KINEMATICS OF A PARTICLE
1
Introduction:
In this chapter, we study the kinematics of a particle - recall that a particle has a mass but negligible
size and shape. Therefore, we limit the discussion to those objects that have dimensions that are of no
consequence in the analysis of the motion. Such objects may be considered as particles, provided motion
of the body is characterized by motion of its center of mass and any rotation of the body is neglected.
By Curvilinear motion we refer to the motion of an object along a curve. The first step in describing the
motion of a particle is to identify its position at any time t. This can be done by specifying its rectangular
coordinates (x, y, z), or in the case of plane motion (x, y) coordinates, its cylindrical coordinates (r, θ, z),
or in the case of plane motion polar coordinates (r, θ). The motion of the particle can also be described by
measurement along the tangent t and normal n to the curve as shown in the figure below. The directions
t and n lie in the local plane of the curve. In the next few paragraphs we will discuss the kinematics of a
particle motion in all these different frames.
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Rectilinear Kinematics: Continuous Motion:
• Rectilinear kinematics refers to straight line motion. The kinematics of a particle is characterized by
specifying at any given time t, the particle’s position, velocity, and acceleration.
– Position. The position of the particle is represented by a position vector ~r starting from the
origin O of the axis of the motion. If s is the algebraic scalar that represents the position
coordinate of the particle, then
~r = s û
where û is a unit vector pointing toward the positive direction of the axis of the motion. The
magnitude of ~r is (r =| s |). The direction of ~r is the same as û if s > 0 and opposite to û if
s < 0. For analytical work it is therefore convenient to represent ~r by the algebraic scalar s.
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– Displacement. The displacement of a particle is a vector ∆~r defined as the change in he
particle’s position vector ~r. Algebraically the displacement of a particle from P of coordinate s
to point P 0 of coordinate s0 is represented by
∆s = s0 − s
– Velocity. The velocity of a particle is a vector
~av = ∆~r/∆t or algebraically
∗ The average velocity is the displacement divided by time V
as
∆s
Vav =
∆t
∆~r
d~r
~
~
∗ The instantaneous velocity (or velocity) is V = lim Vav = lim
=
or algebraically
∆t→0 ∆t
∆t→0
dt
as
ds
= ṡ
V =
dt
∗ Instantaneous speed (or speed) is a positive scalar and refers to the magnitude of the
instantaneous velocity
ds ~
v =| V |= dt
∗ The average speed is a scalar and refers to the total distance divided by the total time
!
~
~
dV
∆V
=
or algebraically
– Acceleration. The acceleration of a particle is a vector ~a = lim
∆t→0 ∆t
dt
as
dV
d2 s
= V̇ = 2 = s̈
dt
dt
In rectilinear kinematics, the acceleration has the same direction as the velocity when the
particle is speeding and opposite when the particle is slowing down.
A particle can have an acceleration and yet have zero velocity
One additional thing is that
dV
dV
or dt =
a=
dt
a
ds
ds
V =
or dt =
dt
V
ds
dV
=
or a ds = V dV
V
a
The last equality can be integrated to solve for the speed as an example if one can express
a as a function of s.
a=
∗
∗
∗
∗
2
2.1
Constant acceleration
• Let a = ac = constant. Assume that V = Vo and s = so at time t = 0. Then
– Velocity as a function of time
Z
V
t
Z
ac dt or V = Vo + ac t
dV =
o
Vo
– Position as a function of time
Z s
Z t
1
ds =
(Vo + ac t) dt or s = so + Vo t + ac t2
2
so
o
– Speed as a function of position
Z V
Z s
V dV =
(ac ) ds or v 2 = vo2 + 2 ac (s − so )
Vo
2.2
so
Procedure for Solving Problems
The equations of rectilinear kinematics should be applied as follow
• Coordinate System
– Establish a position coordinate s along the linear path and specify its fixed origin and positive
direction
– Since the motion os along a straight line, the particle’s position, velocity, and acceleration can be
represented as algebraic scalars. For analytical work, the sense of s, V , and a is then determined
from their algebraic signs
• Kinematic Equations
– If a relationship is known between any two or the four variables a,V , s, and
third
t,
then a dV
ds
variable can be obtained by using one of the kinematic equations a =
, V =
, or
dt
dt
(a ds = V dV ), which relates all the variables.
