App Mechanics Complete1

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CHAPTER- 1:

INTRODUCTION TO DYNAMICS

Mechanics as the origin of Dynamics:Mechanics is defined as that science which describe and predicts the conditions of rest or

motion of bodies under the action of forces. It is the foundation of most engineering sciences. It can

be divided and subdivided as below:

(i) Newtonian Mechanics(Engineering Mechancis)

(ii) Relativistic Mechanics(It deals with the conditions

involving seed of bodiesclose to the speed of light )

(iii) Quantum Mechanics(It deals with the conditionsinvolving extremely small

mass and size ie atomic distance)

(a) Mechancis of rigid bodies (b) Mechanics of deformable bodies (c) Mechanics of fluids

Statics Dynamics

Kinematics Kinetics

Mechanics of Compressiblefluids

Mechanics ofIncompressible fluids

Dynamics:

It is which of Newtonian Mechanics which deals with the forces and their effects, while

acting upon the bodies in motion. When we talk about the motion of the planets in our solar system,

motion of a space craft, the acceleration of an automobile, the motion of a charged particle in an

electric field, swinging of a pendulum, we are talking about Dynamics.

Kinematics:It is that branch of Dynamics which deals with the displacement of a particles or rigid body

over time with out reference to the forces that cause or change the motion. It is concerned with the

position, velocity and acceleration of moving bodies as functions of time.

Kinetics:It is that branch of Dynamics which deals with the motion of a particle or rigid body, with

the reference to the forces and other factor that cause or influence the motion. For the study of

motion Newtons Second Law is widely used.


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-By Er. Biraj Singh Thapa (Lecturer, Eastern College of Engineering, Biratnagar)/ -2

Chapter:- 2

Determination of motion of particles:

In general motion of particles (position, velocity and acceleration ) is expressed in terms offunction as,

X = f(x) , [ x = 6t2+t3]

* But in practice the relation of motion may be defined by any other equation with function of x,v,& t .

a = f(t)

a = f(x)

a = f(v) etc.

so these given relation are integrated to get the general relation of motion x = f(t) .

Case-I: When acceleration is given as function of time [i.e a = f(t) ] [ a = 6t2+t3]

We know,

a = dv/dt dv = adt

or, dv = f(t) dtNow integrating both sides taking limit as time varies from 0 to t and velocity varies form voto v.

).......()(

)(

)(0

idttfvv

dttfvv

dttfdv

t

oo

t

oo

tv

vo

+=

=

=

Again, velocity is given by,

V = dx/dt dx = vdtAgain integration both sides of equation similarly form time 0 to t and position xoto x.

We get,

=x

x

t

o

vdtdx0

x xo=

+

t

o

t

dtdttfv0

0 )( Putting value of V form equation (i)

++=

t t

oo iidtdttfvxx0 0

).........()(

Thus position is obtained from equation of a = f(t)

# Find the velocity and position of a particles after its 5 sec from Rest, which moves with equation of

a = 6t2-4t.

Solution:

Given equation a = f(t) a = 6t2 4t

xo= 0 , vo= 0 and t = 5.

We know,

V0=

===

t

o

t

o

ttdtttdttfdttf 0

55

0

232

2

4

3

6)46()()(

[ ] smttv /20022 0523 == Again,


3/92


X = xo+ dtdttfv

tt

o

+

00

)(

= 0 + [ ] 500

5

0]200[2002000 tdtdt

t

==+ = 100 m

Therefore, x = 100m and v = 200m/s after 5 second of motion.

Case-II When the acceleration is a given function of position [ i.e a = f(x) eg. x2+4x]We know,

a = dv/dt = dv/dx . dx/dt = v.dv/dx

or, vdv = adx

or, vdv = f(x)dx [ a = f(x)]

Now , Integrating both sides of above equation , taking limit as velocity varies from Vo to v as

position p varies form xoto x.

i.e =

=

x

xv

vx

x

t

v ooo

dxxfv

dxxfvdv )(2

)(0

2

or, =x

xdxxf

vv

0

)(22

2

02

)1.......()(22

1

2

00

+=

x

xdxxfvv

Again We know,

V = dx/dt dx = vdt.Integrating both sides with limits as time varies from 0 to t and position from x oto x .

i.e )1........( =t

o

x

x

vdtdxo

Putting value of v varies from equation (1) we get,

x xo=

+ dtdxxfv

x

xo

2

1

2

0 )(2

++=t x

xo dtdxxfvxx

0

2

1

2

00

)(2

Case III : When acceleration is a given function of velocity (i.e a =f(v) eg. a = v2+v)

We know, a = v dv/dx f(v) = v dv/dxOr , dx = v dv/f(v)Integrating both sides taking limit as velocity varies form voto v and position varies from xoto x

==v

v

v

v

x

x vf

dvvxx

vf

dvvdx

000 )()(0

+= vv vfdv

vxx0 )(

0 e.g The acceleration of a particle is defined as a = -0.0125v

2, the particle is given as initial velocity

v0, find the distance traveled before its velocity drops to half.

Solution:Given, a = -0.0125v

2i.e a = f(v) ,

Initial velocity vo, final velocity vo/2For motion a = f(v) xo= 0, x = ?


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x = xo+ v

v vf

dvv

0 )(

=x

=

2 22

0

0

0

0

1

0125.0

1

0125.0

v

v

v

vdv

vdv

v

v

[ ]

==

0

02

2

ln

0125.0

1ln

0125.0

1 0

0

v

vv

v

v

Or, x = 24.08 m ans.

2.2 Uniform Rectilinear motion:

* Uniform motion means covering equal distance over equal intervals of time. ie velocity = constant.

We have,

V = dx/dt = v [ v = constant velocity of body]

dx = udt =tx

xvdtdx

00

[ Integrating both sides under limits as position varies from xoto x and

time 0 to t]

x xo= vt

x = xo+vt

Change in position (or displacement) is equal to uniform velocity x change in time [ i.e s = vt]

2.3 Uniform Accelerated Rectilinear motion:

If constant acceleration be a then,

dv/dt = a = constant dv = adtIntegrating both sides with limit v0to v and 0 to t .

We get,

atvvdtadvtv

v== 000

)1........(0 atvv +=

Again for position , we have

v = dx/dt ..(2) from 1 and 2.

dx = (vo+at) dt ,

Integrating both sides overthe limits

( ) 2000

02

1

0

attvxxatdtvdxtx

x+=+=

2

00 2

1attvxx ++=

Also, a =dx

dvv

Or, vdv = adx

Integrating both sides under limits

=x

x

v

vdxavdv

00

( ) ( )02022

1xxavv =

( )02

0

2 2 xxavv +=

2.4 Motion of several particles:Two or more particles moving in straight line.

Equations of motion may be written for each particles as:

xo= Initial position

x = Final position

v0= Initial velocity.

v = Final velocity

0 = Initial time

t = Final time


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xA+2xB= Constant ..(iv) [ AC+DE+FG = Constant ]

If A block A is given xA motion it will produce xB= (-xA/2) as the motion of Block B.

Differentiating equation (iv) w.r.t time , we get,

VA+2VB= 0 , (v)

Or, VB= -(VA/2)Similarly differentiating equation (v) w.r.t time we get,

AA+2aB= 0 (vi)

Or, aB= - (aA/2) [ Negative sign denotes opposite in direction ]

* In this case displacement, velocity and acceleration of one body gives the displacement, velocity and

acceleration of other body. This arrangement is called 1-degree freedom .

# Derive the equation of motion of the given pulley system.

Solution:

2xA+2xB+xc= Constant .(1)

2vA+2vB+vc= 0 ..(2)

2aA+2aB+ac= 0 .(3)

A

xA

B

C

G

xB

xC

IM N o

J LK

D E F

2.6 Graphical solution of Rectilinear motion problems:* Graphical solution are very helpful to simply and solve the problems of Dynamics.

* Using the motion graphs (i.e x t , v-t and a-t ) the missing value at any point can be obtained .

* If any on equation of motion is known all the three graphs can be obtained as follows.

If equation of displacement x = f(t) ..(i) is known,

Then , V = dx/dt .(ii)

a = dv/dt .(ii)

i.e velocity is slope of x-t curve and acceleration is slope of v-t curve.

x1

t1

x

t

x= f(t)

Sloe= dx/dt = v1

V1

t1t

V

Sloe= dx/dt = a1

V= f(t) a1

t1t

a

a= f(t)


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If x-t curve is given, then computing slope at each point of x-t curve corresponding v-t curve can be

generated and computing slope at each point of v-t curve a-t curve can be generated.

Again,

From equation (ii)

vdtdx=

==

2

1

2

1

2

1

).(..........12x

x

t

t

t

t

ivvdtxxvdtdx

And from equation (iii) ,

dv = adt

).(..........22

1

2

1 112 vadtvvadtdv

t

t

t

t

v

v ==

This means change in position in given by the area under curve v-t and change in velocity is given byarea under the curve a-dt.

a1

t

a

t1 t2

V2

t1t

V

V1vdt = x2- x1

t2

t1

t2

x

tt1

x1

x2

Tutorial Examples:

1) The motion of a particles is defined by the position vector kt

jtitr 4

463

2 ++= )r

where r in meter and t in

second. At the instant when t = 3 sec, find the unit position vector, velocity and acceleration.

