Download Engg-Mechanics-Kinematics-P

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Introduction
Mechanics is that branch of science which deals with
the state of rest or motion of bodies under the action
of forces. The subject of Mechanics is logically
divided into two parts.Statics which concerns the
equilibrium of bodies under the action of forces,and
Dynamics concerns the motion of Bodies. Dynamics
is again subdivided into
a.
Kinematics
b.
Kinetics
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Study of Kinematics concerns with the motion of a
body, without referring to the forces causing the motion
of that body.
Study of Kinetics concerns with the motion of the body
considering the forces causing the motion.
Terms and definitions
space: is the geometric region occupied by bodies
whose positions are described by linear and angular
measurements relative to a coordinate system.
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Time is the measure of succession of events and is
basic quantity in dynamics.
Mass is a measure of the inertia of a body, which is its
resistance to a change of velocity
Particle. A body of negligible dimensions is called a
particle. In the mathematical sense a particle is a body
whose dimensions approach to zero so that it may be
analyzed as point mass.
Rigid body. A body is considered rigid when the
relative movements between its parts are negligible for
the purpose at hand.
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Newton’s Laws of Motion
First Law . A particle remains at rest or continues to
move in a straight line with a uniform velocity if there
is no unbalanced force acting on it.
Second Law. The acceleration of a particle is
proportional to the resultant force acting on it and is
in the direction of this force.
Third Law. The forces of action and reaction between
interacting bodies are equal in magnitude, opposite in
direction, and collinear.
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Newton’s second law forms the basis for most of the
analysis in dynamics. As applied to a particle of mass m, it
may be stated as
F=ma Where F is the resultant force acting
on the particle and a is the resulting acceleration. This
equation is a vector equation.
Rectilinear Motion: It is the motion of a particle along a
straight line.
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Rectilinear motion with uniform acceleration:
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O
M
X
N
∆s
t
∆t
A particle moves along a straight line OX as shown in
the figure. The moving particle is in position M at any time
‘t’ and it covers a distance ‘s’ from ‘O’. The particle moves
to N, through a distance ∆s in a small interval of time ∆t.
The velocity v at the instant when the particle is at certain
point M, at time ‘t’ is the rate of change of displacement ∆s
as the increment of time ∆t approaches to zero as limit is
known INSTANTANEOUS VELOCITY and is given by
Lt
v = ∆t  0 ∆s/∆t = dS/dt
S
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Average Velocity : It is the uniform velocity with which the
particle may be considered to be moving in order to cover the
total distance s in a total time t.
(i) When the particle moves with uniform velocity
s = u.t
(ii) When particle moves with variable velocity
vav = Total distance covered (s)
Total time (t)
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(iii) When particle moves with initial velocity u and constant
acceleration a, its velocity changes to v, then
vav = (u+v)/2
s = (u+v)/2 x t
s = distance covered in time ‘t’
(iv) If the distances moved by the particle from start are s1 in
t1, s2 in t2, then average velocity may also be found by
vavg = (s2 - s1) / (t2-t1)
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Acceleration : k9
If the velocity of a particle is v at M and it changes by
∆v, in a small interval of time ∆t then the acceleration of
moving particle, a at the instant at which particle is at M
i. e. the instantaneous acceleration is given by
a  Lt t 0
v dv d  ds  d 2 s

   2
t dt dt  dt  dt
A particle may move in a straight line with constant
acceleration or with variable acceleration.
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Uniform Acceleration :
If the velocity of a body changes by equal amounts in
equal intervals of time, the body is said to move with
uniform acceleration.
Variable Acceleration:
If the velocity of a body changes by unequal amounts in
equal intervals of time, the body is said to move with
variable acceleration.
Note: When the velocity is increasing the acceleration
is reckoned as positive, when decreasing as
negative (retardation or deceleration) .
