF PARTICLES •We consider a system consisting of n particles •One can treat individual...

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CHAPTER 4

DYNAMICS OF A SYSTEM OF PARTICLES

• We consider a system consisting of n particles

• One can treat individual particles, as before; i.e.,one can draw FBD for each particle, define a coordinate system and obtain an expression of the absolute acceleration for the particle. One can then use Newton’s second law and proceed to get n second-order coupled ODEs.

• Focus here is on overall motion of the system-also a precursor to rigid body dynamics.

2

4.1 Equations of Motion:

Consider a system with:

• n particles

• masses -

• positions -

There are two types

of forces acting:

• External forces -

• Internal forces -

r i

Fi ;

m i

f ij

Z

X

Y

O

r1

F1

rC

ri

F2

Fi

mi

m2m1

mC

f12 f21

f2i

fi2

i

3

- force on the particle due to its

interaction with the particle

• Newton’s 3rd law

Also

• Newton’s 2nd law for particle:

f ij i th

jth

i th

internal forcesareequaland opposi( )

0 when , . .

te

0

ij ji

ij ii

f f

f i j i e f

1

, 1, 2,3,. . . .,n

i i i ij

j

m r F f i n

4

Now, for 3-dimensional motions, the position

of each particle (in Cartesian coordinates) is:

Thus, each equation in Newton’s second law

has 3 scalar second-order ordinary diff.

equations. 3n scalar second-order o.d.e.’s

for the system

In order to solve for the motion, one needs to

know:

• external forces on each of the particles

• nature of internal forces

F i

f ij

, 1,2,3, . . . ,i i i ir x i y j z k i n

5

e.g., Newton’s law of gravitation:

We also need:

• initial conditions:

The general solutions to these nonlinear

ODEs are unknown; they are difficult to

solve except for in some very simple cases

and small n.

2

3

( )

, ( ) /

j ii j

ij

j ij i

ij i j i j j i

r rm mf G

r rr r

or f Gm m r r r r

(0), (0), 1,2, . . . . ,i ir r i n

6

Suppose we would like to get overall motion of

the system, not those of individual particles.

Adding the n equations:

Now, (net interaction force is zero)

• - total mass

• - defines center of mass;

note that it is a function of

time since the particles move.

1 1 1 1

n n n n

i i i ij

i i i j

m r F f

1 1

0n n

ij

i j

f

1

n

i

i

m m

1

( ) ( )n

C i i

i

mr t m r t

7

• - total external force

• Equation of motion for

the center of mass

Internal forces do not affect the motion of

the center of mass.

1 1

Thus,addition of Eqns.n n

i i i C

i i

F m r mr

1

Letn

i

i

F F

1

n

i C

i

F F mr

8

4.2 Work and Kinetic Energy

• The motion of individual particle is defined by

• The motion of center of mass C is defined by

where the total mass is

1

, 1,2,3,. . . .,n

i i i ij

j

m r F f i n

1

n

i C

i

F F mr

1

n

i

i

m m

9

Consider a motion of

the system. The initial

state is A, and the final

state is B. Let AC and BC

denote the positions of

the CM.

• Now, for the CM

• work-energy statement for the CM

2( / 2)C C

C

C

C C

C

B BB

C C C C AA A

F mr

F dr mr dr mv

Y

Z

X

OrC

ri

mi

m

Ai

Bi

AC

BC

10

Note that is only the work done by

external forces, and it is related to the change

in translational kinetic energy associated with

the CM

• Let work done on the particle by all

the forces acting on it in moving from Wi i th

A to Bi i

C

C

B

C

A

F dr

1

( )i

i

B n

i i ij i

jA

W F f dr

11

Now:

where - position of particle relative to

the CM of the system

• Total work done=sum of the work done on all

particles:

i thi C ir r

i

1

1 1

1 1 1 1 1

( ) ( )

, ( ) ( )

i

i

i C

i C

n

i

i

Bn n

i ij C i

i jA

B Bn n n n n

i ij C i ij C

i j i i jA A

W W

W F f dr d

Now F f dr F f dr

12

work done by summation of the

total ext. forces work done on all the

through the displ. particles through their

of the CM displacements relative

to the CM

• For each particle, the work done is:

1 1

( )C i

C i

B Bn n

C i ij i

i jA A

W F dr F f d

1 1( ) ( )

2 2

i i

i i

B B

i i i i i C i C i

A A

W m r r m r r

13

- the sum of increase/change in

KE of the system.