Solution:

We have , kt

jtitr 4

463

2 ++= )r

At time t = 3 sec. kjir 4273618 ++=r

( ) ( ) mrr 81.404

273618

2

22=

++==

r

Now unit position vector at t = 3 sec.

81.40

4

273618

kji

r

rr

++== r

r

Anskjir 165.088.044.0 ++=

Again,


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++== k

tjtit

dt

d

dt

rdv

446

32

rr

ktjtiv 4

3862++=

r

At time t = 3 sec.

++= kjiv 4

24246

r

Velocity (v) = smv /64.257

24246

2

12

22 =

++=

r

V = 25.64 m/s

Again, acceleration ( )

++== ktjti

dt

d

dt

vda 2

4

386

rr

ktja 2

38 +=r

At t = 3 sec , kja 5.48 +=r

Acceleration, (a) = ( )[ ] Anssma 221

22 /18.95.48 =+=r

2) A ball is thrown vertically upward with a velocity of 9.15m/s. After 1s another ball is thrown with the

same velocity. Find the height at which the two ball pass each other?Solution:

Let the initial velocity of both balls

V01= v02= vo= 9.15 m/s

h be the height at which two balls pass each other t 1be the time elapsed by the first ball before passing second

and t2be the time elapsed by second.

From the given condition:

t1 t2 = 1 ..(i)

for 1stball , n = v0t1 9t1

2 ..(ii)

For 2nd

ball n = v0t2 9t22(iii)

Substituting equation (iii) form equation (ii) , we get,

( ) ( )210222192

1ttvtt =

( )( ) ( )210122192

1ttvtttt =+

( )

( ) ).(..........865.1

865.181.9

15.92

9

2

21

021

ivtt

vtt

=+

=

==+

Adding equation (i) and (ii) , we get t1= 1.43 sec and t2= 0.43 sec

mttvh 05.392

1 2110 ==

h = 3.05 m

Hence , two balls pass each other at 3.05m above the ground.

3) In the following pulley system, Block 2 has velocity 2m/s upward and its acceleration is 3m/s2downward

while block 3 has velocity and acceleration 2m/s up ward and 4m/s2 downward respectively. Find the

velocity and acceleration of block 1.


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Solution:

Given, V2= 2m/s ()

a2= 3m/s2()

v3= 2m/s ()

a3= 4m/s2()

Here ,

AB +CF = constantGI +HJ = Constant

DE = Constant

Portion of rope around the pulley is also constant.

Now,

X1= AB + constant ..(i)

X2= GI +CF+ constant .(ii)

X3= HJ+CF+ Constant (iii)

Multiplying equation (i) by (2) and adding

(i), (ii) and (iii) , we get

2x1+x2+x3= 2AB+CF+CF+GI+HJ+Const.

2x1+x2+x3= const. ..(iv)

Differentiating equation (iv) w.r.t time.

2v1+v2+v3= 0 (v)

2a1+a2+a3= 0 ..(vi)

From equation v

smvv

v /22

22

2

321 =

=

=

Therefore, velocity of block 1 (v1) = 2 m/s ()

From equation (vi)

2321 /5.3

2

43

2sm

aaa =

++=

=

Therefore, acceleration of block 1 (a1) = 3.5 m/s2()

1

3

B

H

x1

2

x2

C

G

x3

I

J

A

F

E


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Chapter 3

Curvilinear Motion of Particles

3.1 Position vector, Velocity and Acceleration:When a particle moves along a curve path other than a straight line, it is said to be in

curvilinear motion. Vector analysis is used to analyze the change in position and direction of motion

of particles.

Let, at time t the position vector of particle be rand at another time )( tt + the particle takes a new

position p' and its position be r!. Then r represents the change in directoin as well as magnitude

of the position vector r. (fig. a) The average velocity of the particles at time intervalt

rt

= ( in magnitude and direction of r )

Instantaneous Velocity,dt

rd

t

rv

t=

=

0lim

As r and t becomes shorter, PP & gets closer and v is tangent to the path of the particle. (fig c)

And, As t decreases, length of PP ( r ) equals to length of arc s (fig b)

dt

ds

t

s

t

PP

tt=

=

=

00limlim

Change in position ( r ) can be resolved into two components,

i One parallel to x-axis and ( PP )

ii Other parallel to y-axis ( PP )

PPPPr +=

t

PP

t

PP

t

ror ttt

+

=

000 limlimlim,

O X

Y

r

r

r

p

p

(b)

r s

r

t

O X

Y

r

r

r

p

p

s

(a) o X

Y

r

p

(c)

s

Y

yP p

p

r

V

X

Y

Ox x

Y

XO

Pv

vy

vx

v


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jt

yi

t

xor

tt

limlim,00

+

=

jdt

dyi

dt

dxor , +=

jvivor yx, +=

jyixor , +=

Then,22

yx vvV += [Magnitude of Velocity]

x

y

v

v=tan

x

y

v

v1tan = [Direction of Velocity]

Similarly, acceleration by curvilinear motion can be computed as:

If v and v be the velocities at time t 4 )( tt + i.e. tangents at Pand P , then the average

acceleration of the particle over the timer interval is given by

t

vat

= )(

dt

vd

t

vaor

t=

=

0lim,

Again, v can be resolved into QQQQ & parallel to x & y-axes respectively. Then,

QQQQv +=

t

QQ

t

QQ

t

vor

ttt

+

=

000

limlimlim,

jt

vi

t

vaor

y

t

x

t

limlim,00

+

=

jdt

dvi

dt

dvor

yx +

jaiaor yx, +

jyixaor , +=

magnitudeinonacceleratiaaaor yx +=22 )()(,

Positive Value of xv Right Direction

Positive Value of yv Upward Direction

ydt

dyvx

dt

dxv yx ==== ;

O X

Y

r

r

r

v

v

s

(a)O X

Y

v

v

(b)

Q

Q

v

vy

vx

Q

X(c)

q

q

Y

O

ay

axq

dt

dva

dtdva

y

y

xx

=

=


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du

Qd

du

PdQP

du

d+=+ )(

du

PdfP

du

df

du

Pdf+=

directioninaaa

axy

x

y == )(tantan 1

3.2 Derivatives of a vector function:

O X

Y

(a)

p

O X

Y

(b)

dpdu

p= f(u) =2u2+4u+3

Let, )(uP be a vector function of scalar variable u. If value of u is varied, P will trace a

curve in space. Considering change of vector Pcorresponding to the values u 4 ( )uu + as shown infigure(a). Then

)()( upuupp +=

- (1)

As ,0u p becomes tangent to the curve. Thusdu

pdis tangent to the curve as shown in

figure(b).

Again,

Considering the sum of two vector functions )(&)( uQup of the same scalar variable u. Then

the derivative of the vector )( QP+ is given by:

u

Q

u

P

u

Q

u

P

u

QPQP

du

d

uuuu

+

=

+

=

+=+

0000limlim

){){lim

)(lim)(

- (2)

Again, product of scalar function f(u) and pf a vector function )(uP of the same scalar variable u.

Then, derivative of fPis given by:

( )u

PfPPff

du

pdf

u

++=

0lim

.

+

=

u

PfP

u

f

u 0lim

- (3)

+=

=

u

upuup

u

p

du

pdei

uu

)()(limlim..

00


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du

QdPQ

du

PdQP

du

d..).( +=

du

QdPQ

du

PdQP

du

d+= )(

kPjPiPP zyx ++=

kdu

dP

jdu

dP

idu

dP

du

Pd yyx

++=

kdt

dPj

dt

dPi

dt

dP

dt

Pd yyx ++=

kzjyixr &&& ++=

kvjvivV zyx ++= )(

222

zyx vvvV ++=

zyx

zyx

azayax

vzvyvx

===

===

&&&&&&

&&&

&,

&,

kajaiaa zyx ++=

)( 222 zyx aaaa ++=

o X

Y

Z

vz

vx

vy

v

Similarly, scalar product and vector product of two vector functions )()( uQanduP may be

obtained as:

- (4) [Scalar Product]

- (5) [Vector Product]

Again,

- (6)

where, zyx PPP &, are the rectangular scalar components of vector P & kji,, are the unit vector.

