chapter 1 part1

1

EME1076

Applied Dynamics

Trimester 2, 2014/15

2

Topic 1 (Chapter 12):

Kinematics of a particle

Part I

EME1076 Applied Dynamics

Trimester2, 2014/15

This lecture note is taken and modified from ‘Engineering Mechanics-Dynamics, R.C.Hibbeler, Prentice Hall which

copyright belongs to Pearson Education South Asia Pte Ltd.

3

Chapter Objectives • To introduce the concepts of position, displacement,

velocity, and acceleration.

• To study particle motion along a straight line and

represent this motion graphically.

• To investigate particle motion along a curved path

using different coordinate systems.

• To present an analysis of dependent motion of two

particles.

• To examine the principles of relative motion of two

particles using translating axes.

4

Chapter Outline 1.1 Introduction

1.2 Rectilinear Kinematics : Continuous Motion

1.3 Rectilinear Kinematics : Erratic Motion

1.4 General Curvilinear Motion

1.5 Curvilinear Motion : Rectangular Components

1.6 Motion of a Projectile

1.7 Curvilinear Motion : Normal and Tangential Components

1.8 Curvilinear Motion : Cylindrical Components

1.9 Absolute Dependent Motion Analysis of Two Particles

1.10 Relative Motion Analysis of Two Particles Using

Translating Axes

5

1.1 Introduction

• Mechanics – the state of rest or motion of

bodies subjected to the action of forces

• Static – equilibrium of a body that is either at

rest or moves with constant velocity

• Dynamics – deals with accelerated motion of a

body

1) Kinematics – treats with geometric aspects of

the motion

2) Kinetics – analysis of the forces causing the

motion

6

1.2 Rectilinear Kinematics: Continuous Motion

• Rectilinear – straight-line path

• Kinematics – characterized by specifying, at any

given instant(t), the particle’s position(s),

velocity(v), and acceleration(a).

7

Position

1) Single coordinate axis, s

2) Origin, O is the fixed point on the path

3) Magnitude of s = distance from O to P

4) The sense of direction is defined by algebraic sign

on s :

=> +ve = right of origin, -ve = left of origin


8

Displacement – change in its position, vector quantity

If particle moves from P to P’ ,

displacement :

is +ve if particle’s position is right of its initial

position

is -ve if particle’s position is left of its initial

position

**total distance travel is a +ve scalar, sT =total

length of path over which the particle travels.

sss

s

s


9

• Velocity

Average velocity :

Instantaneous velocity is defined as :

t

sv

avg

tsvt

ins

/lim

0

dt

dsv

ins


10

Representing as an algebraic scalar,

Velocity is +ve = particle moving to the right

Velocity is –ve = Particle moving to the left

Magnitude of velocity is the speed (m/s)

insv

dt

dsv


11

Average speed is defined as total distance traveled by a particle, sT, divided by the elapsed time .

The particle travels along the path of length sT in time

=>

t

t

sv T

avgsp

t

t

sv T

avgsp


12

• Acceleration – velocity of particle is known at

points P and P’ during time interval Δt, average

acceleration is

• Δv represents difference in the velocity during the

time interval Δt, ie

t

vaavg

vvv '

t

vaavg


13

Instantaneous acceleration at time t is found by

taking smaller and smaller values of Δt and

corresponding smaller and smaller values of Δv,

tvat

/lim0

2

2

dt

sda

dt

dva


14

• Particle is slowing down, its speed is decreasing

=> decelerating => will be negative

• Consequently, a will also be negative, therefore it

will act to the left, in the opposite sense to v

• If velocity is constant,

acceleration is zero

vvv '


15

• Velocity as a Function of Time

Integrate ac = dv/dt,

assuming that initially v = v0 when t = 0.

t

c

v

vdtadv

00

tavv c 0

Constant Acceleration


16

• Position as a Function of Time

Integrate v = ds/dt = v0 + act, assuming that

initially s = s0 when t = 0

200

00

2

1

0

tatvss

dttavds

c

t

c

s

s



17

• Velocity as a Function of Position

Integrate v dv = ac ds, assuming that initially v =

v0 at s = s0

0

2

0

2 2

00

ssavv

dsavdv

c

s

sc

v

v

v= ds/dt

dt=ds/v

a=dv/dt

a=dv/(ds/v)

a ds= vdv



18

The car moves in a straight line such that for a

short time its velocity is defined by :

v = (0.9t2 + 0.6t) m/s where t is in sec.

