Chapter 12 kinematics_of_a_particle

CHAPTER-12
KINEMATICS OF A PARTICLE

KINEMATICS OF A PARTICLE
• Rectilinear Kinematics: Continuous Motion
• Rectilinear Kinematics: Erratic Motion
• General Curvilinear Motion
• Curvilinear Motion: Rectangular Components
• Motion of a projectile
• Curvilinear Motion: Normal & Tangential Components
• Curvilinear Motion: Cylindrical Components
• Absolute Dependent Motion Analysis of Two Particles
• Relative Motion Analysis of Two Particles

INTRODUCTION
MECHANICS
(concerned with the state of rest or
motion of bodies subjected to the
action of forces )
STATICS
(concerned with equilibrium of
a body that is either at rest or
moves with constant velocity)
DYNAMICS
(Deals with the accelerated
motion of a body)
KINEMATICS
(deals only with the geometric aspects of
motion without regard to the forces which
produce that motion)
KINETICS
(analysis of forces causing the
motion)
In this chapter we will study the geometric aspects of the motion of a
particle

RECTILINEAR KINEMATICS: CONTINUOUS MOTION
• A particle has a mass but negligible size and shape
• Thus our applications will be limited to those objects that have
dimensions that are of no consequence in the analysis of motion
• In most problems, one is interested in bodies of finite size, such as
rockets, projectiles or vehicles.
• Such objects may be considered as particles provided motion of the
body is characterized by motion of its mass center and any rotation of
the body is neglected.
• A particle can move along either a straight or a curved path.
• Rectilinear motion is the motion of a body along a straight line
• The kinematics of this motion is characterized by specifying, at any
given instant, the particle’s position, velocity and acceleration.

POSITION:
• The straight line path of particle will be defined using a single coordinate
axis, s
• Origin ‘O’ on the path is a fixed point, and from this fixed point the
position vector r is used to specify the location of the particle P at any
given instant.
• Since r is always along the s-axis and so its direction never changes, only
its magnitude changes, so for analytical work it is often convenient to
represent r by an algebraic scalar s, representing the position coordinate
of the particle
• The magnitude of s (and r) is the distance from O to P, and the sense (or
arrowhead direction of r) is defined by the algebraic sign of s.

DISPLACEMENT:
• The displacement of the particle is defined as the change in its position.
e.g. if the particle moves from P to P’ the displacement is ∆r = r’ – r.
Using algebraic scalars to represent ∆r, we have : ∆s = s’ – s
• Here ∆s is positive since the particle’s final position is to the right of its
initial position i.e. s’>s, likewise, if the final position is to the left of its
initial position, ∆s is negative
• Displacement of a particle must be distinguished from the distance
traveled by the particle, which is always a positive scalar quantity that
represents the total length of path over which a particle travels.

VELOCITY:
• If the particle moves through a displacement ∆r from P to P’ during time
interval ∆t, the average velocity of the particle is:
vavg = ∆r/∆t
• If we take smaller and smaller values of ∆t, the magnitude of ∆r becomes
smaller and smaller. Consequently the instantaneous velocity is defined by :
v = lim∆t-0 (∆r/∆t) or v = dr/dt
• Representing v as an algebraic scalar, we can also write:
v = ds/dt ------------------------------(1)
• Since ∆t or dt is always positive, the sign used to define the sense of velocity
is the same as that of ∆s or ds. For example, if the particle is moving to the
right, the velocity is positive, whereas if it is moving to the left, the velocity
is negative.

ACCELERATION:
• The average acceleration of a particle during the time interval ∆t is
defined by: aavg = ∆v/∆t
• Here ∆v represents the difference in the velocity during the time interval
∆t i.e. ∆v = v’ – v
• If we take smaller and smaller values of ∆t, the magnitude of ∆v becomes
smaller and smaller. Consequently the instantaneous acceleration is
defined by :
a = lim∆t-0 (∆v/∆t) or a = dv/dt or a = d2
s / dt2
------------------(2)
• Both the average and instantaneous acceleration can be either positive or
negative.