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Displacement – Time Variations:
Time
Fig. 1
In fig (1)The graph is parallel to the time axis indicating that the
displacement is not changing with time. The slope of the graph is
zero. The body has no velocity and is at rest.
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∆x
∆t
Time
Fig.2
In fig (2)The displacement increases linearly with time. The
displacement increases by equal amount in equal intervals of
time. The slope of the graph is constant.
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∆t2
∆t1
∆x2
∆x1
Time
Fig. 3
In fig(3) The displacement is not changing by equal amounts
in equal intervals of time. The slope of the graph is different
at different times.The velocity of the body is changing with
time. The motion of the body is accelerated.
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Velocity – Time Variations:
∆v
∆t
O
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Constant velocity
Time
Fig. (a)
In fig. (a): The velocity of the body increases linearly with
time. The slope of the graph is constant i.e. the velocity changes
by equal amounts in equal intervals of time. (acceleration of the
body is constant and at t = 0, the velocity is finite. Thus the body
moving with a finite initial velocity,
and has constant
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acceleration).
Velocity – Time Variations:
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Uniform acceleration
∆v
∆t
O
Time
Fig. (b)
In fig. (b): The body has a finite initial velocity. As time passes,
the velocity decreases linearly with time until its final velocity
becomes zero i.e it comes to rest. Thus the body at a constant
deceleration , since the slope of the graph is negative.
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Variable acceleration
∆v2
∆t2
O
∆t1
∆v1
Time
Fig. (c)
In fig. (c): The velocity –Time graph is a curve. The slope is,
therefore, different at different times. In other words, the
velocity is not changing at constant rate. The body does not have
a uniform acceleration since acceleration is changing with time.
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Equations of Motion
Under Uniform Acceleration
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1. Equation of motion
(Relation between v,u, a & t)
If we assume a body starts with an initial velocity u and
uniform acceleration a. After time t, it attains a velocity v.
Therefore the change in velocity in t seconds is (v – u)
Change in velocity / sec. = v – u / t = a v = u + at -----(1)
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2. Equation of Motion:
(Relation between s, u, a and t)
Let a body moving with an initial uniform velocity u be
accelerated with a uniform acceleration a for time t. If v is
the final velocity, the distance s which the body travels in
time t is determined as follows.
Now since acceleration is uniform it is obvious that the
average velocity = (u + v) /2
 Distance traveled = vav x t
= (u + v)/2 x t
= (u + (u + at))/2 x t (Substituted from 1)
s = ut + ½ at2-----(2)
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3. Equation of motion:
(Relation between u, v, a and s)
s = average velocity x time
= (u + v)/2 x t
= (u + v)/2 x (v - u)/a for t = (v – u)/a
therefore
s = (v2 - u2)/2a
v2 = u2 +2as-----------(3)
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Motion under Gravity
It has been observed that bodies falling to the earth
(through distances which are small as compared to the
radius of the earth) experience entirely unrestricted increase
in their velocity by 9.81 m/s for every second during their
fall. This acceleration is called the acceleration due to
gravity and as conventionally denoted by g.
For downward motion:
a = +g
v = u + gt
h = ut + ½ gt2
v2 = u2 + 2gh
For upward motion:
a=–g
v = u – gt
h = ut – ½ gt2
v2 = u2 – 2gh
h = Distance moved in vertical direction
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Comparison between equations of motion under
uniform acceleration and variable acceleration: k21
Acceleration
Uniform
Variable
Equations
2
1
v = u + at
3
S = ut + ½ at2
v2 = u2 + 2as
t
t
s
0
0
0
v = u+∫ a(t)dt S = ut + 1/2∫ a (t2)dt v2 – u2 = 2 ∫ a (s)ds
Graphical Representation:
v = ds/dt
s-t curve
S
ds
dt
t1
t2
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The slope ds/dtwww.bookspar.com
at any point
gives the velocity v at that point.
t
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v
dv
dt
v–t
curve
v
t1
t
dt
t2
The slope dv/dt at any point gives the acceleration a at that
point. The shaded area under v – t curve shown above gives
the incremental displacement ds during the small interval of
time dt.