Given when t = 0, s = 0. Determine it position

and acceleration when t = 3s.

EXAMPLE 12.1

19

Solution:

Coordinate System. The position coordinate

extends from the fixed origin O to the car,

positive to the right.

Position. Since v = f(t), the car’s position can be

determined from v = ds/dt, since this equation

relates v, s and t. Noting that s = 0 when t = 0, we

have

ttdt

dsv 6.09.0 2

EXAMPLE 12.1

20

23

0

23

0

0

2

0

3.03.0

3.03.0

6.09.0

tts

tts

dtttds

ts

ts

When t = 3s,

s = 10.8m

EXAMPLE 12.1

21

Acceleration.

Knowing v = f(t), the acceleration is determined by

a = dv/dt, since this equation relates a, v and t.

6.08.1

6.09.0 2

t

ttdt

d

dt

dva

When t = 3s,

a = 6m/s2

EXAMPLE 12.1

22

A small projectile is forced downward into a fluid medium

with an initial velocity of 60m/s.

Due to the resistance of the fluid the projectile experiences

a deceleration equal to a =(-0.4v3)m/s2, where v is in

m/s.

Determine the projectile’s velocity and position 4s after it is

fired.

EXAMPLE 12.2

23

Solution:

Coordinate System. Since the motion is downward,

the position coordinate is downwards positive,

with the origin located at O.

Velocity. Here a = f(v), velocity is a function of

time using a = dv/dt, since this equation relates

v, a and t. 34.0 v

dt

dva

EXAMPLE 12.2

24

smtv

tv

dtv

dtv

dv

tv

tv

sm

/8.060

1

60

11

8.0

1

1

2

1

4.0

1

4.0

2/1

2

22

0602

0/60 3

When t = 4s,

v = 0.559 m/s

EXAMPLE 12.2

34.0 vdt

dv

25

Position. Since v = f(t), the projectile’s position can

be determined from v = ds/dt, since this equation

relates v, s and t. Noting that s = 0 when t = 0, we

have

2/1

28.0

60

1

t

dt

dsv

t

ts

ts

dttds

0

2/1

2

0

2/1

20

8.060

1

8.0

2

8.060

1

EXAMPLE 12.2

26

When t = 4s,

s = 4.43m

mts

60

18.0

60

1

4.0

12/1

2

EXAMPLE 12.2

** Homework : try to use : a ds= vdv to find s.

27

A rocket travel upward at 75

m/s. When it is 40 m from

the ground, the engine fails.

Determine max height sB

reached by the rocket and

its speed just before it hits

the ground.

EXAMPLE 12.3

28

Solution:

Coordinate System. Origin O for the position

coordinate at ground level with positive upward.

Maximum Height. Rocket traveling upward, vA =

+75m/s when t = 0. s = sB when vB = 0 at max ht. For

entire motion, acceleration aC = -9.81m/s2 (negative

since it act opposite sense to positive velocity or

positive displacement)

EXAMPLE 12.3

29

)(222

ABCAB ssavv

sB = 327 m

Velocity.

2 2 2 ( )

80.1 / 80.1 /

C B C C B

C

v v a s s

v m s m s

The negative root was chosen since the rocket is

moving downward

EXAMPLE 12.3

30

A particle moves along a horizontal path with a velocity

of v = (3t2 – 6t) m/s. if it is initially located at the origin O,

determine the distance traveled in 3.5s and the particle’s

average velocity and speed during the time interval.

EXAMPLE 12.5

31

Solution:

Coordinate System. Assuming positive motion to

the right, measured from the origin, O

Distance traveled. Since v = f(t), the position as a

function of time may be found integrating v = ds/dt

with t = 0, s = 0.