- (7) [where, )](ufP=

And,

- (8) [where, )](tfP=

3.3 Rectangular Components of Velocity and Acceleration:

When, the position of a particle is defined by at any instant by its rectangular co-ordinates x, y,

z as:

- (i)

Then, differentiating both sides w.r.t. time, we get,

kzjyix &&&&&& ++==dt

rd

where,

So,


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ax

ay

Vo= 100m/s

nmax

80mHill

P

A

B

When the motion in each axis can be represented independent with each other then the use of

rectangular components to describe the position, velocity and acceleration of a particle is effective

i.e. motion in each axis can be considered separately.For e.g. for projectile motion, neglecting air resistance, the components of acceleration are:

gaanda yx == 0

gvandvor yx == 0,

gdtdyanddvor x == 0,

On Integrating both sides under the limits,

==tv

v

v

vx dtgdyanddv

x

x

y

y 00

o o

gtvvandvv yyxx == oo 0

)(vgtvvandvv yyxx === oo

gtvyandvxor yx == oo &&,

gtvdt

dyandv

dt

dxor yx ==

oo

,

gtdtdtvdyanddtvdxor yx == oo,

Integrating both sides under limits considering motion starts from origin by co-ordinates i.e. x at t=0

; x=0 ; y=0 and at t=t0, x=x0and y=y0

==ty t

y

tx

x tdtgdtvdyanddtvdxo

00 000 0

0

0

)(21 200 00 vigttvyandtvx yx ==

Thus motion under projectile can be represented by 2-independent rectilinear motion.Y

XO

Vx

XoVxo

youy

uyo

Vxo

V

Problems:

1. A bullet is fired upward at an angle of 30 to the horizontal from point P on a hill and it strkies a

target which is 80m lower than the level of projection as shown in figure. The initial velocity of the

bullet is 100m/s. Calculate:

a. The maximum height to which the bullet will rise above the horizontal.

b. The actual velocity with which it will strike the target

c. The total time required for the flight of the bullet.

Solution:

smVV

smVVsmV

y

x

/5030sin

/60.8630cos/100

0

0

0

0

0

==

===


15/92


nmax

80mB

AVo= A

P

a)

( )Ansmhg

Vh

42.127

42.1272

30sin

max

22

0max

=

==

b) Let, =yV1 vertical component of velocity at highest point A = 0

=yV2 vertical component of velocity striking target

H = vertical distance between point A and target = 127.42+80 = 207.42mThen, ]0[2 1

2

1

2

2 == yyy VgHVV

smVor y /79.63, 2 =

]tan[/60.860012

tconsVsmVVV xxxx ====

smVVV yx /55.1072

2

2

22 =+= (Ans)

& == 37.36tan 1

xy

zy

V

V

c) 222021)( gttVh y =

)(sec60.11

,

08050905.4,

8.92

15080,

2

2

2

2

2

22

Anstxv

Solving

ttor

ttor

==

=

=

2) The motion of a vibrating particle is defined by the equation x=100sin t and y=25cos2 t , where

x & y are expanded in mm & t in sec.

a) Determine the velocity and acceleration when t=15

b) Show that the path of the particle is parabolic.

Solution:

a) We have,

x=100sin txVt x cos100== &

txax sin1002== &&

Again, y=25cos2 t tyVy 2sin50== &

tyay 2cos1002

== &&

Then, for t=2sec,

]1sin50[:]1cos100[ 2 == yx VV

22

yx VVVV +==

22 )2sin50()cos100( += t

0)(tan/100 1 ===

x

y

V

VsmmV

Total time of flight is the sum of time to

reach B from A & to C from BT=t1+t2

t1=g

V sin2 0 & PB=Range=g

V sin20


16/92



And, For t=1sec

==

=+==

==

270)(tan

/100

12cos100:]sin100[(

1

2222

22

x

y

yx

yx

a

a

smmaaaa

aa

b) Since, x=100sin t

tx

sin100

=

)(sin)100

(22 it

x=

Again, ty 2cos25=

)(cos50

25

cos2125

,

1cos225

,

2

2

2

iity

ty

or

ty

or

=+

=

+

=

Adding equations (i) and (ii), we get ;

][parabolaofequationtheiswhich,5000200,

100005000200,

1cossin50

25

10000

22

2

222

cbyaxyxor

yxor

ttyx

=+=++

=++

=+=+

+

3) The motion of a particle is given by the relation V x=2cost & Vy=sint. It is known that initially

both x & y co-ordinates are zero. Determine

a) Total acceleration at the instant of 25b) Equation of the parabola

a) Here, Vx=2 cost & Vy =sint

Then, ax= tdt

dvx sin2=

and, tdt

dva

y

y cos==

jtita

jaiaa yx

cossin2

+=

+=

For t = 2

sec

At t=2sec

ax=-2sinz=-1.82

ay=cos2=-0.42

222 /865.1 smaa yx =+=

== 193tan 1y

a

a

( )

( )

parabolaofequationrequiredtheiswhich,084,

1)1(4,)(

)(cos11cos,

22

22

22

=+

=+

=+=

yyxor

yxiiandiAdding

iitytyor


17/92


[ ]

)(19382.1

42.0tantan

/865.1)42.0()82.1(

onacceleratiTotal

42.082.1

radianinis2,2cos2sin2

11

222

Ansa

a

sm

jia

whereiia

x

y=

==

=+==

=

=

+=

Q

b) ===x t

x dtdxtdt

dxtV

0 0cos2cos2cos2

)(sin4

,

sin2,

22

itx

or

txor

=

=

Again, ===ty

y tdtdytdx

dyV

00sinsin

3.4 Motion Relative to a Frame in Translation:

Let A and B be the particles moving is a same plane with BA rr & be their position with respect to

XY axis.

Considering New axes (X'-Y') centered at A and parallel to original axes X-Y, the motion of

particle B can be defined with respect to motion of particle A such that:

From vector triangle OAB

)(/ irrr ABAB +=

,similarlyand

ABAB

ABAB

YYY

XXX

/

/

+=

+=

Differentiating equ(i) w.r.t. time, we get:

)(iiiVVV BAB +=

In scalar form:

ABBB

ABAB

YYY

XXX

/

/

&&&

&&&

+=

+=

OR

( )

( ) yyy

xxx

ABAB

ABAB

VVV

VVV

)/()(

)/()(

+=

+=

Again,

differentiating (iii) with respect to time, we get:

)(/ vaaa ABAB += In scalar,

- ii

where,

XA,YA& XB& YBare co-ordinates of A & Bw.r.t. XY axes

XB/A, YB/Aare co-ordinates of B .r.t. X'-Y' axis

- iv axis-yinmotionofEquationY

axis-in xmotionofEquation

axesyandboth xinmotionofEquation

X

r

where,

ABABAB

BB

AA

YX

YX

YX

///

B

A

VofcomponentsY&Xare

VofcomponensY&Xare,

VofcomponentsY&Xare,

&&

&&

&&

( ) ( )

Bx

By

ByBxB

a

a

aaa

1

22

tan=

+=


18/92



ABAB

ABAB

YYY

XXX

/

/

&&&&&&

&&&&&&

+=

+= OR

yAByAyB

xABxAxB

aaa

aaa

)/()()(

)/()()(

+=

+= - (vi)

3.4 Tangential and Normal Components:

The velocity of particle is vector tangent to the path of particle. But acceleration may not be

tangent in curvilinear motion.

The acceleration vector may be resolved into two components perpendicular with each other

in directions

i First component along the tangent of path of particle (a t)

ii Second component along the normal of path of particle (an)

Let nt ee and be the unit vectors directed along the tangent and normal of the path

respectively. Then, in curvilinear motion, nt ee and would change the direction as particle

moves from one point to another.

x

yy

pp

xoo

p

(a)

(b)

rr r

et=dr dsr/ s

From fig (a)

lim&'0s

srrrr ==

Then,ds

dr

s

re

st =

=

0lim

)( ids

dret =

Again, [ ]srsssr

tess ==

=

= 00 lim1lim

Therefore, path.otangetn tthealongrunit vectotheisds

rdte =

Let, be the radius of curvature of the path at the point P and '& tt ee be the tangent unit vectors at

P and P'. te be the change in unit vector while the particle moves from P to P'.


19/92


x

y

o

(a)

en p

per

ex

o(b)

et

exex

y

xo (c)

p

at=dVet dt

a =a t et + an en

Now, from fig,== 'PPs

[ ]magnitudeineeeeee nntttt 1e0sAs t' ===

)(

lim0

iid

ede

ee tn

t

sn

=

=

Similarly,

(iii)

and1

== n

t ed

ed

ds

d

Also,

dtd

tts

dtdsV

tt ==== 00

limlim

)(ivdt

dV

==

tt evvedt

ds

ds

rd

dt

rdV . ====

)( veVV t =

And,

dt

ds

ds

d

d

edve

dt

dv

dt

edve

dt

dveV

dt

d

dt

vda t

t

t

tt..

)(

+=+===

)(,

)1

)((,

2

vieV

eVaor

eVeVaor

nt

nt

+=

+=

&

&

which can be represented as in fig(c).

where, nntt eaeaa +=

=ta Tangential component of acceleration = vdt

dv

&=

=na Normal component of acceleration =2

2

&=V

Notes:

For increasing velocity at will be in the direction of velocity and fxdecreasing velocity atwill

be in opposite to the direction of velocity.