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dv=adt
a
t2
t1
t
dt
The shaded area under a – t curve shown above gives the
incremental velocity dv during the small interval of time dt.
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PROJECTILES
INTRODUCTION
Assumptions:
1. Mass of the projectile is not considered.
2. Air resistance is neglected.
3. The trajectory of the particle is in the vertical plane.
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A Projectile is a particle moving in space under the
action of gravity.
The velocity with which the particle is projected into
space has horizontal and vertical components. The combined
effect of both components is to move the particle along a
parabolic path. The parabolic path traced by the projectile
is known as Trajectory of the Projectile.
The horizontal component remains constant ( as air
resistance is ignored) while the vertical component of
motion is always subjected to acceleration due to gravity.
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DEFINITIONS:
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Velocity of Projection is the velocity with which a body is
projected into space.
Angle of projection is the angle which the initial velocity
vector makes with the horizontal, or the angle at which a
projectile is projected with respect to horizontal.
Range is the distance along the reference plane between the
point of projection and the point at which it strikes the
plane.
Time of Flight is the total time during which the particle
remains in motion.
Maximum Height is the maximum vertical distance
covered by the projectile from the point of projection.
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MOTION OF A PROJECTILE
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y
u
P(x,y) Q

O
trajectory
h
x
R
M
Consider a particle thrown upwards from a point O, with an
initial velocity u, at an angle  with the horizontal as shown
in the figure above. After attaining maximum height h, it
descends and finally hits the reference plane.
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Equation of the Trajectory k28
From the figure above
x = ux . t = u (Cos ) t
t = x / (u Cos ) --------------(1)
y = uy. t – ½ gt2
=u (Sin  )t – ½ gt2------------------(2)
Sub (1) in (2) we get
y = x tan  - gx2/ (2u2 Cos2 )
This is an equation for a parabola. Hence the path of the
projectile is a parabola.
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The horizontal distance covered by the projectile is known as
Range of Projectile denoted by R.
Resolving the initial velocity u into horizontal and vertical
components
ux = u Cos 
uy = uSin 
(constant)
Time of Flight:
We know that
At Q vy = 0
vy = uy + at
0 = uy - gt
tm = uy / g = u Sin  / g
 Time of flight T = 2tm = 2 (u Sin  /g)
Where tm is the time taken in seconds to reach maximum
height.
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Maximum height attained (h):
vy2 - uy2 =- 2gh
vy = 0
(upward motion)
h = uy2/2g ;
h = u2(Sin2) / 2g
Range:
R= ux x T
(time of flight)
= 2u2(SinCos) / g
R = u2Sin2/g
(Sin2 = 2SinCos)
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From the above equation it is clear that range will be maximum
if Sin2 = 1
i.e. 2 = 90 or  = 45
Rmax = u2/g
Projectile will cover a maximum range when it is directed at an
angle of 45°.
Two angles of Projections for a given range:
We know that
Range R = u2Sin2/g
Sin = Sin ( - )
u2Sin ( - 2)/g = u2Sin (2 1)/g (say)
 where 21= ( - 2)
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Thus for same range
R = u2Sin2/g = u2Sin ( - 2)/g = u2Sin21/g
Which shows that the horizontal range remains the same
when  is replaced by 1.