T TB A

1 1

1

1 1

1

/ 2

/ 2

, 0 0

/ 2 / 2

i

C

C

i

i

i

i

C

C

i

Bn n

B

i C C C i iAi i A

Bn

i i i

i A

n n

i i i i

i i

Bn

B

C C i i iAi A

W W mr r r m

m

Now m m

W mr r m

14

work-energy principle for

the system of particles

T = K.E. at any instant

Recalling the work-energy principle for the CM:

Work done by all forces (external as well as

internal) in relative motion KE for relative

motion

A B B AW T T

2

1

/ 2 / 2n

C i i i

i

mv m

1 1 1

( ) / 2

ii

i i

BBn n n

i ij i i i i

i j iA A

F f d m

15

Important: In general, internal forces do

work in any motion of the system. Sometimes,

net work (that on the whole system) may be

zero even though there is work done on

individual particles.

Ex: Consider the force in a spring

connecting two moving bodies - there is

net work done by the spring force -

evaluated by potential function .

f ij

sp

16

Ex 1: Consider two particles connected by

a massless rigid (inextensible) rod, and

acted upon by a force F.

FBDs for individual

particles are:

F

1

2r1

r2

rC

C

O

m1

m2

m1

m2

F

Cf12

f21Note: f12=- f21

17

• Work done in relative motion by internal

forces:

• constraint

• Differentiate:

Now:

12 1 21 2 12 1 2( )dW f d f d f d d

22

12 2 1 2 1( ) ( )r l

2

12 2 1 2 1( ) ( ) ( ) 0d r d d

12 12 1 2 1 2

12 1 2

( ) /

Thus, ( ) 0

f f

f d d

18

Ex 2:

Consider the system

shown here. A slider

moves on a rough

guide, and a pendulum

is attached to it at A.

• connected by a massless rigid link.

• Coulomb friction between and the

horizontal guide. Force P acts on the block A.

m m1 2,

m1

m

1

m

2

x

l

xyA

P

rough guide

19

The FBDs are:

The positions of the two particles can now

be defined: 1

2

( ) ( )

( ) { ( ) sin } cos

r t x t i

r t x t l i l j

m1g

PTf

N

yx

T

m2g

m2

20

The equations of motion for the individual particles are:

• Try to write the equation of motion for the

CM of the system.

1 1 1

2

2 2

2

2

: ( sin ) ( cos )

where (sgn( ))

: {( cos sin )

( sin cos ) } sin

( cos )

m m x i P f T i N m g T j

f N x

m m x l l i

l l j T i

T m g j

21

4.3 Conservation of Mechanical Energy

• Suppose that the External forces are conservative, that is, are conservative.

for the CM of the system

Total energy conserved for motion of the CM

• Suppose that Internal forces also conservative:

E = T + V

Total energy conserved for the whole system.

1

n

i

i

F F

C C CE T V

22

Ex. 3 (4.2): Consider the system shown.

• connected by a massless spring.

• A constant force F applied to at t = 0.

• No friction between the floor and the blocks

Find:

IC (t = 0), spring

unstretched

1 2andm m

m1

1 1 2( ) ; when massesareequal :x t m m m

1 2 1 2 0;x x x x

m1 m2

k

xC

x1

F Cx2

23

Motion of the CM:

Motion of the block m1:

FBD:

1 2

1 2

2

( ) / ( ) / 2

Newton'sSecond law :

( 2 )

/ 2 ; Init.Conds. are : (0) 0

( / 2 ) ; ( / 2 ) / 2

C i i C

x C

C C C

C C

r m r m x x x

F F mx m m m m

x F m x x

x F m t x F m t

F

xk(x2- x1)m1

24

Newton’s law for block m1:

2 1 1

2 1

1 1

1 1

1 1

1

( ) ; ICs.: (0) 0

Also,note that 2

2 ( ) (1)

Also: / 2 (2)

(1) (

Har

2)

m

( ) 2 ( ) / 2

ICs: [ (0) (0)] [ (0) (0)] 0

: ( ) {1 cos 2 / }/Sol

o

n 4

(

x C C

C

C

C

C C

C C

C

F F k x x mx x x

x x x

F k x x mx

mx F

m x x k x x F

x x x x

x x F k m k

nicoscillation)

25

Aside (steps involved in the solution):

1Theeqn.is : 2 / 2 where ( )

Thesolution is ( ) ( ) ( )

: 2 / 2 / 4

: ( ) cos sin , where 2 /

(0) 0 / 4 0 / 4

(0) 0 0 0

: ( ) {Sol

Harmonicoscill

1 s }

(

n co / 4

C

h p

p p p

h h n n n

n

n

my ky F y x x

y t y t y t

y ky F y F k

y y t A t B t k m

y A F k A F k

y B B

y t F t k

ation)

26

Or

Energy considerations: (verification)

Recall that

2

1( ) ( / 4 ) {1 cos 2 / }/ 4x t F m t F k mt k

2 2 2

2

CWork done on CM (W )=changein K.