VELOCITY: v = ds/dt ------------------------------(1)
ACCELERATION: a = dv/dt ------------------------------(2)
• A differential relation involving the displacement, velocity and acceleration
along the path may be obtained by eliminating the time differential dt
between equation (1) and (2).
a ds = v dv ------------------------------(3)

CONSTANT ACCELERATION:
When the acceleration is constant, each of the three kinematic equations
v = ds/dt, a = dv/dt, a ds = v dv, may be integrated to obtain formulas that
relate ac, v, s and t.
Velocity as a function of Time:
v = vo + act --------------------------------(4)
Position as a function of Time:
s = so + vot + (1/2)act2
---------------------(5)
Velocity as a function of Position:
v2
= vo
2
+ 2ac(s – so) -----------------------(6)
• It is important to remember that equations (4), (5) and (6) are useful only
when the acceleration is constant and when t=0, s=so, v = vo

Examples:
12.1, 12.2, 12.3, 12.4 and 12.5
Fundamental Problems:
F12-1, F12-6
Practice Problems:
12.7, 12.17, 12.21, 12.29, 12.31

EXAMPLE 12-1
During a test, the car shown below 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 seconds. Determine its
position and acceleration when t = 3s. When t=0, s=0.

EXAMPLE 12-3
During a test, a rocket is
traveling upward at 75m/s, and
when it is 40m from the ground
its engine fails. Determine the
maximum 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 9.81m/s2
due to
gravity. Neglect the effect of air
resistance.

RECTILINEAR KINEMATICS: ERRATIC MOTION
• When a particle’s motion during a time period is erratic (inconsistent,
irregular) it may be difficult to obtain a continuous mathematical
function to describe its position, velocity or acceleration.
• Instead, the motion may best be described graphically using a series of
curves that can be generated experimentally from computer output.
• If the resulting graph describes the relationship between any two of the
variables, a, v, s, t, a graph describing the relationship between the other
variables can be established by using the kinematic equations 1, 2 and 3.
• The concept is explained with the help of the following examples:

EXAMPLE 12-6
(Given the s-t graph, construct the v-t and a-t graphs)
A bicycle moves along a
straight road such that
its position is described
by the graph shown in
figure. Construct the v-t
and a-t graphs for
st 300 ≤≤

EXAMPLE 12-6 contd.
(Given the s-t graph, construct the v-t and a-t graphs)

EXAMPLE 12-7
(Given the a-t graph, construct the v-t and s-t graphs)
The rocket sled shown in
figure starts from rest and
travels along a straight line
such that it accelerates at a
constant rate for 10s and
then decelerates at a
constant rate. Draw the v-t
and s-t graphs and
determine the time t’ needed
to stop the sled. How far has
the sled traveled?

EXAMPLE 12-7 contd.
(Given the a-t graph, construct the v-t and s-t graphs)

EXAMPLE 12-8
(Given the a-s graph, construct the v-s graph)
or (Given the v-s graph, construct the a-s graph)
The v-s graph describing
the motion of a
motorcycle is shown.
Construct the a-s graph
of the motion and
determine the time
needed for the
motorcycle to reach the
position s = 400 ft.

EXAMPLE 12-8 contd.
(Given the a-s graph, construct the v-s graph)
or (Given the v-s graph, construct the a-s graph)

RECTILINEAR KINEMATICS: ERRATIC MOTION
Examples:
12.6, 12.7, 12.8
F12-9, F12-14
Practice Problems:
12.47, 12.53, 12.57, 12.65,
12.67

GENERAL CURVILINEAR MOTION
• Curvilinear motion occurs when the particle moves along a curved path
• Since this path is often described in three dimensions vector analysis
will be used to formulate the particle’s position, velocity and
acceleration
POSITION:
• Consider a particle located at point P on a
space curve defined by the path function s
• The position of the particle, measured
form a fixed point O, will be designated
by the position vector r = r(t)
• This vector is a function of time since, in
general, both its magnitude and direction
changes as the particle moves along the
curve.