If the speed is constant at=0 but an0. [an=0 fxRecti = ]

anis always directed towards the centre of curvature

For higher velocity and smaller radius higher is an.

3.6 Radial and Transverse Components:


20/92



For the motion described by polar co-ordinates.

Position of particles P is defined by the co-ordinates r & , where r is the length and is the

angle in radians.

The unit vectors in radial and transverse direction are denoted by eer and respectively along

radius and 90 clockwise to the radius in direction. fig (a)

x

y

(a)

e

x

y

(b)

e

P(r, )r = rer er

er

eo

r

r eo

As the particle moves from '.PtoP The unit vectors eer &, change to '&' ee r by

eer and respectively.

Here,

===

eofdirectioninis

d

edofdirection(i)-----.

r&& e

dt

d

d

ed

dt

ede rrr

=== re-ofdirectioninis

d

edofdirection(ii)-----.

&&re

dt

d

d

ed

d

ede

Now,

rerr =

Then, rrr

rr ererdt

edre

dt

drer

dt

d

dt

rdV &&

)( +=+===

&&&& eererV r e(iii)- r=+=

which can expressed as

eVeVV rr += , where

Vr= Radial component of velocity = r&

And, V = Transverse component of velocity = r&

Similarly,

( )ererdt

d

dt

vda r

&& +==

ererererer rr&&&&&&&&&& ++++=

&&&&&&&&&&&&&

rrrr eeeeererererer &2 ==+++=

( ) ( ) )(22 verrerra r ++= &&&&&&& which can be represented as,

eaeaa rr +=

where,

=ra Radial component of acceleration =2

&

&& rr and, =a Transverse component of acceleration = ( ) &&&& rr 2+


21/92


In case of a particle moving along a circular path with its centre at the origin O, we have

r=constant

or, 0&0 == rr &&& Then,

)( vierv = &

)(2 viierera r += &&&

Problems:

1) The motion of a particle is defined by the position vector, ,543 432 ktjtitr ++= where r is in m

and t is in sec. At instant when t=4 sec, find the normal and tangential component of acceleration and

the radius of curvature.

Solution, we have

ktjtidt

vda

ktjtitdt

rdV

ktjtitr

60246&

20126

543

2

32

432

++==

++==

++=

Again,

( ) )(40014436 21

642 itttVV ++==

( ) ( )[ ] )(602436 21

222 iittaa ++==

Now,

At t = 4sec

V=1294.54m/s [putting t=4 in equ-(i)]

a=964.81m/s2[putting t=4 in equ-(ii)]

Again,

Tangential component of acceleration,

( )

( )

( )53

2

1642

2

1642

240057672

40014436

1.

2

1

40014436

ttt

ttt

tttdt

d

dt

dvat

++

++

=

++==

At time t=4 sec, at=963.56m/s2(Ans)

Now,

( ) ( )

)(/1.49

56.96381.964

2

2222

Anssma

aaa

n

tn

=

==

Again,

( ))(03.34131

1.49

54.129422

Ansma

V

n

===

2. A car is traveling on a curved section of the road of radius 915m at the speed of 50km/hr. Brakesare suddenly applied causing the car to slow down to the 32 km/hr after 6 sec. Calculate the

acceleration of the car immediately after the brake have been applied.


22/92



Solution: Given,

m/sec8.88km/hr32V

m/sec13.8850km/hrV

915m

1

0

==

==

=

At the instant when the brake is applied,

( )

tn

ttnn

t

n

eea

eaeaa

smt

VVa

smVa

833.0210.0

/833.0

/210.0915

88.13

201

2

22

=

+=

=

=

===

( ) ( )

)(2.1483.0

21.0

tantan

)(/856.083.021.0

11

222

Ansa

a

Anssmaa

t

n

=

==

=+==

3. The plane curvilinear motion of the particle is defined in polar co-ordinates by r=t3/4+3t and

=0.5t2where r is in m,is in radian and t is in second. At the instant when t=4 sec, determine the

magnitude of velocity, acceleration and radius of curvature of the path.

Solution: We have,

2/334

33

4

23

trt

rtt

r =+=+= &&&

Again, 15.0

2 ===

&&&

tt Now, we have

)(34

34

3

32

iettt

et

ererv rr

++

+=+=

&&

Again, at t = 4 sec

( ) ( ) )(/13311215

11215

222 Anssmvv

eev r

=+==

+=

Again,

( )

ett

tt

etttt

errerra

r

r

34

3213

43

42

3

)2(

232

3

2

++

++

+=

++= &&&&&&&

At t = 4 sec,

( ) ( ) ( )Anssmaa

er

ea

2/12.4662

121482442

148442

=

+==

+=

Again, from equ(i) [for ]


23/92


++

+++

=

+++==

++

+==

ttt

ttt

a

ttt

dt

d

dt

dv

ta

ttt

vv

274

333

8

53

2

1

92

227

16

433

16

6

1

2

1

2

1

92

227

16

433

16

6

2

12

234

42

34

23

Q

Q

At t = 4 sec,

( )[ ] ( ) ( )[ ]2

1

222

1

22

2

005.1612.466

/055.16

==

=

tn

t

aaa

sma

Q

( )m

a

v

sma

n

n

41.2784.465

113

/84.465

22

2

===

=

Hence, Radius of curvature = 27.41m (Ans)


24/92



Chapter 4

KINETICS OF PARTICLES

NEWTONS SECOND LAW

4.1 Newtons Second Law of Motion:

Newton has given his understanding of motion of particles and their causes and effects in 3

laws.

The first and third law of motion deals with the bodies at rest or moving with uniform velocity

i.e. without any acceleration.

For the bodies under the motion with acceleration the analysis of motion and forces producing

it is done by the application of Newtons Second Law.

Statement of Newtons 2nd

Law:

If the resultant force acting on a particle is not zero, the particle will have an accelerationproportional to the magnitude of the resultant and in the direction of this resultant force.

a1

F1

a2F2

a2F2

321 F,F,FIf , etc be the resultant forces of different magnitude and direction acting on the particle.

Each time the particle moves in the direction of the force acting on it and if 321 ,, aaa , etc be the

magnitude of the accelerations produced by the resultant forces. Then,

(m)particleofmassconstant..........FFF

........F,F,F

3

3

2

2

1

1

332211

=====

aaa

etcaaa

So, when a particle of mass m is acted upon by a force ,nacceleatioandF a they must satisfy the

relation,

same]are&Fofdirectionwhere[)(amF ai=

i.e. kajaiakji zyx mFFF zyx ++=++ which is Newtons Second Law.

When a particle is subjected simultaneously to several forces equation(i) is modified as:

( ) ( ) ++=++= kajaiamkjieia zyx FFF..mF zyx

where, F = sum of resultant of all forces acting.

zyx aaa ,,,F,F,F zyx are x, y and z component of the forces and acceleration acting onthe particle respectively.

Notes:(i) When the resultant force is zero, the acceleration of the particle is zero.

(ii)When V0=0 and 0F= , Then particle would remain at rest.


25/92


(iii)When V0=V and 0F= , Then particle would move with constant velocity, V along thestraight line.

(iv)All the above cases defines the first law, hence the Newtons 1stLaw of Motion is a particular

case of Newtons 2ndLaw of Motion.

4.2 Linear Momentum and Rate of Change [Impulse Momentum Theorem}:From Newtons 2ndLaw, amF=

or,dt

vdmF=

( )ivmF = )(dt

d

Multiplying both sides by dt and integrating under the limits, we get:

)(12212

1

2

1iivmvmIvdmFdt

v

v

t

t==

The term 2

1

t

tdtF is called the impulse (I) of the force during time interval (t2-t1) whereas vm is

the linear momentum vector of the particle.

So, equation (ii) states that

The impulse (I) over the time interval (t2-t1) equal the change in linear momentum of a

particle during that interval. [Impulse Momentum Theorem]

The impulse of force is known even when the force itself may not be known.

Again, from equatin(ii)

)(2112 iiiIvmvm +=

i.e. Final momentum 2vm of the particle may be obtained by adding vectorically its initial

momentum 1vm and the impulse of the force Fduring the time interval considered.

Or, showing in vector form.

mV1

mV1

mV2

I1-2

I1-2

When several forces act on a particle, the impulse produced by each of the forces should be

considered.

i.e. )(2211 ivvmIvm =+

where, ( ) ( ) ++=++==2

1

2

121

2

1321

2

121..........

t

t

t

t

t

t

t

tdtFdtFdtFFFdtFI

Improper Path Function


26/92



a1

F1

F2F2

m mF1

F

m

=0

(ma)rev

4.3 System of Unit:

Units of measurement should be consistent and one of the standards should be followed.

Generally 2 standard units are taken:

a) System de' International Unit (SI unit)

b) U.S. Customary Units (used by American Engineers)

SI Units:

SI stands for System de International. SI units are the world-wide standards for the measuringsystem. SI units are fundamental or derived.