– Whenever integration is performed, it is important that the 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 limit of integration if a definite integral is used.
– The equation developed in section 2-1 apply only when the acceleration is constant
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3
Rectilinear Kinematics: Erratic Motion
When a particle’s motion during a time period is erratic or there is a discontinuity in the motion, it may
be difficult to obtain a continuous 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 by
computer. There are several frequently occurring situations:
• Given s-t graph, construct V-t graph. The position of a particle can be plotted over time (s-t
graph), the particle’s velocity as a function of time (V-t graph)can be obtained by measuring the
slope of the s-t graph.
ds
=V
dt
slope of s-t graph = velocity
• Given v-t graph, construct a-t graph. When the particle’s V-t graph is known, the particle’s
acceleration as a function of time (a-t graph)can be obtained by measuring the slope of the V-t graph.
dV
=a
dt
slope of V-t graph = acceleration
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• Given a-t graph, construct V-t graph. When the particle’s a-t graph is given, the particle’s
velocity as a function of time (V-t graph)may be constructed by:
Z
∆V =
a dt
change in velocity = area under a-t graph
• Given V-t graph, construct s-t graph. When the particle’s V-t graph is given, the particle’s
position as a function of time (s-t graph)may be constructed by:
Z
∆s =
V dt
displacement = area under V-t graph
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• Given a-s graph, construct V-s graph. When the particle’s a-s graph is given, the particle’s
velocity as a function of position (V-s graph)may be constructed by:
Z
V2
V1
1
V dV = (V2 2 − V1 2 ) =
2
Z
s1
a ds
so
1
(V2 2 − V1 2 ) = area under a-s graph
2
6
• Given V-s graph, construct a-s graph. When the particle’s V-s graph is given, the particle’s
acceleration as a function of position (a-s graph)may be constructed by:
dV
ds
acceleration=Velocity × slope of V-s graph
a=V
4
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.
In this section the general aspects of curvilinear motion are discussed, and in the subsequent sections we
will consider three types of coordinate systems often used to analyze this motion: rectangular, Normal and
Tangential, and Cylindrial.
• Position. The position of the particle is measured from a fixed point and described by the position
vector ~r(t). This vector is a function of time since both its magnitude and direction change as the
particle moves along its path (described by the path function s(t)).
• displacement. The displacement ∆~r = ~r0 − ~r represents the change in the particle’s position.
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• Velocity.
~av = ∆~r
– The average velocity of the particle is defined as V
∆t
~ (t) = lim ∆~r = d~r = ~r˙ . The velocity
– The velocity of the particle is described by the vector V
∆t→0 ∆t
dt
is tangent to the curve.
d~r ds
– As ∆t ≈ 0, | d~r |≈| ds | and the speed v = =
= ṡ: the speed can be obtained by
dt
dt
differentiating the path function s(t) with respect to time.
• Acceleration.
– The Average acceleration ~aav is defined as ~aav =
8
~
∆V
∆t
~
~
∆V
dV
d2~r
=
= 2
∆t→0 ∆t
dt
dt
– The acceleration of the particle is described by the vector ~a = lim
~ and consequently ~a must account for changes made in both the magnitude and direction
– ∆V
~ . as the particle moves from P1 to P2 . Just a magnitude change increases (or
of the velocity V
~ , and this in itself would allow ~a to remain tangent to the path.
decreases) the ”length” of V
However, in order for the particle to follow the path, the directional change always ”swings” the
velocity vector toward the ”inside” or ”concave side” of the path, and therefore ~a cannot remain
tangent to the path. In fact ~a points toward the concave side of the path.
4.1
Curvilinear Motion: Rectangular Components
When the position of a particle P is defined at any instant by its rectangular coordinates x, y, z, it is
~ and acceleration ~a of the particle into rectangular components
convenient to resolve the velocity V
• Position.
~r(t) = x(t)î + y(t)ĵ + z(t)k̂
M agnitude = r =
p
x2 + y 2 + z 2
Direction given by =
9
~r
r
• Velocity.
~ (t) = d~r = dx î + dy ĵ + dz k̂ = Vx î + Vy ĵ + Vz k̂
V
dt
dt
dt
dt
q
M agnitude = v = Vx 2 + Vy 2 + Vz 2
Direction: =
~
V
always tangent to path
v
• Acceleration.