( / 2 )

K.E.of CM (2 ) / 2 ( / 4 )

Work doneon CM ( / 4 )

(for a constant forc

E.of C

)

M

e

C C

C C

C C

x v F m t

T m v F m t

W Fx F F m t

27

Now, consider for the whole system:

2 2 2

1 2

2 2 2 2

2 2

1 2 1

2 2

2 2 2

[(2 ) {( ) ( ) }] / 2

or ( / 4 ) ( / 8 )sin ( 2 / )

( ) / 2 2 ( )

(Work done byinternal forces)

or ( / 8 ){1 cos( 2 / )}

( /

TotalKE:

PotentialEnerg

4 ) ( / 4 )

:

co

y

{1

C C C

C

T m x m x x x x

T F m t F k k mt

V k x x k x x

V F k k mt

T V E F m t F k

s( 2 / )}k mt

28

W = work done by the external force2 2 2

1 ( / 4 ) ( / 4 ){1 cos( 2 / )}

(0 0)

Work done byall forces (externaland internal)

(final totalenergy) (initial totalenergy)

Fx F m t F k k mt

W T V

29

4.4 Linear Impulse and Momentum

Let, - lin. impulse of external forces

Considering Newton’s laws for motion of CM:

Let - total linear momentum of the

system at a given instantThen

2

1

ˆ ( )

t

t

F F d

2 2

1 1

2 1ˆ ( ) ( ) ( )

t t

C C C

t t

F F d mr d m v v

1

( ) ( )n

i i C

i

p t m v t mv

2 1 2 1ˆ ( ) ( ) ( )C CF m v v p t p t

30

4.5 Angular Momentum:

The key point to consider here is the point about

which the moment can be taken.

• Moment about a

fixed reference point:

Z

X

Y

O

r1

F1

rC

ri

F2

Fi

mi

m2m1

mC

f12 f21

f2i

fi2

i

(angular momentum

of theith particleabout

point O)

iO i i iH r m r

31

1 1

1 1 1

1

Rateof chan

Totalangular mome

geof angular momen :

Now,using Newton'second law for a part

ntum of the syst

icl

t

em:

e :

um

n n

O iO i i i

i i

n n n

O i i i i i i i i i

i i i

n

i i i ij O

j

H H r m r

H r m r r m r r m r

m r F f H r

1 1

1

( )

or

n n

i i ij

i j

n

O i i O

i

F f

H r F M

32

• Reference point as the center of mass:

Let

1

1 1

1

1

( ) ( )

( )

n

O i C i C i

i

n n

C C C i i i i C

i i

n

i i i

i

n

O C C i i i

i

H m r r

mr r r m m r

m

H mr r m

i C ir r

33

(Ang. momentum with respect to the CM, as

viewed by a nonrotating observer moving

with the CM)

Now, differentiating:

1

Thus,

where

O C C C

n

C i i i

i

H r mr H

H m

1

1

n

O C C i i i

i

C n

i ii

r F

F

H r mr m

34

• (very convenient for rigid bodies)

1

1

for motion of CMNow ( )

about fixed pointO

ab

Reviewing : ( )

out C, the( )CM

n

CO C i i C C

i

C C C

n

CC i i i

i

OO

OO

CM

M r F F H r mr

r F r mr

M H m

M H

M H

35

• About an arbitrary reference point P:

Let P be an arbitrary

point (could be moving).

Let

Then, one can show

that

- angular momentum about P

Z

X

Y

O

rP

rC

ri

Fi

mi

mC fi2

iC

P

and

i P i

C P C

r r

r r

P C C CH m H

36

And

Choosing an arbitrary point for moments

of forces results in an additional term in the

moment equation.

• If P is a fixed point

• If P is the center of mass

• If P is such that and are parallel

throughout the motion

P C P PM mr H

( 0)P P PM H r

( 0)P P CM H

Pr C

P PM H

37

• Computation of Kinetic energy using P as

a reference point:

The kinetic energy is: 1

22

1

22

1

/ 2

Now, ,

[ 2 ] / 2

If : [ ] / 2 (as before)

n

i i i

i

i P i i P i

n

P i i P C

i

n

C i i

i

T m r r

r r r r

T m r m r m

P C T m r m

38

Ex. 4 (4.7):Consider a particle traveling at a speed ‘v’ to

the right. It strikes a stationary dumbbell (two

particles connected by a massless rigid rod).

The masses are:

Assumption:

• Perfectly elastic impact in (e=1).

Find: motion of the particles just after impact.

m m m m1 2 3

m m1 2,

x

y

C

Ovm1

m2

m3

l/2

l/2

45º

39

FBDs:

Observe that during impact:

• Net force on the whole system = 0

linear momentum conserved for the system

• Resultant moment about O (a fixed point)= 0

angular momentum about O conserved for the system

x

y

C

O

m2

m3

l/2

l/2

45ºF̂m1 F̂

40

Set up of the problem:

Motion before impact:

Motion after impact: It is convenient to think

in terms of the motion

of the CM, and rotational

motion about CM. Use

the triad

to define the motion of

the CM and the particles.

