Kinetics of Particles

Chapter 3

1

Kinetics of Particles It is the study of the relations existing between the forces acting

on body, the mass of the body, and the motion of the body. It is the study of the relation between unbalanced forces and the

resulting motion. The three general approaches to the solution of kinetics problems

Direct application of Newton’s second law (called the forcemass-acceleration method),

Use of work and energy principles, and Solution by impulse and momentum methods.

2

Force, Mass and Acceleration Newton’s second law states that F=ma The mass m is used as a quantitative measure of inertia (The

resistance of any physical object to change in its state ofmotion or rest)

are two vectors having identical direction with respectto an internal frame of reference.

When a particle of mass m acted upon by several forces. TheNewton’s second law can be expressed by the equation.

F and a

∑ = maF

3

To determine the acceleration we must use the analysis used in kinematics, i.e Rectilinear motion Curvilinear motion

Rectilinear Motion If we choose the x-direction, as the direction of the rectilinear motion of

a particle of mass m, the acceleration in the y and z direction will bezero, i.e

Generally, Resultant force are given by

∑ = maF

∑∑∑

=

=

=

0

0

z

y

xx

F

FmaF

ZZ

yy

xx

maFmaFmaF

=

=

=

∑∑∑

( )2 2 2( ) ( )x y zF F F F= + +∑ ∑ ∑ ∑

4

Curvilinear motion In applying Newton's second law, we shall make use of the three coordinate

descriptions of acceleration in curvilinear motion. Rectangular coordinates

Where

Normal and tangential coordinate Where

Polar coordinates Where

∑∑

=

=

yy

xx

maFmaF ••

= xax••

= yay

∑∑

=

=

tt

nn

maF

maF 2 2

,n

t

va

a v

ρ βρ

•

•

= =

=

∑∑

=

=

θθ maF

maF rr

2•••

−= θrrar

••••

+= θθ rran 2

5

Example6

1.

7

2.

8

3.

9

4.

10

6.

11

7.

12

8.

13

Work and Energy

14

Work and Kinetic Energy There are two general classes of problems in which

the cumulative effects of unbalanced forces acting ona particle over an interval of motion. These requires integration of the forces with respect to the

displacement. OR integration of the forces with respect to the time they

are applied. The first one will result in Work-Energy equation The second will give Impulse-momentum equation

15

Work Consider the following figure

The work done by the force over the displacement is

The magnitude of the dot product is

and is the angle between Or

- the component of resultant of resultant displacement in the direction of force.

- the component of resultant force in the direction of displacement

F

dr

dU F dr=

dr ds=

cos dU Fds where dr dsα= =

α & F dr

( cos )dU F dsα=

cosds α

cosF α

-

F component in the displacement

dUdirection of displacement

=

16

- has no work because it can notdisplace the particles along itsdirection, it is called reactive force.

- has work besause there isdisplacement in its direction and itscalled active force.

If the force and displacement are in the same direction the workdone by is +ve, otherwise it is –ve.

Example of negative work Work by spring force Work by gravitational force

tdU F ds=

nF

tF

F

17

Consider the work done on a particle ofmass m, moving along a curved pathunder the action force F as shownbelow.

2

1

2 2

1,21 1

s

ts

U du F dr F ds= = =∫ ∫ ∫

t tF ma=2

1

1,2

s

ts

U ma ds= ∫ta ds vdv=

2

1

2

1,21

v

v

U F dr mvdv= =∫ ∫

2 21,2 2 1

1 ( )2

U m v v= −18

Kinetic Energy of a Particle (T) K.E of a particle is the work done on the particle to bring the particle from state of rest to

a velocity V.

K.E is a scalar quantity with the unit of Nm or Joule (J). K.E is always positive.

-the total work done by all external acting forces on the particle during an interval of its motion from condition 1 to 2.