EXAMPLE 12.5

32

mtts

tdtdttds

dttt

vdtds

s tt

23

0 00

2

2

3

63

63

0 ≤ t < 2 s

-> -ve velocity -> the particle is moving to the left,

t > 2s

-> +ve velocity -> the particle is moving to the right

msst

125.65.3

ms

st0.4

2

0

0

ts

EXAMPLE 12.5

33

The distance traveled in 3.5s is

sT = 4.0 + 4.0 + 6.125 = 14.125m

Velocity. The displacement from t = 0 to t = 3.5s is Δs

= 6.125 – 0 = 6.125m

And so the average velocity is

sm

t

sv

avg/75.1

05.3

125.6

Average speed, smt

sv T

avgsp /04.405.3

125.14

EXAMPLE 12.5

34

1.3 Rectilinear Kinematics: Erratic Motion

• When particle’s motion is erratic, it is best described

graphically using a series of curves that can be generated

experimentally from computer output.

• a graph can be established describing the relationship with

any two of the variables, a, v, s, t

• using the kinematics equations

•a = dv/dt,

•v = ds/dt,

•a ds = v dv

35

Given the s-t Graph, construct the v-t Graph

•The s-t graph can be plotted if the position of the

particle can be determined experimentally during a

period of time t.

•To determine the particle’s velocity as a function of

time, the v-t Graph, use v = ds/dt

•Velocity as any instant is determined by

measuring the slope of the s-t graph


36

vdt

ds

Slope of s-t graph =

velocity


37

Given the v-t Graph, construct the a-t Graph

•When the particle’s v-t graph is known, the

acceleration as a function of time, the a-t graph can

be determined using a = dv/dt

•Acceleration as any instant is determined by

measuring the slope of the v-t graph


38

adt

dv

Slope of v-t graph =

acceleration


39

• Since differentiation reduces a polynomial of

degree n to that of degree n-1, then if the s-t graph

is parabolic (2nd degree curve), the v-t graph will be

sloping line (1st degree curve), and the a-t graph

will be a constant or horizontal line (zero degree

curve)


40

EXAMPLE 12.6

A bicycle moves along a

straight road such that it

position is described by

the graph as shown.

Construct the v-t and a-t

graphs for 0 ≤ t ≤ 30s.

41

Solution:

v-t Graph. The v-t graph can be determined by

differentiating the eqns defining the s-t graph

6306;3010

6.03.0;100 2

dt

dsvtssts

tdt

dsvtsst

The results are plotted.

EXAMPLE 12.6

42

We obtain specify values of v by measuring the

slope of the s-t graph at a given time instant.

smt

sv /6

1030

30150

a-t Graph. The a-t graph can be determined by

differentiating the eqns defining the lines of the

v-t graph.

EXAMPLE 12.6

43

06;3010

6.06.0;100

dt

dvavst

dt

dvatvst

The results are plotted.

EXAMPLE 12.6

44

Given the a-t Graph, construct the v-t Graph

• When the a-t graph is known, the v-t graph may

be constructed using a = dv/dt

dtav

Change in

velocity Area under

a-t graph =


45

• Knowing particle’s initial velocity

v0, and add to this small increments

of area (Δv)

• Successive points v1 = v0 + Δv, for

the v-t graph

• Each eqn for each segment of the

a-t graph may be integrated to yield

eqns for corresponding segments

of the v-t graph


46

Given the v-t Graph, construct the s-t Graph

• When the v-t graph is known, the s-t graph may

be constructed using v = ds/dt

dtvs

Displacement Area under

v-t graph =


47

• knowing the initial position s0,

and add to this area increments

Δs determined from v-t graph.

• describe each of there

segments of the v-t graph by a

series of eqns, each of these

eqns may be integrated to yield

eqns that describe

corresponding segments of the

s-t graph


48

EXAMPLE 12.7

A test car starts from rest

and travels along a

straight track such that it

accelerates at a constant

rate for 10 s and then

decelerates at a constant

rate. Draw the v-t and s-t

graphs and determine the

time t’ needed to stop the

car. How far has the car

traveled?

49

Solution:

v-t Graph. The v-t graph can be determined by

integrating the straight-line segments of the a-t

graph. Using initial condition v = 0 when t = 0,

tvdtdvasttv

10,10;1010000

When t = 10s, v = 100m/s, using this as initial

condition for the next time period, we have

1202,2;2;1010100

tvdtdvattstv

EXAMPLE 12.7

50

When t = t’ we require v = 0. This yield t’ = 60 s

s-t Graph. Integrating the eqns of the v-t graph

yields the corresponding eqns of the s-t graph.