1 2 3; 0v v i v v

( , , )t a be e e

x

y

C

O

m2

m3

l/2

l/2

45ºet

ea

41

Expressing velocities

in terms of

x

y

C

O

m2

m3

l/2

l/2

45ºet

ea

( , , )t a be e e

2

2

3 3

3

( / 2)

( / 2)

( / 2)

Similarly, ( / 2)

( / 2)

C a a t t

C CO

C a

C t

a a t t

C C C t

a a t t

v v e v e

v v k r

v k le

v le

v v e v l e

v v k r v l e

v v e v l e

42

linear momentum conserved for the system:

(a vector equation 2 scalar equations)

angular momentum conserved for the system:

1 2 3

1 2 2 (1)a a t t

mvi mv i mv mv

or v i v i v e v e

3 30 [ ( / 2) ]

( / 2) / 2 (2)

O a a a t t

t t

r v le v e v l e

l v l k v l

43

Note: The vectors can be expressed

in terms of the unit vectors i and j as:

In equations (1) and (2),

are unknowns but there are only 3 equations.

Thus one more relation is required.

• coefficient of restitution:

1 , , ,t av v v

andt ae e

cos 45 sin 45 ( ) / 2

cos 45 sin 45 ( ) / 2

t

a

e i j i j

e i j i j

vAx vBx

A B x

44

Aside: central impact: Consider two particles

A and B that collide with each other. The

geometry and definitions of terms are:

FBDs: No y-comp of force

vA1 vB1

A

B

vA2 vB2 line of impact

plane of impact

A Bx

F̂

F̂

45

1 1 2 2

Let and

be velocities; , before,and , after.

The i

the ration

s then defined as

of the relative velocity after impact

to the relative velocity b

coefficient of resti

efore i

:tution

m

A Ax Ay B Bx By

A B A B

v v i v j v v i v j

v v v v

2 2 2

1 1 1

, for

velocities along the line of impact:

( ) ( )

pact

( ) ( )

Bx Ax Bx Ax

Bx Ax Bx Ax

v v v ve

v v v v

46

• For the system at hand, elastic impact: e = 1.

Also,

1 1 2 1 2 1

2 1 2

1

1

1

, =0, 0, ,

( 2 ) / 2, 0. Thus

(( 2 ) / 2 )1

(0 )

/ 2 2 (3)

Solving (1), (2), and (3)

(2 2 / 7) , (2 2 / 7) , / 7

Ax Ay Ay Bx Ax

Bx a t By By

a t

a t

a t

v v v v v v v

v v v v v

v v ve

v

v v v v

v v v v v v

47

Ex. (Problem 3.19)

Consider a cylinder rotating

at a constant rate .

• A thin, flexible and

massless rope goes

around the drum.

• There is no gravity,

and the rope does not

slip relative to the drum ;

At

Find: Tension in the rope as a function of time.

t r0 0 0 0, ( ) , ( )

O

Pm

r

l

48

Setup:

Consider a triad

for the moving reference

frame , with coordinate

system located at

• Let be angular

velocity of the moving

reference frame or triad.

• The fixed reference

frame is with origin O.

e e et n b, ,b g

O .

Y

X

O

P

m

r

O’

R

l

l

49

- triad for moving coordinate system

Let be angular velocity of the moving frame.

e e et n b, ,b g

Y

X

O

P

r

l

O’

R

Schematic:

50

•Use the general formulation to express aP:

Let us now consider the various terms:

( ) ( ) 2 ( )P r ra R

( ) and

/

( / ) , ( / )

Position : ,

( / ) ( / )

Also ( / ) ( / )

b

b b

t

t t

b t n

n n

e l r

l r l r

l r e l r e

deR re R r r e

dt

R r l r e e r l r e

R r l r e r l r e

51

2

2

2

2

Thus ( ) ( / )

Now, ( ) ( )

Als

The

o =( / ) ( ) ( / )

( ) ( / )

2 ( ) 2 ( / )

[ ( / ) / 2 ( / )]

( / )

mp:I

n t

n r n r n

b n t

n

r t

P t

n

R l e r l r e

le le le

l r e le ll r e

l l r e

l l r e

a r l r ll r l l r e

l l r e

is acceleration relative to

52

Now, applying Newton’s Second Law:

Initial conditions:

Integration 2 2r t

( ) , ( )0 0 0 r

2

2 2 2

2 2 2 2

: ( / ) / 2 ( / ) 0

or 0

( ) ( )

n

t

F ma Te ma

e r l r l l r l l r

l l l r

d l l dt r d l l r dt

53

Integrating once again,

2 2 2 2 2 2

0 0

l t

d r d l r t l r t

2 2

3

: ( / ) 4

or 4

ne T ml l r ml

T mr t