Fundamental and Derived Units:

Fundamental and Derived units are the SI units. Fundamental units are independent of any

other measuring units and are the basic units for all other system whereas Derived units are the units

which are expressed in terms of powers of one or more fundamental units.

Fundamental Units Derived Units

Length = metre (m) Velocity = L/T = m/s

Mass = kilogram (kg) Acceleration = V/T = L/T2= m/s2

Time = second (s) Force = ma = ML/T2= kgm/s2(N)

SI units are the absolute system of units and results are independent upon the location of

measurement.

US Customary Units:

This system is not absolute system of unit. They are gravitational system of units.

Base Unitslength = foot(ft)

force = pound (lb)

time = second (s)

Conversion from US Customary Units to SI Units:

length : 1 ft = 0.3048 m

force : 1 lb = 4.448 N

mass : 1 slug = 14.59 kg

: 1 pound = 0.4536 kg

4.4 Equations of Motion and Dynamic Equilibrium


27/92


yyxx

yx

avav

vyvx

==

==

&&

&&Q

&

&

TheoremsVarignon'

-pointthe

aboutmomentofsumis

pointaaboutmomentumTotal

Y

Xo

x

y

mV y

mV y

mV y

mV

Considering a particle mass m acted upon by several forces. Then from second law,

( )kajaiamFiamF zyx )( ++== Using rectangular components, the equation of motions are

)(,, iimaFmaFamF zzyyxx ===

(i) and (ii) gives the equation of motion of particle under the force F

or, zmFymFxmF zyx &&&&&& === ,,Integrating these equation as done in 3.3, the equation of motion can be obtained.

Again, the equation(i) may be expressed as

0= amF

i.e., if we add vector am to the resultant force in opposite direction, the system comes under the

equilibrium state. This force ( am ) opposite to the resultant force is called Inertial Force or Inertia

Vector. This equilibrium state of a particle under the given forces and the inertia vector is said to be

dynamic equilibrium.

At the dynamic equilibrium,

=== 0&0,0 zyx FFF Inertia vector measure the resistance that particles offer when we try to set them in motion or when

we try to change the condition of their motion.

Angular Momentum and Rate of Change (Angular Momentum Theorem)

Statement

The rate of change of angular momentum of the particle about any point at any instant is equalto the moment of the force F acting on that particle about the same point.

Let a particle of mass m moving in the XY-plane and the linear momentum of the particle is

equal to the vector .vm

The moment about O of the vector .vm (linear

momentum) is called angular momentum of the particle about O

at that instant and is denoted by 0H

Now, mvxand mvyare components of .vm in x & y

direction.

Then, from definition,

(+ H0=x(mvy)-y(mvx)

)()( iyvxvmH xyo =

Differentiating equ(i) with respect to time, we get:

( )xy yvxvdt

dmH =0&


28/92



F1

F2 m

m

ma

F3

p

Fy

Fx = 0

W

xxyy vyvyvxvxm &&&& +=

= xy yaxam

OaboutForceofmomentH

yFxFH

ymaxmaH

xy

xy

=

=

=

&

&

&Q

( )iimH = 0&

Thus, the rate of change of angular momentum of the particle about any point to any instant is equal

to the moment of force F acting on that particle about the same point.

i.5 Equation of Motion(a) Rectilinear motion of particles:

If a particle of mass m is moving in a straight line under the action of coplanar forces

etcFFF ,,, 321 Then the motion of particle can be written as

++== 321),( FFFFwhereiamF For Rectilinear motion, motion is only along the single co- ordinate,

i.e. ax=a & ay=0

Equn-(i) may be written as

0=

=

y

xx

F

maF

- (ii)

These are the equation of motion for the particle moving in the straight line.

(b) Curvilinear motion of particles:

i Rectangular components

ii Tangential and Normal components

iii Radial and Transverse Components

i. Rectangular components

From Newtons second law,

== yyxx maFmaF ;For Projectile motion, neglecting air resistance

==== 000 xxxx amaFF === mgwmaF yy

gm

mgaor y ==,

ga

a

y

x

=

= 0 (i)

These are the equations of motion.


29/92


Fnat= dV dt

FT

m

an= V2

a

F = ma

m

Fr = mar

ii. Tangential and Normal components:

From Newtons 2ndlaw,

==

==

2vmamF

dt

dvmamF

nn

tt

+= xt FFF These are the equations of motion.

iii. Radial and Transverse components:

From Newtons 2ndlaw,

+=

+==

==

FFF

rrmmaF

rrmmaF

r

rr

)2(

)( 2

&&&&

&&&

`

These are the equations of motion.

Note:

In case of Dynamic Equilibrium all the components of forces are balanced by Inertial Vector or

Inertia force. So, for dynamic equilibrium condition, the equation of motion becomes

==

====

==

==

==

0

0

0

0

0

0

maF

maF

maF

maF

maF

maF

rr

nn

tt

yy

xr

- (ii)

i.6 Motion due to Central Force-Conservation of Angular MomentumWhen the force Facting on a particle P is directed towards or away from the fixed point O, the

particle is said to be moving under a central force. The fixed point O is called the center of force.

As shown in the figure, particle P moves along the curve path.

O = origin of co-ordinates

Now,

Fr= Radial component of force F

F = Transverse component of force F

For central motion F = 0

( )

( ) ( ) 00,

021

,

02

22

2

==

=+

=+=

&&

&&&&

&&&&Q

rdrdt

dor

rrr

r

or

rrF

- (i)

- (iii)


30/92



Y

X

Z

F

dArd

Po

F

d rP

r

Integrating both sides we get

r2& =constant=h - (i)

Now, if elementary section are swept in time dt be dA

rdrdA .2

1=Q [ ]01;/ == dforSRS

drdAor 221, =

[ ]dtbysidesbothDividing2

1

2

1, 22

&r

dt

dr

dt

dAor ==

Here, .).( VAVelocityAreaorAreasweepingofchangeofRatedt

dA=

)(..2

2

1

2

1.. 2

iiVAh

hrVA

=

== &Q

Thus, when a particle moves under the central force, the areal velocity is constant. This is also called

Keplers Law

Again, Angular momenum = momentum of linear momentum about the fixed point.

rmvH = 0

Now,

&rv =

rmrH &= 0

)(20 iiimrH = &

[ ][ ][ ]tconsmtconsVAcetconsH

VAhVAmHor

hrmhHor

tan&tan..,sintan

..2.).(2,

,

0

0

2

0

===

==

==

Q

&Q

Hence, when a particle is moving under a central force, Angular momentum is always conserved.

i.7 Newtons Universal Law of Gravitation:Statement:

Every particle in the universe attracts every other particle with a force, which is directly

proportional to the product of their masses and inversely proportional to the square of the distance

between their centers.

Mathematically,

( )

=

mindistancetheisd

kginmassesareMandmwhere,2

2

id

GMmF

d

MmF

Also, G is the Universal Gravitation constant with its value

6.67310-11Nm2/kg2and

F is the force of attraction between them.


31/92


For a body of mass m located on or near the surface of earth, force exerted by the earth

on a body equals to the weight of the body i.e. F = mg and d = R (radius of the earth).

F = mg =2R

GMm

( )iiR

GMg =

2Q

where, g is the acceleration due to gravity with its standard value 9.81 m/s2

at the sea level.Since, earth is not perfectly spherical so the value of R is different and hence g varies according

to the variation of altitude and latitude.

i.8 Application in space mechanics:Earth satellite and space vehicles are subjected only to the gravitational pull of the earth after

crossing the atmosphere. The gravitation force acts as a central force on them and hence their

motions can be predicted as follows:

From central force motion,

( )ihr =&2

Trajectory of a particle under a central force:

Considering a particle P under central force F(i.e. directed towards center O)

Then we have

Radial component of force,

( ) )(2 iFrrmmaF rr === &&& And, Transverse component of force

)(0)2( iirrmmaF =+==

&&&&

From equ(ii) since m0

02 =+ &&&& rr

( ) 01, 2 =&rdt

d

ror

On integrating,

)(tan2 iiihtconsr ==&

2

r

h

dt

d==

&

Again,

)(1

..2

ivrd

dh

d

dr

r

h

dt

d

d

dr

dt

drr

====

&

====

rd

dh

d

d

r

h

r

h

dt

d

d

rd

dt

rdr

1.

d

rd.

22

&&&&&

)(1

2

2

2

2

vrd

d

r

hr

=

&&

Putting values of getweiequinr )(& &&&


32/92



Fr

hr

rd

d

r

hm =

=

4

2

2

2

2

2 1

Putting getwer

v ,1

=

Fuhd

ud

uhm =

+

22

2

222

( )viumh

Fu

d

ud=+

222

2

This is second order differential equation, which is the trajectory followd by the particle which is

moving under a central force F.

Note:

i. Fis directed towards Oii.