~a(t) =
~
dV
dVx
dVy
dVz
=
î +
ĵ +
k̂ = ax î + ay ĵ + az k̂ = ẍî + ÿ ĵ + z̈ k̂
dt
dt
dt
dt
p
M agnitude = a = ax 2 + ay 2 + az 2
Direction: =
~a
toward the concave part of the path
a
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4.1.1
Application: Motion of a Projectile
The free-flight motion of a projectile is often studied in terms of its rectangular components, since the
projectile’s acceleration always acts in the vertical direction. Free flight projectile motion can be solved
using the following procedure:
• Coordinate system
– Establish the fixed x, y coordinate axes and sketch the trajectory of the particle. Between
any two points on the path specify the problem data and the three unknowns. In all cases
the acceleration of gravity acts downward. The particle’s initial and final velocities should be
represented in terms of their x and y components
– Remember that positive and negative position, velocity and acceleration components always act
in accordance with their associated coordinate directions.
• Kinematic Equations
– Depending upon the known data and what is to be determined, a choice should be made as to
which three of the following five equations should be applied between the two points on the path
to obtain the most direct solution to the problem.
∗ Horizontal Motion
Vx = (Vo )x
x = xo + (Vo )x t
∗ Vertical Motion
Vy = (Vo )y − g t
y = yo + (Vo )y t −
1
g t2
2
Vy2 = (Vo )2y − 2g (y − yo )
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4.2
Curvilinear Motion: Normal and Tangential Components
When the path along which the particle is moving is known, then it is often convenient to describe the
motion using n and t coordinates which act normal and tangent to the path, respectively, and at the instant
considered have their origin located at the particle.
• Coordinate system
– The positive tangent axis always acts in the direction of motion and the positive normal axis is
directed toward the path’s center of curvature. Geometrically the curve is constructed from a
series of differential arc segments ds. Each segment ds is formed from the arc of an associated
circle having a radius of curvature ρ and center of curvature O0 .
• Velocity and acceleration
– The n and t axes are particulary advantageous for studying the velocity and acceleration of the
~ and the acceleration ~a are expressed by the equations:
particle, because the velocity V
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~ = v û = ṡ û
V
~a = at ût + an ûn
Consider a small displacement of the particle in a time dt. Assume ds to be small so that the
radius of curvature ρ stays the same.
We have the following relations
~a =
~
dV
d(vû)
dv
dû
=
=
û + v
dt
dt
dt
dt
dût =| ût | dθ ûn or û˙t = θ̇ ûn
replacing û˙t with its present expression in the acceleration found earlier.
~a = v̇û + v θ̇ ûn
on the other hand we have
ds = ρ dθ or ṡ = ρ θ̇ and θ̇ =
ṡ
ρ
so
~a = v̇û +
ṡ2
ûn
ρ
and
at = v̇ = s̈
an =
ṡ2
ρ
∗ The velocity is always tangent to the path, its magnitude (speed) is the time derivative of
the path function v = ṡ
∗ The tangential component of the acceleration is he result of the time rate of change of the
speed. This component acts in the positive s-direction if the particle’s speed is increasing
or in the opposite direction if the speed is decreasing.
∗ The relation between at , v and s are the same as for rectilinear motion
at = v̇ =
dv
dv ds
=
dt
ds dt
at ds = v dv
· If at is constant at = (at )c , the above equations, when integrated yield
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s = so + vo t +
1
(at )c t2
2
V = Vo + (at )c t
v 2 = vo2 + 2 (at )c (s − so )
∗ The normal component of the acceleration is the result of change in the direction of the
particle velocity. This component is always directed toward the center of curvature of the
path
v2
· The magnitude of this component is determined from an =
ρ
• As a result of these interpretations, a particle moving along a curved path will have accelerations
directed as shown in the figure below
• A special case of the plane curvilinear motion is the circular motion where the radius of curvature
becomes constant and equal to the radius of the circle ρ = r
• 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:
3/2
1 + (dy/dx)2
ρ=
d2 y dx2 14
4.3
Curvilinear Motion: Cylindrical Components
Some engineering problems involve angular position and a radial distance. for these cases it is often
convenient to express the path of motion in terms of cylindrical coordinates. r, θ, z. If the motion is
restricted to a plane, The polar coordinates r, θ are used.
• Coordinate system
– Polar coordinates are particularly suitable for solving problems for which data regarding the
angular motion of the radial coordinate r is given to describe the path’s motion.