DISPLACEMENT:
• Suppose during a small time interval ∆t the particle moves a distance ∆s
along the curve to a new position P’, defined by r’ = r + ∆r
• The displacement ∆r represents the change in the particle’s position and is
determined by vector subtraction:
∆r = r’ – r

VELOCITY:
• During the time interval ∆t, the average velocity of the particle is defined by:
vavg = ∆r / ∆t
• The instantaneous velocity is determined from this equation by letting ∆t – 0,
and consequently the direction of ∆r approaches the tangent to the curve at
point P. Hence,
v = lim∆t-0 (∆r/∆t) or v = dr/dt
• Since dr will be tangent to the curve at P, the direction of v is also tangent to the
curve

ACCELERATION:
• If the particle has a velocity v at time t and a velocity v’ = v + ∆v at t + ∆t, then the
average acceleration of the particle during the time interval ∆t is
aavg = ∆v / ∆t where ∆v = v’ - v
• The two velocity vectors v’ and v are plotted in figure such that their tails are
located at the fixed point O’ and their arrowheads touch points on the curve. This
curve is called a Hodograph, and when constructed, it describes the locus of points
for the arrowhead of the velocity vector in the same manner as the path s describes
the locus of points for the arrowhead of the position vector.
• To obtain the instantaneous acceleration, let ∆t – 0, so
a = lim∆t-0 (∆v/∆t) or a = dv/dt or a = d2
r / dt2
• By definition of the derivative, a acts tangent to the hodograph, and therefore, in
general, a is not tangent to the path of motion.

CURVILINEAR MOTION: RECTANGULAR COMPONENTS
222
zyxr ++=
POSITION:
• If at a given instant, the particle P is at a point (x,y,z) on the curved path s, its
location is then defined by the position vector
r = xi +yj + zk
• The magnitude of r is always positive and is defined as:
• The direction of r is specified by the components of the unit vector, ur = r/r

VELOCITY:
• The first time derivative of r yields the velocity v of the particle, Hence
)()()( zk
dt
d
yj
dt
d
xi
dt
d
dt
dr
v ++==
000
,, zvyvxv zyx ===
kvjviv
dt
dr
v zyx ++==
• The velocity has a magnitude defined as the positive value of:
v = √(vx
2
+vy
2
+vz
2
)
• The direction of v is specified by the components of the unit vector,
uv = v/v
• This direction is always tangent to the path.

ACCELERATION:
• The first time derivative of v yields the acceleration a of the particle, Hence
000000000
,, zvayvaxva zzyyxx ======
kajaia
dt
dv
a zyx ++==
• The acceleration has a magnitude defined as the positive value of:
a = √(ax
2
+ay
2
+az
2
)
• The direction of a is specified by the components of the unit vector,
ua = a/a

EXAMPLE 12-9
At any instant the horizontal position of the weather balloon is
defined by x=(8t) ft where t is given in seconds. If the equation of the
path is y=x2
/10, determine:
(a) the distance of the balloon from the station at A when t=2s,
(b) the magnitude and direction of the velocity when t=2s,
(c) the magnitude and direction of the acceleration when t=2s

MOTION OF A PROJECTILE
• The free flight motion of a projectile is often studied in terms of its rectangular
components, since projectile’s acceleration always acts in the vertical direction
• Consider a projectile launched at point (x0,y0)
• The path is defined in the x-y plane such that the initial velocity is v0, having
components (vx)0 and (vy)0
• When air resistance is neglected, the only force acting on the projectile is its
weight, which causes the projectile to have a constant downward acceleration of
9.8m/s2
(32.3ft/s2
)

