Chapter 12 kinematics of a particle part-i

Course Instructor: Engr. Sajid Yasin
Department of Mechanical Technology
MNS UET Multan

Lecture Schedule
Lecture Timings
Tuesday (Theory)
Room F5
07:30 PM-:09:30 PM
Wednesday (Lab)
D2
06:30 PM-:09:30 PM

Textbooks
“Engineering Mechanics: Dynamics”
R.C. Hibbeler, 12th Edition
“Vector Mechanics for Engineers (Dynamics)”
Beer and Johnston (Latest Edition)
“Engineering Mechanics (Dynamics)”
J. L. Meriam (Latest Edition)

Method of Assessment
Type Marks
Mid-semester Exam 30
End-semester Exam 40
Quiz 10
Assignments 10
Attendance 10
Total 100
*75%Attendance Mandatory

Mechanics
• Mechanics : A branch of physical science which deals
with ( the states of rest or motion of ) bodies under
action of forces
• Mechanics can be divided into 3 branches:
- Rigid-body Mechanics
- Deformable-body Mechanics
- Fluid Mechanics
• Rigid-body Mechanics deals with
- Statics
- Dynamics

Mechanics
• Statics – Equilibrium of bodies
 At rest
 Move with constant velocity
• Dynamics – Accelerated motion of bodies

Application of Mechanics
Mechanics
Statics
Dynamics
Mech of Materials
Fluid Mechanics
Vibration
Fracture Mechanics
Etc.
Structures
Automotives
Robotics
Spacecrafts
MEMs
Etc.

Space: Collection of points whose relative positions
can be described using “a coordinate system”
Time : Relative occurrence of events
or Measure of succession of events
Mass Measure of inertia of a body
(Its resistance to change invelocity)
position,
velocity,
acceleration
r
r
1.2 Fundamentals Concepts

Force: Vector quantity
 An agent which produces or tends to produce
motion, destroy or tends to destroy the motion of
the body.
 A force acting on the body may
a) Change the motion of the body b) Retard the motion of the body
c) Balance the forces already acting d) Give rise to internal stresses
 To determine the effect of forces acting on the body
i. Magnitude of the force
ii. Line of the action of the force
iii. Nature of the force i.e. Push or Pull
iv. Point of the application of force
• The SI unit = newton ( N= (kg·m/s2)

Idealizations
 Concentrated Force: Effect of a loading which is
assumed to act at a point (CG) on a body.
• Provided that, the area over which the load is applied is
very small compared to the overall size of the body.
Example:
Contact Force between
a wheel
• and ground.
40 kN
160 kN

Particle: Body of negligible dimensions
Rigid body: Body with negligible deformations
Non-rigid body: Body which can deform
In Statics, bodies are considered rigid unless stated otherwise.
Fundamentals Concepts

Point: Exact dimension, No size, Only Position
Line: No Thickness, Extends in both directions infinitely
Ray: Line with just one end
Line Segment: Line with both ends
○
Fundamentals Concepts

NEWTON’S LAWS OF MOTION (1st Law)
The study of rigid body mechanics is
formulated on the basis of Newton’s laws
of motion.

v v
0F
First Law:
An object at rest tends to stay at rest and an object in
motion tends to stay in motion with the same speed and in the
same direction, unless acted upon by an unbalanced force.

NEWTON’S LAWS OF MOTION (2nd Law)
Second Law:
The acceleration of an object as produced by a net force is
directly proportional to the magnitude of the net force, in the
same direction as the net force, and inversely proportional to
the mass of the object.
m
F

a
r
amF



NEWTON’S LAWS OF MOTION (3rd Law)
Third Law:
The mutual forces of action and reaction between
two particles are equal in magnitude, opposite in
direction, and collinear.
F
r
F
r
F
r
F
r
Confusing? Concept of FBD (Free Body Diagram)
Point: Isolate the body
Forces always occur in pairs – equal and
opposite action-reaction force pairs.