Power Time rate of doing work. Scalar quantity and units of Nm/s=J/s=watt(W)

Efficiency:- the ratio of work done by a machine to the work done on the machine during the same of interval

212

T mv=

1 2 2 1U T T T→ = − = ∆1 2U →

dU F drP F vdt dt

•= = = •

m

Work done by the machine Power out putework done on the machine Power in put

= =19

Potential Energy Gravitational potential energy

Elastic potential energy

Work energy equation If stands works done of all external force other than gravitational potential

energy and elastic potential energy.

If there is no external forces =0

2 1( )g

g

U mghU mg h h=

∆ = −

2 22 1

1 ( )2eU k x x= −

1 2U →

1 2 e gU U U T→ − − = ∆

1 2U →

0 e gU U T= ∆ + ∆ + ∆2 2 2 22 1 2 1 2 1

1 10 ( ) ( ) ( )2 2

k x x mg h h m v v= − + − + −

2 1 2 1 2 10 e e g gU U U U T T= − + − + −

1 1 2 21 2e g e gU U T U U T+ + ∆ = + + 20

Example

21

1.

22

2.

23

3.

24

4.

25

5.

26

Impulse and Momentum

27

Integrating the equation of motion F=ma with respect to time leadsto the equations of Impulse and Momentum.

Linear impulse and momentum Consider the general curvilinear motion in space of a particle of mass m,

where

where Linear momentum of the Particle.

Momentum:-capacity for progressive development. The power to increase or develop at an every going Forward movement

F ma=∑

dvadt

=

( )dv dF m mvdt dt

= =∑

mv G= →

dF Gdt

=∑

Fdt dG→ =∑

2

1

t

t

Fdt G= ∆∫

2 1G G= −

2 1mv mv= −

28

The scalar components of equation are:

The total linear impulse on a mass m equals the corresponding change in linear momentum of m.

If then

G mv=

∑∑∑ === zzyyxxGFGFGF ,,

( ) ( )

( ) ( )

( ) ( )

2

1

2

1

2

1

2 1

2 1

2 1

t

x x xt

t

y y yt

t

z z zt

F dt mV mV

F dt mV mV

F dt mV mV

= −

= −

= −

∑∫

∑∫

∑∫

0F =∑

2 1

2 1

0

0

G

G G

G G

∆ =

− =

=

The SI units of linear momentum is Kgm/s or Ns.

29

Angular Impulse and Angular Momentum Consider the following

Angular momentum is defined as the moment of linear momentum vector ( ) about the origin o.

And it is given by:-

The angular momentum is a vector perpendicular to plane A defined by r and v.

The direction is clearly defined by the right hand rule for cross product.

The scalar component of angular momentum obtained by cross product.

oH

mv

0H r mv r G= × = ×

oH

( )

o

x y z

H r v m r xi yj zk

v v i v j v k

= × = + +

= + +

x y z

i j km x y z

v v voH =

( ) ( ) ( )z y x z y xr mv m v y v z i m v z v x j m v x v y k= × = − + − + −

oH

( ), ( ) ( )x z y y x z z y xH m v y v z H m v z v x H m v x v y= − = − = −

30

In order to analyze the components of angular momentumconsider following figure

The magnitude of

Where is the angle between and

SI unit of angular momentum is kg(m/s)m=kgm2/s=Nm.s

oH

sinoH r mv θ=

θ r v

( sin ) sin

( sin ) sino

o

H m r v mr

H mr v mr

θ θ

θ θ

= =

= =

31

If represents the resultant of all forces acting on the particles P, the moment Mo about the origin O is the vector cross product.

Angular momentum Differentiate with respect to time

but Since the cross product of parallel vector is zero.

Since

From equation (1) and (2)

F∑

(1)oM r F r mv= × = ×∑ ∑

( )oH r mv= ×

( )od dH r mvdt dt

= ×

( )odr dH mv r mvdt dt

= × + ×

0dr mv v mvdt

× = × =

( )dr mv r Fdt

× = ×∑

( )dF mvdt

=∑

(2)oH r F= ×∑

o o o

o o

dM H Hdt

M dt dH

= =

=

∑∑

32

The moment about the fixed point O of all forces acting on Mequals the time rate of change of angular momentum of M about O.