Using the initial conditions s = 0 when t = 0,

2

005,10;10;100 tsdttdstvst

ts

1202,2;2;1010100

tvdtdvattstv

EXAMPLE 12.7

51

When t’ = 60s, the position is s = 3000m

When t = 10s, s = 500m. Using this initial condition,

600120

1202;1202;6010

2

10500

tts

dttdstvststs

EXAMPLE 12.7

52

Given the a-s Graph, construct the v-s Graph

• v-s graph can be determined by using v dv = a ds,

integrating this eqn between the limit v = v0 at s =

s0 and v = v1 at s = s1

1

0

2

0

2

12

1 s

sdsavv

Area under

a-s graph


53

• determine the eqns which define the segments of

the a-s graph

• corresponding eqns defining the segments of the

v-s graph can be obtained from integration, using

vdv = a ds


54

Given the v-s Graph, construct the a-s Graph

• v-s graph is known, the acceleration a at any

position s can be determined using a ds = v dv

ds

dvva

Acceleration = velocity times slope of v-s graph


55

• At any point (s,v), the slope dv/ds of the v-s graph

is measured

• Since v and dv/ds are known, the value of a can

be calculated


56

EXAMPLE 12.8

The v-s graph describing the motion of a motorcycle

is shown in Fig 12-15a. Construct the a-s graph of

the motion and determine the time needed for the

motorcycle to reach the position s = 120 m.

57

Solution:

a-s Graph. Since the eqns for the segments of the

v-s graph are given, a-s graph can be determined

using a ds = v dv.

0

;15;12060

6.004.0

32.0;600

ds

dvva

vmsm

sds

dvva

svms

EXAMPLE 12.8

58

0

0 60 m; 0.2 3; ;0.2 3

0.2 3

5ln(0.2 3) 5ln 3

t s

o

ds ds dss v s v dt

dt v

dsdt

s

t s

At s = 60 m, t = 8.05 s

Time. The time can be obtained using v-s

graph and v = ds/dt. For the first segment of

motion, s = 0 at t = 0,

EXAMPLE 12.8

59

For second segment of motion,

05.415

15

15;15;12060

6005.8

s

t

dsdt

ds

v

dsdtvms

st

At s = 120 m, t = 12.0 s

EXAMPLE 12.8

60


Curvilinear motion occurs when the particle

moves along a curved path

Position. The position of the particle,

measured from a fixed point O, is designated by

the position vector r = r(t).

61

Displacement. Suppose during a small time interval

Δt the particle moves a distance Δs along the curve

to a new position P`, defined by r` = r + Δr. The

displacement Δr represents the change in the

particle’s position.


62

Velocity. During the time Δt, the average velocity

of the particle is defined as :

tavg

rv

The instantaneous velocity is determined from

this equation by letting Δt 0, and consequently

the direction of Δr approaches the tangent to the

curve at point P. Hence,

dt

dins

rv


Direction of vins is tangent to the curve

63

Magnitude of vins is the speed, which is obtained

by realizing the ength of the straight line segment

Δr approaches the arc length Δs as Δt 0.

Hence,

dt

dsv


64

Acceleration. If the particle has a velocity v at time t

and a velocity v` = v + Δv at time t` = t + Δt.

The average acceleration during the time interval

Δt is :

tavg

va

2

2

dt

d

dt

d rva


65

a acts tangent to the hodograph, therefore it is

not tangent to the path


66

1.5 Curvilinear Motion: Rectangular Components

Position. Position vector is defined by

r = xi + yj + zk

The magnitude of r is always positive and defined

as : 222 zyxr

The direction of r is

specified by the

components of the unit

vector ur = r/r

67

Velocity.

zvyvxv

kvjvivdt

drv

zyx

zyx

where

The velocity has a magnitude

defined as the positive value of

222

zyx vvvv

and a direction that is specified by the components

of the unit vector uv=v/v and is always tangent to

the path.


68

Acceleration.

zva

yva

xva

kajaiadt

dva

zz

yy

xx

zyx

The acceleration has a magnitude defined as the

positive value of : 222

zyx aaaa

where


69

• The acceleration has a direction specified by

the components of the unit vector ua = a/a.

• Since a represents the time rate of change in

velocity, a will not be tangent to the path.