Newton’s Law of Gravitation
2
r
GMm
F 
- M & m are particle masses
- G is the universal constant of gravitation,
6.673 x 10-11 m3/kg-s2
- r is the distance between the particles.
For Gravity on earth (at sea level)
where
- m is the mass of the body in question
- g = GM/R2 = 9.81 m/s2 (32.2 ft/s2)
m
M
W=mg

v r
W mg
M
m
r
F

Units of Measurement
W  mg
F  ma 1 N=kg.m/s2
1 N=kg.m/s2
1 newton is the force required
to give a mass of 1 kg an
acceleration of 1 m/s2
Quantities Dimensional Symbol SI Units Symbols
Mass M Kilogram Kg
Length L Meter m
Time T Second s
Force F newton N

The International System of Units
Prefixes
• For a very large or small numerical quantity, units can
be modified by using a prefix
• Each represent a multiple or sub-multiple of a unit

Significant Figures
 In any measurement, the accurately known digits
and the first doubtful digit are called significant.
• Nonzero digits are always significant e.g. 38.57 (4) 283 (3)
• Zeroes are sometimes significant and sometimes not
– Zeroes at the beginning: never significant 0.052 (2)
– Zeroes between: always 6.08 (3)
– Zeroes at the end after decimal: always 39.0 (3)
– Zeroes at the end with no decimal may or may not:
– 8,000 kg (three, four, five) depending on the accuracy
of measuring instrument )

Rounding Off Numbers
 The process of removing insignificant figures from
measured value till last significant figure to be retained.
a) If the first digit dropped is less than 5, the last digit
retained should remain unchanged.
b) If the first digit dropped is more than 5, the digit to be
retained is increased by one.
c) If the digit to be dropped is 5, the previous digit which is
increased by one if it is odd retained as such if it is even.

Numerical Calculations
• Accuracy obtained would never be better than the
accuracy of the problem data
• Calculators or computers involve more figures in the
answer than the number of significant figures in the data
• Calculated results should always be “rounded off” to an
appropriate number of significant figures
Calculations
• Retain a greater number of digits for accuracy
• Work out computations so that numbers that are
approximately equal
• Round off final answers to three significant figures

General Procedure for Analysis
• To solve problems, it is important to present work in a
logical and orderly way as suggested:
1. Correlate actual physical situation with theory
2. Draw any diagrams and tabulate the problem data
3. Apply principles in mathematics forms
4. Solve equations which are
dimensionally homogenous
5. Report the answer with
significance figures
6. Technical judgment
and common sense

IPE Approach (Problem Solving Strategy)

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 a long 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.

Chapter Outline
• Introduction
• Rectilinear Kinematics: Continuous Motion
• Rectilinear Kinematics: Erratic Motion
• Curvilinear Motion: Rectangular Components
• Motion of a Projectile
• Curvilinear Motion: Normal and Tangential Components
• Curvilinear Motion: Cylindrical Components
• Absolute Dependent Motion Analysis of Two Particles
• Relative Motion Analysis of Two Particles Using Translating
Axes

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 dealing with s, v, a, & t.
2) Kinetics – analysis of the forces causing the
motion

Rectilinear Kinematics: Continuous Motion
• Rectilinear Kinematics – To identify at any given
instant, the particle’s position, velocity, and
acceleration
• Position
• Location of a particle at any given instant with
respect to the origin
r : Displacement ( Vector )
s : Distance ( Scalar )
+ve = right of origin,
-ve = left of origin

Displacement – change in
its position, vector quantity
• r : Displacement ( 3 km )
• s : Distance ( 8 km )
Total length
• If particle moves from P to P’
=>
Vector is direction oriented
D s positive
D s negative
QUT
City
My Place
X
3km
River
8 km
N
sss D
rrr D

• Velocity
Average velocity,
Instantaneous velocity is defined as
t
r
vavg
D
D

 trv
t
ins DD
D
/lim
0
dt
dr
vins 

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
ds
v  








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
=>
tD
 
t
s
v T
avgsp
D

tD
 
t
s
v
t
s
v
avg
T
avgsp
D
D

D


• 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, i.e.
t
v
aavg
D
D

vvv D '
t
v
aavg
D
D


Instantaneous Acceleration at time t is found by
taking smaller and smaller values of Δt and
corresponding smaller and smaller values of Δv,
 tva
t
DD
D
/lim
0
















2
2
dt
sd
a
dt
dv
a



• 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 D '

Velocity as a Function of Time
Integrate ac = dv/dt,
Assuming that initially v = v0 when t = 0.
 
t
c
v
v
dtadv
00
tavv c 0








Constant Acceleration

Position as a Function of Time
Integrate v = ds/dt = v0 + act,
Assuming that initially s = s0 when t = 0
 
2
00
0 0
2
1
0
tatvss
dttavds
c
t
c
s
s

 









Relation involving s, v, and s
No time t
dvvdsa 
• Position s
• Velocity
• Acceleration
dt
dv
a 
v
ds
dt 
dt
ds
v 
a
dv
dt 
a
dv
v
ds


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
s
c
v
v

 









PROCEDURE FOR ANALYSIS
1) Coordinate System
• Establish a position coordinate s along the path and
specify its fixed origin and positive direction.
• The particle’s position, velocity, and acceleration,
can be represented as s, v and a respectively and
their sense is then determined from their
algebraic signs.