Or 1=  - 2  =(  / 2 )- 
2
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Projection on an Inclined Plane
u
Q
B

P

O

R
P1
R(Sin)
R(cos)
R= Range along incline; α=angle of projection;
β =angle of inclined plane
Sx=u x+ ½ axt2
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Projection on an Inclined Plane
R(cosβ)=u(cosα)(t)+(0) ;
ax =0
t= Rcosβ / ucosα…..(1)
Sy=uy + ½ ayt2
R sinβ=u sinα (t)- ½ gt2
Substituting eqn.(1) and simplifying we get
The Range along the inclined plane
R={2u2cos2 α [sin(α- β)] } / gcos2 β ..(2)
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Projection on an Inclined Plane:
To get maximum range on the incline,
Differentiating R w.r.t α and equating it to zero
we get α= π/4 + β/4
Substituting this value of α in eqn.(2)
We get maximum Range
Rmax= u2/g(1+sin β)
To find the time of flight: using the relation
v=u + at
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Projection on an Inclined Plane:
0=u x sin(α- β)-gx(cos β) x t
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
t= {usin(α- β)}/ g(cos β)
a=(g cos β) is the acceleration due to gravity along
inclined plane
t=time taken by the projectile to reach Q where QP is
perpendicular distance to the incline plane
Time of Flight T= 2xt
T={2 u sin(α- β)} / g(cos β)
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Projection on an Inclined Plane:
Maximum Height attained h, (PQ)
using the relation
v2-u2=2as
0-u2 sin2(α- β)=-2 gcos β x h
h={ u2 sin2(α- β)} / 2gcos β
Vertical height P1Q=h/ cos β
={u2 sin2(α- β)} / 2gcos2 β
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Practice Problems:
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1. On turning a corner, a motorist rushing at 15m/s, finds a
child on the road 40m ahead. He instantly stops the engine
and applies brakes,so as to stop the car within 5m of the
child, calculate: (a) retardation (b) time required to stop
the car
Ans: (a) -3.21m/s2,
(b) 4.67s
2. A stone is dropped from the top of a tower 100m high.
Another stone is projected upward at the same time from
the foot of the tower, and meets the first stone at a height
of 40m. Find the velocity,with which the second stone is
projected upwards.
Ans: u=28.6 m/s
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Practice Problems:
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3. A train starting from rest is accelerated and acceleration at
any instant is 3/(v+1) m/s. where v is the velocity of the body in
meters per second at any instant. Find the distance in which the
train attains velocity of 48kmph.
Ans: S=293m
4. A projectile is fired from the edge of a 150m high cliff with
an initial velocity of 180m/s at an angle of elevation of 30o with
the horizontal. Neglecting air resistance, find (a) the horizontal
distance from the gun to the point where the projectile strikes
the ground,and (b) the greatest elevation above the ground
reached by the projectile.
(a) 3125m (b) 563m
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Practice Problems:
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3. A train starting from rest is accelerated and acceleration at
any instant is 3/(v+1) m/s. where v is the velocity of the body in
meters per second at any instant. Find the distance in which the
train attains velocity of 48kmph.
Ans: S=293m
4. A projectile is fired from the edge of a 150m high cliff with
an initial velocity of 180m/s at an angle of elevation of 30o with
the horizontal. Neglecting air resistance, find (a) the horizontal
distance from the gun to the point where the projectile strikes
the ground,and (b) the greatest elevation above the ground
reached by the projectile.
(a) 3125m (b) 563m
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Practice Problems:
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5. An aero plane is flying at a height of 300m,with a
velocity of 360kmph. A Shell is fired from the ground
exactly when the aeroplane is above the gun.what should
be the minimum initial velocity of the shell and the angle
of inclination in order to hit the aeroplane ?
Ans:( 126m/s, 37.47o)
6. A projectile is fired from a point ‘o’ at a velocity of
125m/s has to strike a point located on the top of a tower
of 200m high. The horizontal distance betweem the point
‘o’ and the tower is 1000m. Neglect the air resistance and
take g= 9.8m/s2
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Practice Problems:
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Contd:
Calculate:
a) The angle, to the horizontal,at which the projectile must
be fired in order to strike the point on tower in minimum
time.
b) the time taken for flight
c) The maximum height above ‘o’ reached by the projectile.
Ans: (a) 68.550 or 32.990 ,( b) 9.54 s, (c) 233.8m
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