70

EXAMPLE 12.9

At any instant the horizontal

position of the weather balloon

is defined by x = (9t) m, where

t is in second. If the equation

of the path is y = x2/30,

determine the distance of the

balloon from the station at A,

the magnitude and direction of

the both the velocity and

acceleration when t = 2 s.

71

Solution:

Position. When t = 2 s, x = 9(2) m = 18 m and

y = (18)2/30 = 10.8 m

The straight-line distance from A to B is

218.101822r m

Velocity.

smxdt

dyv

smtdt

dxv

y

x

/8.1030/

/99

2

EXAMPLE 12.9

72

When t = 2 s, the magnitude of velocity is

smv /1.148.10922

The direction is tangent to the path, where

2.50tan 1

x

y

vv

v

Acceleration.

2/4.5

0

smva

va

yy

xx

EXAMPLE 12.9

73

222/4.54.50 sma

The direction of a is

900

4.5tan 1

a

EXAMPLE 12.9

74

The motion of box B is defined

by the position vector r =

{0.5sin(2t)i + 0.5cos(2t)j –

0.2tk} m, where t is in seconds

and the arguments for sine and

cosine are in radians (π rad =

180°). Determine the location

of box when t = 0.75 s and the

magnitude of its velocity and

acceleration at this instant.

EXAMPLE 12.10

75

Solution:

Position. Evaluating r when t = 0.75 s yields

mkjradiradr st })75.0(2.0)5.1cos(5.0)5.1sin(5.0{75.0

mkji }150.00354.0499.0{

The distance of the box from the origin is

mr 522.0)150.0()0354.0()499.0( 222

EXAMPLE 12.10

76

The direction of r is obtained from the

components of the unit vector,

107

1.86

2.17)955.0(cos

287.00678.0955.0

522.0

150.0

522.0

0352.0

522.0

499.0

1

kji

kjir

rur

EXAMPLE 12.10

77

Velocity.

smkjtitdt

rdv /}2.0)2sin(1)2cos(1{

Hence at t = 0.75 s, the magnitude of velocity, is

smvvvv zyx /02.1222

Acceleration. The acceleration is not tangent to

the path. 2/})2cos(2)2sin(2{ smjtit

dt

vda

At t = 0.75 s, a = 2 m/s2

EXAMPLE 12.10

78

• the free-flight motion of a projectile is often

studied in terms of rectangular components

• consider a projectile launched at (x0, y0), with

initial velocity vo having components (vo)x , (vo)y

• path defined in the x-y plane

• air resistance neglected

• only force acting on the projectile is its weight,

resulting in constant downwards acceleration,

ac = g = 9.81 m/s2


79


80

Horizontal Motion Since ax = 0,

);(2

;2

1

;

0

2

0

2

2

00

0

ssavv

tatvxx

tavv

c

c

c

xx

x

xx

vv

tvxx

vv

)(

)(

)(

0

00

0

Horizontal component of velocity, (vo)x remain

constant during the motion


81

Vertical. Positive y axis is directed upward,

then ay = - g

);(2

;2

1

;

0

2

0

2

2

00

0

yyavv

tatvyy

tavv

c

c

c

)(2)()(

2

1)(

)(

0

2

0

2

2

00

0

yygvv

gttvyy

gtvv

yy

y

yy


82

• Problems involving the motion of a projectile

have at most three unknowns since only three

independent equations can be written :

- one in the horizontal direction

- two in the vertical direction

• Velocity in the horizontal and vertical direction

are used to obtain the resultant velocity

• Resultant velocity is always tangent to the path


83

EXAMPLE 12.11

A sack slides off the

ramp with a horizontal

velocity of 12 m/s. If the

height of the ramp is 6 m

from the floor, determine

the time needed for the

sack to strike the floor

and the range R where

the sacks begin to pile

up.

84

Coordinate System. Origin of the coordinates is

established at the beginning of the path, point A.

Initial velocity of a sack has components (vA)x = 12

m/s and (vA)y = 0

Acceleration between point A and B ay = -9.81 m/s2

Since (vB)x = (vA)x = 12 m/s, the three unknown are

(vB)y, R and the time of flight tAB

EXAMPLE 12.11

85

Vertical Motion. Vertical distance from A to B is

known

st

tatvyy

AB

ABcABy

11.1

2

1)( 2

00

The above calculations also indicate that if a sack

is released from rest at A, it would take the same

amount of time to strike the floor at C

EXAMPLE 12.11

Horizontal Motion.

mR

tvxx ABx

3.13

)( 00

86

The chipping machine is designed to eject wood at

chips vO = 7.5 m/s. If the tube is oriented at 30°

from the horizontal, determine how high, h, the

chips strike the pile if they land on the pile 6 m

from the tube.