Magnitude of Fis +ve if Fis actually towards O (i.e. attractive force)

iii. F should be -ve if Fis directed away from O.The trajectory of a particle under a central force is

( )iiumh

Fu

d

ud=+

222

2

Again,

( )iiiGMmur

GMmF == 2

2

where, M = mass of earth

m = mass of the space vehicle

r = distance from the centre of earth to the space vehicle, u =r

1

From equ (ii) & (iii)

)(constant222

2

2

2

ivh

GM

umh

GMmuu

d

ud===+

This equn(iv) is second order differential equation with constant co-efficient .2

h

GM The general

solution of the differential equation is equal to the sum of the complementary i.e.

U = Uc+Up

wherem,

Uc= A sin+ B cos

Up= 2h

GM

Again,

Uc= A sin+ B cos= C (coscos0+ sinsin0) = C cos(-0)

( )20

cosh

GMCU +=

Uc= complementy solution i.e. for tangient condition

Up= particular solution i.e. for steady state condition


33/92


earth

satellite

trajectory of motion

e1

Now,

choosing 0= 0 and [ ]symmetryofaxisislineinital..1

eir

U=

we, get:

( )vch

GM

r+= cos

12

Again, we have the equation of conic section,

cosel

lr

+=

( )vil

e

lr+= cos

11

Comparing equn(v) & (vi), we get:

clel

ec ==

Again,

GM

hl

h

GM

l

2

2

1==

GM

che

2

= which is eccentricity of the conic section.

So, three cases may arise:

a) If e>1 (i.e. conic is a hyperbola)

i.e.2

2

,1h

GMcor

GM

ch>>

b) If e=1 (i.e. conic is a parabola)

i.e.2

2

,1h

GMcor

GM

ch==

c) If e


34/92



o rA

Vofree flight

powered flight

At vertex A, =0, r=roand v=vo

( ) ( ) [ ]10cos

12

000

== viiivr

GM

rc

For parabolic trajectory,2h

GMc=

)(22

0

ixvr

GMc =

From equn(viii) & (ix)

( )xrGMv

rvr

GMor

vr

GM

rvr

GM

=

=

=

0

0

0

2

0

2

0

2

0

2

00

2

0

2

0

2

12,

1

This velocity v0 is called the escape velocity. Since, this is the minimum velocity required for the

vehicle so that it does not return to its starting point.

=

===

2

2

0

2

0

22

R

GMmmg

gRGM

r

gR

r

GMVesc

( )xir

gRVesc =

0

22

Note:

If Vo>Vesc, trajectory will be hyperbolic

Vo=Vesc, trajectory will be parabolic

Vo


35/92


AAA

Vo=Vese

Vo=Vcir


36/92



A

700kg

c

300kg

D

T1

T2

T1

T2

WA400kg

F.B.D of A

mBaB

300kg

WB

Altitude gained by satellite (H) = r'-R = 23659.44 km

Again, to calculate time period:

When the satellite covers 180, it will make

kmr

r

07.71255

10576.61098.71

1

88

1

=

+=

Then, kmrr

a o 54.390472

07.712256870

2

1 =+

=+

=

kmrrb o 49.221201 ==

QTime period of the satellite,10

63

1006.7

1049.221201054.3904722

==

h

ab

secmin2618hrs12

sec10670647.4 4

=

=

2. The two blocks shown in the figure start from rest. The horizontal plane and the pulleys are

frictionless, and the pulley is assumed to be of negligible mass. Determine the acceleration of each

block and the tension in each cord.

Soln: Let, tension in the cord ACD be T1 & cord BC be T2. From

figure, if block A moves through distance SAthen block B moves

through SA/2.

( )ia

aV

VS

S ABA

BA

B ===222

Q

Using Newtons 2ndlaw for Block A, Block B and Pulley C

Block A:

( )iiaT

amF

A

AAx

=

=1001

Block B:

( )iiiTTaT

aTW

amF

A

BB

BBy

=

=

=

=

022/30081.9300

300

12

2

2

Pulleys

Since mass of pulley is considered zero, we have:

( )ivTT

amF ccy

=

==02

0

12

Putting values of T1& T2in equn(iv), we get:

2943-1500A-2100aA=0

2

/41.8 smaA =Q

2/42052

sma

a AB ==


37/92


NTT

NaT A

16822

841100

12

1

==

==Q

3. The bob of a 3 m pendulum describes an arc of a circle, in a vertical plane. If the tension is twice

of the weight of the bob for the position when it is displaced through an angle of 30 from its mean

position, then find the velocity and acceleration of the bob.

Soln:

Applying Newtons Second Law,

= tt maF

2/9.430sin

30sin,

smga

mamgor

t

t

==

=

Again,

= xx maF

xmamgmg = 30cos2

[ ] 221

22

2

/15.12

/12.1130cos2

smaaa

smgga

rt

x

=+=

==Q

== 22.36tan 1

t

n

a

a

Velocity of Bob ( )

==

2v

aav xx

chordthetoperpv

v

.sec/78.5

12.113

=

=Q

4. The motion of a 500 gm Block B in a horizontal plane is defined by the relation r=2(1+cos2 t)

and =2t, where r is expressed in meters, t in seconds and in radians. Determine the radial and

transverse component of the force exerted on the block when t=0 & t=0.75 sec.

Soln:

Here, m = 500 gm = 0.5 kg

r = 2(1+cos2t) - (i)

= 2t - (ii)

Differentiating with respect to time, we getr& = -4sin2t 2=&

02cos8 == &&&& tr

Now,

( )( )

+==

==

&&&&

&&&

rrmmaF

rrmmaF rr

2

2

When, t = 0, r = 4, 28&0 == rr &&&

0&20 ===

&&&


38/92



( )

( )

0

0202045.0

43.118

4485.0 22

=

==

=

=

F

F

NF

F

r

r

Q

Q

Similarly for t = 0.75 sec,

NFNFr 0.79,5.39 == 025.1

204

===

===

&&&

&&& rrr


39/92


F

A

AA

FFsin

Fcos

dr

ds

Fcos

U1-2

S1 S2

F

A2

A1x

Y1 Y

Y2A1

w

A2

Chapter 5

Kinetics of Particle : Energy and Momentum Method

5.1 Work done by a Force:When a particle moves by the application of force Fproducing the displacement ds, then the

work done by the force during the displacement ds is defined by:

du = component of force along the direction of motion distance travelled.

( )iFdsdudsFdu

=

=

cos.

.cosQ

where, rdds=

[ ]motionofdirectionandforceebetween thangletheis

Particular cases:

(a) When Fis along the direction of rd , then

[ ]10coscos === Fdsdu

(b)If Fis perpendicular to the direction of rd , then

[ ]090coscos0 === du

(c) For finite work done from s1to s2,

Integrating (i), we get:

( ) ( )iidsFUs

s=

2

1

cos21

s-Fcoscurveunder theArea21 =U

5.1.1Work of a const Force in Rectilinear Motion( ) xFU = cos21

[ ]Motionctilinear

Ax

Re

toAfromntDisplaceme 21=

5.1.2Work of a weight (or Force of gravity)The work du of the weight is equal to the product

of weight (w) and the vertical displacement of the center of gravity G of the body.

i.e. du = -wdy

( )ywU

yywU

wdyduy

y

=

=

=

21

1221

21

2

1

U1-2is -ve when work is done on the body

U1-2is +ve when work is done by the body.


40/92



x1

x2Ao

A1

A2

5.1.3Work of a force exerted by a springWhen a spring is deformed, the magnitude of force Fexerted by it on the body is proportional

to the elongation of the spring.

i.e. F = kx - (i)

where, k = spring constant

x = elongation lengthAgain, Elementary work

du = -Fdx = -kdx

finite work done during elongation from x1to x2

( )212221

21

2

1

2

1

xxkU

dxkUx

x

=

=

work is positive, when x

==

=

RrHere

wRmgRGMmivwR

rrU

,

1122

2

12

21

Q

5.2 Kinetic Energy of a Particle:For a mass m acted upon by a force Fand moving along the curve path, the component of

force along the direction of motion is given by:

=

=

====

=

2

1

2

1

cos

getwelimitstakingsides,bothgIntegratin

cos,

cos,

v

v

s

s

t

tt

vdvmdsF

mvdvdsFor

ds

dvv

dt

ds

ds

dv

dt

dva

ds

dvmvFor

maF

Q


41/92


)(,

2

1

2

1,

2

1

2

1,

1221

2

1

2

221

2

1

2

221

iTTUor

mvmvUor

mvmvUor

==

=

=

where, T2and T1is final and initial K.E. of the particle.

Hence, the work of the force Fis equal to the change of K.E. of the particle. This is also called asprinciple of work and energy.