2) Kinematic Equation
• If a relationship is known between any two of the four
variables a, v, s and t, then a third variable can be
obtained by using one of the three the kinematic
equations
• The positive sense for each scalar can be indicated by
an arrow shown alongside each kinematics equation as
it is applied

• When integration is performed, it is important that
position and velocity be known at a given instant in order
to evaluate either the constant of integration if an
indefinite integral is used, or the limits of integration if a
definite integral is used
• Remember that the three kinematics equations can only
be applied to situation where the acceleration of the
particle is constant.

The car moves in a straight line such that for a
short time its velocity is defined by v = (3t2 + 2t)
ft/s where t is in sec. Determine it position and
acceleration when t = 3s. When t = 0, s = 0.
EXAMPLE 12.1

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
 tt
dt
ds
v 23 2








EXAMPLE 12.1

 
 
23
0
23
0
0
2
0
23
tts
tts
dtttds
ts
ts


 
When t = 3s,
s = (3)3 + (3)2= 36ft
EXAMPLE 12.1

Acceleration. Knowing v = f(t), the acceleration
is determined from a = dv/dt, since this equation
relates a, v and t.








 
26
23 2


t
tt
dt
d
dt
dv
a
When t = 3s,
a = 6(3) + 2
= 20ft/s2  
EXAMPLE 12.1

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/s2.
Determine the projectile’s
velocity and position 4s
after it is fired.
EXAMPLE 12.2

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.
3
4.0 v
dt
dv
a   
EXAMPLE 12.2

 
 
smtv
t
v
dt
v
dt
v
dv
tv
tv
sm
/8.0
60
1
60
11
8.0
1
1
2
1
4.0
1
4.0
2/1
2
22
060
2
0/60 3


































When t = 4s,
v = 0.559 m/s  
EXAMPLE 12.2

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
2
8.0
60
1







 t
dt
ds
v
 
 
t
ts
ts
dttds
0
2/1
2
0
2/1
20
8.0
60
1
8.0
2
8.0
60
1





















 
EXAMPLE 12.2

When t = 4s,
s = 4.43m
 
mts
















60
1
8.0
60
1
4.0
1
2/1
2
 
EXAMPLE 12.2

During a test, a rocket travel
upward at 75m/s. When it is
40m from the ground, the
engine fails. Determine max
height sB reached by the rocket
and its speed just before it hits
the ground.
While in motion the rocket is
subjected to a constant
downward acceleration of 9.81
m/s2 due to gravity. Neglect the
effect of air resistance.
EXAMPLE 12.3

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

)(222
ABCAB ssavv 
sB = 327 m
Velocity.


smsmv
ssavv
C
BCCBC
/1.80/1.80
)(2
2
22
The negative root was chosen since the rocket is
moving downward
EXAMPLE 12.3

A metallic particle travels
downward through a fluid that
extends from plate A and plate B
under the influence of magnetic
field. If particle is released from
rest at midpoint C, s = 100 mm,
and acceleration, a = (4s) m/s2,
where s in meters, determine
velocity when it reaches plate B
and time need to travel from C to B
EXAMPLE 12.4

Solution:
Coordinate System. It is shown that s is taken
positive downward, measured from plate A
Velocity. Since a = f(s), velocity as a function of
position can be obtained by using v dv = a ds.
Realizing v = 0 at s = 100mm = 0.1m
EXAMPLE 12.4

 2
1
2
1.0
2
0
2
1.00
01.02
2
4
2
1
4





sv
sv
dssdvv
dsadvv
Sv
sv
 
At s = 200mm = 0.2m,
 smmsmvB /346/346.0
EXAMPLE 12.4

 
 
 
  tss
tss
dt
s
ds
dts
dtvds
ts
ts
233.201.0ln
201.0ln
2
01.0
01.02
2
01.0
2
01.0 5.02
5.02