EXAMPLE 12.12

87

Coordinate System. Three unknown h, time of

flight, tOA and the vertical component of velocity

(vB)y. Taking origin at O, for initial velocity of a chip,

(vA)x = (vO)x = 6.5 m/s and ay = -9.81 m/s2

smv

smv

yO

xO

/75.3)30sin5.7()(

/5.6)30cos5.7()(

EXAMPLE 12.12

88

Horizontal Motion.

st

tvxx

OA

OAxA

9231.0

)( 00

Vertical Motion.

Relating tOA to initial and final elevation of the chips,

mh

tatvyhy OAcOAyOA

38.1

2

1)(1.2 2

0

EXAMPLE 12.12

89

The track for this racing event

was designed so that the

riders jump off the slope at

30°, from a height of 1m.

During the race, it was

observed that the rider

remained in mid air for 1.5 s.

Determine the speed at which he was traveling off

the slope, the horizontal distance he travels before

striking the ground, and the maximum height he

attains. Neglect the size of the bike and rider.

EXAMPLE 12.13 (self-study)

90

Coordinate System. Origin is established at point A.

Three unknown are initial speed vA, range R and

the vertical component of velocity vB.

Vertical Motion. Since time of flight and the vertical

distance between the ends of the paths are known,

smv

tatvss

A

ABCAByAyAyB

/4.13

2

1)()()( 2


91

Horizontal Motion

•For maximum height h, we consider path AC

•Three unknown are time of flight, tAC, horizontal

distance from A to C and the height h

•At maximum height (vC)y = 0

m

R

tvss ABAxAxB

4.17

)5.1(30cos38.130

)()()(


92

•Since vA known, determine h using the following

equations

•Show that the bike will strike the ground at B with

velocity having components of

smvsmv

mh

h

ssavv

yBxB

yAyCcAc yy

/02.8)(,/6.11)(

28.3

]0)1)[(81.9(2)30sin38.13()0(

])()[(2)()(

22

22


93

• When the path of motion of a particle is

known, describe the path using n and t

coordinates which act normal and tangent to

the path

Planar Motion

• Consider particle P which is moving in a

plane along a fixed curve, such that at a

given instant it is at position s, measured

from point O

1.7 Curvilinear Motion :

Normal and Tangential Components

94

• Consider a coordinate

system that has origin at a

fixed point on the curve, and

at the instant, considered

this origin happen to coincide

with the location of the

particle

• t axis is tangent to the curve at P and is positive in

the direction of increasing s



95

• Designate this positive position direction with unit

vector ut

• For normal axis, note that geometrically, the curve

is constructed from series differential arc

segments

• Each segment ds is

formed from the arc of an

associated circle having

a radius of curvature ρ

(rho) and center of

curvature O’



96

• Normal axis n is perpendicular to the t axis and

is directed from P towards the center of

curvature O’

• Positive direction is always on the concave

side of the curve, designed by un

• Plane containing both the n and t axes is

known as the oscillating plane and is fixed on

the plane of motion



97

Velocity.

• Since the particle is moving, s is a function of time

• Particle’s velocity v has direction that is always

tangent to the path and a magnitude that is

determined by taking the time derivative of the

path function s = s(t)

sv

uvv t

where



98

Acceleration

• Acceleration of the particle is the time rate of

change of velocity

tt uvuvva



99

• As the particle moves along the arc ds in time dt, ut

preserves its magnitude of unity

• When particle changes direction, it becomes ut’

ut’ = ut + dut

• dut stretches between the arrowhead of ut and ut’, which

lie on an infinitesimal arc of radius ut = 1

nnnt uv

us

uu

ddudu tt )1(

nt udud

100

2

or

va

vdvdsava

uauaa

n

tt

nntt

• Magnitude of acceleration is the positive value

of 22nt aaa

where

and



101

Consider two special cases of motion

• If the particle moves along a straight line, then ρ → ∞ and

an = 0. Thus , we can conclude that the

tangential component of acceleration represents the time

rate of change in the magnitude of velocity.