5.3 Applications of Principle of work and energy:With the help of work energy principle, solution of problems, involving force, displacement

and velocity can be obtained in simple form,

e.g. Analysis of Pendulum

To determine the velocity of bob as it falls freely from A1to A2,

weve

wLU =21

Again, at KE at A1

]0[0 11 == VT Q

KE at A2

2

222

1mvT =

Now, using principle of work and energy,

2112 = UTT

[ ]pointreferencefrombobofheightverticaltheis22

1,

2

2

2212

LgLv

mgLwLmvUTor

=

===

Advantages of this method:

To find v2it is not necessary to find a2

Equation is in the form of scalar, hence it is easy to handle.

Forces which do not work (e.g. Tension on strings), etc are eliminated.

5.4 Power and Efficiency:Power is defined as rate of change of work.

( ) ( )

dt

dsFP

dsFdui

tt

uPavg

cos

cosPutting,dt

duP

getwe0aslimitTaking,

=

==

=

( )

== V

dt

dsiiFVPor cos,

where, V is magnitude of the velocity at the point of application of force F.


42/92



( )workInput

workOutputEfficiency =

[ ]frictiontoduelossestodue1, (Vg)1then work is ve (i.e. PE increases)

If (Vg)2


43/92


44/92



TA+ VA= wL - (iii)

From equation (i), (ii) and (iii), the total mechanical energy of pendulum at any position is same and

is equal to wL.

At A1, Total energy is entirely due to PE

At A2, Total energy is entirely due to KE

At A, Total energy is sum of PE + KENote:

For the system interacting with other forms of energy as electrical, frictional, etc all the forms

of energy should be considered. In that case as well the total energy of system is always conserved.

Hence, energy is conserved in all the cases.

5.7 Principle of Impulse and Momentum:Considering a particle of mass m acted upon by a force F.

Then from Newtons 2ndLaw,

amF=

In x & y components,

dt

dvmF

dt

dvmF

maFmaF

y

yx

x

yyxx

==

==

&

&

Since mass m is constant

( ) ( )imvdt

dFmv

dt

dF yyxx == )(&

Vectorically, we have

( ) ( )iivmdt

dF =

This equation states, Force Facting on the particle is equal to the rate of change of momentum

vm of the particle.

Multiplying equation (i) by dt and integrating on both sides, we get:

( ) ( )12

2

1

2

1xx

v

vx

t

tx mvmvdvmdtF ==

( ) ( ) ( )iiimvdtFmv xt

txx =+ 21

2

1

Similarly,

( ) ( ) ( )ivmvdtFmvt

tyyy =+

21

2

1

In vector form,

( )

+=

+=

+=

=+ yx

yx

yx

t

t

FFF

vvv

vvv

vmvdtFvm22

11

2

12

1

21

where,


45/92


Line of contact

Line of impact

A BBA

VA

CA CB

CA CB

VA VB

VB

Impact/actionLine of

(a) direction central impact (b) oblique central impact

( )21

forceofImpulse2

1

== mpt

tIdtF

Given, by area under the curve F-t

Hence,

( ) ( )vivmIvm mp =+ 2211

From vector diagram,

When several forces are acting on a particle,

( ) ( )viivmIvm mp =+ 2211

5.8 Impulsive motion and Impact(1) Impulsive Motion:

When a very large force is acted during a very short time interval on a particle and produce a definite

change in momentum, such a force is called as impulsive force and the resulting motion is called

impulsive motion.

Example of Impulsive motion:

Striking the ball with a cricket bat, large force F is applied in a small time ( )t , the resulting

impulse tF is large enough to change the direction of motion of ball.

+ =

mV1 F tmV2

From impulse momentum principle,

( )ivmtFvm =+ 21 Here non-impulsive forces (like weight of ball, bat, etc) are not included.

2. Impact

A collision between two bodies, which occurs in very short interval of time and during which

the two bodies exert relatively large forces on each other is called an Impact.

The common normal to the surfaces in contact during the impact is called the line of impact or

line of action.

Types of Impact:

If the mass centers of the two colliding bodies are located on this line of impact, the impact is

central impact otherwise eccentric impact.

If the velocities of the two particles are directed along the line of impact, it is said to be direct

impact. If either or both particle moves along the line other than the line of impact, the impact is

said to be an oblique impact.

Hence, four types of impact may

occur. They are:

a) Direct Central Impactb) Oblique Central Impact

c) Direct Eccentric Impact


46/92



A A

CA CA

CB

CB VA

VA VB

VB

BB

Line ofimpact

(c) direct ecentric impact (d) oblique ecentric impact

+ =

mAvA pdt mAv

+ =mAvARdtmAv

d) Oblique Eccentric Impact

5.9 Direct Central Impact: Two particles A and B of mass mAand mBare moving in a straight line with velocities vA& vB.

If vA>vBthe particle A strikes B.

Under the impact, they deform and at the end of period of deformation they will move with the

same velocity u.

After the impact the particles may gain their original shape or are permanently deformed,

depending upon the magnitude of impact and material involved which is called restitution.

After the impact and separation the particles move with '' and BA vv velocities.

The duration of time of impact when the particles comes under the deformation and restitution

during impact is called deformation period and restitution period respectively.

U

VA VB

A A AB BB

VA VB

Considering that only impulsive forces are acting, the total momentum of the system is conserved.

i.e. ( )ivmvmvmvm BBAABBAA +=+''

In scalar form,

( )iivmvmvmvm BBAABBAA +=+''

+ve value is for +ve axis and ve value is for ve axis.

Analysis during Impact

Following phenomena will occur for the particle A.

( )

)(' ivvmRdtum

iiiumpdtvm

AAA

AAA

=

=

where, dtRdtp and are the impulses during the period of deformation and restitution respectively.Then the co-efficient of restitution is defined as:

( )vPdt

Rdte =


47/92


Value of e depends upon

Materials of particles

Impact velocity

Shape & size of colliding bodies

Generally, 0


48/92



( ) ( )2'2'222

1

2

1

2

1

2

1, BBAABBAA vmvmvmvmor +=+

i.e. Initial KE of system = Final KE of system

Hence, for perfectly elastic condition, KE of the system is conserved.

When e, there is loss of KE and this lost energy is converted into heat, sound and other forms ofenergy.

5.10 Oblique Central Impact:When the velocities of the two colliding bodies are not directed along the line of impact, then it

is called oblique impact as shown in the figure.

V1

V1m1

m2V2

V2

m1

m2

1

1

1

2

XLine of contact

Here, line of impact is along y-axis and line of contact is along x-axis. Then the following

phenomena occur.

(a) x-component of the momentum of the particle 1stis conserved

i.e. ( )ixvxvvmvm xx =='

11

'

1111

(b) x-component of the momentum of 2ndparticle is conserved

i.e. ( )iivxvvmvm xxx =='

22

'

2222

From (a) and (b) [ ] [ ]'2'121 xx vvxvxv = (c) From (a) and (b), the total momentum of the particles in x-direction is also conserved

i.e. ( )iiivmvmvmvm xxxx +=+'

22

'

112211

(d) y-component of total momentum of the particle is conserved.

i.e. ( )ivvmvmvmvm yyyy =+'

22

'

112211

(e) y-component of relative velocity after impact is obtained by multiplying y-component of relative

velocity before impact by co-efficient of restitution.

i.e.( ) ( )

( ) ( ) ( )vvvevvvvevv

yyyy

yyyy

+=+

=

21

'

1

'

2

21

'

1

'

2

The above five equation are applied for the analysis of the problems related to oblique impact.

Remember:

(a)Along the line of contact, momentum of each particle is conserved.

(b)Along the line of impact, the total momentum of particles is conserved.

Tutorials:


49/92


200mm

1Datum

150mm

2

0.5kg

0.4kg

B

A

M1

M2

V

30o

(a) A 10 kg collar slides without friction along a vertical rod. The spring attached to the collar has an

undeformed length of 100 mm and a constant of 500 N/m. If the collar is released from rest in

position 1, determine its velocity after it has moved 150mm to position2.

Solution:

Given, K = 500 N/m

Undeformed length of spring = 100mm = 0.1m

We have from conservation of energy, KE + PE at 1 = KE +PE at 2

i.e. )(2211 iVTVT +=+

[ ]

( )

NmV

V

kxVVV

VT

ge

5.2

1.05002

1

2

1

00

1

2

1

2

11

11

11

=

=

=+=

==

Again,

( ) ( )

Nmv

wykx

vvv

vvY

ge

09.9

15.081.91015.05002

1

2

1

5102

1

2

2

2

2

2

2

2

2

22

22

=

+=

+=

+=

==

Putting all the values in equ(1), we get

smvv /52.109.955.20 22

2 ==+

2) A particle having mass 0.5 kg is released from rest and strikes. The stationary particle of mass 0.4

kg as shown in the figure. Assume the impact is direct and elastic. If the horizontal surface has a

dynamic co-efficient of friction 3.0= , locate the final position of each mass from the origin of the

axis.