– To use coordinates, the origin is established at a fixed point, and the radial line r is directed to
the particle
– The transverse coordinate θ is measured counterclockwise from a fixed reference line to the
radial line
• Position, Velocity and Acceleration
– The position of the particle is defined by the position vector ~r = r ûr
– The velocity of the particle is given by
~ = ~r˙ = r ˙ûr = ṙ ûr + r û˙r = ṙ ûr + r dθ ûθ
V
dt
∆ûr
∆θ
û˙r = lim
= lim
ûθ
∆t→0 ∆t
∆t→0 ∆t
– The acceleration of the particle is given by
~˙ = d ṙ ûr + r θ̇ûθ = r̈ ûr + ṙ û˙ r + ṙθ̇ ûθ + rθ̈ ûθ + rθ̇ û˙ θ
~a = V
dt
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∆ûθ
∆θ
û˙θ = lim
= − lim
ûr
∆t→0 ∆t
∆t→0 ∆t
~a = (r̈ − rθ̇2 ) ûr + (rθ̈ + 2ṙθ̇) ûθ
• If the particle travels in a circular path, s = constant so that ṙ = r̈ = 0 the previous equations
simplify considerably
• If the particle moves along a space curve as shown in teh figure below, then its location may be
specified by the three cylindrical coordinates r, θ, z. The z coordinate is identical to that used
for rectangular coordinates. Since the unit vector defining its direction, ûr is constant, the time
derivatives of this vector are zero and
~rp = r ûr + z ûz
~ = ṙ ûr + rθ̇ ûθ + ż ûz
V
~a = (r̈ − rθ̇2 ) ûr + (rθ̈ + 2ṙθ̇) ûθ + z̈ ûz
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5
Absolute dependent motion. Analysis of two particles
In some 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. When each particle moves along a rectilinear path, the magnitude of the velocity
and acceleration of the particles change, not their line of direction.
The following procedure can be used:
• Position-Coordinate equations
– Establish position coordinates which have their origin located at fixed or datums
– The coordinates are directed along the path of motion and extend to a point having the same
motion as each of the particles
– it is not necessary that the origin be the same for each of the coordinates; however, it is important
that each coordinate axis selected be directed along the path of motion of the particle
– Using geometry or trigonometry, relate the coordinates to the total length of the cord, lT or
to that portion of the cord, l, which excludes the segments that do not change length as the
particles move, such as arc segments wrapped over pulleys.
– If a problem involves a system of two or more cords wrapped around pulleys, then the position
of a point one one cord must be related to the position of a point on another cord. Separate
equations are written for a fixed length of each cord of the system and teh position of the two
particles are then related by these equations
Example:
In the figure above the position coordinate equations are:
sA + lCD + sB = lT
• Time derivatives
– Two successive time derivatives of the position-coordinate equations yield the required velocity,
and acceleration equations which relate the motions of the particles
– The signs of the terms in these equations will be consistent with those that specify the positive
and negative sense of the position coordinates
Example:
In the figure above the position coordinate equations are:
dsA
dsB
+
= 0 or VA = −VB
dt
dt
d2 sA
d2 sB
+
= 0 or aA = −aB
dt2
dt2
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6
Relative motion. Analysis of two particles using translating
axes
Throughout this chapter the absolute motion of a particle has been determined using a single fixed reference frame for measurement. There are many cases, however, where the path of motion for a particle
is complicated, so that it may be feasible to analyze the motion in parts by using two or more frames of
reference. In this section translating frames of reference will be considered for the analysis.
• When applying the relative-position equation ~rB = ~rA + ~rB/A , it is first necessary to specify the
location of the fixed frame x, y, z and the translating frame x0 , y 0 , z 0
• Usually the origin A of the translating axes is located at a point having known position, ~rA .
• Since vector addition form a triangle, there can be at most two unknowns.
• these unknowns can be solved for either graphically, using trigonometry, or by resolving each of the
three vectors ~rA , ~rB and ~rb/A into rectangular or cartesian components, thereby generating a set of
scalar equations.
~B = V
~A + V
~B/A and ~aB = ~aA +~aB/A are applied in teh same manner
• The relative-motion equations V
as mentioned above except that in this case teh origin of the fixed x, y, z axes does not have to be
specified.
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