• If the particle moves along the curve with a

constant speed, then and

vaa t

0 vat /2vaa n

2vava

uauaa

nt

nntt

or



102

• Normal component of acceleration represents the time

rate of change in the direction of the velocity. Since an

always acts towards the center of curvature, this component

is sometimes referred to as the centripetal acceleration

• As a result, a particle moving along the curved path will

have accelerations directed as shown



103

Three Dimensional Motion (space motion)

• If the particle is moving along a space curve, at a

given instant, t axis is uniquely specified however,

an infinite number of straight lines can be

constructed normal to tangent axis.



104

• For planar motion,

- choose positive n axis directed from P towards

path’s center of curvature O’

-The above axis also referred as principle normal

to curve

- ut and un are always perpendicular to one

another and lies in the osculating plane



105

• For spatial motion,

a third unit vector ub, defines a binormal axis b

which is perpendicular to ut and un

Three unit vectors are related by vector cross

product

ub = ut x un

where un is always on the concave side



106

EXAMPLE 12.14

When the skier reaches the

point A along the parabolic

path, he has a speed of 6m/s

which is increasing at 2m/s2.

Determine the direction of his

velocity and the direction and

magnitude of this

acceleration at this instant.

Neglect the size of the skier

in the calculation.

107

Coordinate System. Establish the origin of the n, t

axes at the fixed point A on the path and

determine the components of v and a along these

axes.

Velocity. The velocity is directed tangent to the

path. 2

10

1, 1

20 x

dyy x

dx

v make an angle of θ = tan-1(1) = 45° with the x

axis smvA /6

EXAMPLE 12.14

108

Acceleration. Determined from nt uvuva

)/( 2

mdxyd

dxdy28.28

/

])/(1[22

2/32

The acceleration becomes

2

2

/}273.12{ smuu

uv

uva

nt

ntA

EXAMPLE 12.14

109

5.57327.1

2tan

/37.2237.12

1

222

sma

Thus, 57.5° – 45 ° = 12.5 °

a = 2.37 m/s2

EXAMPLE 12.14

110

Race car C travels round the horizontal circular

track that has a radius of 90 m. If the car

increases its speed at a constant rate of 2.1 m/s2,

starting from rest, determine the time needed for it

to reach an acceleration of 2.4 m/s2. What is its

speed at this instant?

EXAMPLE 12.15

111

Coordinate System. The origin of the n and t axes

is coincident with the car at the instant. The t axis

is in the direction of the motion, and the positive n

axis is directed toward the center of the circle.

Acceleration. The magnitude of acceleration can

be related to its components using 22nt aaa

t

tavv ct

1.2

)(0

EXAMPLE 12.15

112

222

/049.0 smtv

an

Thus,

The time needed for the acceleration to reach

2.4m/s2 is 22nt aaa

Solving for t = 4.87 s

Velocity. The speed at time t = 4.87 s is

smtv /2.101.2

EXAMPLE 12.15

113

The boxes travels alone

the industrial conveyor. If

a box starts from rest at A

and increases its speed

such that at = (0.2t) m/s2,

determine the magnitude

of its acceleration when it

arrives at point B.

EXAMPLE 12.16

114

Coordinate System. The position of the box at any

instant is defined by s, from the fixed point A. The

acceleration is to be determined at B, so the origin

of the n, t axes is at this point.

Acceleration. Since vA= 0 when t = 0

2

00

1.0

2.0

2.0

tv

dttdv

tva

tv

t

(1)

(2)

EXAMPLE 12.16

115

The time needed for the box to reach point B can

be determined by realizing that the position of B is

sB = 3 + 2π(2)/4 = 6.142 m, since sA = 0 when t = 0

st

dttds

tdt

dsv

B

tB

690.5

1.0

1.0

0

2142.6

0

2

EXAMPLE 12.16

116

Substituting into eqn (1) and (2),

smv

smva

B

BtB

/238.3)69.5(1.0

/138.1)690.5(2.0)(

2

2

At B, ρB = 2 m,

22

/242.5)( smv

aB

BnB

2

22

/36.5

)242.5()138.1(

sm

aB

EXAMPLE 12.16

chapter 1 part1

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