Solution:

Applying conservation of energy at Pt. A & B

Lost of energy = work done against friction

Now for mass m1,

KE at A (TA1) = 0 [vA=0]

PE at A (VA1) = mghA= 0.59.81(0.25-0.25sin30)

(VA1) = 0.613 J

KE at B (TB1) =2

1

2

1

2

1 25.025.02

1

2

1BBB vvmv ==

PE at B (VB1) = 0 [B is datum]Now, from conservation of energy,

datumat0

1.01.02.0

1atspringofelongation

1

2

1

1

=

==

=

gv

x

x

At point 2, the total length of spring is

( ) ( )

15.01.025.0

25.015.02.0

2

22

==

=+

x


50/92



smvvor

VTVT

BB

BBAA

/56.1025.0613.00,1

1111

2 =+=+

+=+

sm /56.1vBpt.atMatVelocity1B1

=

Now, at the point of impact

Velocity of m1before impact (v1) =1B

v = 1.56 m/s

Velocity of m1after impact ( ) '1'1 vv = Velocity of m2before impact (v2) = 0

Velocity of m2after impact'

2

'

2 )( vv =

Now, for direct impact

( )ivvor

vvor

vmvmvmvm

=+

+=+

+=+

78.04.051.0,

4.05.0056.15.0,

2

'

1

'

2

'

1

'

22

'

112211

Again, we have:

( )iivv

vv

vv

vve

=

==

=

56.1

056.11

'

1

'

2

'

1

'

2

21

'

1

'

2

Solving equ(i) and (ii), we get:

smv

smv

/733.1

/173.0

'

2

'

1

=

=

Now, using work energy relation to find the distance travelled by the particle

For M1:

Work done due to friction = ( )iiiVT +

(Energy lost due to friction)

( ) J

T

00748.0173.05.02

10

KEInitial-KEFinal

2

1

==

=

[ ]001 == hV

work done due to friction for mass m1 = JVT 00748.0=+

For M2:

( )

JVT

V

JT

60065.0

0

60065.0733.14.02

10

2

2

2

=+

=

==

Now,

Work done due to friction = travelleddistance mg

mg

frictiontoduedonework-(x)travelleddistance =

For mass 1,


51/92


line of contact

VA= 9m/s VB= 9m/s

m m

A B

30

o60

o

( )

( )

mx

x

mx

mx

510.0

81.94.03.0

6065.0

00508.0

00508.081.95.03.0

00748.0

2

2

1

1

=

=

=

=

=

3. The magnitude and direction of the velocities of two identical frictionless balls before they strike

each other as shown. Assuming e=0.90 determine the magnitude and direction of the velocity of each

ball after the impact.

Solution:

smVV

smVV

smVV

smVV

BBy

BBx

AAy

AAx

/6.1060sin

/1.660cos

/5.430sin

/8.730cos

==

==

==

==

Now, in oblique impact

For motion along the line of contact,

smVVAnd

smVV

ByBy

AyAy

/6.10,

/5.4

'

'

==

==

For motion along the line of impact,

[ ]

)(7.1

1.68.7''

''

''

''

iVV

VV

VVVV

mmVmVmVmVm

BxAx

BxAx

BxAxBxAx

BABxBAxABxBAxA

=+

+=

+=+

=+=+

Again,AxBx

AxBx

VV

VVe

=

''

( )

( )25.12

5.12)1.6(8.790.0

''

''

=

==

AxBx

AxBx

VV

VVQ

Solving (1) and (2), we get:

4.5&1.7

'

== VVBx

Resultant Motion:

Adding components of velocities after impact, we get:

( ) ( )

( ) ( ) smVVV

smVVV

AyAxB

AyAxA

/8.12

/7

2'2''

2'2''

=+=

=+=


52/92


53/92


For all n particles the equation (i) can be written as:

( )=

===

+=n

i

n

jij

ij

n

i

i

n

i

ii iiFF

dt

rdm

1 1112

2

For considering all particles the summation of internal forces is zero. Hence, equ(ii) modifies as :

==

==

FF

dt

rdmF

n

i

i

n

i

ii

11 2

2

Q

( )iiirmdt

dFor

n

i

ii = =1

2

2

,

If, crbe position vector of mass centre of system of particles and M is the total mass of particles,

then from principle of first moment of inertia (moment due to entire mass = sum of moments due to

individual mass),

( )ivrMrM iic =

From (iii) and (iv), we get

( )

( )vdt

rdMF

rMdt

dF

c

c

=

=

2

2

2

2

6.2Linear and Angular Momentum for a system of particles(1) Linear Momentum for a system of particles:

For a system of particles, applying Newtons 2ndlaw to any jthparticle, we have:

( )= =n

j

jj idt

vdMF

1

Multiplying (i) by dt and integrating from t1to t2, we get:

( )iiVMVMIdtFt

n

j

jj

t

n

j

jjext

t

t

==

==12

2

1 11

This shows, The impulse of the total external force on the system of particles during a time interval

equals to the sum of the changes of the linear momentum vector of the particles during the same time

interval.

From the concept of mass center,

( )=

=n

j

jjc iiirMrM1

Differentiating with respect to time, we get

( )=

=n

j

jjc ivVMVM1

From (ii) and (iv),

( )vVMVMdtFIt

tccext == 1

2

12 )()(

Thus, the total external impulse of a system particles is equal to the change in linear momentum of

the particles, moving with the mass center velocity.


54/92



z

x

y

fij

o

oo

o

o

o

M

r1 rc

2. Angular Momentum For a system of particles:

Angular Momentum of ithparticle,

( )

( ) iiiiiii

iiiii

marPrdt

drvm

Prdt

drvm

===

==

&

For system of particles the angular momentum equation for the

ithparticle about origin O is given by:

[ ] ( )iPrdt

dFrFr ii

n

j

ijiii =

+

=1

where, particleofmomentumLinear=iP

For the system of n particles, equn(i) becomes:

( )

==+

====

n

i

oii

n

i

n

jj

iji

n

i

ii Fr

dt

dFrFr

11111

&

Since, internal force vanishes for all particles, the moment also become zero for all particles. Hence,

[ ] ( )iiMMFr ooon

i

iio === =

&&1

Similarly for any other fixed point A,

( )iiiM AA =&

Thus, the total moment of external forces acting on an aggregate of particles about a fixed point A

in an inertial reference equals the rate of change of the angular momentum relative to the same point

A and same inertial reference.

Again, considering center of mass of the aggregate of particles

For ithparticle, [ CGw.r.t.iofpositionci ]

( )ivrr cici +=

Now, the angular momentum for aggregate particle about O is then,

[ ]momentumlinearppr in

i

iio == =

Q

1

( ) ( ){ }[ ] ( ) += =++= = ciciiiiii

n

i

ciiicico

rrvmvmpvvmvor

&&&

&&&&

,1

Carrying out the cross-product and extracting cr from the summation we get:

[ ]

( ) ( ) ( ) ( )

====

=

+++=

+++=

n

i

ciicic

n

i

cii

n

i

ciic

n

i

cic

n

i

ciiciciciciiccico

mrmmrrmr

mrmmrrmr

1111

1

&&&&

&&&

We know that the sum of the first moment of mass about the centroid is zero i.e.

Thenmm ciicii ,0henceand0 == &


55/92


( ) [ ] =

=+=n

i

iciicicco MmmrMr1

&&

( ) ciicucccco mvrMr &Q& =+=

where, c be the sum of the angular momentum about the center of mass .Similarly for any point

A, we get:

( )virMr cACACA += &

where, ACr& is the velocity of center of mass relative to fixed point A = ACV

Now, differentiating equn(vi) with respect to time, we get:

[ ] ( ) ];[, ACACAAcACACAcACACA

aVMviiaMrMor

VMr

==+=

+=

&&

&&&

6.3Motion of mass center of a system of particle:Center of mass for particles is not the center of mass of system.

We know that (from 6.0) for system of particles

( ) ( )

( ) ( )

( ) ( )

=

=

=

zextz

yexty

xextx

maF

maF

maF

(i)

If mass center of the system of particles is considered with co-ordinates ( ),,, zyxG Then we have:

( ) ( )

( ) ( )( ) ( )

=

=

=

mzzm

myym

mxxm

(ii)

Differentiating equ(ii) twice with respect to time, we get:

( ) ( )( ) ( )( ) ( )

=

=

=

zmzm

ymym

xmxm

&&&&

&&&&

&&&&

( ) ( )( ) ( )( ) ( )

=

=

=

zz

yy

xx

maam

maam

maam

- (iii)

where, zyx aaa &, are the components of acceleration a of G (i.e. center of mass) of the system,

From equ(i) & equ(iii), we have:( ) ( )( ) ( )

( ) ( ) zextzyexty

xextx

amF

amF

amF

=

=

=

- (iv)

which defines the motion of center of mass of system. It shows that the center of a system of

particles move as if the entire mass of the system and all the external forces were concentrated at that

point G.

6.4 Conservation of Momentum:We know that the final momentum of the particle is obtained by adding vectorically its initial

momentum and the impulse of the force Fduring the time interval considered i.e.

i.e. component of moment due to entire mass

= component of sum of the moment due to

individual mass

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App Mechanics Complete1