→+
→+
Horizontal Motion:
Since ax=0, application of the constant acceleration equation yields:
v = v0 + act; vx = (vx)0
x = x0 + v0t +(1/2)act2
; x = x0 + (vx)0 t
v2
= v0
2
+ 2ac(s-s0); vx = (vx)0
Vertical Motion:
Since positive y-axis is directed upward, then ax= -g, application of the constant acceleration
equation yields:
v = v0 + act; vy = (vy)0 - gt
y = y0 + v0t +(1/2)act2
; y = y0 + (vy)0 t – (1/2)gt2
v2
= v0
2
+ 2ac(y-y0); vy
2
= (vy)0
2
– 2g(y-y0)
→+

Examples:
12.9, 12.10, 12.11, 12.12, 12.13
F12-17, F12-22
Practice Problems:
12.97, 12.98, 12.105, 12.107,
12.109

EXAMPLE 12-11
A sack slides off the ramp, as shown in figure, with a horizontal
velocity of 12m/s. If the height of the ramp is 6m from the floor,
determine the time needed for the sack to strike the floor and the
range R where sacks begin to pile up.

EXAMPLE 12-13
The track for the racing event is designed so that the rider jumps off
the slope at 30º, from a height of 1m. During a 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 the rider.

PROBLEM 12-105
The boy at A attempts to throw a ball over the roof of a barn with an
initial speed of vA = 15 m/s. Determine the angle ӨA at which the ball
must be thrown so that it reaches its maximum height at C. Also,
find the distance d where he should stand to make the throw.

PROBLEM 12-109
Determine the horizontal velocity vA of a tennis ball at A so that it
just clears the net at B. Also, find the distance s where the ball
strikes the ground.

CURVILINEAR MOTION: Normal & Tangential Coordinates
When the path along which a particle is moving is known, it is often convenient to
describe the motion using n and t coordinates which act normal and tangent to the path.
Planar Motion:
• Consider the particle P shown, which is moving in a plane along a fixed curve,
such that at a given instant it is at the position s, measured from point O.
• The t axis is tangent to the curve at P and is positive in the direction of increasing
s. This positive direction is designated by the unit vector ut.
• The normal axis n is perpendicular to the t-axis and is directed from P towards the
center of curvature O’. The positive direction, which is always on the concave side
of the curve, will be designated by the unit vector un.
• The plane which contains the n and t axis is referred to a s the osculating plane.

Velocity:
• Since the particle is moving, s is a function of time.
• The particle’s velocity v has a direction that is always tangent to the path, and a
magnitude that is determined by taking the time derivative of the path function
s = f(t), i.e. v = ds/dt. Hence:
v = v ut where v =
0
s

Acceleration:
• The acceleration of the particle is the time rate of change of velocity. Thus
------------------(1)
• In order to determine the time derivative of unit vector ut note that as the particle
moves along the arc ds in time dt, ut preserves its magnitude of unity, however, its
direction changes and becomes ut’

• As shown in the figure, ut’ = ut + dut
• dut has a magnitude of dut = (1) dӨ, and its direction is defined by un
• So dut = dӨ un
• Therefore the time derivative becomes

SPECIAL CASES:
• If the particle moves along a straight line, then ρ = ∞ , thus
an = 0. Thus a = at, and we can conclude that the tangential
component of acceleration represents the time rate of change
in the magnitude of the velocity
• If the particle moves along the curve with a constant speed,
then at= 0 and a = an= v2
/ρ. Therefore, the normal component
of acceleration represents the time rate of change in the
direction of the velocity. Since an always acts towards the
center of the curvature, this component is sometimes referred
to as the centripetal acceleration.

Three Dimensional Motion
• If the path is expressed as y=f(x), the radius of curvature ρ at any point on the
path is determined from the equation:
22
2/32
/
])/(1[
dxyd
dxdy+
=ρ

Examples:
12.14, 12.15, 12.16
F12-28, F12-31
Practice Problems:
12.116, 12.118, 12.138,
12.140, 12.141

EXAMPLE 12-15
A race car C travels around the horizontal circular track that has a
radius of 300ft. If the car increases its speed at a constant rate of
7ft/s, starting from rest, determine the time needed for it to reach an
acceleration of 8 ft/s2
. What is its speed at this instant?