The product of momentum and time is defined as angular impulse.

Where

If the resultant moment about a fixed point O of all forces acting on a particle is zero

o o odM H Hdt

= =∑

o oM dt dH=∑

2

2 1

1

t

o o o ot

M dt H H H= − = ∆∑∫

2

1

2 2

1 1

o

o

H r mv

H r mv

= ×

= ×

2

2 1

1

0t

o o ot

M dt H H= = −∑∫

1 20 o o oH or H H∆ = =

33

Example34

1.

35

2.

36

3.

37

4.

38

Impact39

Impact refers to the collision between two bodies and ischaracterized by the generation of relatively large contact forcesthat act over a very short interval of time.

Direct central impact Consider the collinear motion of two spheres of masses m1 and m2

travelling with velocities V1 & V2. If V1 is greater than V2, collision occurswith the contact forces directed along the line of centers.

a. Before impact

b. Maximum deformation during impact

c. After impact

40

As far as the contact forces are equal and opposite during impact, the linearmomentum of the system is conserved.

In this equation we have two unknowns and so we need anotherrelation or equation.

During collision, there are two kinds of forces experienced by the particles Force of deformation: for Force of restoration: for

Particle 1

Before collision

During collision

After collision41

' '1 1 2 2 1 1 2 2m v m v m v m v+ = +

'1v '

2v

0 ot t≤ ≤'

ot t t≤ ≤

0 ot t≤ ≤ 'ot t t≤ ≤

Particle 1Before collision During collision

After collision

42

0 ot t≤ ≤ 'ot t t≤ ≤

Applying the principle of linear momentum and impulse(state of deformation) (State of restoration)

Coefficient of restoration (e)It is the ratio of magnitude of the restoration impulse to the magnitude of the deformation impulse.

0 ot t≤ ≤

1 1( )ot

d oo

F dt m v v= −∫

'ot t t≤ ≤

'

'1 1( ( ))

o

t

r ot

F dt m v v= − − −∫'

'1 1( )

o

t

r ot

F dt m v v= −∫

'

'1 1

1 1

( )( )

o

o

t

rt ot

od

o

F dtm v vem v v

F dt

−= =

−

∫

∫'1

1

( ) (1)( )

o

o

v vev v−

=−

1 1( ( ))ot

d oo

F dt m v v= − − −∫

43

Particle 2Before collision During collision

After collision

0 ot t≤ ≤ 'ot t t≤ ≤

Applying the principle of linear momentum and impulse(state of deformation) (State of restoration)

Coefficient of restoration (e)It is the ratio of magnitude of the restoration impulse to the magnitude of the

deformation impulse.

0 ot t≤ ≤ 'ot t t≤ ≤

2 2( )ot

d oo

F dt m v v= −∫'

'1 2( )

o

t

r ot

F dt m v v= −∫

'

'1 2

1 2

( )( )

o

o

t

rt ot

od

o

F dtm v vem v v

F dt

−= =

−

∫

∫

'2

2

( ) (2)( )

o

o

v vev v−

=−

From equation (1) and (2)

If e=1 then the collision is perfectly elastic collision (no energy loss) If e=0 then the collision is perfectly plastic collision (maximum energy loss)

44

' '2 1

2o

v vev v−

=−

Oblique Central Impact Occurs when the direction of motion of the mass centers of the

colliding particles are not the line of impact.

45

' '1 1 2 2 1 1 2 2n n n nm v m v m v m v+ = +

'1 1 1 1

'2 2 2 2

t t

t t

m v m v

m v m v

=

=

' '2 1

1 2

n n

n n

v vev v

−=

−

Example46

1.

47

2.

48