 
At s = 200mm = 0.2m, t = 0.658s
EXAMPLE 12.4

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

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

 
 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 > 2a -> +ve velocity -> the particle is
moving to the right
ms
st
125.6
5.3


ms
st
0.4
2


0
0

t
s
EXAMPLE 12.5

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



D
D
 sm
t
s
aavg /75.1
05.3
125.6
Average speed,   sm
t
s
v T
avgsp /04.4
05.3
125.14



D

EXAMPLE 12.5

Home-Work Problems
• Exercise Problems 12.1 to 12.25
• Selected problems will be solved in Class-Room
• Rest of problems will be submitted as assignment
within one-week time.

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).
Example :
r = {sin (2t) i + cos (2t) j – 0.5 t k}
General Curvilinear Motion

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
change in particle’s position.
* S is a path function

Velocity. During the time Δt, the average velocity
of the particle is defined as
t
r
vavg
D
D

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
dr
vins 

• Since Δr is tangent to the curve at P, therefore,
Direction of vins is tangent to the curve
• Magnitude of vins is the speed, which may be
obtained by noting the magnitude of the
displacement Δr is the length of the straight-line
segment from P to P`.
dt
ds
v 

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
t
v
aavg
D
D

2
2
dt
rd
dt
dv
a 

• a acts tangent to the hodograph, therefore it is
not tangent to the path
• a is not tangent to the path of motion
• a directed toward the inside or concave side

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
Curvilinear Motion: Rectangular Components

Velocity.
zvyvxv
kvjviv
dt
dr
v
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.

Acceleration.
zva
yva
xva
kajaia
dt
dv
a
zz
yy
xx
zyx








The acceleration has a magnitude defined as the
positive value of
222
zyx aaaa 
where

• 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.

PROCEDURE FOR ANALYSIS
Coordinate System
• A rectangular coordinate system can be used to
solve problems for which the motion can
conveniently be expressed in terms of its x, y and z
components.

Kinematic Quantities
• Since the rectilinear motion occurs along each
coordinate axis, the motion of each component is
found using v = ds/dt and a = dv/dt, or a ds = v ds
• Once the x, y, z components of v and a have
been determined. The magnitudes of these vectors
are found from the Pythagorean theorem and their
directions from the components of their unit
vectors.

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.
EXAMPLE 12.9

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
    218.1018
22
r m
Velocity.
 
  smx
dt
d
yv
smt
dt
d
xv
y
x
/8.1030/
/99
2




EXAMPLE 12.9

When t = 2 s, the magnitude of velocity is
    smv /1.148.109
22

The direction is tangent to the path, where

2.50tan 1
 
x
y
v
v
v

Acceleration.
2
/4.5
0
smva
va
yy
xx




EXAMPLE 12.9

    222
/4.54.50 sma 
The direction of a is

 90
0
4.5
tan 1
a
EXAMPLE 12.9

For a short time, the path of the plane is described by
y = (0.001x2) m. If the plane is rising with a constant
velocity of 10 m/s, Determine the magnitudes of the
velocity and acceleration of the plane when it is at
y=100 m.
EXAMPLE 12.10

Solution:
Position.
sec10t
)/10(100m,
also
2.316
001.0100
then,10myWhen
2





tsmtvy
mx
x
y
EXAMPLE 12.10

Velocity.
smvvv
smv
vm
xvxx.
x
dt
d
yv
yx
x
x
x
y
/7.18)10()81.15(
Therefore,
/81.15
)2.316(002.010m/s
Thus,
002.0)020(
)001.0(
RuleChainUsing
2222
2







EXAMPLE 12.10

Acceleration.
sma
aaa
sma
am
avsm
xav.
vxvx
xx
dt
d
va
yx
x
x
yy
xx
xx
yy
/791.0
)0()791.0(
MagnitudeTherefore,
/791.0
))(2.316)81.15((002.00
0,/81.15v316.2m,When x
)(0020(
002.0002.0
)002.0(
RuleChainUsing
2222
2
2
x
2











EXAMPLE 12.10

Home-Work Problems
• Exercise Problems 12.71 to 12.86
• Selected problems will be solved in Class-Room

Chapter 12 kinematics of a particle part-i

More Related Content

What's hot (20)

Similar to Chapter 12 kinematics of a particle part-i (20)

Recently uploaded (20)

Chapter 12 kinematics of a particle part-i