EXAMPLE 12-16
The box travels along the industrial conveyor as shown. If it starts
from rest at A and increases its speed such that at = (0.2t) m/s2
, where
t is in seconds, determine the magnitude of its acceleration when it
arrives at point B.

Problem 12-141
The truck travels along a circular road that has a radius of 50m at a
speed of 4m/s. For a short distance when t=0, its speed is then
increased by v0
=(0.4t) m/s2
, where t is in seconds. Determine the speed
and the magnitude of the truck’s acceleration when t=4s.

CURVILINEAR MOTION: Cylindrical Components
Path of motion of a particle can sometimes be expressed in terms of cylindrical
components (r, Ө, z). If the motion is restricted to the plane, the polar coordinates r and Ө
are used.
Polar Coordinates:
• The location of the particle P can be specified by using both the radial
coordinate r, which extends outward from the fixed origin O to the particle, and
a transverse coordinate θ, which is the counterclockwise angle between a fixed
reference line and the r axis.
• The angle is generally measured in degrees or radians
• The positive directions of r and θ coordinates are defined by the unit vectors ur
and uθ respectively.

Position:
• At any instant the position of the particle is defined by the position vector,
r = r ur
Velocity:
• The instantaneous velocity v is obtained by taking the time derivative of
position vector r. Thus we have

• To evaluate the time derivative of ur, it must be noticed that ur changes only its
direction with respect to time, since by definition the magnitude of this vector is
always one unit.
• During the time Δt, a change Δθ will cause ur to become ur’ where ur’
= ur + Δur
• The time change in ur is then Δur
• For small angles Δθ this vector has a magnitude Δur ≈ 1(Δθ) and acts in the uθ
direction.
• Hence Δur = Δθ uθ
• Thus,

• Therefore the velocity vector v can be written as:
where:
• These components are shown graphically in the above figure
• Since vr and vθ are mutually perpendicular, the magnitude of velocity is defined by:
• And the direction of v is tangent to the path at P.

Acceleration:
• Taking the time derivative of velocity vector, the particle’s instantaneous acceleration is obtained, which is:
• To evaluate the time derivative of uθ, it must be noticed that uθ changes only its direction with respect to time, since
by definition the magnitude of this vector is always one unit.
• During the time Δt, a change Δθ will cause uθ to become uθ’ where uθ’ = uθ + Δ uθ. The time change in uθ
is then Δ uθ
• For small angles Δθ this vector has a magnitude Δuθ ≈ 1(Δθ) and acts in the -ur direction. Hence Δuθ = -Δθ ur
• Thus,

• Therefore the acceleration vector a can be written as:
where:
• These components are shown graphically in the above figure
• Since ar and aθare mutually perpendicular, the magnitude of acceleration is
defined by:

Cylindrical Coordinates:
• If the particle P moves along a space curve as shown, then its location may be specified
by the three cylindrical coordinates, r, θ and z.
• The z coordinate is identical to that used for rectangular coordinates.

Examples:
12.17, 12.18, 12.19, 12.20
F12-34, F12-36
Practice Problems:
12.165, 12.167, 12.169, 12-186
12.187

Example 12-18
The rod OA is rotating in the horizontal plane such that θ=(t3
) rad.
At the same time, the collar B is sliding outward along OA so that
r=(100t2
) mm. If in both cases t is in seconds, determine the velocity
and acceleration of the collar when t=1s.

Example 12-20
Due to the rotation of the forked rod, the cylindrical peg A travels
around the slotted path, a portion of which is in the shape of a
cardioid, r=0.5(1-Cosθ) ft, where θ is in radians. If the peg’s velocity
is v=4 ft/s and its acceleration is a=30 ft/s2
at the instant θ=180º,
determine the angular velocity and angular acceleration of the fork.

Absolute Dependent Motion Analysis of Two Particles
• In some types of problems the motion of one particle will depend on the
corresponding motion of another particle.
• This dependency commonly occurs if the particles are interconnected by
inextensible cords which are wrapped around pulleys.
• For example, the movement of block A downward along the inclined plane
will cause a corresponding movement of block B up the other incline.
• We can show this mathematically by first specifying the location of the blocks
using position coordinates sA and sB
• Each of the coordinate axes is:
i) referenced from a fixed point (O) or fixed datum line
ii) measured along each incline plane in the direction of motion of block A
and block B
iii) has a positive sense from C to A and D to B.

• If the total cord length is lT, the position coordinates are related by the equation:
sA + lCD + sB = lT
• Taking the time derivative of this expression and noting that lCD and lT reamin
constant, we have:
(d sA/dt) + (d sB/dt) = 0
or vB = -vA
• The negative sign indicates that when block A has a velocity downward, i.e. in
the direction of positive sA, it causes a corresponding upward velocity of block B;
i.e. B moves in negative sB direction.
• Similarly, time differentiation of the velocities yields the relation between the
accelerations, i.e.
aB = -aA

• Consider a more complicated example
involving dependent motion of two blocks.
• In this case the position of A is specified by
sA, and the position of the end of the cord
from which block B is suspended is defined
by sB
• Coordinate axes are chosen such that they
are:
i) referenced from fixed points or datums
ii) measured in the direction of motion of
each block, and
iii) positive to the right (sA) and positive
downward (sB)
• If l represents the total length of the cord,
then:
2sB+h+sA= l
2vB= -vA
2aB= -aA

• Same example can also be worked out
by defining the position of block B
from the center of the bottom pulley.
• In this case:
2(h-sB) + h + sA = l
2vB = vA
2aB = aA

Example 12.21
Determine the speed of block
A, if block B has an upward
speed of 6ft/s.

Example 12.23
Determine the speed with
which block B rises if the end
of the cord at A is pulled
down with a speed of 2 m/s.

Relative Motion Analysis of Two Particles Using
Translating Axes
• Uptil now, absolute motion of a particle has been determined using a
single fixed reference frame for measurement
• However, when the path of motion for a particle is complicated,
sometimes it is feasible to analyze the motion in parts by using two or
more frames of reference.
• At this instant, only translating frames of reference will be considered for
analysis.

Translating Axes
Position:
• Consider particles A and B which move along the
arbitrary paths aa and bb respectively.
• The absolute position of each particle rA and rB is
measured from the common origin O of the fixed x, y,
z reference frame.
• The origin of a second frame of reference x’, y’, z’is
attached to and moves with particle A.
• The relative position of “B with respect to A” is
designated by a relative position vector rB/A, such that:
rB = rA + rB/A
Velocity:
• An equation that relates the velocities of the particles
can be determined by taking the time derivative of
above equation:
vB = vA + vB/A
Acceleration:
• The time derivative of velocity equation yields a
similar vector relationship between the absolute and
relative accelerations of particles.
aB = aA + aB/A

Examples:
12.21, 12.22, 12.23, 12.24,
12.25, 12.26, 12.27
F12-39, F12-42
Practice Problems:
12.202, 12.207, 12.217,
12.223, 12.225
Translating Axes

Example 12-25
A train traveling at a constant speed of 60 mi/h, crosses over a road
as shown. If the automobile A is traveling at 45 mi/h along the road,
determine the magnitude and direction of the relative velocity of the
train with respect to the automobile.

Example 12-26
Plane A is flying along a straight line path, whereas plane B is flying
along a circular path having a radius of curvature of ρB=400 km.
Determine the velocity and acceleration of B as measured by the
pilot of A.

Example 12-27
At the instant shown cars A and B are traveling at speeds of 18 m/s
and 12 m/s respectively. Also at this instant, A has a decrease in
speed of 2 m/s2
and B has an increase in speed of 3 m/s2
. Determine
the velocity and acceleration of B with respect to A.

Chapter 12 kinematics_of_a_particle

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Chapter 12 kinematics_of_a_particle